QM: Linear momentum of angular momentum eigenstate

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
center o bass
Messages
545
Reaction score
2

Homework Statement


Find [Lz, Px] and [Lz,Py] and use this to show that [tex]\langle l'm'|P_x|lm\rangle = 0[/tex] for [tex]m' \neq m \pm 1.[/tex]


Homework Equations


[tex]L_z|lm\rangle = \hbar m |lm\rangle[/tex]
[tex]L^2|lm\rangle = \hbar^2 l(l+1)|lm\rangle[/tex]
[tex]L_{\pm}|lm\rangle = \hbar \sqrt{l(l+1)-m(m\pm 1)}|l,m\pm 1 \rangle[/tex]


The Attempt at a Solution


It was easy to show that [tex][L_z,P_x] = i \hbar P_y[/tex] and that [tex][L_z,P_y] = - i\hbar P_x[/tex] but how might I use this to show that [tex]\langle l'm'|P_x|lm\rangle = 0[/tex] for [tex]m' \neq m \pm 1?[/tex]
 
Physics news on Phys.org