tag:blogger.com,1999:blog-3933248450119789432024-03-13T13:46:00.975-04:00Behind the GuessesSolutions from the brain, not the buttEli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.comBlogger11125tag:blogger.com,1999:blog-393324845011978943.post-56318622079988261462009-11-30T23:42:00.001-05:002009-11-30T23:44:40.685-05:00Quaternions -- Part 1: How many?[<a href="http://s87572736.onlinehome.us/btg/btg07-quaternions1.pdf">Click here for a PDF of this post with nicer formatting</a> or <a href="http://behindtheguesses.blogspot.com/2009/11/quaternions-part-1-how-many.html#more">see below</a>]<br />
<span style="font-weight: bold;">The Setup</span><br />
Quaternions seem to be one of the least understood mathematical things amongst physicists. I have sat in countless lectures where at some point the lecturer pointed out that a particular topic could be understood or explained using quaternions, but, when pressed, could not really explain what, precisely, one of these quaternion thingies actually is.<br />
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The first encounter people have with quaternions is generally after they learn about the complex plane and its relationship to the regular 2D Cartesian plane. After seeing all sorts of nifty properties and uses of this relationship (we'll see one shortly) it's only natural to ask if there's a 3D complex analogue to the 2D complex plane. And, therefore, most books ask precisely this question. However, they usually give less-than-satisfactory attempts at generalizing, highlighting the mysterious algebraic problem of ``closure'' or something to that effect.<br />
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Then, often retelling the story of Hamilton and a bridge<sup>1</sup> they pull some strange, ``4D'' quaternions out of a hat and show how they happily resolve all the algebraic problems. This, it seems, should be enough to placate even the most thoughtful reader, and stands in place of an actual explanation. And even though there is a lot of information about these buggers out there on the intertubes, all that I've seen is of the same approach.<br />
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So, it doesn't surprise me, honestly, that ``quaternion'' is also one of the most popular searches on this blog. The topic of quaternions is really too big to handle fully in one post (and, for full disclosure, I do not completely understand them myself), so this post will deal primarily with a rationale for the initial guess of a ``4D'' quaternion.<br />
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This post assumes you have read, and thoroughly grokked my discussion of dot and cross products,[1] and have a solid understanding of traditional complex numbers.<br />
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<span style="font-weight: bold;">Complex Numbers and 2D Vectors</span><br />
My approach in this section is based on the fantastic book, Visual Complex Analysis by Tristan Needham.[2] If something here isn't clear (and it's not the fault of my writing), or is different from the way you learned complex numbers, read this book. Even if everything is perfectly clear, read this book. What I'm trying to say is: Read this book.<sup>2</sup><br />
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Recall that a complex number <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z=x+iy" /> can be represented by a vector <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7Bz%7D=x%5Chat%5Cimath+y%5Chat%5Cjmath" /> in a 2D <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?xy" />-plane. Also, if <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?r=%5Csqrt%7Bx%5E2+y%5E2%7D" /> -- i.e. the ``modulus'' of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z" /> or the length of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%20z" /> -- and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Ctheta" /> is the ``argument'' of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z" /> or the angle between <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%20z" /> and the <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?x" />-axis, we can also write <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z=re%5E%7Bi%5Ctheta%7D" /> in ``polar form.'' See [2] for pictures. We also have Euler's identity<br />
<div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?e%5E%7Bi%5Ctheta%7D=%5Ccos%5Ctheta+i%5Csin%5Ctheta" /><br />
</div><div style="text-align: right;">(1)<br />
</div><br />
Furthermore, recall that multiplying two complex numbers together effects a rotation and scaling. For example, multiplying a complex number <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z=re%5E%7Bi%5Cphi%7D" /> -- graphically, a vector of length <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?r" /> making angle <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cphi" /> with respect to the real axis (<img align="middle" border="0" src="http://www.codecogs.com/gif.latex?x" />-axis) -- by <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cxi=%5Crho%20e%5E%7Bi%5Ctheta%7D" /> gives <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cxi%20z=%28%5Crho%20r%29e%5E%7Bi%28%5Cphi+%5Ctheta%29%7D" /> . This can be understood graphically as a scaling of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?r" /> by <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Crho" /> and a rotation of the direction of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z" /> by the angle <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Ctheta" />. Finally, the complex conjugate <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z%5E*" /> of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z" /> is given either by <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z%5E*=x-iy" /> or <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z=re%5E%7B-i%5Ctheta%7D" />.<br />
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<span style="font-weight: bold;">From Complex Multiplication to Vector Products</span><br />
For two complex numbers <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?A=a%20e%5E%7Bi%5Calpha%7D" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?B=b%20e%5E%7Bi%5Cbeta%7D" /> let's see what <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?A%5E*%20B" /> is. This demonstration (at least initially) is based on [2].<sup>3</sup> Anyway,<br />
<div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?A%5E*%20B%20=%20a%20b%20e%5E%7Bi%20%28%5Cbeta%20-%20%5Calpha%29%7D" /><br />
</div><div style="text-align: center;"></div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?=a%20b%20e%5E%7Bi%5Ctheta%7D." /><br />
</div><div style="text-align: right;">(2)<br />
</div><br />
Graphically this is a vector with length <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?ab" /> at an angle <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Ctheta=%28%5Cbeta-%5Calpha%29" /> from the <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?x" />-axis. Expanding this into a real and complex part using Euler's identity (1) gives:<br />
<div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?A%5E*%20B=ab%5Ccos%5Ctheta%20+%20i%5C,ab%5Csin%5Ctheta." /><br />
</div><div style="text-align: right;">(3)<br />
</div>We now note that the real part of this expression corresponds to the dot product between the two vectors<br />
<img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%20A" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%20B" />. But should we do with the imaginary part?<br />
<br />
Well, the magnitude of the imaginary part certainly corresponds to the magnitude of <span style="font-style: italic;">one dimension</span> of the cross product between the two vectors. That is, if we relate the complex plane to the Cartesian <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?xy" />-plane then the imaginary part of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?A%5E*%20B" /> is the <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z" />-component of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D%5Ctimes%5Cvec%7BB%7D" />. This important point is often lost in passing, and thus this property of complex multiplication is relegated to the realm of ``cool trick.'' However, we'll make good use of this detail.<br />
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<span style="font-weight: bold;">Rethinking complex numbers</span><br />
Now we are ready for the conceptual jump. Although we got to the representation of dot and cross products through use of a 2D complex plane, we're going to distance ourselves from this wonderful visualization for the moment and note that <span style="font-weight: bold;">an arbitrary complex number has two parts: One corresponds to a dot product, the other corresponds to <span style="font-style: italic;">one dimension</span> of a cross product of two vectors.</span>[3]<sup>4</sup> If we want to find a relationship between complex numbers and 3D vectors we need to pick one of these parts to generalize.<br />
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Now, recall that the dot product yields a scalar quantity equal to the amount that two vectors point in the same direction. Since there is no directionality or dimensionality inherent in this quantity -- it's just a length -- there's really no way to add extra bits here. Length stays a scalar in <span style="font-style: italic;">any</span> dimension.<br />
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So, instead we turn to the cross-product part. In the preceding section I repeatedly stressed that the imaginary part of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?A%5E*B" /> corresponds to <span style="font-style: italic;">one dimension</span> of a 3D cross product. However, <span style="font-style: italic;">which</span> single dimension of the cross product we choose is completely arbitrary: Just as with the calculation of area for the cross product, the 2D Cartesian plane we choose to map to the complex plane could just as easily be the <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?xy" />-, <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?zx" />- or the <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?yz" />-planes.<br />
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Recall, that to resolve this ambiguity in in cross-product land we chose to identify which plane we were talking about by a right-hand rule normal vector to the plane. However, here we're attempting to generalize complex numbers, not cross products per se. So, instead of assigning different normal vectors to each cross product term, let's assign a different complex number to each term. That is, <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?i%5E2=-1" /> <span style="font-style: italic;">and</span> <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?j%5E2=-1" />, but <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?i%5Cneq%20j" /> for example. Then, we assign <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?i" /> to the cross product of two vectors in the <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?yz" />-plane and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?j" /> to the cross product of two vectors in the <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?zx" />-plane.<br />
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The one question remaining, though, as we generalize our complex plane, is how many additional complex numbers do we need? Maybe, naively, we can try adding just one extra cross-product dimension. That is, <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?i" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?j" /> only. The problem, though, can be seen in cross-product land.<br />
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<span style="font-weight: bold;">Closure</span><br />
Remember, that a cross product resultant vector is a normal vector to an <span style="font-style: italic;">arbitrary</span> plane in 3D Cartesian space, and thus <span style="font-style: italic;">always</span> requires <span style="font-style: italic;">all three</span> unit vectors <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Chat%7Bx%7D" />, <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Chat%20y" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Chat%7Bz%7D" />. For example, the cross product <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Chat%7Bx%7D%5Ctimes%5Chat%20y" /> is <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Chat%20z" />. That is, in order to make sense of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Chat%20x" /> cross <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Chat%20y" /> which can exist in 2D, you must already have a third unit vector <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Chat%20z" />.<br />
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Physically, in Cartesian vector space, it means that you must be able to add any arbitrary 3D cross product resultant vectors and still get a 3D vector. In fact, if this wasn't true, there'd be no way to even write the 3D cross product in the first place since you need to project the arbitrary vectors to three (independent) 2D planes and then add the resulting normal vectors. You can't have just two cross-product parts and get a result that <span style="font-style: italic;">always</span> makes sense. This is the requirement of ``closure.''<br />
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The reason there's no problem in the 2D plane version is simply because there's only one possible normal vector, so we only look at the <span style="font-style: italic;">magnitude</span> of the cross product -- i.e. the amount of area -- and the sign. And that is just a scalar! In 2D land nothing is preventing you from adding the cross product to the dot product -- they're both scalars -- so you can write a two-element complex number <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?x+i%5C,y" /> combination with no trouble.<br />
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However, in 3D we can't simply add a vector to a scalar, and therefore we need all three parts of the cross product. So too, then, if we want a generalized complex number to have a dot-product part and a cross-product part that makes physical sense, we need <span style="font-style: italic;">three</span> complex numbers: <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?i" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?j" /> from above, <span style="font-style: italic;">plus</span> a <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?k%5E2=-1" />, <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?k%5Cneq%20i,j" /> corresponding to the cross product of the projection of vectors in the <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?xy" />-plane.<br />
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Thus, we now have a generalized complex number -- quaternion -- of the form<br />
<div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z=%5Cell+i%5C,x+j%5C,y+k%5C,z." /><br />
</div><div style="text-align: right;">(4)<br />
</div><br />
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<span style="font-weight: bold;">References</span><br />
[1] E. Lansey. The dot product and cross products [online]. April 2009. Available from:<a href="http://behindtheguesses.blogspot.com/2009/04/dot-and-cross-products.html"> http://behindtheguesses.blogspot.com/2009/04/dot-and-cross-products.html</a>.<br />
[2] Tristan Needham. <span style="font-style: italic;">Visual Complex Analysis</span>. Oxford University Press, Oxford, UK, 1997.<br />
[3] C. Doran and A.N. Lasenby. <span style="font-style: italic;">Geometric algebra for physicists</span>. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.<br />
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<hr /><br />
<sup>1</sup> Just Google it, it's not really worth retelling, in my opinion.<br />
<sup>2</sup> If I was stranded on an island forever but could bring only one math book, this would be it.<br />
<sup>3</sup> Just go out and get that book already! What are you waiting for?<br />
<sup>4</sup>Thanks, <a href="http://peeterjoot.wordpress.com/">Peeter</a>, for recommending the book ([3]) which highlighted this point.<br />
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<a name='more'></a><br />
<iframe height="910" src="http://docs.google.com/viewer?url=http%3A%2F%2Fs87572736.onlinehome.us%2Fbtg%2Fbtg07-quaternions1.pdf&embedded=true" width="700"></iframe>Eli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.com7tag:blogger.com,1999:blog-393324845011978943.post-35835375459884278092009-09-30T09:02:00.004-04:002009-09-30T09:16:55.969-04:00Nothing new<div>Unless I miraculously complete 3 problems sets, find the errors in the calculations I've been working on for 4 months and the bugs in the code based on those calculations, and mow the lawn by a reasonable hour today, there will likely not be a new Behind the Guesses post of substance this month.</div><div><br /></div><div>However, I am considering a new method of posting, using <a href="https://docs.google.com/viewer">Google Document Viewer</a>, rather than converting LaTeX into pictures, etc. This is the <a href="http://behindtheguesses.blogspot.com/2009/08/noncommuting-rotation-and-angular.html">last post</a>, Noncommuting Rotation and Angular Momentum Operators, using the new method. Would you prefer this, or the old way? Or do you just download the PDF, and it makes no difference?</div><br /><iframe src="http://docs.google.com/viewer?url=http%3A%2F%2Fs87572736.onlinehome.us%2Fbtg%2Fbtg06-noncommuting.pdf&embedded=true" width="700" height="910"></iframe><br /><div><br /><script type="text/javascript" charset="utf-8" src="http://static.polldaddy.com/p/2059107.js"></script><noscript><br /><a href="http://answers.polldaddy.com/poll/2059107/">Should I use Google Docs Viewer?</a><span style="font-size:9px;">(<a href="http://answers.polldaddy.com/">polling</a>)</span><br /></noscript><br /></div>Eli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.com0tag:blogger.com,1999:blog-393324845011978943.post-50842636233489486502009-08-31T16:49:00.020-04:002009-08-31T17:30:24.522-04:00Noncommuting Rotation and Angular Momentum Operators[<a href="http://s87572736.onlinehome.us/btg/btg06-noncommuting.pdf">Click here for a PDF of this post with nicer formatting</a>]<br /><span style="font-weight: bold;">The Setup</span><br />Avi Ziskind<sup>1</sup> asked me to cover non-commuting operators in quantum mechanics, specifically why angular momentum operators do not commute. He pointed out that Griffiths [1] gives an intuitive argument for understanding why position and momentum operators do not commute but does not present any rationale given for why the different components of angular momentum have the commutation relation<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5B%5Cmathcal%7BL%7D_i,%5Cmathcal%7BL%7D_j%5D=i%5Chbar%5Cepsilon_%7Bijk%7D%5Cmathcal%7BL%7D_k" align="middle" border="0" /><br /></div><div style="text-align: right;">(1)<br /></div>Additionally, Schwabl [2], for example, defines the angular momentum operator, presents the commutation relations, and at least attempts (I think) to show (in a post-facto way) why they should have such relations. Likewise, in a related (as we'll see) problem, Goldstein, et. al. [3] discuss the commutation relations of generators of rotation without any physical argument.<br /><br />However, both Sakurai [4] and Landau and Lifshitz [5], to some degree, present physical rationales for these relations. Landau and Lifshitz derive the notion of angular momentum in quantum theory quite nicely, and succinctly, but do not argue for why the commutation relations should hold. Sakurai develops a set of commutation relations independently of QM (as I will, shortly), but, I feel, bridges the gap to angular momentum rather poorly.<br /><br />This post assumes familiarity with the ``generator of transformation'' ideas in [<a href="http://behindtheguesses.blogspot.com/2009/06/schrodinger-equation-corrections.html">6</a>].<br /><br /><br /><span style="font-weight: bold;">The Generator of Rotation</span><br /><div style="text-align: center;"><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_D1sP-NndkqU/Spw5m0of5kI/AAAAAAAAP90/d3-ZZ7YdGeY/s1600-h/3D.gif"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 231px; height: 238px;" src="http://1.bp.blogspot.com/_D1sP-NndkqU/Spw5m0of5kI/AAAAAAAAP90/d3-ZZ7YdGeY/s400/3D.gif" alt="" id="BLOGGER_PHOTO_ID_5376235394299848258" border="0" /></a>(a) In 3D.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_D1sP-NndkqU/Spw5nF_sLKI/AAAAAAAAP98/l1uQEzTclRA/s1600-h/2D.gif"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 360px; height: 359px;" src="http://3.bp.blogspot.com/_D1sP-NndkqU/Spw5nF_sLKI/AAAAAAAAP98/l1uQEzTclRA/s400/2D.gif" alt="" id="BLOGGER_PHOTO_ID_5376235398960524450" border="0" /></a>(b) The projection of (a) onto the <img src="http://www.codecogs.com/gif.latex?xy" align="middle" border="0" />-plane.</div>Figure 1: The rotation of a vector around the <img src="http://www.codecogs.com/gif.latex?z" align="middle" border="0" />-axis.<br /><br />In a previous post I covered the notion of ``generators of transformations,''[6] and claimed, as an example, that the ``generator of rotation'' is the angular momentum. Actually, I was getting ahead of myself there, and the statement in that context was not entirely correct. As I did not derive this result in that post, I will now, and will hopefully clear things up.<br /><br />Suppose we have a function <img src="http://www.codecogs.com/gif.latex?f%28x,y,z%29" align="middle" border="0" /> and we want to rotate it in space around the <img src="http://www.codecogs.com/gif.latex?z" align="middle" border="0" /> axis through some angle <img src="http://www.codecogs.com/gif.latex?%5CDelta%5Ctheta" align="middle" border="0" /> to <img src="http://www.codecogs.com/gif.latex?f%28x%5Ccos%28%5CDelta%5Ctheta%29-y%5Csin%28%5CDelta%5Ctheta%29,x%5Csin%28%5CDelta%5Ctheta%29+y%5Ccos%28%5CDelta%5Ctheta%29,z%29" align="middle" border="0" />. To do this, we'll find an ``angle rotation'' operator <img src="http://www.codecogs.com/gif.latex?R%5Ez_%7B%5CDelta%5Ctheta%7D" align="middle" border="0" />, which, when applied to <img src="http://www.codecogs.com/gif.latex?f%28x,y,z%29" align="middle" border="0" />, gives <img src="http://www.codecogs.com/gif.latex?f%28x%5Ccos%28%5CDelta%5Ctheta%29-y%5Csin%28%5CDelta%5Ctheta%29,x%5Csin%28%5CDelta%5Ctheta%29+y%5Ccos%28%5CDelta%5Ctheta%29,z%29" align="middle" border="0" />. That is,<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?f%28x%5Ccos%28%5CDelta%5Ctheta%29-y%5Csin%28%5CDelta%5Ctheta%29,x%5Csin%28%5CDelta%5Ctheta%29+y%5Ccos%28%5CDelta%5Ctheta%29,z%29=R%5Ez_%7B%5CDelta%5Ctheta%7D%20f%28x,y,z%29." align="middle" border="0" /><br /></div><div style="text-align: right;">(2)<br /></div>The shift in coordinates can be derived from regular vector analysis, see Fig. 1 and Ref. [7], applied inside the arguments of the function.<br /><br />Now the tricky part -- the Taylor expansion. Unlike the last time where the translated function had a simple <img src="http://www.codecogs.com/gif.latex?%28x+%5CDelta%20x%29" align="middle" border="0" /> argument, here we have <img src="http://www.codecogs.com/gif.latex?%5CDelta%5Ctheta" align="middle" border="0" /> inside sines and cosines. Since I'm really too lazy to do this expansion by hand I had Mathematica do it for me (click to see full-size):<br /><div style="text-align: right;"><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_D1sP-NndkqU/Spw6WgDwvTI/AAAAAAAAP-E/6OxRxnfnFjw/s1600-h/series.gif"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 400px; height: 54px;" src="http://1.bp.blogspot.com/_D1sP-NndkqU/Spw6WgDwvTI/AAAAAAAAP-E/6OxRxnfnFjw/s400/series.gif" alt="" id="BLOGGER_PHOTO_ID_5376236213410774322" border="0" /></a>(3)</div>In Mathematica's notation, <img src="http://www.codecogs.com/gif.latex?f" align="middle" border="0" /> raised to those parenthetical powers denotes partial derivatives. Say, <img src="http://www.codecogs.com/gif.latex?f%5E%7B%280,1,0%29%7D%28x,y,z%29" align="middle" border="0" /> means <img src="http://www.codecogs.com/gif.latex?%5Cfrac%7B%5Cpartial%20f%28x,y,z%29%7D%7B%5Cpartial%20y%7D" align="middle" border="0" />, for example. This expression is a bit of a mess, but we are not completely lost. From our discussion in the beginning of [6], we know that at least one similar operator takes an exponential form. So, we'll guess that here, as well, our operator will take an exponential form. We just need to process the mess of (3) to find that hidden exponential.<br /><br />The first two terms in the series give us hope. They can be written as<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cleft%5B1+%5CDelta%5Ctheta%5Cleft%28x%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D-y%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D%5Cright%29%5Cright%5Df%28x,y,z%29," align="middle" border="0" /><br /></div><div style="text-align: right;">(4)<br /></div>which are, indeed, what we would expect to see at the beginning of an exponential expansion, where <img src="http://www.codecogs.com/gif.latex?%5Cleft%28x%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D-y%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D%5Cright%29" align="middle" border="0" /> is the generator. Now we check that this keeps up for higher powers.<br /><br />Continuing with the quadratic term, let's see if we can write <img src="http://www.codecogs.com/gif.latex?x%5E2%20f%5E%7B%280,2,0%29%7D%28x,y,z%29+y%5E2%20f%5E%7B%282,0,0%29%7D%28x,y,z%29-x%20f%5E%7B%281,0,0%29%7D%28x,y,z%29-2%20x%20y%20f%5E%7B%281,1,0%29%7D%28x,y,z%29-y%20f%5E%7B%280,1,0%29%7D%28x,y,z%29" align="middle" border="0" /> as <img src="http://www.codecogs.com/gif.latex?%5Cleft%28x%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D-y%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D%5Cright%29%5E2f%28x,y,z%29" align="middle" border="0" /> which would be the next term in an exponential. We check:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cleft%5Bx%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D-y%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D%5Cright%5D%5E2f%28x,y,z%29%20=%20%5Cleft%5Bx%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D-y%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D%5Cright%5D%5Cleft%5Bx%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D-y%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D%5Cright%5Df%28x,y,z%29" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=%5Cleft%5Bx%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7Dx%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D-x%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7Dy%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D-y%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7Dx%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D+y%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7Dy%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D%5Cright%5Df%28x,y,z%29" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=%5Cleft%5Bx%5E2%5Cfrac%7B%5Cpartial%5E2%20%7D%7B%5Cpartial%20y%5E2%7D+y%5E2%5Cfrac%7B%5Cpartial%5E2%20%7D%7B%5Cpartial%20x%5E2%7D-x%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7Dy%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D-y%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7Dx%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D%5Cright%5Df%28x,y,z%29" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=%5Cleft%5Bx%5E2%5Cfrac%7B%5Cpartial%5E2%20%7D%7B%5Cpartial%20y%5E2%7D+y%5E2%5Cfrac%7B%5Cpartial%5E2%20%7D%7B%5Cpartial%20x%5E2%7D%20-x%5Cleft%28y%5Cfrac%7B%5Cpartial%5E2%20%7D%7B%5Cpartial%20x%5Cpartial%20y%7D+%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D%5Cright%29%20-y%5Cleft%28x%5Cfrac%7B%5Cpartial%5E2%20%7D%7B%5Cpartial%20x%5Cpartial%20y%7D+%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D%5Cright%29%20%5Cright%5Df%28x,y,z%29" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=%5Cleft%5Bx%5E2%5Cfrac%7B%5Cpartial%5E2%20%7D%7B%5Cpartial%20y%5E2%7D+y%5E2%5Cfrac%7B%5Cpartial%5E2%20%7D%7B%5Cpartial%20x%5E2%7D%20-x%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D%20-2xy%5Cfrac%7B%5Cpartial%5E2%20%7D%7B%5Cpartial%20x%5Cpartial%20y%7D-%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D%20%5Cright%5Df%28x,y,z%29" align="middle" border="0" /><br /></div><div style="text-align: right;">(5)<br /></div>which does match the mess for the <img src="http://www.codecogs.com/gif.latex?%28%5CDelta%5Ctheta%29%5E2" align="middle" border="0" /> term in the expansion (3). You can verify on your own that this pattern continues in the higher powers.<br /><br />Thus we conclude that<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?R%5Ez_%7B%5CDelta%5Ctheta%7D=%5Cexp%5Cleft%5B%5CDelta%5Ctheta%5Cleft%28%20x%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D-y%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D%5Cright%29%5Cright%5D," align="middle" border="0" /><br /></div><div style="text-align: right;">(6)<br /></div>where we now identify<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?L_z%5Cequiv%5Cleft%28%20x%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D-y%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D%5Cright%29" align="middle" border="0" /><br /></div><div style="text-align: right;">(7)<br /></div>as the generator of the rotation.<br /><br />This generator is the <img src="http://www.codecogs.com/gif.latex?z" align="middle" border="0" />-component of the cross product<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cvec%7Br%7D%5Ctimes%5Cnabla," align="middle" border="0" /><br /></div>where <img src="http://www.codecogs.com/gif.latex?%5Cvec%7Br%7D=%5Chat%7B%5Cimath%7Dx+%5Chat%7B%5Cjmath%7Dy+%5Chat%7Bk%7Dz" align="middle" border="0" /> and <img src="http://www.codecogs.com/gif.latex?%5Cnabla=%5Chat%7B%5Cimath%7D%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D+%5Chat%7B%5Cjmath%7D%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D+%5Chat%7Bk%7D%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20z%7D" align="middle" border="0" />. Thus, we can simplify<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?L_z=%28%5Cvec%7Br%7D%5Ctimes%5Cnabla%29_z." align="middle" border="0" /><br /></div><div style="text-align: right;">(8)<br /></div>If we carry through these same calculations for rotations around the <img src="http://www.codecogs.com/gif.latex?x" align="middle" border="0" /> or <img src="http://www.codecogs.com/gif.latex?y" align="middle" border="0" /> axes (try it yourself!) we get similar generators<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?L_x=%28%5Cvec%7Br%7D%5Ctimes%5Cnabla%29_x," align="middle" border="0" /><br /></div><div style="text-align: right;">(9)<br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?L_y&=%28%5Cvec%7Br%7D%5Ctimes%5Cnabla%29_y" align="middle" border="0" /><br /></div><div style="text-align: right;">(10)<br /></div>This allows us to write the rotation operator for a rotation around an arbitrary axis <img src="http://www.codecogs.com/gif.latex?%5Chat%7Bn%7D" align="middle" border="0" />, as<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?R_%7B%5CDelta%5Ctheta%7D=e%5E%7B%5CDelta%5Ctheta%5C,%5Chat%7Bn%7D%5Ccdot%20%5Cvec%7BL%7D%7D," align="middle" border="0" /><br /></div><div style="text-align: right;">(11)<br /></div>where <img src="http://www.codecogs.com/gif.latex?%5Chat%7Bn%7D%5Ccdot%5Cvec%7BL%7D" align="middle" border="0" /> for<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cvec%7BL%7D%5Cequiv%20%5Cvec%7Br%7D%5Ctimes%5Cnabla" align="middle" border="0" /><br /></div><div style="text-align: right;">(12)<br /></div>is the generator of the transformation.<br /><br /><br /><span style="font-weight: bold;">Commutators in general</span><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_D1sP-NndkqU/Spw9nOiG1ZI/AAAAAAAAP_c/YUTWezmEm9g/s1600-h/fig2.GIF"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 400px; height: 281px;" src="http://1.bp.blogspot.com/_D1sP-NndkqU/Spw9nOiG1ZI/AAAAAAAAP_c/YUTWezmEm9g/s400/fig2.GIF" alt="" id="BLOGGER_PHOTO_ID_5376239799298872722" border="0" /></a>In general, rotations do not commute. That is, rotating an object first around the <img src="http://www.codecogs.com/gif.latex?x" align="middle" border="0" />-axis and then around the <img src="http://www.codecogs.com/gif.latex?y" align="middle" border="0" />-axis will give a different result than rotating in the opposite order. You can convince yourself of this by the ultimate hand-waving argument<sup>2</sup> -- twist your hand around different axes in different orders. Or see Fig. 2.<br /><br />We'd like to find a way to quantify the difference between applying the rotations in different order, but, for the sake of generality, we'll discuss this for any two arbitrary operators <img src="http://www.codecogs.com/gif.latex?A" align="middle" border="0" /> and <img src="http://www.codecogs.com/gif.latex?B" align="middle" border="0" />. The most natural way to quantify a difference is to look at, well, the difference. That is, if these operators act on a vector <img src="http://www.codecogs.com/gif.latex?%5Cvec%7Bv%7D" align="middle" border="0" />, we'd like to know what<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?BA%5Cvec%7Bv%7D-AB%5Cvec%7Bv%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(13)<br /></div>is. This difference (for linear operators) does not depend on the particular vector <img src="http://www.codecogs.com/gif.latex?%5Cvec%7Bv%7D" align="middle" border="0" />, so we'll define the commutator of two operators as<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cleft%5BB,A%5Cright%5D%5Cequiv%20BA-AB." align="middle" border="0" /><br /></div><div style="text-align: right;">(14)<br /></div>Thus, a <span style="font-weight: bold;">commutator of two operators is another operator which enacts this difference</span>. If the order of operator application does not affect the end result the commutator is 0, and the operators are said to ``commute.''<br /><br />In quantum mechanics, the issue of non-commuting operators is closely tied to the problem of measurement and the uncertainty principle. For example, if I have a state <img src="http://www.codecogs.com/gif.latex?%5Cpsi" align="middle" border="0" /> and I want to measure the position I apply the position operator <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BX%7D" align="middle" border="0" />. Likewise, if I want to measure the momentum I apply the momentum operator <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BP%7D" align="middle" border="0" />. However, in quantum mechanics, the order of taking these measurements affects the results, such that <img src="http://www.codecogs.com/gif.latex?%5Cleft%5B%5Cmathcal%7BX%7D,%5Cmathcal%7BP%7D%5Cright%5D%5Cpsi=i%5Chbar%5Cpsi" align="middle" border="0" />, for example. However, the applicability of commutators is not relegated only to quantum mechanics.<br /><br /><br /><span style="font-weight: bold;">Commutators for rotation</span><br />This brings us back to our original question of the commutator of rotations. Because any two rotations through arbitrary angles, done in opposite orders give drastically different results depending on the angles, we'll consider rotations through small angles <img src="http://www.codecogs.com/gif.latex?%5Cdelta%5Ctheta" align="middle" border="0" />, such that we can approximate (11) by the first two terms in the expansion:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?R%5Capprox%201+%5Cdelta%5Ctheta%5C,%5Chat%7Bn%7D%5Ccdot%20%5Cvec%7BL%7D." align="middle" border="0" /><br /></div><div style="text-align: right;">(15)<br /></div>This simpler expression makes calculating the commutator much simpler. For rotations around <img src="http://www.codecogs.com/gif.latex?%5Chat%20n_1" align="middle" border="0" /> and <img src="http://www.codecogs.com/gif.latex?%5Chat%20n_2" align="middle" border="0" />, the commutator <img src="http://www.codecogs.com/gif.latex?%5Cleft%5BR_2,R_1%5Cright%5D" align="middle" border="0" /> depends only on the commutator of the generators <img src="http://www.codecogs.com/gif.latex?%5Cleft%5B%5Chat%7Bn%7D_2%20%5Ccdot%20L_2,%5Chat%7Bn%7D_1%20%5Ccdot%20L_1%5Cright%5D" align="middle" border="0" />.<sup>3</sup> This commutator is the generator of the transformation for ``the difference between the order of the rotations.'' That is<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?C%5Capprox%201+%5Cdelta%5Cphi%5C,%5Cleft%5B%5Chat%7Bn%7D_2%20%5Ccdot%20L_2,%5Chat%7Bn%7D_1%20%5Ccdot%20L_1%5Cright%5D," align="middle" border="0" /><br /></div><div style="text-align: right;">(16)<br /></div>where <img src="http://www.codecogs.com/gif.latex?%5Cdelta%5Cphi" align="middle" border="0" /> is the parameter for this transformation. Then, just as any rotation can then be built up from repeated applications of the generator (as in that exponential), the commutators for larger angles can be built up from repeated applications of the commutators of the generators.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_D1sP-NndkqU/Spw-nSrhYeI/AAAAAAAAP_k/YNnSri1T9Gs/s1600-h/coms.gif"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 400px; height: 326px;" src="http://1.bp.blogspot.com/_D1sP-NndkqU/Spw-nSrhYeI/AAAAAAAAP_k/YNnSri1T9Gs/s400/coms.gif" alt="" id="BLOGGER_PHOTO_ID_5376240899923730914" border="0" /></a>Figure 3: Graphical commutator of <img src="http://www.codecogs.com/gif.latex?%5Cleft%5BL_y,L_x%5Cright%5D" align="middle" border="0" />. Blue vector is application of either <img src="http://www.codecogs.com/gif.latex?R_x" align="middle" border="0" /> or <img src="http://www.codecogs.com/gif.latex?R_y" align="middle" border="0" />. Red is further application of <img src="http://www.codecogs.com/gif.latex?R_y" align="middle" border="0" /> to <img src="http://www.codecogs.com/gif.latex?R_x" align="middle" border="0" /> and green is further application of <img src="http://www.codecogs.com/gif.latex?R_x" align="middle" border="0" /> to <img src="http://www.codecogs.com/gif.latex?R_y" align="middle" border="0" />. Brown is difference between the two.<br /><br />For ease of illustration, we'll consider small rotations around the <img src="http://www.codecogs.com/gif.latex?x" align="middle" border="0" />- and <img src="http://www.codecogs.com/gif.latex?y" align="middle" border="0" />-axes (i.e. <img src="http://www.codecogs.com/gif.latex?%5Chat%7Bn%7D_1=%5Chat%7B%5Cimath%7D" align="middle" border="0" /> and <img src="http://www.codecogs.com/gif.latex?%5Chat%7Bn%7D_2=%5Chat%7B%5Cjmath%7D" align="middle" border="0" />). There are two ways to find the commutator <img src="http://www.codecogs.com/gif.latex?%5Cleft%5BL_y,L_x%5Cright%5D" align="middle" border="0" />. One way is by brute force calculation which I encourage you to try on your own (use the expressions for <img src="http://www.codecogs.com/gif.latex?L_x" align="middle" border="0" /> (9) and <img src="http://www.codecogs.com/gif.latex?L_y" align="middle" border="0" /> (10)). However, I prefer showing it graphically, see Fig. 3. Starting with a vector in the <img src="http://www.codecogs.com/gif.latex?xy" align="middle" border="0" />-plane, we apply a small rotation around <img src="http://www.codecogs.com/gif.latex?x" align="middle" border="0" />. This directs the vector upwards (blue in the picture). Then we apply another small rotation around <img src="http://www.codecogs.com/gif.latex?y" align="middle" border="0" />, which directs the vector along the red line.<br /><br />If we start with the same vector, and apply a small rotation around <img src="http://www.codecogs.com/gif.latex?y" align="middle" border="0" />, the vector follows the blue line again. However, when we then rotate around <img src="http://www.codecogs.com/gif.latex?x" align="middle" border="0" />, the vector veers off in the opposite direction at the same rate. The difference between the red and green vectors, as well as that difference added to the initial vector is shown in brown. The picture illustrates that<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cleft%5BL_y,L_x%5Cright%5D=L_z," align="middle" border="0" /><br /></div><div style="text-align: right;">(17)<br /></div>i.e. the generator of rotation around the <img src="http://www.codecogs.com/gif.latex?z" align="middle" border="0" />-axis. Similar relationships<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5BL_i,L_j%5D=%5Cepsilon_%7Bijk%7DL_k" align="middle" border="0" /><br /></div><div style="text-align: right;">(18)<br /></div>hold for other permutations of <img src="http://www.codecogs.com/gif.latex?xyz" align="middle" border="0" />.<br /><br /><br /><span style="font-weight: bold;">Angular momentum</span><br />Looking back at the expression for the generator of rotations (12), we see that we can re-write this in terms of the momentum operator<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cvec%7Bp%7D=-i%5Chbar%5Cnabla" align="middle" border="0" /><br /></div><div style="text-align: right;">(19)<br /></div>in quantum mechanics:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cvec%7BL%7D=%5Cfrac%7Bi%7D%7B%5Chbar%7D%5Cvec%7Br%7D%5Ctimes%5Cvec%7Bp%7D" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=%5Cfrac%7Bi%7D%7B%5Chbar%7D%5Cmathcal%7BL%7D," align="middle" border="0" /><br /></div><div style="text-align: right;">(20)<br /></div>where we call <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BL%7D" align="middle" border="0" /> the ``quantum mechanical angular momentum'' operator.<sup>4</sup> Flipping this around to solve for <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BL%7D" align="middle" border="0" /> in terms of <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BL%7D" align="middle" border="0" />:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BL%7D=-i%5Chbar%5Cvec%7BL%7D." align="middle" border="0" /><br /></div><div style="text-align: right;">(21)<br /></div>In other words, the quantum mechanical angular momentum is the same (up to a constant) as the generator of rotations. Thus, the reason that quantum angular momentum has commutation relations (1) is due to the fact that it's simply a generator of rotation masquerading as a quantum mechanical operator.<br /><br /><br /><span style="font-weight: bold;">References</span><br />[1] D.J. Griffths. <span style="font-style: italic;">Introduction to Electrodynamics</span>. Pearson Prentice Hall, 3rd edition, 1999.<br />[2] F. Schwabl. <span style="font-style: italic;">Quantum Mechanics</span>. Springer, 3rd edition, 2005.<br />[3] H. Goldstein, C. Poole, and J. Safko. <span style="font-style: italic;">Classical Mechanics</span>. Cambridge University Press, San Francisco, CA, 3rd edition, 2002.<br />[4] J.J. Sakurai. <span style="font-style: italic;">Modern Quantum Mechanics</span>. Addison-Wesley, San Francisco, CA, revised edition, 1993.<br />[5] L.D. Landau and E.M. Lifshitz. <span style="font-style: italic;">Quantum Mechanics</span>. Butterworth-Heinemann, Oxford, UK, 3rd edition, 1977.<br />[6] E. Lansey. The Schrodinger Equation -- Corrections [online]. June 2009. Available from: http://behindtheguesses.blogspot.com/2009/06/schrodinger-equation-corrections.html.<br />[7] D.C. Lay. <span style="font-style: italic;">Linear Algebra and Its Applications</span>. Addison-Wesley, Reading, MA, 3rd edition, 2003.<br />[8] C.T.J. Dodson and T. Poston. <span style="font-style: italic;">Tensor Geometry: The Geometric Viewpoint and its Uses</span>. Springer, 2nd edition, 1997.<br /><br /><hr /><br /><sup>1</sup> Everyone congratulate him on the birth of a son!<br /><sup>2</sup> Borrowing a joke from Dodson and Poston, [8]<br /><sup>3</sup> If this isn't obvious, work it out for yourself. Hint: The identity operator 1 commutes with everything.<br /><sup>4</sup> There are better arguments (see [5]) using symmetry for why <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BL%7D" align="middle" border="0" /> should <span style="font-style: italic;">actually</span> be the angular momentum, not just <span style="font-style: italic;">called</span> it, as I've argued, but they require much more talking. And this post is long enough already.Eli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.com3tag:blogger.com,1999:blog-393324845011978943.post-27588270888623185562009-07-29T13:32:00.022-04:002009-07-29T15:49:54.205-04:00Transverse Electric and Magnetic Fields in a Waveguide[<a href="http://s87572736.onlinehome.us/btg/btg05-waveguide.pdf">Click here for a PDF of this post with nicer formatting</a>]<br /><span style="font-weight: bold;">The Setup</span><br /><div style="text-align: center;"><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_D1sP-NndkqU/SnCH4EM4kQI/AAAAAAAAOKM/abbQFLxbwJg/s1600-h/guide.gif"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 360px; height: 205px;" src="http://3.bp.blogspot.com/_D1sP-NndkqU/SnCH4EM4kQI/AAAAAAAAOKM/abbQFLxbwJg/s400/guide.gif" alt="" id="BLOGGER_PHOTO_ID_5363936553468465410" border="0" /></a><br />Figure 1: An example of a section cylindrical waveguide with embedded coordinate axes.</div><br />A conducting waveguide is a metal tube -- think pipe or air conditioning duct, for example -- through which electromagnetic waves can propagate. If you want to know what real-life waveguides look like, just do a quick <a href="http://images.google.com/images?q=waveguide">internet image search</a>. We'll assume the length of the tube is oriented along the <img src="http://www.codecogs.com/gif.latex?z" border="0" align="middle" />-direction, see Fig 1. There is no loss of generality in doing this, since we can always choose a coordinate system as we like. So really, we're picking a coordinate system such that the <img src="http://www.codecogs.com/gif.latex?z" border="0" align="middle" />-axis points along the tube.<br /><br />Now, we can decompose the electric field <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BE%7D" border="0" align="middle" /> and magnetic (inductance) field <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" border="0" align="middle" /> vectors into two parts each. One part points along the <img src="http://www.codecogs.com/gif.latex?z" border="0" align="middle" /> (normal) direction while the other is pointing somewhere in the <img src="http://www.codecogs.com/gif.latex?xy" border="0" align="middle" /> (transverse) plane. Explicitly:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cvec%7BE%7D=%20E_z%5Chat%7Bz%7D%20+%5Cvec%7BE%7D_t" border="0" align="middle" /><br /></div><div style="text-align: right;">(1a)<br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D=%20B_z%5Chat%7Bz%7D%20+%5Cvec%7BB%7D_t" border="0" align="middle" /><br /></div><div style="text-align: right;">(1b)<br /></div><br />In the first([1], Eq. (8.24)) and third[2], Eq. (8.26)) editions of <span style="font-style: italic;">Classical Electrodynamics</span>, J.D. Jackson gives the transverse fields in terms of the <img src="http://www.codecogs.com/gif.latex?z" border="0" align="middle" />-components of the fields. (I have no idea why he left the complete expression out of the second edition.) In the third edition, for example, he assumes plane wave propagation in the positive <img src="http://www.codecogs.com/gif.latex?z" border="0" align="middle" /> direction -- that is an <img src="http://www.codecogs.com/gif.latex?e%5E%7B+ikz%7D" border="0" align="middle" /> dependance -- and simply states, without any real explanation:<br /><blockquote>the transverse fields are<br /><img src="http://www.codecogs.com/gif.latex?%5Cvec%7BE%7D_t=%5Cfrac%7Bi%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-k%5E2%5Cright%29%7D%5Cleft%5Bk%5Cnabla_tE_z-%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%20%5Ctimes%5Cnabla_t%20B_z%5Cright%5D" border="0" align="middle" /><br /><img src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D_t=%5Cfrac%7Bi%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-k%5E2%5Cright%29%7D%5Cleft%5Bk%5Cnabla_tB_z+%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%20%5Ctimes%5Cnabla_t%20E_z%5Cright%5D" border="0" align="middle" /></blockquote>where I've converted his new choice of MKSA units back into the clearer CGS units. However, back in the first edition he does not insist on the assumption of positive <img src="http://www.codecogs.com/gif.latex?z" border="0" align="middle" /> propagation. Moreover, he does not just state the fields; he suggests a method for getting them -- namely, manipulation of the curl equations in Maxwell's equations. However, in that edition, he does not expand the curl equations in light of the separation of the fields into transverse and parallel components as he does in the second and third editions.<br /><br />Because of all this confusion, I'm going to derive the cavity modes fully, starting from Maxwell's equations, once and for all. This derivation is based on a combination of all three editions of Jackson's book. This is a tedious, although not completely trivial exercise. Brace yourselves for quite a bit of algebra.<br /><br /><br /><span style="font-weight: bold;">Maxwell's Equations - The Curls</span><br />Here we'll deal with the two curl equations in Maxwell's equations:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cnabla%5Ctimes%5Cvec%7BE%7D=-%5Cfrac%7B1%7D%7Bc%7D%5Cfrac%7B%5Cpartial%5Cvec%7BB%7D%7D%7B%5Cpartial%20t%7D" border="0" align="middle" /><br /></div><div style="text-align: right;">(2a)<br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cnabla%5Ctimes%5Cvec%7BH%7D=%5Cfrac%7B1%7D%7Bc%7D%5Cfrac%7B%5Cpartial%5Cvec%7BD%7D%7D%7B%5Cpartial%20t%7D+%5Cfrac%7B4%5Cpi%7D%7Bc%7D%5Cvec%7BJ%7D" border="0" align="middle" /><br /></div><div style="text-align: right;">(2b)<br /></div>where <img src="http://www.codecogs.com/gif.latex?H" border="0" align="middle" /> is the magnetic field and <img src="http://www.codecogs.com/gif.latex?D" border="0" align="middle" /> is the electric displacement field. We will assume the inside of the waveguide has uniform permittivity and permeability, so <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BD%7D=%5Cepsilon%5Cvec%7BE%7D" border="0" align="middle" /> and <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D=%5Cmu%20%5Cvec%7BH%7D" border="0" align="middle" />. Also, we'll assume the absence of any currents, so <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BJ%7D=0" border="0" align="middle" /> and we'll drop it from here on. Additionally, we'll assume the same sinusoidal time dependance <img src="http://www.codecogs.com/gif.latex?e%5E%7B-i%5Comega%20t%7D" border="0" align="middle" /> for both the fields. Thus, the time derivatives ``bring down'' a factor of <img src="http://www.codecogs.com/gif.latex?-i%5Comega" border="0" align="middle" />.<br /><br />Furthermore, since we're splitting up <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BE%7D" border="0" align="middle" /> and <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" border="0" align="middle" /> into normal and transverse parts, we'll do the same with the gradient operator <img src="http://www.codecogs.com/gif.latex?%5Cnabla" border="0" align="middle" />:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cnabla=%5Cunderbrace%7B%5Chat%7Bx%7D%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20x%7D+%5Chat%7By%7D%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20y%7D%7D_%7B%5Cequiv%5Cnabla_t%7D+%5Chat%7Bz%7D%20%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20z%7D=%5Cnabla_t+%5Chat%7Bz%7D%20%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20z%7D" border="0" align="middle" /><br /></div><br />Because curl equations are annoying, and because we're ultimately looking for an equation for the transverse fields, I'm going to try and get rid of the <img src="http://www.codecogs.com/gif.latex?%5Cnabla%5Ctimes" border="0" align="middle" />'s. The symmetry of form in (2) means that we'll only need to do these calculations once; I will use <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" border="0" align="middle" /> in place of either <img src="http://www.codecogs.com/gif.latex?%5Cvec%20E" border="0" align="middle" /> or <img src="http://www.codecogs.com/gif.latex?%5Cvec%20B" border="0" align="middle" />.<br />First, we'll expand <img src="http://www.codecogs.com/gif.latex?%5Cnabla%5Ctimes%5Cvec%7BA%7D" border="0" align="middle" />:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cnabla%5Ctimes%5Cvec%7BA%7D%20=%20%5Cleft%5B%5Cnabla_t+%5Chat%7Bz%7D%20%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20z%7D%5Cright%5D%5Ctime%5Cleft%5B%5Cvec%7BA%7D_t+%5Chat%7Bz%7D%20A_z%5Cright%5D" border="0" align="middle" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=%5Cnabla_t%5Ctimes%5Cat+%5Cnabla_t%5Ctimes%5Chat%7Bz%7D%20A_z+%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20z%7D%5Cleft%5B%5Chat%7Bz%7D%20%5Ctimes%5Cat%5Cright%5D+%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20z%7D%5Cleft%5B%5Ccancel%7B%5Chat%7Bz%7D%20%5Ctimes%5Chat%7Bz%7D%20%7D%5E%7B%5C,0%7DA_z%5Cright%5D" border="0" align="middle" /><br /></div><div style="text-align: right;">(3)<br /></div>We've killed one term through this expansion. However, the leftmost cross product term <img src="http://www.codecogs.com/gif.latex?%5Cnabla_t%5Ctimes%5Cvec%7BA%7D_t" border="0" align="middle" /> gives a quantity with only a <img src="http://www.codecogs.com/gif.latex?z" border="0" align="middle" /> component. The righthand side of these equations also have a <img src="http://www.codecogs.com/gif.latex?%5Chat%7Bz%7D" border="0" align="middle" /> term. We can get rid of both by multiplying the entire equation(s) by <img src="http://www.codecogs.com/gif.latex?%5Chat%7Bz%7D%20%5Ctimes" border="0" align="middle" />:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Chat%7Bz%7D%20%5Ctimes%5Cnabla%5Ctimes%5Cvec%7BA%7D%20=%20%5Ccancel%7B%5Chat%7Bz%7D%5Ctimes%5Cnabla_t%5Ctimes%5Cvec%7BA%7D_t%7D%5E%7B%5C,0%7D+%5Chat%7Bz%7D%5Ctimes%5Cnabla_t%5Ctimes%5Chat%7Bz%7D%20A_z+%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20z%7D%5Cleft%5B%5Cunderbrace%7B%5Chat%7Bz%7D%20%5Ctimes%5Chat%7Bz%7D%5Ctimes%5Cvec%7BA%7D_t%7D_%7B-%5Cvec%7BA%7D_t%7D%5Cright%5D" border="0" align="middle" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=%20%5Chat%7Bz%7D%20%5Ctimes%5Cnabla_t%5Ctimes%5Chat%7Bz%7D%20%20A_z-%5Cfrac%7B%5Cpartial%20%5Cvec%7BA%7D_t%7D%7B%5Cpartial%20z%7D" border="0" align="middle" /><br /></div><div style="text-align: right;">(4)<br /></div><br /><div style="text-align: center;"><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_D1sP-NndkqU/SnCX44ZdQPI/AAAAAAAAOKc/2nXPhItEJ1g/s1600-h/zcross.gif"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 352px; height: 252px;" src="http://3.bp.blogspot.com/_D1sP-NndkqU/SnCX44ZdQPI/AAAAAAAAOKc/2nXPhItEJ1g/s400/zcross.gif" alt="" id="BLOGGER_PHOTO_ID_5363954159665889522" border="0" /></a>Figure 2: Vectors <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D_t" border="0" align="middle" />, <img src="http://www.codecogs.com/gif.latex?%5Chat%7Bz%7D%20%5Ctimes%5Cvec%7BA%7D_t" border="0" align="middle" /> and <img src="http://www.codecogs.com/gif.latex?%5Chat%7Bz%7D%20%5Ctimes%5Chat%7Bz%7D%20%5Ctimes%5Cvec%7BA%7D_t" border="0" align="middle" />.<br /></div><br />For why <img src="http://www.codecogs.com/gif.latex?%5Chat%7Bz%7D%20%5Ctimes%5Chat%7Bz%7D%20%5Ctimes%5Cvec%7BA%7D_t=-%5Cvec%7BA%7D_t" border="0" align="middle" /> see Fig. 2. Also, we note that<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Chat%7Bz%7D%20%5Ctimes%5Cnabla_t%5Ctimes%5Chat%7Bz%7D%20=%5Chat%7Bz%7D%20%5Ctimes%28-%5Chat%7Bz%7D%5Ctimes%5Cnabla_t%29=%5Cnabla_t" border="0" align="middle" /><br /></div><div style="text-align: right;">(5)<br /></div>for the same reason. We could have used the vector multiplication identity<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?a%5Ctimes%20%28b%5Ctimes%20c%29=%28a%5Ccdot%20c%29b-%28a%5Ccdot%20b%29c" border="0" align="middle" /><br /></div>to simplify both of these expressions, or expanded <img src="http://www.codecogs.com/gif.latex?%5Cnabla_t" border="0" align="middle" /> and <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D_t" border="0" align="middle" /> and carried through even more algebra, but I think the picture is clearer.<br /><br />Thus,<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Chat%7Bz%7D%20%5Ctimes%5Cnabla%5Ctimes%5Cvec%7BA%7D%20=%5Cnabla_t%20A_z-%5Cfrac%7B%5Cpartial%5Cvec%7BA%7D_t%7D%7B%5Cpartial%20z%7D," border="0" align="middle" /><br /></div><div style="text-align: right;">(6)<br /></div>and we can write (2) as<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cnabla_t%20E_z-%5Cfrac%7B%5Cpartial%20%5Cvec%7BE%7D_t%7D%7B%5Cpartial%20z%7D&=%5Cfrac%7Bi%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%20%5Ctimes%5Cvec%7BB%7D_t" border="0" align="middle" /><br /></div><div style="text-align: right;">(7a)<br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cnabla_t%20B_z-%5Cfrac%7B%5Cpartial%20%5Cvec%7BB%7D_t%7D%7B%5Cpartial%20z%7D&=-i%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%20%5Ctimes%5Cvec%7BE%7D_t" border="0" align="middle" /><br /></div><div style="text-align: right;">(7b)<br /></div><br />At this point, it's time to introduce the explicit <img src="http://www.codecogs.com/gif.latex?z" border="0" align="middle" /> dependence and process the <img src="http://www.codecogs.com/gif.latex?z" border="0" align="middle" /> derivatives.<br /><br /><br /><span style="font-weight: bold;">Some </span><img style="font-weight: bold;" src="http://www.codecogs.com/gif.latex?%5Cpm" border="0" align="middle" /><span style="font-weight: bold;"> and </span><img style="font-weight: bold;" src="http://www.codecogs.com/gif.latex?%5Cmp" border="0" align="middle" /><span style="font-weight: bold;"> notes</span><br />Unlike Jackson, who works with the assumption of upward propagating waves -- i.e. an <img src="http://www.codecogs.com/gif.latex?e%5E%7B+ikz%7D" border="0" align="middle" /> dependence -- we'll work with an assumed <img src="http://www.codecogs.com/gif.latex?e%5E%7B%5Cpm%20i%20k%20z%7D" border="0" align="middle" /> dependance, thus allowing both upward and downward propagating waves. Thus, the <img src="http://www.codecogs.com/gif.latex?z" border="0" align="middle" /> derivatives ``bring down'' a factor of <img src="http://www.codecogs.com/gif.latex?%5Cpm%20ik" border="0" align="middle" />. Whenever we have <img src="http://www.codecogs.com/gif.latex?%5Cpm" border="0" align="middle" /> or <img src="http://www.codecogs.com/gif.latex?%5Cmp" border="0" align="middle" /> the upper symbol is the sign for upward propagating waves, the lower symbol is for downward propagating. Because we'll be mucking about with these plus-minus guys in some algebra, I want to get a few issues out of the way.<br /><br />The first thing to keep in mind about these plus-minus operators is that an equation like<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?A%20=%20%5Cpm%20C%20+%20D" border="0" align="middle" /><br /></div><div style="text-align: right;">(8)<br /></div><span style="font-weight: bold;">is shorthand for two different equations:</span><br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?A%20=%20+C+D" border="0" align="middle" /><br /></div><div style="text-align: right;">(9a)<br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?A%20=%20-C+D" border="0" align="middle" /><br /></div><div style="text-align: right;">(9b)<br /></div><br />So, there are essentially two ways to approach these things. One way is to carefully trace at the outset what happens to <img src="http://www.codecogs.com/gif.latex?%5Cpm" border="0" align="middle" /> or <img src="http://www.codecogs.com/gif.latex?%5Cmp" border="0" align="middle" /> under various arithmetic operations like addition, multiplication, etc. This has the benefit of being more concise -- you only need to write each equation once -- but is a lot easier to make errors and hides the double-equation nature of the symbol. I'll admit, though, that when I'm writing a paper I'm generally inclined to take this path.<br /><br />However, for the purposes of this blog post, I'll explicitly carry out the calculations in parallel equations. (This really looks <span style="font-style: italic; font-weight: bold;">much</span> better <a href="http://s87572736.onlinehome.us/btg/btg05-waveguide.pdf">in the PDF</a>. If anyone has any suggestions for improving the web version, please, let me know!) The left-hand column corresponds to <img src="http://www.codecogs.com/gif.latex?+" border="0" align="middle" />, the right-hand column to <img src="http://www.codecogs.com/gif.latex?-" border="0" align="middle" />. At the end I will also show what the results looks like in the shorthand notation and I encourage you to work out the rules on your own. Perhaps in another post I'll address the shorthand notation in detail.<br /><br /><br /><span style="font-weight: bold;">Some more algebra</span><br />Now, it's time for some more algebra.<sup>1</sup> Taking the <img src="http://www.codecogs.com/gif.latex?z" border="0" align="middle" /> derivative in (7) gives:<br /><div style="text-align: center;"><table border="0"><br /><tbody><tr><br /><td> <img src="http://www.codecogs.com/gif.latex?%5Cnabla_t%20E_z-i%20k%5Cvec%7BE%7D_t=i%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%5Ctimes%5Cvec%7BB%7D_t" border="0" align="middle" /> </td><br /><td><br /></td><td><br /></td><td> | </td><td><br /></td><td><br /></td><br /><td> <img src="http://www.codecogs.com/gif.latex?%5Cnabla_t%20E_z+%20i%20k%5Cvec%7BE%7D_t=i%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%5Ctimes%5Cvec%7BB%7D_t" border="0" align="middle" /> </td><br /></tr><br /></tbody></table><br /><div style="text-align: right;">(10)<br /></div></div>and<br /><div style="text-align: center;"><table border="0"><br /><tbody><tr><br /><td> <img src="http://www.codecogs.com/gif.latex?%5Cnabla_t%20B_z-i%20k%5Cvec%7BB%7D_t=%20-i%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%5Ctimes%5Cvec%7BE%7D_t" border="0" align="middle" /> </td><br /><td><br /></td><td><br /></td><td> | </td><td><br /></td><td><br /></td><br /><td> <img src="http://www.codecogs.com/gif.latex?%5Cnabla_t%20B_z+i%20k%5Cvec%7BB%7D_t=%20-i%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%5Ctimes%5Cvec%7BE%7D_t" border="0" align="middle" /> </td><br /></tr><br /></tbody></table><br /><div style="text-align: right;">(11)<br /></div></div><br />Solving (10) for <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BE%7D_t" border="0" align="middle" /> gives<br /><div style="text-align: center;"><table border="0"><br /><tbody><tr><br /><td> <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BE%7D_t=-%20%5Cfrac%7Bi%7D%7Bk%7D%5Cnabla_tE_z-%5Cfrac%7B%5Comega%7D%7Bck%7D%5Chat%7Bz%7D%5Ctimes%5Cvec%7BB%7D_t" border="0" align="middle" /> </td><br /><td><br /></td><td><br /></td><td> | </td><td><br /></td><td><br /></td><br /><td> <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BE%7D_t=+%20%5Cfrac%7Bi%7D%7Bk%7D%5Cnabla_tE_z+%5Cfrac%7B%5Comega%7D%7Bck%7D%5Chat%7Bz%7D%5Ctimes%5Cvec%7BB%7D_t" border="0" align="middle" /> </td><br /></tr><br /></tbody></table><br /><div style="text-align: right;">(12)<br /></div></div>Substituting this into (11) and simplifying:<br /><div style="text-align: center;"><table border="0"><br /><tbody><tr><br /><td> <img src="http://www.codecogs.com/gif.latex?%5Cnabla_t%20B_z-i%20k%5Cvec%7BB%7D_t=-i%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%20%5Ctimes%5Cleft%5B-%5Cfrac%7Bi%7D%7Bk%7D%5Cnabla_tE_z-%5Cfrac%7B%5Comega%7D%7Bck%7D%5Chat%7Bz%7D%5Ctimes%5Cvec%7BB%7D_t%5Cright%5D" border="0" align="middle" /> </td><br /><td><br /></td><td><br /></td><td> | </td><td><br /></td><td><br /></td><br /><td> <img src="http://www.codecogs.com/gif.latex?%5Cnabla_t%20B_z+i%20k%5Cvec%7BB%7D_t=-i%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%5Ctimes%5Cleft%5B%5Cfrac%7Bi%7D%7Bk%7D%5Cnabla_tE_z+%5Cfrac%7B%5Comega%7D%7Bck%7D%5Chat%7Bz%7D%5Ctimes%5Cvec%7BB%7D_t%20%5Cright%5D" border="0" align="middle" /> </td><br /></tr><br /><tr><br /><td> <img src="http://www.codecogs.com/gif.latex?=-%5Cfrac%7B%5Cmu%5Cepsilon%7D%7Bk%7D%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%5Ctimes%5Cnabla_tE_z%20+i%5Cfrac%7B%5Cmu%5Cepsilon%7D%7Bk%7D%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D%5Cunderbrace%7B%5Chat%7Bz%7D%5Ctimes%5Chat%7Bz%7D%20%5Ctimes%5Cvec%7BB%7D_t%7D_%7B-%5Cvec%7BB%7D_t%7D" border="0" align="middle" /> </td><br /><td><br /></td><td><br /></td><td> | </td><td><br /></td><td><br /></td><br /><td> <img src="http://www.codecogs.com/gif.latex?=%5Cfrac%7B%5Cmu%5Cepsilon%7D%7Bk%7D%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%5Ctimes%5Cnabla_tE_z%20-i%5Cfrac%7B%5Cmu%5Cepsilon%7D%7Bk%7D%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D%5Cunderbrace%7B%5Chat%7Bz%7D%5Ctimes%5Chat%7Bz%7D%20%5Ctimes%5Cvec%7BB%7D_t%7D_%7B-%5Cvec%7BB%7D_t%7D" border="0" align="middle" /> </td><br /></tr><br /><tr><br /><td> <img src="http://www.codecogs.com/gif.latex?%5Cnabla_tB_z=-%5Cfrac%7Bi%7D%7Bk%7D%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-%20k%5E2%5Cright%29%5Cvec%7BB%7D_t%20-%20%5Cfrac%7B%5Cmu%5Cepsilon%7D%7Bk%7D%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%5Ctimes%5Cnabla_tE_z" border="0" align="middle" /> </td><br /><td><br /></td><td><br /></td><td> | </td><td><br /></td><td><br /></td><br /><td> <img src="http://www.codecogs.com/gif.latex?%5Cnabla_tB_z=+%5Cfrac%7Bi%7D%7Bk%7D%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-k%5E2%5Cright%29%5Cvec%7BB%7D_t+%5Cfrac%7B%5Cmu%5Cepsilon%7D%7Bk%7D%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%20%5Ctimes%5Cnabla_tE_z" border="0" align="middle" /> </td><br /></tr><br /></tbody></table><br /><div style="text-align: right;">(13)<br /></div></div><br />Solving this for <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D_t" border="0" align="middle" /> gives:<br /><div style="text-align: center;"><table border="0"><br /><tbody><tr><br /><td> <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D_t%20=%20%5Cfrac%7B1%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-%20k%5E2%5Cright%29%7D%5Cleft%5B%20ik%5Cnabla_t%20B_z%20+i%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%20%5Ctimes%5Cnabla_tE_z%5Cright%5D" border="0" align="middle" /> </td><br /><td><br /></td><td><br /></td><td> | </td><td><br /></td><td><br /></td><br /><td> <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D_t%20=%20%5Cfrac%7B1%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-%20k%5E2%5Cright%29%7D%5Cleft%5B%20-ik%5Cnabla_t%20B_z%20+i%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%20%5Ctimes%5Cnabla_tE_z%5Cright%5D" border="0" align="middle" /> </td><br /></tr><br /></tbody></table><br /><div style="text-align: right;">(14)<br /></div></div>Or, in <img src="http://www.codecogs.com/gif.latex?%5Cpm" border="0" align="middle" /> form:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cboxed%7B%5Cvec%7BB%7D_t%20=%20%5Cfrac%7B1%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-%20k%5E2%20%5Cright%29%7D%5Cleft%5B%20%5Cpm%20ik%5Cnabla_t%20B_z%20+i%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%20%5Ctimes%5Cnabla_tE_z%5Cright%5D%7D" border="0" align="middle" /><br /></div><div style="text-align: right;">(15)<br /></div><br />In the first edition, Jackson converts the <img src="http://www.codecogs.com/gif.latex?%5Cpm%20ik%20B_z" border="0" align="middle" /> back into <img src="http://www.codecogs.com/gif.latex?%5Cfrac%7B%5Cpartial%20B_z%7D%7B%5Cpartial%20z%7D" border="0" align="middle" /> to get rid of the <img src="http://www.codecogs.com/gif.latex?%5Cpm" border="0" align="middle" />, but I feel this confuses things, as this expression only holds for a plane wave in the <img src="http://www.codecogs.com/gif.latex?z" border="0" align="middle" /> direction. In any case, we now substitute this expression for <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D_t" border="0" align="middle" /> back into (12) and simplify:<br /><div style="text-align: center;"><table border="0"><br /><tbody><tr><br /><td> <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BE%7D_t=-%20%5Cfrac%7Bi%7D%7Bk%7D%5Cnabla_tE_z-%5Cfrac%7B%5Comega%7D%7Bck%7D%5Chat%7Bz%7D%5Ctimes%5Cfrac%7B1%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-%20k%5E2%5Cright%29%7D%5Cleft%5B%20ik%5Cnabla_t%20B_z%20+i%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%20%5Ctimes%5Cnabla_tE_z%5Cright%5D" border="0" align="middle" /> </td><br /><td><br /></td><td><br /></td><td> | </td><td><br /></td><td><br /></td><br /><td> <img src="http://www.codecogs.com/gif.latex?%5Cvec%7BE%7D_t=%20%5Cfrac%7Bi%7D%7Bk%7D%5Cnabla_tE_z+%5Cfrac%7B%5Comega%7D%7Bck%7D%5Chat%7Bz%7D%20%5Ctimes%5Cfrac%7B1%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-%20k%5E2%20%5Cright%29%7D%5Cleft%5B-ik%5Cnabla_t%20B_z%20+%20i%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%5Ctimes%5Cnabla_tE_z%5Cright%5D" border="0" align="middle" /> </td><br /></tr><br /><tr><br /><td> <img src="http://www.codecogs.com/gif.latex?=-%20%5Cfrac%7Bi%7D%7Bk%7D%5Cnabla_tE_z-%20%5Cleft%5Bi%5Cfrac%7B%5Comega%7D%7Bc%7D%5Cfrac%7B1%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-k%5E2%20%5Cright%29%7D%5Chat%7Bz%7D%20%5Ctimes%5Cnabla_t%20B_z%20+%20%5Cfrac%7Bi%7D%7Bk%7D%5Cfrac%7B%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-k%5E2%20%5Cright%29%7D%5Cunderbrace%7B%5Chat%7Bz%7D%20%5Ctimes%5Chat%7Bz%7D%5Ctimes%5Cnabla_tE_z%7D_%7B-%5Cnabla_tE_z%7D%5Cright%5D" border="0" align="middle" /> </td><br /><td><br /></td><td><br /></td><td> | </td><td><br /></td><td><br /></td><br /><td> <img src="http://www.codecogs.com/gif.latex?=%20%5Cfrac%7Bi%7D%7Bk%7D%5Cnabla_tE_z%20+%20%5Cleft%5B-i%5Cfrac%7B%5Comega%7D%7Bc%7D%5Cfrac%7B1%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-k%5E2%20%5Cright%29%7D%5Chat%7Bz%7D%20%5Ctimes%5Cnabla_t%20B_z%20+%20%5Cfrac%7Bi%7D%7Bk%7D%5Cfrac%7B%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-k%5E2%5Cright%29%7D%20%5Cunderbrace%7B%5Chat%7Bz%7D%20%5Ctimes%5Chat%7Bz%7D%5Ctimes%5Cnabla_tE_z%7D_%7B-%5Cnabla_tE_z%7D%5Cright%5D" border="0" align="middle" /> </td><br /></tr><br /><tr><br /><td> <img src="http://www.codecogs.com/gif.latex?=ik%5Cfrac%7B1%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-k%5E2%20%5Cright%29%7D%5Cnabla_tE_z-i%5Cfrac%7B%5Comega%7D%7Bc%7D%5Cfrac%7B1%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-k%5E2%20%5Cright%29%7D%5Chat%7Bz%7D%20%5Ctimes%20%5Cnabla_t%20B_z" border="0" align="middle" /> </td><br /><td><br /></td><td><br /></td><td> | </td><td><br /></td><td><br /></td><br /><td> <img src="http://www.codecogs.com/gif.latex?=-ik%5Cfrac%7B1%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-k%5E2%20%5Cright%29%7D%5Cnabla_tE_z-i%5Cfrac%7B%5Comega%7D%7Bc%7D%5Cfrac%7B1%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-k%5E2%20%5Cright%29%7D%5Chat%7Bz%7D%20%5Ctimes%20%5Cnabla_t%20B_z" border="0" align="middle" /> </td><br /></tr><br /><tr><br /><td> <img src="http://www.codecogs.com/gif.latex?=%5Cfrac%7B1%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-%20k%5E2%5Cright%29%7D%5Cleft%5Bik%5Cnabla_tE_z%20-i%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%20%5Ctimes%5Cnabla_t%20B_z%5Cright%5D" border="0" align="middle" /> </td><br /><td><br /></td><td><br /></td><td> | </td><td><br /></td><td><br /></td><br /><td> <img src="http://www.codecogs.com/gif.latex?=%5Cfrac%7B1%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-%20k%5E2%5Cright%29%7D%5Cleft%5B-ik%5Cnabla_tE_z%20-i%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%20%5Ctimes%5Cnabla_t%20B_z%5Cright%5D" border="0" align="middle" /> </td><br /></tr><br /></tbody></table><br /><div style="text-align: right;">(16)</div></div>Or, in <img src="http://www.codecogs.com/gif.latex?%5Cpm" border="0" align="middle" /> form:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cboxed%7B%5Cvec%7BE%7D_t=%5Cfrac%7B1%7D%7B%5Cleft%28%5Cmu%5Cepsilon%5Cfrac%7B%5Comega%5E2%7D%7Bc%5E2%7D-k%5E2%5Cright%29%7D%5Cleft%5B%5Cpm%20ik%5Cnabla_t%20E_z-i%5Cfrac%7B%5Comega%7D%7Bc%7D%5Chat%7Bz%7D%20%5Ctimes%5Cnabla_t%20B_z%5Cright%5D%7D" border="0" align="middle" /><br /></div><div style="text-align: right;">(17)<br /></div><br />So, we've <span style="font-style: italic;">finally</span> achieved Jackson's result, allowing for both upward and downward propagating waves.<br /><br /><br /><span style="font-weight: bold;">References</span><br />[1] J.D. Jackson. <span style="font-style: italic;">Classical Electrodynamics</span>. John Wiley & Sons, Inc., 1st edition, 1966.<br />[2] J.D. Jackson. <span style="font-style: italic;">Classical Electrodynamics</span>. John Wiley & Sons, Inc., 3rd edition, 1998.<br /><br /><br /><hr /><br /><sup>1</sup> In case you were wondering why Jackson left out the whole calculation...Eli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.com5tag:blogger.com,1999:blog-393324845011978943.post-71509216311617207632009-06-30T10:04:00.011-04:002009-07-01T23:15:55.662-04:00Derivative and Integral of the Heaviside Step Function[<a href="http://s87572736.onlinehome.us/btg/btg04-intstep.pdf">Click here for a PDF of this post with nicer formatting</a>]<br /><span style="font-weight: bold;">The Setup</span><br /><div style="text-align: center;"><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_D1sP-NndkqU/SkohSrSOgDI/AAAAAAAALok/lGxUys5_oMk/s1600-h/hb.gif"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 360px; height: 249px;" src="http://1.bp.blogspot.com/_D1sP-NndkqU/SkohSrSOgDI/AAAAAAAALok/lGxUys5_oMk/s400/hb.gif" alt="" id="BLOGGER_PHOTO_ID_5353127711823790130" border="0" /></a>(a) Large horizontal scale<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_D1sP-NndkqU/SkohScViW0I/AAAAAAAALoc/rpUGS0gcjoQ/s1600-h/hs.gif"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 360px; height: 237px;" src="http://4.bp.blogspot.com/_D1sP-NndkqU/SkohScViW0I/AAAAAAAALoc/rpUGS0gcjoQ/s400/hs.gif" alt="" id="BLOGGER_PHOTO_ID_5353127707811142466" border="0" /></a>(b) ``Zoomed in''<br /><div style="text-align: left;">Figure 1: The Heaviside step function. Note how it doesn't matter how close we get to </div></div><img src="http://www.codecogs.com/gif.latex?x=0" align="middle" border="0" /> the function looks exactly the same.<br /><br />The Heaviside step function <img src="http://www.codecogs.com/gif.latex?H%28x%29" align="middle" border="0" />, sometimes called the Heaviside theta function, appears in many places in physics, see [1] for a brief discussion. Simply put, it is a function whose value is zero for <img src="http://www.codecogs.com/gif.latex?x%3C0" align="middle" border="0" /> and one for <img src="http://www.codecogs.com/gif.latex?x%3E0" align="middle" border="0" />. Explicitly,<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?H%28x%29=%5Cbegin%7Bcases%7D0&%20x%3C0,%5C%5C1&%20x%3E0%5Cend%7Bcases%7D." align="middle" border="0" /><br /><div style="text-align: right;">(1)<br /></div></div>We won't worry about precisely what its value is at zero for now, since it won't effect our discussion, see [2] for a lengthier discussion. Fig. 1 plots <img src="http://www.codecogs.com/gif.latex?H%28x%29" align="middle" border="0" />. The key point is that crossing zero flips the function from 0 to 1.<br /><br /><br /><span style="font-weight: bold;">Derivative -- The Dirac Delta Function</span><br /><div style="text-align: center;"><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_D1sP-NndkqU/SkohS3mqjWI/AAAAAAAALos/Nuc9lWl1zco/s1600-h/delta.gif"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 360px; height: 268px;" src="http://2.bp.blogspot.com/_D1sP-NndkqU/SkohS3mqjWI/AAAAAAAALos/Nuc9lWl1zco/s400/delta.gif" alt="" id="BLOGGER_PHOTO_ID_5353127715130740066" border="0" /></a>(a) Dirac delta function<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_D1sP-NndkqU/SkohS9wF00I/AAAAAAAALo0/DT6AxQuYx3o/s1600-h/ramp.gif"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer; width: 360px; height: 253px;" src="http://1.bp.blogspot.com/_D1sP-NndkqU/SkohS9wF00I/AAAAAAAALo0/DT6AxQuYx3o/s400/ramp.gif" alt="" id="BLOGGER_PHOTO_ID_5353127716780888898" border="0" /></a>(b) Ramp function<br /><div style="text-align: left;">Figure 2: The derivative (a), and the integral (b) of the Heaviside step function.<br /></div> </div>Say we wanted to take the derivative of <img src="http://www.codecogs.com/gif.latex?H" align="middle" border="0" />. Recall that a derivative is the slope of the curve at at point. One way of formulating this is<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cfrac%7BdH%7D%7Bdx%7D=%5Clim_%7B%5CDelta%20x%20%5Crightarrow%200%7D%20%5Cfrac%7B%5CDelta%20H%7D%7B%5CDelta%20x%7D." align="middle" border="0" /><br /></div><div style="text-align: right;">(2)<br /></div>Now, for any points <img src="http://www.codecogs.com/gif.latex?x%3C0" align="middle" border="0" /> or <img src="http://www.codecogs.com/gif.latex?x%3E0" align="middle" border="0" />, graphically, the derivative is very clear: <img src="http://www.codecogs.com/gif.latex?H" align="middle" border="0" /> is a flat line in those regions, and the slope of a flat line is zero. In terms of (2), <img src="http://www.codecogs.com/gif.latex?H" align="middle" border="0" /> does not change, so <img src="http://www.codecogs.com/gif.latex?%5CDelta%20H=0" align="middle" border="0" /> and <img src="http://www.codecogs.com/gif.latex?dH/dx=0" align="middle" border="0" />. But if we pick two points, equally spaced on opposite sides of <img src="http://www.codecogs.com/gif.latex?x=0" align="middle" border="0" />, say <img src="http://www.codecogs.com/gif.latex?x_-=-a/2" align="middle" border="0" /> and <img src="http://www.codecogs.com/gif.latex?x_+=a/2" align="middle" border="0" />, then <img src="http://www.codecogs.com/gif.latex?%5CDelta%20H=1" align="middle" border="0" /> and <img src="http://www.codecogs.com/gif.latex?%5CDelta%20x=a" align="middle" border="0" />. It doesn't matter how small we make <img src="http://www.codecogs.com/gif.latex?a" align="middle" border="0" />, <img src="http://www.codecogs.com/gif.latex?%5CDelta%20H" align="middle" border="0" /> stays the same. Thus, the fraction in (2) is<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cfrac%7BdH%7D%7Bdx%7D=%5Clim_%7Ba%20%5Crightarrow%200%7D%20%5Cfrac%7B1%7D%7Ba%7D" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=%5Cinfty." align="middle" border="0" /><br /></div><div style="text-align: right;">(3)<br /></div>Graphically, again, this is very clear: <img src="http://www.codecogs.com/gif.latex?H" align="middle" border="0" /> <span style="font-style: italic;">jumps</span> from 0 to 1 at zero, so it's slope is essentially vertical, i.e. infinite. So basically, we have<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cdelta%28x%29%5Cequiv%5Cfrac%7BdH%7D%7Bdx%7D=%5Cbegin%7Bcases%7D0&%20x%3C0%5C%5C%5Cinfty&%20x=0%5C%5C0&%20x%3E0%5Cend%7Bcases%7D." align="middle" border="0" /><br /></div><div style="text-align: right;">(4)<br /></div>This function is, loosely speaking, a ``Dirac Delta'' function, usually written as <img src="http://www.codecogs.com/gif.latex?%5Cdelta%28x%29" align="middle" border="0" />, which has seemingly endless uses in physics.<br /><br />We'll note a few properties of the delta function that we can derive from (4). First, integrating it from <img src="http://www.codecogs.com/gif.latex?-%5Cinfty" align="middle" border="0" /> to any <img src="http://www.codecogs.com/gif.latex?x_-%3C0" align="middle" border="0" />:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cint_%7B-%5Cinfty%7D%5E%7Bx_-%7D%5Cdelta%28x%29dx=%5Cint_%7B-%5Cinfty%7D%5E%7Bx_-%7D%5Cleft%28%5Cfrac%7BdH%7D%7Bdx%7D%5Cright%29dx" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=H%28x_-%29-H%28-%5Cinfty%29" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=0" align="middle" border="0" /><br /></div><div style="text-align: right;">(5)<br /></div>since <img src="http://www.codecogs.com/gif.latex?H%28x_-%29=H%28-%5Cinfty%29=0" align="middle" border="0" />. On the other hand, integrating the delta function to any point greater than <img src="http://www.codecogs.com/gif.latex?x=0" align="middle" border="0" />:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cint_%7B-%5Cinfty%7D%5E%7Bx_+%7D%5Cdelta%28x%29dx=%5Cint_%7B-%5Cinfty%7D%5E%7Bx_+%7D%5Cleft%28%5Cfrac%7BdH%7D%7Bdx%7D%5Cright%29dx" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=H%28x_+%29-H%28-%5Cinfty%29" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=1" align="middle" border="0" /><br /></div><div style="text-align: right;">(6)<br /></div>since <img src="http://www.codecogs.com/gif.latex?H%28x_+%29=1" align="middle" border="0" />.<br /><br />At this point, I should point out that although the delta function blows up to infinity at <img src="http://www.codecogs.com/gif.latex?x=0" align="middle" border="0" />, it still has a finite integral. An easy way of seeing how this is possible is shown in Fig. 2(a). If the width of the box is <img src="http://www.codecogs.com/gif.latex?1/a" align="middle" border="0" /> and the height is <img src="http://www.codecogs.com/gif.latex?a" align="middle" border="0" />, the area of the box (i.e. its integral) is <img src="http://www.codecogs.com/gif.latex?1" align="middle" border="0" />, no matter how large <img src="http://www.codecogs.com/gif.latex?a" align="middle" border="0" /> is. By letting <img src="http://www.codecogs.com/gif.latex?a" align="middle" border="0" /> go to infinity we have a box with infinite height, yet, when integrated, has finite area.<br /><br /><br /><span style="font-weight: bold;">Integral -- The Ramp Function</span><br />Now that we know about the derivative, it's time to evaluate the integral. I have two methods of doing this. The most straightforward way, which I first saw from Prof. T.H. Boyer, is to integrate <img src="http://www.codecogs.com/gif.latex?H" align="middle" border="0" /> piece by piece. The integral of a function is the area under the curve,<sup>1</sup> and when <img src="http://www.codecogs.com/gif.latex?x%3C0" align="middle" border="0" /> there is no area, so the integral from <img src="http://www.codecogs.com/gif.latex?-%5Cinfty" align="middle" border="0" /> to any point less than zero is zero. On the right side, the integral to a point <img src="http://www.codecogs.com/gif.latex?x" align="middle" border="0" /> is the area of a rectangle of height 1 and length <img src="http://www.codecogs.com/gif.latex?x" align="middle" border="0" />, see Fig. 1(a). So, we have<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cint_%7B-%5Cinfty%7D%5Ex%20H%5C,dx=%5Cbegin%7Bcases%7D0&%20x%3C0,%5C%5Cx&%20x%3E0%5Cend%7Bcases%7D." align="middle" border="0" /><br /></div><div style="text-align: right;">(7)<br /></div>We'll call this function a ``ramp function,'' <img src="http://www.codecogs.com/gif.latex?R%28x%29" align="middle" border="0" />. We can actually make use of the definition of <img src="http://www.codecogs.com/gif.latex?H" align="middle" border="0" /> and simplify the notation:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?R%28x%29%5Cequiv%5Cint%20H%5C,dx=xH%28x%29" align="middle" border="0" /><br /></div><div style="text-align: right;">(8)<br /></div>since <img src="http://www.codecogs.com/gif.latex?0%5Ctimes%20x=0" align="middle" border="0" /> and <img src="http://www.codecogs.com/gif.latex?1%5Ctimes%20x=x" align="middle" border="0" />. See Fig. 2(b) for a graph -- and the reason for calling this a ``ramp'' function.<br /><br />But I have another way of doing this which makes use of a trick that's often used by physicists: <span style="font-weight: bold;">We can always add zero for free</span>, since <img src="http://www.codecogs.com/gif.latex?%7Banything%7D+0=%7Banything%7D" align="middle" border="0" />. Often we do this by adding and subtracting the same thing,<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?A=%28A+B%29-B," align="middle" border="0" /><br /></div><div style="text-align: right;">(9)<br /></div>for example. But we can use the delta function (4) to add zero in the form<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?0=x%5C,%5Cdelta%28x%29." align="middle" border="0" /><br /></div><div style="text-align: right;">(10)<br /></div>Since <img src="http://www.codecogs.com/gif.latex?%5Cdelta%28x%29" align="middle" border="0" /> is zero for <img src="http://www.codecogs.com/gif.latex?x%5Cneq%200" align="middle" border="0" />, the <img src="http://www.codecogs.com/gif.latex?x" align="middle" border="0" /> part doesn't do anything in those regions and this expression is zero. And, although <img src="http://www.codecogs.com/gif.latex?%5Cdelta%28x%29=%5Cinfty" align="middle" border="0" /> at <img src="http://www.codecogs.com/gif.latex?x=0" align="middle" border="0" />, <img src="http://www.codecogs.com/gif.latex?x=0" align="middle" border="0" /> at <img src="http://www.codecogs.com/gif.latex?x=0" align="middle" border="0" />, so the expression is still zero.<br /><br />So we'll add this on to <img src="http://www.codecogs.com/gif.latex?H" align="middle" border="0" />:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?H=H+0" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=H+x%5C,%5Cdelta%28x%29" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=H+x%5Cfrac%7BdH%7D%7Bdx%7D%20%5Cquad%20%5Cquad%20%5Cquad%20%5Ctext%7Bby%20%284%29%7D" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=%5Cfrac%7Bdx%7D%7Bdx%7DH+x%5Cfrac%7BdH%7D%7Bdx%7D" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5BxH%28x%29%5Cright%5D," align="middle" border="0" /><br /></div><div style="text-align: right;">(11)<br /></div>where the last step follows from the ``product rule'' for differentiation. At this point, to take the integral of a full differential is trivial, and we get (8).<br /><br /><br /><span style="font-weight: bold;">References</span><br />[1] M. Springer. Sunday function [online]. February 2009. Available from: <a href="http://scienceblogs.com/builtonfacts/2009/02/sunday_function_22.php">http://scienceblogs.com/builtonfacts/2009/02/sunday_function_22.php</a> [cited 30 June 2009].<br />[2] E.W. Weisstein. Heaviside step function [online]. Available from: <a href="http://mathworld.wolfram.com/HeavisideStepFunction.html">http://mathworld.wolfram.com/HeavisideStepFunction.html</a> [cited 30 June 2009].<br /><br /><hr /><br /><sup>1</sup> To be completely precise, it's the (signed) area between the curve and the line <img src="http://www.codecogs.com/gif.latex?x=0" align="middle" border="0" />.Eli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.com13tag:blogger.com,1999:blog-393324845011978943.post-57693335208157734592009-06-04T09:43:00.010-04:002009-06-04T14:43:37.939-04:00The Schrödinger Equation - Corrections[<a href="http://s87572736.onlinehome.us/btg/btg03a-schroed.pdf">Click here for a PDF of this post with nicer formatting</a>]<br />In my <a href="http://behindtheguesses.blogspot.com/2009/05/schrodinger-equation.html">last post</a>, I claimed<br /><blockquote>Additionally, we can extend from here that any quantum operator <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BG%7D" align="middle" border="0" /> is written in terms of its classical counterpart <img src="http://www.codecogs.com/gif.latex?G" align="middle" border="0" /> by <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BG%7D=-i%20G/%5Chbar." align="middle" border="0" /></blockquote>Peeter Joot correctly <a href="http://behindtheguesses.blogspot.com/2009/05/schrodinger-equation.html?showComment=1243377144823#c3781014951742869278">pointed out</a> that this result does not follow from the argument involving the Hamiltonian. While it is true that<br /><blockquote>any arbitrary unitary transformation, <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BU%7D" align="middle" border="0" />, can be written as <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BU%7D=e%5E%7B-i%20%5Cfrac%7B%5Calpha%7D%7B%5Chbar%7D%20G%7D," align="middle" border="0" /><br />where <img src="http://www.codecogs.com/gif.latex?G" align="middle" border="0" /> is an Hermitian operator,</blockquote>the relationship between a classical <img src="http://www.codecogs.com/gif.latex?G" align="middle" border="0" /> and its quantum counterpart is not as straightforward as I claimed. In reality, we can only relate the classical Poisson brackets to the quantum mechanical commutators, and we must work from there. Perhaps I will discuss this further in a later post.<br /><br />In any case, though, the derivation of the Schrödinger equation only makes use of the relationship between the classical and quantum mechanical Hamiltonians, so the remainder of the derivation still holds. I am leaving the original post up as reference, but the corrected, restructured version (with some additional, although slight, notation changes) is below.<br /><br /><br /><span style="font-weight: bold;">A brief walk through classical mechanics</span><br />Say we have a function of <img src="http://www.codecogs.com/gif.latex?f%28x%29" align="middle" border="0" /> and we want to translate it in space to a point <img src="http://www.codecogs.com/gif.latex?%28x+%5CDelta%20x%29" align="middle" border="0" />, where <img src="http://www.codecogs.com/gif.latex?%5CDelta%20x" align="middle" border="0" /> need not be small. To do this, we'll find a ``space translation'' operator <img src="http://www.codecogs.com/gif.latex?S_%7B%5CDelta%20x%7D" align="middle" border="0" /> which, when applied to <img src="http://www.codecogs.com/gif.latex?f%28x%29" align="middle" border="0" />, gives <img src="http://www.codecogs.com/gif.latex?f%28x+%5CDelta%20x%29" align="middle" border="0" />. That is,<br /><div style="text-align: center;"></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?f%28x+%5CDelta%20x%29=S_%7B%5CDelta%20x%7D%20f%28x%29" align="middle" border="0" /><br /></div><div style="text-align: right;">(1)<br /></div>We'll expand <img src="http://www.codecogs.com/gif.latex?f%28x+%5CDelta%20x%29" align="middle" border="0" /> in a Taylor series:<br /><div style="text-align: center;"></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?f%28x+%5CDelta%20x%29=f%28x%29+%5CDelta%20x%20%5C,%5Cfrac%7Bdf%28x%29%7D%7Bdx%7D+%5Cfrac%7B%28%5CDelta%20x%29%5E2%7D%7B2%21%7D%5Cfrac%7Bd%5E2f%28x%29%7D%7Bdx%5E2%7D+%5Cldots" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=%5Cleft%5B1+%5CDelta%20x%5C,%5Cfrac%7Bd%7D%7Bdx%7D+%5Cfrac%7B%28%5CDelta%20x%29%5E2%7D%7B2%21%7D%5Cfrac%7Bd%5E2%7D%7Bdx%5E2%7D+%5Cldots%5Cright%5Df%28x%29" align="middle" border="0" /><br /></div><div style="text-align: right;">(2)<br /></div>which can be simplified using the series expansion of the exponential<sup>1</sup> to<br /><div style="text-align: center;"></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?e%5E%7B%5Cleft%5B%5CDelta%20x%5C,%5Cfrac%7Bd%7D%7Bdx%7D%5Cright%5D%7Df%28x%29" align="middle" border="0" /><br /></div><div style="text-align: right;">(3)<br /></div>from which we can conclude that<br /><div style="text-align: center;"></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?S_%7B%5CDelta%20x%7D=e%5E%7B%5Cleft%5B%5CDelta%20x%5C,%5Cfrac%7Bd%7D%7Bdx%7D%5Cright%5D%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(4)<br /></div>If you do a similar thing with rotations around the <img src="http://www.codecogs.com/gif.latex?z" align="middle" border="0" />-axis, you'll find that the rotation operator is<br /><div style="text-align: center;"></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?R_%7B%5CDelta%20%5Ctheta%7D=e%5E%7B%5CDelta%5Ctheta%5C,%20L_z%7D," align="middle" border="0" /><br /></div><div style="text-align: right;">(5)<br /></div>where <img src="http://www.codecogs.com/gif.latex?L_z" align="middle" border="0" /> is the <img src="http://www.codecogs.com/gif.latex?z" align="middle" border="0" />-component of the angular momentum.<br /><br />Comparing (4) and (5), we see that both have an exponential with a parameter (distance or angle) multiplied by something (<img src="http://www.codecogs.com/gif.latex?%5Cfrac%7Bd%7D%7Bdx%7D" align="middle" border="0" /> or <img src="http://www.codecogs.com/gif.latex?L" align="middle" border="0" />). We'll call the something the ``generator of the transformation.'' So, the generator of space translation is <img src="http://www.codecogs.com/gif.latex?%5Cfrac%7Bd%7D%7Bdx%7D" align="middle" border="0" /> and the generator of rotation is <img src="http://www.codecogs.com/gif.latex?L" align="middle" border="0" />. So, we'll write an arbitrary transformation operator <img src="http://www.codecogs.com/gif.latex?O" align="middle" border="0" /> through a parameter <img src="http://www.codecogs.com/gif.latex?%5CDelta%5Calpha" align="middle" border="0" /><br /><div style="text-align: center;"></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?O_%7B%5CDelta%20%5Calpha%7D=e%5E%7B%5CDelta%5Calpha%20%5C,G%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(6)<br /></div>where <img src="http://www.codecogs.com/gif.latex?G" align="middle" border="0" /> is the generator of this particular transformation.<sup>2</sup> See [1] for an example with Lorentz transformations.<br /><br /><br /><span style="font-weight: bold;">From classical to quantum</span><br />Generalizing (6), we'll postulate that any arbitrary quantum mechanical (unitary) transformation operator <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BO%7D" align="middle" border="0" /> through a parameter <img src="http://www.codecogs.com/gif.latex?%5CDelta%5Calpha" align="middle" border="0" /> can be written as<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BO%7D_%7B%5CDelta%20%5Calpha%7D=e%5E%7B%5CDelta%5Calpha%20%5C,%5Cmathcal%7BG%7D%7D," align="middle" border="0" /><br /></div><div style="text-align: right;">(7)<br /></div>where <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BG%7D" align="middle" border="0" /> is the quantum mechanical version of the classical operator <img src="http://www.codecogs.com/gif.latex?G" align="middle" border="0" />. We'll call this the ``quantum mechanical generator of the transformation.'' If we have a way of relating a classical generator to a quantum mechanical one, then we have a way of finding a quantum mechanical transformation operator.<br /><br />For example, in classical dynamics, the time derivative of a quantity <img src="http://www.codecogs.com/gif.latex?f" align="middle" border="0" /> is given by the Poisson bracket:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cfrac%7Bdf%7D%7Bdt%7D=%5C%7Bf,H%5C%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(8)<br /></div>where <img src="http://www.codecogs.com/gif.latex?H" align="middle" border="0" /> is the classical Hamiltonian of the system and <img src="http://www.codecogs.com/gif.latex?%5C%7B%5C,%5C,,%5C,%5C%7D" align="middle" border="0" /> is shorthand for a messy equation.[2] In quantum mechanics this equation is replaced with<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cfrac%7Bdf%7D%7Bdt%7D=i%5Chbar%5Bf,%5Cmathcal%7BH%7D%5D" align="middle" border="0" /><br /></div><div style="text-align: right;">(9)<br /></div>where the square brackets signify a commutation relation and <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BH%7D" align="middle" border="0" /> is the quantum mechanical Hamiltonian.[3] This holds true for any quantity <img src="http://www.codecogs.com/gif.latex?f" align="middle" border="0" />, and <img src="http://www.codecogs.com/gif.latex?i%5Chbar" align="middle" border="0" /> is a number which commutes with everything, so we can argue that the quantum mechanical Hamiltonian operator is related to the classical Hamiltonian by<br /><div style="text-align: center;"></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?H=i%5Chbar%5Cmathcal%7BH%7D%20%5CRightarrow%20%5Cmathcal%7BH%7D=-i%20H/%5Chbar" align="middle" border="0" /><br /></div><div style="text-align: right;">(10)<br /></div><br /><br /><span style="font-weight: bold;">Time translation of a quantum state</span><br />Consider a quantum state at time <img src="http://www.codecogs.com/gif.latex?t" align="middle" border="0" /> described by the wavefunction <img src="http://www.codecogs.com/gif.latex?%5Cpsi%28%5Cvec%7Br%7D,t%29" align="middle" border="0" />. To see how the state changes with time, we want to find a ``time-translation'' operator <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BT%7D_%7B%5CDelta%20t%7D" align="middle" border="0" /> which, when applied to the state <img src="http://www.codecogs.com/gif.latex?%5Cpsi%28%5Cvec%7Br%7D,t%29" align="middle" border="0" />, will give <img src="http://www.codecogs.com/gif.latex?%5Cpsi%28%5Cvec%7Br%7D,t+%5CDelta%20t%29" align="middle" border="0" />. That is,<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cpsi%28%5Cvec%7Br%7D,t+%5CDelta%20t%29=%5Cmathcal%7BT%7D_%7B%5CDelta%20t%7D%5Cpsi%28%5Cvec%7Br%7D,t%29." align="middle" border="0" /><br /></div><div style="text-align: right;">(11)<br /></div>From our previous discussion we know that if we know the classical generator of time translation we can write <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BT%7D" align="middle" border="0" /> using (7). Classically, the generator of time translations is the Hamiltonian![4] So we can write<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BT%7D_%7B%5CDelta%20t%7D=e%5E%7B%5CDelta%20t%20%5C,%5Cmathcal%7BH%7D%7D" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=e%5E%7B-i%20%5Cfrac%7B%5CDelta%20t%7D%7B%5Chbar%7D%20H%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(12)<br /></div>where we've made the substitution from (10). Then (11) becomes<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cpsi%28%5Cvec%7Br%7D,t+%5CDelta%20t%29=e%5E%7B-i%20%5Cfrac%7B%5CDelta%20t%7D%7B%5Chbar%7D%20H%7D%5C,%5Cpsi%28%5Cvec%7Br%7D,t%29." align="middle" border="0" /><br /></div><div style="text-align: right;">(13)<br /></div><br />This holds true for any time translation, so we'll consider a small time translation and expand (13) using a Taylor expansion<sup>3</sup> dropping all quadratic and higher terms:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cpsi%28%5Cvec%7Br%7D,t+%5CDelta%20t%29%5Capprox%5Cleft%5B1-i%5Cfrac%7B%5CDelta%20t%7D%7B%5Chbar%7DH+%5Cldots%20%5Cright%5D%5Cpsi%28%5Cvec%7Br%7D,t%29" align="middle" border="0" /><br /></div><div style="text-align: right;">(14)<br /></div>Moving things around gives<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?H%5Cpsi%28%5Cvec%7Br%7D,t%29=i%5Chbar%5Cleft%5B%5Cfrac%7B%5Cpsi%28%5Cvec%7Br%7D,t+%5CDelta%20t%29-%5Cpsi%28%5Cvec%7Br%7D,t%29%7D%7B%5CDelta%20t%7D%5Cright%5D" align="middle" border="0" /><br /></div><div style="text-align: right;">(15)<br /></div>In the limit <img src="http://www.codecogs.com/gif.latex?%5CDelta%20t%5Crightarrow%200" align="middle" border="0" /> the right-hand side becomes a partial derivative giving the Schrödinger equation<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?H%5Cpsi%28%5Cvec%7Br%7D,t%29=i%5Chbar%5Cfrac%7B%5Cpartial%20%5Cpsi%28%5Cvec%7Br%7D,t%29%7D%7B%5Cpartial%20t%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(16)<br /></div><br />For a system with conserved total energy, the classical Hamiltonian is the total energy<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?H=%5Cfrac%7B%5Cvec%7Bp%7D%5C,%5E2%7D%7B2m%7D+V" align="middle" border="0" /><br /></div><div style="text-align: right;">(17)<br /></div>which, making the substitution for quantum mechanical momentum <img src="http://www.codecogs.com/gif.latex?%5Cvec%7Bp%7D=i%5Chbar%5Cnabla" align="middle" border="0" /> and substituting into (19) gives the familiar differential equation form of the Schrödinger equation<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?-%5Cfrac%7B%5Chbar%5E2%7D%7B2m%7D%5Cnabla%5E2%5Cpsi%28%5Cvec%7Br%7D,t%29+V%5Cpsi%28%5Cvec%7Br%7D,t%29=i%5Chbar%5Cfrac%7B%5Cpartial%20%5Cpsi%28%5Cvec%7Br%7D,t%29%7D%7B%5Cpartial%20t%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(18)<br /></div><br /><br /><span style="font-weight: bold;">References</span><br />[1] J.D. Jackson. <span style="font-style: italic;">Classical Electrodynamics</span>. John Wiley & Sons, Inc., 3rd edition, 1998.<br />[2] L.D. Landau and E.M. Lifshitz. <span style="font-style: italic;">Mechanics</span>. Pergamon Press, Oxford, UK.<br />[3] L.D. Landau and E.M. Lifshitz. <span style="font-style: italic;">Quantum Mechanics</span>. Butterworth-Heinemann, Oxford, UK.<br />[4] H. Goldstein, C. Poole, and J. Safko. <span style="font-style: italic;">Classical Mechanics</span>. Cambridge University Press, San Francisco, CA, 3rd edition, 2002.<br /><br /><hr /><br /><sup>1</sup> <img src="http://www.codecogs.com/gif.latex?e%5Ex=%5Csum_%7Bn=0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7Bx%5En%7D%7Bn%21%7D=1+x+%5Cfrac%7Bx%5E2%7D%7B2%21%7D+%5Cldots" align="middle" border="0" /><br /><sup>2</sup> There are other ways to do this, differing by factors of <img src="http://www.codecogs.com/gif.latex?i" align="middle" border="0" /> in the definition of the generators and in the construction of the exponential, but I'm sticking with this one for now.<br /><sup>3 </sup>Kind of the reverse of how we got to this whole exponential notation in the first place...Eli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.com7tag:blogger.com,1999:blog-393324845011978943.post-70906949661177828282009-05-26T11:43:00.006-04:002009-06-04T10:56:41.431-04:00The Schrödinger Equation<span style="font-weight: bold;">Update:</span> A corrected and improved version of this post is now up: <a href="http://behindtheguesses.blogspot.com/2009/06/schrodinger-equation-corrections.html">http://behindtheguesses.blogspot.com/2009/06/schrodinger-equation-corrections.html</a><br /><br />[<a href="http://s87572736.onlinehome.us/btg/btg03-schroed.pdf">Click here for a PDF of this post with nicer formatting</a>]<br />notElon <a href="http://behindtheguesses.blogspot.com/2009/03/behind-guesses.html?showComment=1237865400000#c1353998693559438166">asked</a> me to discuss, and to try and derive the Schrödinger equation, so I'll give it a shot. This derivation is partially based on Sakurai,[1] with some differences.<br /><br /><span style="font-weight: bold;">A brief walk through classical mechanics</span><br />Say we have a function of <img src="http://www.codecogs.com/gif.latex?f%28x%29" align="middle" border="0" /> and we want to translate it in space to a point <img src="http://www.codecogs.com/gif.latex?%28x+a%29" align="middle" border="0" />. To do this, we'll find a ``space translation'' operator <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BS%7D_a" align="middle" border="0" /> which, when applied to <img src="http://www.codecogs.com/gif.latex?f%28x%29" align="middle" border="0" />, gives <img src="http://www.codecogs.com/gif.latex?f%28x+a%29" align="middle" border="0" />. That is,<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?f%28x+a%29=%5Cmathcal%7BS%7D_%7Ba%7D%20f%28x%29" align="middle" border="0" /><br /></div><div style="text-align: right;">(1)<br /></div>We'll expand <img src="http://www.codecogs.com/gif.latex?f%28x+a%29" align="middle" border="0" /> in a Taylor series:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?f%28x+a%29&=f%28x%29+a%5Cfrac%7Bdf%28x%29%7D%7Bdx%7D+%5Cfrac%7Ba%5E2%7D%7B2%21%7D%5Cfrac%7Bd%5E2f%28x%29%7D%7Bdx%5E2%7D+%5Cldots" align="middle" border="0" /><br /></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=%5Cleft%5B1+a%5Cfrac%7Bd%7D%7Bdx%7D+%5Cfrac%7Ba%5E2%7D%7B2%21%7D%5Cfrac%7Bd%5E2%7D%7Bdx%5E2%7D+%5Cldots%5Cright%5Df%28x%29" align="middle" border="0" /><br /></div><div style="text-align: right;">(2)<br /></div>which can be simplified using the series expansion of the exponential<sup>1</sup> to<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?e%5E%7B%5Cleft%5Ba%5Cfrac%7Bd%7D%7Bdx%7D%5Cright%5D%7Df%28x%29" align="middle" border="0" /><br /></div><div style="text-align: right;">(3)<br /></div>from which we can conclude that<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BS%7D_a=e%5E%7B%5Cleft%5Ba%5Cfrac%7Bd%7D%7Bdx%7D%5Cright%5D%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(4)<br /></div>If you do a similar thing with rotations around the <img src="http://www.codecogs.com/gif.latex?z" align="middle" border="0" />-axis, you'll find that the rotation operator is<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BR%7D_%5Ctheta=e%5E%7B%5Ctheta%20L_z%7D," align="middle" border="0" /><br /></div><div style="text-align: right;">(5)<br /></div>where <img src="http://www.codecogs.com/gif.latex?L_z" align="middle" border="0" /> is the <img src="http://www.codecogs.com/gif.latex?z" align="middle" border="0" />-component of the angular momentum.<br /><br />Comparing (4) and (5), we see that both have an exponential with a parameter (distance or angle) multiplied by something (<img src="http://www.codecogs.com/gif.latex?%5Cfrac%7Bd%7D%7Bdx%7D" align="middle" border="0" /> or <img src="http://www.codecogs.com/gif.latex?L" align="middle" border="0" />). We'll call the something the ``generator of the transformation.'' So, the generator of space translation is <img src="http://www.codecogs.com/gif.latex?%5Cfrac%7Bd%7D%7Bdx%7D" align="middle" border="0" /> and the generator of rotation is <img src="http://www.codecogs.com/gif.latex?L" align="middle" border="0" />. So, we'll write an arbitrary transformation operator <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BO%7D" align="middle" border="0" /> through a parameter <img src="http://www.codecogs.com/gif.latex?%5Calpha" align="middle" border="0" /> as<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BO%7D_a=e%5E%7B%5Calpha%20G%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(6)<br /></div>where <img src="http://www.codecogs.com/gif.latex?G" align="middle" border="0" /> is the generator of this particular transformation.<sup>2</sup> See [2] for an example with Lorentz transformations.<br /><br /><br /><span style="font-weight: bold;">From classical to quantum</span><br />In classical dynamics, the time derivative of a quantity <img src="http://www.codecogs.com/gif.latex?f" align="middle" border="0" /> is given by the Poisson bracket:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cfrac%7Bdf%7D%7Bdt%7D=%5C%7Bf,H%5C%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(7)<br /></div>where <img src="http://www.codecogs.com/gif.latex?H" align="middle" border="0" /> is the classical Hamiltonian of the system and <img src="http://www.codecogs.com/gif.latex?%5C%7B%5C,%5C,,%5C,%5C%7D" align="middle" border="0" /> is shorthand for a messy equation.[3] In quantum mechanics this equation is replaced with<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cfrac%7Bdf%7D%7Bdt%7D=i%5Chbar%5Bf,%5Cmathcal%7BH%7D%5D" align="middle" border="0" /><br /></div><div style="text-align: right;">(8)<br /></div>where the square brackets signify a commutation relation and <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BH%7D" align="middle" border="0" /> is the quantum mechanical Hamiltonian.[4] This holds true for any quantity <img src="http://www.codecogs.com/gif.latex?f" align="middle" border="0" />, and <img src="http://www.codecogs.com/gif.latex?i%5Chbar" align="middle" border="0" /> is a number which commutes with everything, so we can argue that the quantum mechanical Hamiltonian operator is related to the classical Hamiltonian by<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?H=i%5Chbar%5Cmathcal%7BH%7D%20%5CRightarrow%20%5Cmathcal%7BH%7D=-i%20H/%5Chbar" align="middle" border="0" /><br /></div><div style="text-align: right;">(9)<br /></div>specifically.<br /><br />Additionally, we can extend from here that any quantum operator <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BG%7D" align="middle" border="0" /> is written in terms of its classical counterpart <img src="http://www.codecogs.com/gif.latex?G" align="middle" border="0" /> by<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BG%7D=-i%20G/%5Chbar." align="middle" border="0" /><br /></div><div style="text-align: right;">(10)<br /></div><br />So, using (4) the quantum mechanical space translation operator is given by<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BS%7D_a=e%5E%7B%5Cleft%5B-i%5Cfrac%7Ba%7D%7B%5Chbar%7D%5Cfrac%7Bd%7D%7Bdx%7D%5Cright%5D%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(11)<br /></div>and, using (5), the rotation operator by<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BR%7D_%5Ctheta=e%5E%7B-i%5Cfrac%7B%5Ctheta%7D%7B%5Chbar%7D%20L_z%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(12)<br /></div>or, from (6) any arbitrary (unitary) transformation, <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BU%7D" align="middle" border="0" />, can be written as<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BU%7D=e%5E%7B-i%20%5Cfrac%7B%5Calpha%7D%7B%5Chbar%7D%20G%7D," align="middle" border="0" /><br /></div><div style="text-align: right;">(13)<br /></div>where <img src="http://www.codecogs.com/gif.latex?G" align="middle" border="0" /> is (an Hermitian operator and is) the classical generator of the transformation.<br /><br /><br /><span style="font-weight: bold;">Time translation of a quantum state</span><br />Consider a quantum state at time <img src="http://www.codecogs.com/gif.latex?t" align="middle" border="0" /> described by the wavefunction <img src="http://www.codecogs.com/gif.latex?%5Cpsi%28%5Cvec%7Br%7D,t%29" align="middle" border="0" />. To see how the state changes with time, we want to find a ``time-translation'' operator <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BT%7D_%7B%5CDelta%20t%7D" align="middle" border="0" /> which, when applied to the state <img src="http://www.codecogs.com/gif.latex?%5Cpsi%28%5Cvec%7Br%7D,t%29" align="middle" border="0" />, will give <img src="http://www.codecogs.com/gif.latex?%5Cpsi%28%5Cvec%7Br%7D,t+%5CDelta%20t%29" align="middle" border="0" />. That is,<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cpsi%28%5Cvec%7Br%7D,t+%5CDelta%20t%29=%5Cmathcal%7BT%7D_%7B%5CDelta%20t%7D%5Cpsi%28%5Cvec%7Br%7D,t%29." align="middle" border="0" /><br /></div><div style="text-align: right;">(14)<br /></div>From our previous discussion we know that if we know the classical generator of time translation we can write <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BT%7D" align="middle" border="0" /> using (13). Well, classically, the generator of time translations is the Hamiltonian![5] So we can write<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BT%7D_%7B%5CDelta%20t%7D=e%5E%7B-i%20%5Cfrac%7B%5CDelta%20t%7D%7B%5Chbar%7D%20H%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(15)<br /></div>and (14) becomes<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cpsi%28%5Cvec%7Br%7D,t+%5CDelta%20t%29=e%5E%7B-i%20%5Cfrac%7B%5CDelta%20t%7D%7B%5Chbar%7D%20H%7D%5C,%5Cpsi%28%5Cvec%7Br%7D,t%29." align="middle" border="0" /><br /></div><div style="text-align: right;">(16)<br /></div><br />This holds true for any time translation, so we'll consider a small time translation and expand (16) using a Taylor expansion<sup>3</sup> dropping all quadratic and higher terms:<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?%5Cpsi%28%5Cvec%7Br%7D,t+%5CDelta%20t%29%5Capprox%5Cleft%5B1-i%5Cfrac%7B%5CDelta%20t%7D%7B%5Chbar%7DH+%5Cldots%20%5Cright%5D%5Cpsi%28%5Cvec%7Br%7D,t%29" align="middle" border="0" /><br /></div><div style="text-align: right;">(17)<br /></div>Moving things around gives<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?H%5Cpsi%28%5Cvec%7Br%7D,t%29=i%5Chbar%5Cleft%5B%5Cfrac%7B%5Cpsi%28%5Cvec%7Br%7D,t+%5CDelta%20t%29-%5Cpsi%28%5Cvec%7Br%7D,t%29%7D%7B%5CDelta%20t%7D%5Cright%5D" align="middle" border="0" /><br /></div><div style="text-align: right;">(18)<br /></div>In the limit <img src="http://www.codecogs.com/gif.latex?%5CDelta%20t%5Crightarrow%200" align="middle" border="0" /> the righthand side becomes a partial derivative giving the Schrödinger equation<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?H%5Cpsi%28%5Cvec%7Br%7D,t%29=i%5Chbar%5Cfrac%7B%5Cpartial%20%5Cpsi%28%5Cvec%7Br%7D,t%29%7D%7B%5Cpartial%20t%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(19)<br /></div><br />For a system with conserved total energy, the classical Hamiltonian is the total energy<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?H=%5Cfrac%7B%5Cvec%7Bp%7D%5C,%5E2%7D%7B2m%7D+V" align="middle" border="0" /><br /></div><div style="text-align: right;">(20)<br /></div>which, making the substitution for quantum mechanical momentum <img src="http://www.codecogs.com/gif.latex?%5Cvec%7Bp%7D=i%5Chbar%5Cnabla" align="middle" border="0" /> and substituting into (19) gives the familiar differential equation form of the Schrödinger equation<br /><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?-%5Cfrac%7B%5Chbar%5E2%7D%7B2m%7D%5Cnabla%5E2%5Cpsi%28%5Cvec%7Br%7D,t%29+V%5Cpsi%28%5Cvec%7Br%7D,t%29=i%5Chbar%5Cfrac%7B%5Cpartial%20%5Cpsi%28%5Cvec%7Br%7D,t%29%7D%7B%5Cpartial%20t%7D" align="middle" border="0" /><br /></div><div style="text-align: right;">(21)<br /></div><br /><br /><span style="font-weight: bold;">References</span><br />[1] J.J. Sakurai. <span style="font-style: italic;">Modern Quantum Mechanics</span>. Addison-Wesley, San Francisco, CA, revised edition, 1993.<br />[2] J.D. Jackson. <span style="font-style: italic;">Classical Electrodynamics</span>. John Wiley & Sons, Inc., 3rd edition, 1998.<br />[3] L.D. Landau and E.M. Lifshitz. <span style="font-style: italic;">Mechanics</span>. Pergamon Press, Oxford, UK.<br />[4] L.D. Landau and E.M. Lifshitz. <span style="font-style: italic;">Quantum Mechanics</span>. Butterworth-Heinemann, Oxford, UK.<br />[5] H. Goldstein, C. Poole, and J. Safko. <span style="font-style: italic;">Classical Mechanics</span>. Cambridge University Press, San Francisco, CA, 3rd edition, 2002.<br /><br /><hr /><br /><sup>1</sup> <img src="http://www.codecogs.com/gif.latex?e%5Ex=%5Csum_%7Bn=0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7Bx%5En%7D%7Bn%21%7D=1+x+%5Cfrac%7Bx%5E2%7D%7B2%21%7D+%5Cldots" align="middle" border="0" /><br /><sup>2</sup> There are other ways to do this, differing by factors of <img src="http://www.codecogs.com/gif.latex?i" align="middle" border="0" /> in the definition of the generators and in the construction of the exponential, but I'm sticking with this one for now.<br /><sup>3 </sup>Kind of the reverse of how we got to this whole exponential notation in the first place...Eli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.com5tag:blogger.com,1999:blog-393324845011978943.post-4908734618764744732009-04-27T06:30:00.003-04:002010-05-13T09:35:33.591-04:00The Dot and Cross Products[<a href="http://elansey.googlepages.com/btg02-dtcrprd.pdf">Click here for a PDF of this post with nicer formatting</a>]<br />
<div><span style="font-weight: bold;">A bad way</span></div><div>The dot product and cross product of two vectors are tools which are heavily used in physics. As such, they are typically introduced at the beginning of first semester physics courses, just after vector addition, subtraction, etc. Although they are not strictly <i>required</i> for these intro courses (see [1], for example), they make the development and computations of work and energy, torque, and electromagnetism far simpler.<br />
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Unfortunately, they are consistently introduced in an awful way: by straight definition. That is, using the dot product for example, for two vectors <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D=A_x%5Chat%7Bx%7D+A_y%5Chat%7By%7D+A_z%5Chat%7Bz%7D" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D=B_x%5Chat%7Bx%7D+B_y%5Chat%7By%7D+B_z%5Chat%7Bz%7D" /> they say something like</div><blockquote><div>We <i>define</i> the dot product between <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /> as:</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D%5Ccdot%5Cvec%7BB%7D=A_xB_x+A_yB_y+A_zB_z," /></div><div>or,</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D%5Ccdot%5Cvec%7BB%7D=%7C%5Cvec%7BA%7D%7C%5C,%7C%5Cvec%7BB%7D%7C%5Ccos%5Ctheta," /></div><div>where <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Ctheta" /> is the angle between them.</div></blockquote><div>Then, for the cross product, either they use an equation like the latter of the above two equations coupled with the ``right-hand rule,'' or a strange algebraic combination of the components of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" />, often ``simplified'' with help of a startling determinant.<sup>1</sup> See [2], [3], [4], [5] and [6] as a few examples. Although a few of these give a geometric interpretation after the fact, it is usually in passing, and does not really contribute to their discussion. These approaches are not limited to textbooks, either. See [7] for an in-class lecture example.</div><div></div><div><br />
In these examples, the dot product is introduced first and then the cross product. From one standpoint this makes some sense -- the dot product is definitionally simpler and usually easier to calculate. However, from a conceptual standpoint, I think this order is backwards. Furthermore, in my experience, students, by and large, miss the <i>physical</i> and <i>graphical</i> significance of these definitions, and upon encountering the concepts of work or torque later on, take the resulting expressions purely as definitions as well.<sup>2</sup> This is yet another example of the fact that definition <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cneq" /> explanation.</div><div></div><div><br />
Personally, it is my inclination to wait to introduce these products until they're needed, thus motivating the discussion in the first place. However, I do understand the notion of ``getting it over with,'' and, it's possible that introducing them as abstract concepts lends to easier application of the concepts to general problems. In any case, my discussion follows the latter approach (for better insertion into standard texts) and presupposes understanding of vector basics: addition, decomposition, etc..</div><div></div><div><br />
<br />
<b>A better way</b></div><div><i><b>The Cross Product</b></i><br />
<div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/_D1sP-NndkqU/SfDq13JeaCI/AAAAAAAAHHc/Q_IyEv4AXkY/s1600-h/cross.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/_D1sP-NndkqU/SfDq13JeaCI/AAAAAAAAHHc/Q_IyEv4AXkY/s320/cross.gif" /></a></div><div style="text-align: center;">(a) Geometrical view of the cross product as the parallelogram area.</div><br />
<div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_D1sP-NndkqU/SfDq4A_2oSI/AAAAAAAAHHk/lWBYwv6Mh4s/s1600-h/decomp.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="119" src="http://1.bp.blogspot.com/_D1sP-NndkqU/SfDq4A_2oSI/AAAAAAAAHHk/lWBYwv6Mh4s/s400/decomp.gif" width="400" /></a></div><div style="text-align: center;">(b) Graphical derivation of area for two 2D arbitrary vectors, from [8].</div>Figure 1: The 2D cross product of vectors<img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /> .</div><div><br />
Say we have two vectors<img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /> with lengths <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?A" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?B" />, and we want to find something which is a measure of how much of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /> is perpendicular to <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" />. Looking at Fig. 1(a), we can see that the area of the parallelogram sided by the two vectors is such a measure. The area of a parallelogram is</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Ctext%7Barea%7D=%28%5Ctext%7Bbase%7D%29%5Ctimes%28%5Ctext%7Bheight%7D%29," /></div><div style="text-align: right;">(1)</div><div>which, for our case, is the same as</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Ctext%7Barea%7D=%28%5Ctext%7Blength%20of%20one%20vector%7D%29%5Ctimes%28%5Ctext%7Bamount%20of%20the%20other%20vector%20perpendicular%20to%20the%20first%7D%29." /></div><div style="text-align: right;">(2)</div><div>That is, you can only have an area if you have a ``base'' <i>and </i>a ``height'' perpendicular to the base. Thus area is a good measure of perpendicularity.<sup>3</sup><br />
<span style="font-size: small;"><span style="font-size: 13px;"><br />
</span></span></div><div></div><div>There are two different ways of calculating this area. If the angle between the two vectors is <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Ctheta" />, as in Fig. 1(a), we see that, choosing <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> as the ``base'' we can write the ``height'' as <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?B%20%5Csin%5Ctheta" />. Alternatively, choosing <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /> as the base, we write the perpendicular part as <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?A%20%5Csin%5Ctheta" />. Then the area is</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Ctext%7Barea%7D=%28A%29%5C,%28B%20%5Csin%5Ctheta%29=%28B%29%5C,%28A%5Csin%5Ctheta%29" />.</div><div style="text-align: right;">(3)</div><div>However, if we don't know angle between them, we're not completely out of luck. If you look at Fig. 1(b), you can see that for a simple, two-dimensional case, we can express the area in terms of the <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?x" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?y" /> components of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" />:</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Ctext%7Barea%7D=A_xB_y-B_xA_y" /></div><div style="text-align: right;">(4a)</div><div>Of course, I could just have easily labeled the axes <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?y" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z" /> which would give a different area</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Ctext%7Barea%7D%27=A_yB_z-B_yA_z," /></div><div style="text-align: right;">(4b)</div><div>or <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?x" />, which would give yet another area</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Ctext%7Barea%7D%27%27=A_zB_x-B_zA_x" /></div><div style="text-align: right;">(4c)</div><div></div><div><br />
If all we've done is relabel our axes, keeping <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /> fixed, then we wouldn't expect the size of these areas to be different -- and they're not. However, although the amount of area is the same, in a way the areas <i>are</i> different in that they're facing different directions in each case. So, we need a way to distinguish these three areas from each other, and from an arbitrarily oriented area. What we'll do is pick a vector perpendicular to both <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> to <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /> -- and thus perpendicular to the area of the parallelogram -- with magnitude equal to the area. We'll call this vector</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BC%7D=%5Cvec%7BA%7D%5Ctimes%5Cvec%7BB%7D," /></div><div style="text-align: right;">(5)</div><div>and say it's the result of a ``cross product'' of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" />. However, in principle, we have a choice of two such perpendicular vectors. In Fig. 1, for example, we could choose the vector pointing in either the <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?+z" /> or <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?-z" /> direction. Additionally, this arbitrariness can be seen in choosing whether to measure the angle in (3) from <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> to <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /> or vise-versa.</div><div></div><div><br />
So, as a matter of convention, we'll decide to always measure angles from the first term in the cross product (<img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> in (5)) such that</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Chat%7Bx%7D%5Ctimes%5Chat%7By%7D=+%5Chat%7Bz%7D%20," /></div><div style="text-align: right;">(6)</div><div>so if the fingers in your <i>right</i> hand point along the little arcs we draw for angles, your thumb points in the direction that this vector goes. Thus,</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D%5Ctimes%5Cvec%7BB%7D=-%5Cvec%7BB%7D%5Ctimes%5Cvec%7BA%7D," /></div><div style="text-align: right;">(7)</div><div>since your hand would curl in the other direction. This is called the ``Right-Hand Rule.'' Then, the areas we discussed in equations (4) become</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Ctext%7Barea%7D_%7Bxy%7D=%28A_xB_y-B_xA_y%29%5Chat%7Bz%7D," /></div><div style="text-align: right;">(8a)</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Ctext%7Barea%7D%27_%7Byz%7D=%28A_yB_z-B_yA_z%29%5Chat%7Bx%7D," /></div><div style="text-align: right;">(8b)</div><div>and</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Ctext%7Barea%7D%27%27_%7Bzx%7D=%28A_zB_x-B_zA_x%29%5Chat%7By%7D," /></div><div style="text-align: right;">(8c)</div><div>where the subscripts tell us which coordinate plane the two crossed vectors are in. Thus, the cross product represents <b>how much these two vectors point in perpendicular directions</b>, and is a signed area vector perpendicular to the plane described by <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" />.</div><div><br />
<div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/_D1sP-NndkqU/SfDrt4Z-ZEI/AAAAAAAAHHs/yYqx4O3SCjQ/s1600-h/3dvec.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/_D1sP-NndkqU/SfDrt4Z-ZEI/AAAAAAAAHHs/yYqx4O3SCjQ/s400/3dvec.gif" /></a></div><div style="text-align: center;">(a) Geometrical view of the 3D cross product as the parallelogram area.</div><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/_D1sP-NndkqU/SfDrvfs6TpI/AAAAAAAAHH0/TP2On8aOryI/s1600-h/areas.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/_D1sP-NndkqU/SfDrvfs6TpI/AAAAAAAAHH0/TP2On8aOryI/s400/areas.gif" /></a></div><div style="text-align: center;">(b) Looking at the area from the <i>xy</i>-plane (dashed outline), the <i>yz</i>-plane (shaded) and the <i>zx</i>-plane (solid).</div>Figure 2: The 3D cross product of vectors <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /> and the decomposed area.</div><div></div><br />
So far, though, we've only discussed vectors which have only two coplanar components. But it's fairly straightforward to generalize to arbitrary 3D vectors. See Fig. 2(a), for example. Here the area vector, and hence the cross product vector, is pointing in a complicated direction. However, we know we can decompose <i>any</i> vector into its <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?x" />, <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?y" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z" /> components, and this area vector is no different:<br />
<div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?\vec{A}\times\vec{B}=\text{area}_{3D}=(\text{z%20area})\hat{z}+(\text{x%20area})\hat{x}+(\text{y%20area})\hat{y}" /></div><div style="text-align: right;">(9)</div><div><br />
All we need to do is find out how much area is pointing in each direction. To do that, look at Fig. 2(b). This picture shows what the area between the two vectors looks like if we look only at two coplanar components at a time -- in other words the <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?z" />, <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?x" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?y" /> components of the area. But we already know what each of these areas are from (8)! So, then we can combine these equations and write the cross product</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D%5Ctimes%5Cvec%7BB%7D=%28A_yB_z-B_yA_z%29%5Chat%7Bx%7D+%28A_zB_x-B_zA_x%29%5Chat%7By%7D+%28A_xB_y-B_xA_y%29%5Chat%7Bz%7D" /></div><div style="text-align: right;">(10)</div><br />
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<i><b>The Dot Product</b></i></div><div><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/_D1sP-NndkqU/SfDsaKRrApI/AAAAAAAAHH8/npeXOGApkX0/s1600-h/BonAa.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/_D1sP-NndkqU/SfDsaKRrApI/AAAAAAAAHH8/npeXOGApkX0/s400/BonAa.gif" /></a></div><div style="text-align: center;">(a) When <i>B < A.</i></div><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_D1sP-NndkqU/SfDsazoNi7I/AAAAAAAAHIE/43ikfYEcLFM/s1600-h/BonAb.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/_D1sP-NndkqU/SfDsazoNi7I/AAAAAAAAHIE/43ikfYEcLFM/s400/BonAb.gif" /></a></div><div style="text-align: center;">(b) When <i>B > A</i>.</div>Figure 3: The projection of vector <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /> on to vector <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" />.</div><div><div><br />
Having discussed the perpendicularity of two vectors, it's natural to ask if there's a similar measure of the parallelity of two vectors. There are two ways of doing this. The way I'll do it first is explicitly geometrical, the second way is only implicitly geometrical.</div><div></div><div><br />
Say we have two vectors <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /> again, and we want to know how much of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /> is pointing (projected) along <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" />. From Fig. 3 we see that this is equal to</div></div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?B%20%5Ccos%20%5Ctheta." /></div><div style="text-align: right;">(11)</div><div>Similarly, the amount of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> that is projected along <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /> is</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?A%20%5Ccos%20%5Ctheta." /></div><div style="text-align: right;">(12)</div><div>Now, it would be nice if we could have one statement which somehow combined the these two statements and gave a measure both of how much of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> is pointing along <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /><i>and</i> of how much of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" /> is pointing along <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" />; that is, a measure of <b>how much these two vectors point in the same direction</b>. Additionally, since (2) used a multiplicative combination of the two vectors as a measure of perpendicularity, we'll try a similar multiplicative measure here, as well.</div><br />
<div>If we multiply (11) by <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?A" /> and (12) by <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?B" /> we can write a single, symmetric statement</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?D=%5Cvec%7BA%7D%5Ccdot%5Cvec%7BB%7D=AB%5Ccos%5Ctheta," /></div><div style="text-align: right;">(13)</div><div>and say it's the result of a ``dot product'' of <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> and <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" />, which amounts to multiplying together the parallel parts of two vectors. Here, too, if we don't know the angle between them, we're not out of luck. For a vector written in component form, it's straightforward to multiply the parallel parts together:</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D%5Ccdot%5Cvec%7BB%7D=A_xB_x+A_yB_y+A_zB_z." /></div><div style="text-align: right;">(14)</div><div></div><div><br />
However, unlike the cross product which gave us an actual area with a natural direction, this area-like structure is actually a measure of ``non-area'' and doesn't really have a natural direction. Although we could, completely arbitrarily, define a direction for this dot product,<sup>4</sup> and thus make it a vector as well, to the best of my knowledge such a quantity does not have any uses in physics, so we'll leave it alone and treat it only as a number (scalar).</div><div></div><div><div><br />
Alternatively, we know that the largest area possible between two vectors occurs when they are perpendicular to each other, where the area is <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?AB" /> (you can also see this from (3)). If we are interested in the maximal ``amount perpendicular'' we can write</div></div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%28%5Ctext%7Bamount%20perpendicular%7D%29_%7Bmax%7D%5E2=%28AB%29%5E2," /></div><div style="text-align: right;">(15)</div><div>where they are squared to take care of sign problems. Now, when they are completely parallel there is no area, and we're left only with non-area, which, also, can't be larger than the total maximum area, so</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%28%5Ctext%7Bamount%20parallel%7D%29_%7Bmax%7D%5E2=%28AB%29%5E2," /></div><div style="text-align: right;">(16)</div><div>as well.</div><div></div><div><br />
Then using a rough analogue to the Pythagorean theorem we see that</div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%28%5Ctext%7Bamount%20parallel%7D%29%5E2+%28%5Ctext%7Bamount%20perpendicular%7D%29%5E2=%28%5Ctext%7Bmax%20total%20amount%7D%29%5E2" /></div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%28%5Ctext%7Bamount%20parallel%7D%29%5E2=%28%5Ctext%7Bmax%20total%20amount%7D%29%5E2-%28%5Ctext%7Bamount%20perpendicular%7D%29%5E2" /></div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%28%5Ctext%7Bamount%20parallel%7D%29%5E2=%28AB%29%5E2-%28AB%5Csin%5Ctheta%29%5E2" /></div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%28%5Ctext%7Bamount%20parallel%7D%29%5E2=%28AB%29%5E2%5Cleft%5B1-%5Csin%5E2%5Ctheta%5Cright%5D" /></div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%28%5Ctext%7Bamount%20parallel%7D%29%5E2=%28AB%29%5E2%5Ccos%5E2%5Ctheta" /></div><div style="text-align: center;"><img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%28%5Ctext%7Bamount%20parallel%7D%29=%5Cpm%20AB%5Ccos%5Ctheta," /></div><div style="text-align: right;">(17)</div><div>which, choosing the positive root, is the same as (13).</div><br />
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<div></div><div><b>References</b></div><div><div>[1] F.W. Sears and M.Z. Zemansky. University Physics. Addison-Wesley, Reading, MA, 2nd edition, 1955.</div><div>[2] D. Halliday amd R. Resnick and J. Walker. Fundamentals of Physics. John Wiley & Sons, Inc., 7 edition, 2005.</div><div>[3] G.R. Fowles and G.L. Cassiday. Analytical Mechanics. Thomson Brooks/Cole, Belmont, CA, 7th edition, 2005.</div><div>[4] J.R. Reitz, F.J. Milford, and R.W. Christy. Foundations of Electromagnetic Theory. Addison-Wesley, 4th edition, 1992.</div><div>[5] D.J. Griffths. Introduction to Quantum Mechanics. Pearson Prentice Hall, 2nd edition, 2005.</div><div>[6] J. Stewart. Multivariable Calculus. Brooks/Cole Publishing Company, Pacific Grove, CA, 4th edition, 1999.</div><div>[7] W. Lewin. Lec 3 | 8.01 Physics I: Classical Mechanics, Fall 1999 [online]. Available from:<a href="http://www.youtube.com/watch?v=fwNQKjTj-0w#t=13m45s">http://www.youtube.com/watch?v=fwNQKjTj-0w#t=13m45s</a> [cited 16 March 2009].</div><div>[8] C.T.J. Dodson and T. Poston. Tensor Geometry: The Geometric Viewpoint and its Uses. Springer, 2nd edition, 1997.</div></div><div></div><div><br />
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<sup>1</sup> Of course, not all first semester physics students even know what a determinant is, but that is not my point.</div><div><sup>2</sup> Work <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?W=%5Cint%20%5Cvec%7BF%7D%5Ccdot%20d%5Cvec%7Br%7D" />, and torque <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7B%5Ctau%7D=%5Cvec%7Br%7D%5Ctimes%5Cvec%7Bp%7D" /></div><div><sup>3</sup>Another way to approach this is to start by calculating the area, and then explain that this can also be viewed as a measure of perpendicularity.</div><div><sup>4</sup> i.e. along either <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BA%7D" /> or <img align="middle" border="0" src="http://www.codecogs.com/gif.latex?%5Cvec%7BB%7D" />, or along a line midway between them, or perpendicular to them, or some other arbitrary choice</div>Eli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.com28tag:blogger.com,1999:blog-393324845011978943.post-16365183352821191702009-04-20T16:20:00.000-04:002009-04-20T16:20:28.813-04:00New LayoutI'm trying a new, wider layout for the blog to allow for wider equations. I'm curious what you think about the new layout. Please, let me know in the comments, via email, or via the poll on the top-right of the page. For comparison, this is what the site used to look like:<br />
<div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_D1sP-NndkqU/SezY0NHSJnI/AAAAAAAAHFs/HPbiARGUUSE/s1600-h/old+layout.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/_D1sP-NndkqU/SezY0NHSJnI/AAAAAAAAHFs/HPbiARGUUSE/s400/old+layout.PNG" /></a></div>Eli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.com1tag:blogger.com,1999:blog-393324845011978943.post-17187521028793349162009-03-30T07:00:00.005-04:002009-06-26T08:43:01.442-04:00The Quantum Harmonic Oscillator Ladder Operators<div style="margin: 0px;">[<a href="http://elansey.googlepages.com/btg01-qhm.pdf">Click here for a PDF of this post with nicer formatting</a>]</div><div style="margin: 0px;"><span style="font-weight: bold;">The Setup</span></div><div style="margin: 0px;">As a first example, I'll discuss a particular pet-peeve of mine, which is something covered in many introductory quantum mechanics classes: The algebraic solution to quantum (1D) simple harmonic oscillator.<sup>1 </sup>The one-dimensional, time-independent Schrödinger equation is:</div><div style="text-align: center;"></div><div style="text-align: center;"><div style="margin: 0px;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BH%7D%5CPsi=E%5CPsi," align="middle" border="0" /></div></div><div style="text-align: right;"><div style="margin: 0px;">(1)</div><div style="text-align: left;"><div style="margin: 0px;">where <img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BH%7D" align="middle" border="0" /> is the Hamiltonian of the system. Explicitly, this Hamiltonian is</div></div></div><div style="text-align: center;"></div><div style="text-align: center;"><div style="margin: 0px;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BH%7D=%5Cfrac%7Bp%5E2%7D%7B2m%7D+V%28x%29," align="middle" border="0" /></div></div><div style="text-align: right;"><div style="margin: 0px;">(2)</div></div><div style="margin: 0px;">where <img src="http://www.codecogs.com/gif.latex?p" align="middle" border="0" /> is the particle's momentum, <img src="http://www.codecogs.com/gif.latex?m" align="middle" border="0" /> is it's mass and <img src="http://www.codecogs.com/gif.latex?V%28x%29" align="middle" border="0" /> is the potential the particle is placed into. The potential associated with a classical harmonic oscillator is</div><div style="text-align: center;"><div style="margin: 0px;"><img src="http://www.codecogs.com/gif.latex?V%28x%29&=%5Cfrac%7B1%7D%7B2%7Dkx%5E2" style="cursor: move;" align="middle" border="0" /></div></div><div style="text-align: center;"><div style="margin: 0px;"></div></div><div style="text-align: center;"><img src="http://www.codecogs.com/gif.latex?=%5Cfrac%7B1%7D%7B2%7Dm%5Comega%5E2x%5E2," align="middle" border="0" /></div><div style="text-align: right;"><div style="margin: 0px;">(3)</div></div><div style="margin: 0px;"><img src="http://www.codecogs.com/gif.latex?%5Cbegin%7Bsplit%7D%3Cbr%3EV%28x%29&=%5Cfrac%7B1%7D%7B2%7Dkx%5E2%5C%5C%3Cbr%3E&=%5Cfrac%7Bmx%5E2%7D%7B2%5Comega%5E2%7D,%3Cbr%3E%5Cend%7Bsplit%7D" align="middle" border="0" /></div><div style="margin: 0px;">where <img src="http://www.codecogs.com/gif.latex?%5Comega%5E2%5Cequiv%20k/m" align="middle" border="0" />. For the sake of convenience, so we don't get bogged down with various factors,<sup>2</sup> we'll consider <img src="http://www.codecogs.com/gif.latex?m=%5Comega=%5Chbar=1" align="middle" border="0" />. Then, if we substitute (3) back into (2) we write the Hamiltonian as:</div><div style="text-align: center;"><div style="margin: 0px;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BH%7D=%5Cfrac%7Bp%5E2+x%5E2%7D%7B2%7D." align="middle" border="0" /></div></div><div style="text-align: right;"><div style="margin: 0px;">(4)</div></div><div style="margin: 0px;"><b><br /></b><br /><span style="font-weight: bold;">A bad way</span></div><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_D1sP-NndkqU/ScjYEiPgXcI/AAAAAAAAGP8/rNnbwSws4j8/s1600-h/emc2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img src="http://1.bp.blogspot.com/_D1sP-NndkqU/ScjYEiPgXcI/AAAAAAAAGP8/rNnbwSws4j8/s320/emc2.jpg" border="0" /></a></div><div style="margin: 0px;">Figure 1: How Einstein developed his famous <img src="http://www.codecogs.com/gif.latex?E=mc%5E2" align="middle" border="0" /> expression.</div><div style="margin: 0px;"></div><div style="margin: 0px;"><br />Now, at this point, many texts, (see [1] or [2] for example,) define, with no motivation other than future ``convenience'', two operators</div><div style="text-align: center;"><div style="margin: 0px;"><img src="http://www.codecogs.com/gif.latex?a%20%5Cequiv%20%28x+ip%29/%5Csqrt%7B2%7D" align="middle" border="0" /></div></div><div style="text-align: right;"><div style="margin: 0px;">(5a)</div></div><div style="text-align: center;"><div style="margin: 0px;"><img src="http://www.codecogs.com/gif.latex?a%5E%5Cdag%20%5Cequiv%20%28x-ip%29/%5Csqrt%7B2%7D" align="middle" border="0" /></div></div><div style="text-align: right;"><div style="margin: 0px;">(5b)</div></div><div style="margin: 0px;">and proceed to show how these can be used to simplify the Hamiltonian and easily solve the problem. While it is an elegant and quick solution, this presentation is completely useless. I find it highly unlikely that Dirac sat down to solve this problem and tried a whole series of random operators</div><div style="text-align: center;"><div style="margin: 0px;"><img src="http://www.codecogs.com/gif.latex?c%20%5Cequiv%20%28x%5E2+ip%29/7" align="middle" border="0" /></div><div style="margin: 0px;"><img src="http://www.codecogs.com/gif.latex?d%20%5Cequiv%20%28x%5Ep-e%5E%7Bip%7D%29/%5Cpi" align="middle" border="0" /></div></div><div style="text-align: center;"><div style="margin: 0px;"><img src="http://www.codecogs.com/gif.latex?f%20%5Cequiv%20%28%5Csqrt%7Bx%7De%5E%7B-x%5E2/8%7D+ip%5E3%29/%5Csqrt%5Cpi" align="middle" border="0" /></div><div style="margin: 0px;"><img src="http://www.codecogs.com/gif.latex?g%20&%5Cequiv%20%28%5Csqrt%7Bx%20p%7D-ip/x%29/3" align="middle" border="0" /></div></div><div style="text-align: left;"><div style="margin: 0px;">and so on, along with their complex conjugates, until he lucked out with the solution in (5), see Fig. 1.</div><div style="margin: 0px;"></div><div style="margin: 0px;"></div><div style="margin: 0px;"><b><br /></b><br /><span style="font-weight: bold;">A better way</span></div><div style="margin: 0px;">What is far more likely is the argument given by Griffiths in [3], which I'll loosely follow. He presents a rationale and a <span style="font-style: italic;">method</span> for approaching this problem. Namely, he suggests factoring the Hamiltonian (4) into terms linear in <img src="http://www.codecogs.com/gif.latex?x" align="middle" border="0" /> and <img src="http://www.codecogs.com/gif.latex?p" align="middle" border="0" />. If we ignore the operator properties of <img src="http://www.codecogs.com/gif.latex?x" align="middle" border="0" /> and <img src="http://www.codecogs.com/gif.latex?p" align="middle" border="0" /> momentarily, and consider the classical quantities, we can factor the Hamiltonian</div></div><div style="text-align: center;"><div style="margin: 0px;"><img src="http://www.codecogs.com/gif.latex?%5Cfrac%7B%28x+ip%29%7D%7B%5Csqrt%202%7D%5Cfrac%7B%28x-%20ip%29%7D%7B%5Csqrt%202%7D=%5Cmathcal%7BH%7D_%7B%5Ctextit%7Bclassical%7D%7D=%5Cfrac%7B%28p+%20ix%29%7D%7B%5Csqrt%202%7D%5Cfrac%7B%28p-%20ix%29%7D%7B%5Csqrt%202%7D." align="middle" border="0" /></div></div><div style="text-align: right;"><div style="margin: 0px;">(6)</div></div><div style="margin: 0px;">Now we see a reason why (5) makes sense to try. Each term, either on the right or left side of (6),<sup>3</sup> contains two terms which are complex conjugates of each other. If this were a classical problem, we could, in principle, make a change of variables converting the Hamiltonian to something of the form</div><div style="text-align: center;"><div style="margin: 0px;"><img src="http://www.codecogs.com/gif.latex?%5Cmathcal%7BH%7D=%5Ceta%5E2/2=%5Ceta%5E*%5Ceta/2," align="middle" border="0" /></div></div><div style="text-align: right;"><div style="margin: 0px;">(7)</div></div><div style="margin: 0px;">where <img src="http://www.codecogs.com/gif.latex?%5Ceta" align="middle" border="0" /> is any of the combinations of <img src="http://www.codecogs.com/gif.latex?x" align="middle" border="0" /> and <img src="http://www.codecogs.com/gif.latex?p" align="middle" border="0" /> in (6). Although this is not a ``canonical transformation,'' the symmetric form<sup>4</sup> of the Hamiltonian allows us to reduce the Hamiltonian from a function of two dynamical variables to a function of a single dynamical variable.</div><div style="margin: 0px;"></div><div style="margin: 0px;"><br />Switching back to quantum mechanics, we now see a rationale for choosing <img src="http://www.codecogs.com/gif.latex?a" align="middle" border="0" /> and <img src="http://www.codecogs.com/gif.latex?a%5E%5Cdag" align="middle" border="0" /> as we did.<sup>5</sup> Although deciding which variable to attach the <img src="http://www.codecogs.com/gif.latex?i" align="middle" border="0" /> to and its choice of sign is a guess,<sup>6</sup> we now have a general <span style="font-style: italic;">method</span> for approaching Hamiltonians that look like they might be easily factored classically -- <span style="font-weight: bold;">try using the classical factorizations with quantum quantities and see what happens</span>.<br /><br /><br /><b>References</b><br />[1] J.J. Sakurai. Modern Quantum Mechanics. Addison-Wesley, San Francisco, CA, revised edition, 1993.<br />[2] F. Schwabl. Quantum Mechanics. Springer, 3rd edition, 2005.<br />[3] D.J. Griffiths. Introduction to Electrodynamics. Pearson Prentice Hall, 3rd edition, 1999.<br />[4] P.A.M. Dirac. The Principles of Quantum Mechanics. Oxford University Press, USA.</div><br /><hr /><br /><span style="line-height: 15px; white-space: pre-wrap;font-size:13;" ><sup>1</sup> I'll be discussing this in the boring algebraic sense of symbols and whatnot, leaving off the geometric/visual interpretation of the algebra for another time.</span><br /><span style="line-height: 15px; white-space: pre-wrap;"><span style="line-height: normal; white-space: normal;"><span style="font-family:inherit;"><sup><span style="font-size:small;">2</span></sup><span style="font-size:small;"> Or maybe I'm just lazy</span></span></span></span><br /><span style="font-family:inherit;"><sup><span style="font-size:small;">3</span></sup><span style="font-size:small;"> And, since this is a classical problem, the order does not matter, either.</span></span><br /><span style="font-family:inherit;"><sup><span style="font-size:small;">4</span></sup><span style="font-size:small;"> Yes, it's only up to a constant which I've set to one, but you can still symmetrize things by changing to unitless variables, see [2].</span></span><br /><span style="font-family:inherit;"><sup><span style="font-size:small;">5</span></sup><span style="font-size:small;"> Recall that the </span></span><span style="font-family:inherit;"><span style="font-size:small;"><img src="http://www.codecogs.com/gif.latex?%5Cdag" align="middle" border="0" /> for a quantum-mechanical operator/matrix serves the role of the </span></span><span style="font-family:inherit;"><span style="font-size:small;"><img src="http://www.codecogs.com/gif.latex?*" align="middle" border="0" /> in (7).</span></span><br /><span style="font-family:inherit;"><sup><span style="font-size:small;">6</span></sup><span style="font-size:small;"> It actually does not matter to the solution of the problem. The only change is which operator acts to `step up' the state. Dirac actually defined his operators with the </span></span><span style="font-family:inherit;"><span style="font-size:small;"><img src="http://www.codecogs.com/gif.latex?i" align="middle" border="0" /> attached to the </span></span><span style="font-family:inherit;"><span style="font-size:small;"><img src="http://www.codecogs.com/gif.latex?x" align="middle" border="0" /> variable, see [4].</span></span>Eli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.com7tag:blogger.com,1999:blog-393324845011978943.post-12121308894781369342009-03-23T15:39:00.010-04:002009-03-23T15:45:13.405-04:00Behind the Guesses[<a href="http://elansey.googlepages.com/btg00-intro.pdf">Click here for a PDF of this post with nicer formatting</a>]<br />
<span style="font-weight: bold;">Why I'm starting this blog</span><br />
<div class="separator" style="clear: both; text-align: center;"></div><div style="text-align: auto;"><div style="text-align: center;"><a href="http://2.bp.blogspot.com/_D1sP-NndkqU/ScfgTEwA2AI/AAAAAAAAGPQ/Nf_lziLJ5_g/s1600-h/20090309.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/_D1sP-NndkqU/ScfgTEwA2AI/AAAAAAAAGPQ/Nf_lziLJ5_g/s320/20090309.gif" style="cursor: move;" /></a></div></div><div class="separator" style="clear: both; text-align: center;">Figure 1: How physics does not work. Comic (cropped) from [1].</div><br />
If you sit in enough physics, and to a lesser degree mathematics lectures, and/or read enough books, you're bound to encounter the phrase ``We guess the solution...'' Or, ``If we define ... it turns out that..." Now, this is not inherently problematic. After all, if the goal is only to solve some problem which does not have an obvious solution, or prove a random mathematical statement, these approaches will suffice. Then, the student or reader, assuming they memorize the solution (or remember where to find it), will now be exquisitely equipped to solve exactly the same problem.<br />
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However, this approach does not teach students any skills, tools or methods for approaching new and exciting problems. Additionally, I find, it is difficult to gain any meaningful understanding of the material through the ``Guess and Show'' method. It's possible that for many people definition is the same as explanation, but from my personal experience, and from discussions with other students, it seems that such people are not the majority of physics students.<br />
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So I'm starting this blog to attempt to show some of the reasoning that goes into these guesses. After all, the great physicists didn't just pull their solutions out of their butt, see Fig. 1. Even if getting the correct solution ultimately was a result of guesswork, physicists don't sit around solving problems by randomly throwing darts at variables to see what sticks.<br />
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Now, I am <span style="font-style: italic;">not</span> claiming that science should be taught according to the historical development of the theories. In fact, usually doing that leads to needless confusion. What I am saying, however, is that even if we know the answer already, it is still essential to explain <span style="font-style: italic;">some</span> rationale as to why such an answer makes sense -- even if it's not the original idea behind the guess.<br />
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<span style="font-weight: bold;">What to expect</span><br />
So what can you expect to see on this blog? Well, a lot of quantum mechanics, to be sure. The conceptual difficulty and widespread lack of understanding,<sup>1</sup> of this subject, together with the plethora of ``known solutions'' make it a prime target. But I'll hopefully cover topics in electrodynamics, classical mechanics, thermodynamics, and so on. Maybe even some mathematics, if I'm feeling up for it.<br />
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In general, I will not carry through the solution to the end; you can always follow my references to see the complete thing. Instead I will try to provide some physical or mathematical reasoning behind some so-called ``guesses'' which ``turn out'' the correct solution. Additionally, with each post I will try to highlight a key <span style="font-weight: bold;">methodological point</span> that going behind the guesses gives us. These points will be <span style="font-weight: bold;">written in a different typeface</span> to easily spot.<br />
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<div style="text-align: left;">I do not expect to always get things right. This is a learning experience for me, too. If I make mistakes, please point them out in the comments or email me. If I make corrections, I will leave the original, unedited version up as reference -- sometimes we learn more from mistakes! I will try to post about once a month until I run out of ideas. I suggest you use one of the subscription options on the side so you don't have to keep checking the site to see if I've updated. So, if you have a particular solution (or problem) that's always bothered you, let me know, and I'll see if I can come up with something. And if I can't I'll post it anyway, and we'll see if anyone out there in the vast Intertubes who stumbles across the blog can help out.<br />
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Each blog post will be have a link to a nicely formatted PDF with the same content. Come back here this Monday (March 30, 2009) for my first real post.<br />
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</div><div style="text-align: left;"></div><div style="text-align: left;"><span style="font-weight: bold;">References</span></div><div style="text-align: left;">[1] Z. Weiner. Saturday morning breakfast cereal [online]. March 2009. Available from: <a href="http://www.smbc-comics.com/index.php?db=comics&id=1452">http://www.smbc-comics.com/index.php?db=comics&id=1452</a> [cited 9 March 2009].</div><br />
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<sup><span style="font-size: small;">1</span></sup><span style="font-size: small;"> I don't claim to understand quantum mechanics. But at least I don't pretend to.</span>Eli Lanseyhttp://www.blogger.com/profile/01955234977479398457noreply@blogger.com4