- #1
Mark_L
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Homework Statement
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##:
Homework 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}##
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.