Square of z-component of angular momentum eigenvalues

[tomwilliam2]

Homework Statement

I'm trying to demonstrate that if:

$$\hat{L}_z | l, m \rangle = m \hbar | l, m \rangle$$

Then

$$\hat{L}_z^2 | l, m \rangle = m^2 \hbar^2 | l, m \rangle$$

Homework Equations

$$\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2$$
$$\hat{L}_z = -i\hbar \left [ x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right ]$$

The Attempt at a Solution

I'm not really sure where to start with this. I can apply the operator $\hat{L}_z^2$ to an arbitrary function $f(x,y)$, but that gives me:

$$\hat{L}_z^2 f(x,y) = \hbar^2\left(x^2 \frac{\partial^2}{\partial y^2} - xy \frac{\partial}{\partial y} \frac{\partial}{\partial x} -xy \frac{\partial}{\partial x} \frac{\partial}{\partial y} + y^2 \frac{\partial^2}{\partial x^2} \right)f(x,y) $$
I've no idea if this demonstrates anything at all...

 

[vanhees71]

Don't think too complicated. Just apply \hat{L}_z twice to the eigenstate. Generally one can say that almost any calculation concerning angular-momentum operators is easier in the representation-free Hilbert-space formulation than using the differential operators of the position representation.