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--- classes:2009:fall:phys4101.001:quiz_3_1113 [2009/11/12 23:19] – x500_chap0326
+++ classes:2009:fall:phys4101.001:quiz_3_1113 [2009/11/13 09:23] (current) – x500_maxwe120
@@ Line 45: / Line 45: @@
 ===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.
+==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===