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Commutators of angular momentum, linear momentum squared and H

  1. Mar 20, 2013 #1
    1. The problem statement, all variables and given/known data

    For the linear momentum operator ##\hat{\mathbf{p}}## and angular momentum operator ##\hat{\mathbf{L}} ##, prove that ##\begin{eqnarray}
    [\hat{\mathbf{L}},\hat{\mathbf{p}}^2]&=&0\end{eqnarray}##:

    [Hint: Write ##\hat{\mathbf{L}}## as the ##x##-component of the angular momentum operator and evaluate ##[\hat{L}_{x},\hat{\mathbf{p}}^2]\psi##.
    The results for the other components will follow by symmetry.]

    Hence show that if the potential is given by ##V(|\mathbf{x}|)##, then ##[\hat{\mathbf{L}},\hat{H}] = 0##:

    2. Relevant equations

    ##\begin{eqnarray}\hat{\mathbf{P}}&=&-i\hbar\nabla\\
    \hat{L}_x&=&y\hat{p}_{z}-z\hat{p}_{y}\\
    \hat{H}&=&-\frac{\hbar^2}{2m}\nabla^2+V\\
    \end{eqnarray}##

    3. The attempt at a solution

    After some algebra just using ##[A,B]=AB-BA## and substituting the equations above,

    ##[\hat{L}_{x},\hat{\mathbf{p}}^2]\psi=i\hbar^3y\frac{\partial d}{\partial z}(\nabla^2(\frac{\partial\psi}{\partial z})-i\hbar^3z\nabla^2(\frac{\partial\psi}{\partial y})+i\hbar^3\nabla^2(z\frac{\partial\psi}{\partial y})-i\hbar^3\nabla^2(y\frac{\partial\psi}{\partial z})##

    and doing all the derivatives eventually gave ##[\hat{L}_{x},\hat{\mathbf{p}}^2]\psi=i\hbar^3[\frac{\partial^2 z}{\partial x^2}\frac{\partial\psi}{\partial y}+\frac{\partial^2 z}{\partial y^2}\frac{\partial\psi}{\partial y}+2\frac{\partial^2\psi}{\partial x\partial y}\frac{\partial z}{\partial x}+2\frac{\partial^2\psi}{\partial y^2}\frac{\partial z}{\partial y}-\frac{\partial^2 y}{\partial x^2}\frac{\partial\psi}{\partial z}-\frac{\partial^2 y}{\partial z^2}\frac{\partial\psi}{\partial z}-2\frac{\partial^2\psi}{\partial x\partial z}\frac{\partial y}{\partial x}-2\frac{\partial^2\psi}{\partial z^2}\frac{\partial y}{\partial z}]##

    But this isn't zero so I'm definitely stuck!! I checked all my working and it's definitely all correct so I'm thinking there is a different (hopefully less algebraically intensive) method.

    Any help is greatly appreciated, thanks in advance.
     
  2. jcsd
  3. Mar 20, 2013 #2

    Dick

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    Science Advisor
    Homework Helper

    Yes, it is zero. Things like ##\frac{\partial y}{\partial x}## are zero. Same for all of your second derivatives of coordinates. Only things like ##\frac{\partial x}{\partial x}## aren't zero and that's equal to 1. Aren't they? I think you've been carrying around a lot of terms you could have dropped.
     
  4. Mar 20, 2013 #3
    Why are they zero?
     
  5. Mar 20, 2013 #4

    Dick

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    Science Advisor
    Homework Helper

    Because it's the partial derivative of y taken with respect to x. In taking a partial derivative with respect to x you hold y and z fixed. So it's the derivative of a constant. I'd look back at the definition of partial derivative.
     
  6. Mar 20, 2013 #5
    Okay, so what about the next part?

    We have ##\hat{H}=\frac{1}{2m}\hat{\mathbf{P^2}}+V##
     
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