# Commutators of angular momentum, linear momentum squared and H

1. Mar 20, 2013

### Mark_L

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. Mar 20, 2013

### Dick

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.

3. Mar 20, 2013

### Mark_L

Why are they zero?

4. Mar 20, 2013

### Dick

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.

5. Mar 20, 2013

### Mark_L

Okay, so what about the next part?

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