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classes:2009:fall:phys4101.001:quiz_3_1113 [2009/11/12 23:12] – x500_hruby028classes:2009:fall:phys4101.001:quiz_3_1113 [2009/11/13 09:23] (current) – x500_maxwe120
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 ===chap0326 11/12 13:39=== ===chap0326 11/12 13:39===
-I wouldn't be too worried. The derivations Griffith did, he pulled a bunch of definitions out of nowhere that were pretty arbitrary unless you already knew what you wanted to get. So I would be surprised if we were expected to pull that sort of stuff out of the ether.+I wouldn't be too worried. The derivations Griffiths did, he pulled a bunch of definitions out of nowhere that were pretty arbitrary unless you already knew what you wanted to get. So I would be surprised if we were expected to pull that sort of stuff out of the ether.
  
 ===physics4dummies 11/12 8:30pm=== ===physics4dummies 11/12 8:30pm===
 Been doing the practice test and I need to know if problem 2 is as easy as plugging the change of coordinates into the operator equation.  Been doing the practice test and I need to know if problem 2 is as easy as plugging the change of coordinates into the operator equation. 
 +
 +==Zeno 9am 11/13==
 +Generally, yes, but since the x and y momentum operators involves derivatives of x and y respectively you need to change the derivatives from Cartesian to spherical coordinates. You can do this by exploiting the chain rule for derivatives and obtaining 
 +
 + <math>\frac{\partial}{\partial x}= \frac{\partial \rho}{\partial x} \frac{\partial}{\partial \rho} + 
 + \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta} +
 + \frac{\partial \phi}{\partial x} \frac{\partial}{\partial \phi} </math>
 +
 +Then just use the definitions for the change of coordinates to compute each respective component derivative and you'll be left with spherical coordinate derivative operators multiplied by variables on to the left. If you do it correctly you'll obtain equation 4-129.
  
 ===Ekrpat 1135=== ===Ekrpat 1135===
 Doing problem 3.38 for review, part b for A, I get <math><A> = \lambda (c_1*c_2 + c_2*c_1 + 2|c_3|^2)</math>.  Is it possible to cancel the <math>c_1*c_2</math> and <math>c_2*c_1</math> terms by kroenecker delta leaving <math><A> =  (2\lambda|c_3|^2)</math>?  Thanks! Doing problem 3.38 for review, part b for A, I get <math><A> = \lambda (c_1*c_2 + c_2*c_1 + 2|c_3|^2)</math>.  Is it possible to cancel the <math>c_1*c_2</math> and <math>c_2*c_1</math> terms by kroenecker delta leaving <math><A> =  (2\lambda|c_3|^2)</math>?  Thanks!
 +
 +==chap0326 11/12 23:15==
 +I'm not sure that you can. I think when Griffiths proved that in chapter 2.2, he used two different states of wave function, but the c's {<math>(c_1*c_2 + c_2*c_1 + 2|c_3|^2)</math>} are, I think, more like dimensions of a single state rather than different states.
classes/2009/fall/phys4101.001/quiz_3_1113.1258089161.txt.gz · Last modified: 2009/11/12 23:12 by x500_hruby028

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