Another Question About Angular Momentum in QM

  • Thread starter Bobbo Snap
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  • #1
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Homework Statement


Show that the Hamiltonian [itex]H = (p^2/2m)+V[/itex] commutes with all three components of L, provided that V depends only on r.


Homework Equations


In previous parts of the problem, I've worked out the following relations:
[tex][L_z,x] = i\hbar y, \quad [L_z,y] = -i\hbar x, \quad [L_z, z] = 0,\\
[L_z, p_x] = i\hbar p_y, \quad [L_z, p_y] = -i\hbar p_x, \quad [L_z, p_z] = 0,\\
[L_z, L_x] = i \hbar L_y, \quad [L_z, r^2] = xy(xp_x - yp_y), \quad [L_z, p^2] = (xp_x -yp_y)p_xp_y [/tex]

where
[tex] p_i = \frac{\hbar}{i}\frac{\partial}{\partial i}, \quad r^2 = x^2 + y^2 + z^2,\\
p^2 = p_x^2 + p_y^2 + p_z^2, \quad L_z = yp_z - zp_y [/tex]



The Attempt at a Solution


I started with [itex][L_z, H][/itex], expanded it, cancelled a bunch of stuff and wound up with,
[tex] [L_z, H] = \frac{1}{2m}(xp_yp_x^2 - yp_xp_y^2) + xp_yV - yp_xV + Vxp_y - Vyp_X. [/tex]
Assuming I did everything right, I need this equation to equal 0 to show [itex]L_z[/itex] commutes with [itex]H[/itex] and the problem suggests that this can only be the case if [itex]V[/itex] is a function of [itex]r[/itex], but I don't see how to get there. Any help would be appreciated.
 
Last edited:

Answers and Replies

  • #2
29
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Ok, so I can simplify that last equation to get
[tex][L_z, H] = \frac{1}{2m}[L_z, p^2] + [L_z, V] =\frac{i\hbar}{2m}(p_y^2 -p_x^2) + [L_z, V][/tex]

How does this imply [itex]V[/itex] must be a function of [itex]r[/itex]?
(I guess I should have seen this from the form of the Hamiltonian).
 
Last edited:

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