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QM: Linear momentum of angular momentum eigenstate

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    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]


    2. Relevant 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]


    3. 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]
     
  2. jcsd
  3. Oct 2, 2011 #2

    vela

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    Start with [itex]-i\hbar\langle l' m' \lvert P_x \rvert l m \rangle = \langle l' m' \lvert [L_z,P_y] \rvert l m \rangle[/itex].
     
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