Angular Momentum and Eigenfunctions

1. May 7, 2014

unscientific

1. The problem statement, all variables and given/known data

Part (a): What is momentum operator classically and in quantum?
Part (b): Show the particle has 0 angular momentum.
Part (c): Determine whether angular momentum is present along: (i)z-axis, (ii) x-axis and find expectation values <Lz> and <Lx>.
Part (d): Find the result of finding Lz then Lx.
Part (e): Find the result of finding Lx then Lz.

2. Relevant equations

3. The attempt at a solution

Part (a)
Classically, $\vec L = \vec r x \vec p$.
Quantum mechanically, $\vec L = -i\hbar \vec r x \vec \nabla$.

Part (b)
Since $L^2$ is the angular part of $\nabla^2 = -\frac{1}{sin \theta}\frac{\partial}{\partial \theta}\left(sin \theta \frac{\partial}{\partial \theta}\right) + \frac{1}{sin^2\theta}\frac{\partial^2}{\partial \theta^2}$, it has no radial dependence.

So $\langle \psi|L^2|\psi\rangle = 0$.

Part (c)

Along z-axis, it is zero, since $L_z = -i\hbar \frac{\partial}{\partial \phi}$.

Along x-axis, it is non-zero, since $L_x$ is a function of both $(\theta, \phi)$.

Expectation value $\langle L_z\rangle = 0$.

I'm not sure how to find the expectation value $\langle L_x \rangle$ without using brute force integration. Is there a trick somewhere I've missed?

The rest of the question boggles me very much. Would appreciate any help!

2. May 7, 2014

strangerep

FYI, you can use the "times" macro in Latex instead of your $x$. That makes things look nicer:
$\vec L = \vec r \times \vec p$
$\vec L = -i\hbar \vec r \times \vec \nabla$

You have the expression $\vec L = -i\hbar \vec r \times \vec \nabla$. Can you re-write this in x,y,z component form, i.e.,
$L_x ~=~ \dots\text{what?}\dots$
(Hint: use the definition of the $\times$ cross-product.)

Then apply the operator $L_x$ to your wave function. It's probably best for your education if you work it out by brute force first, then look to see if there's any shortcuts for subsequent parts of the question.

3. May 7, 2014

Simon Bridge

(a) not bad ... in LaTeX the cross-product is provided by "\times" so:
classically: $\vec L = \vec r \times \vec p$
QM: $\hat L = \hat r\times \hat p = -i\hbar(\vec r \times \vec \nabla)$ ... remember that $\hat L$ is an operator. The angular momentum is the result of using the operator.

(b) the statement is kinda correct - but it does not quite answer the question.
note: just because the operator has no radial dependence does not mean that the result of applying the operator to the wavefunction will have no radial dependence,

I suspect you are expected to work the problem in cartesian coordinates - though the spherical argument is a nice one.

(c) No pain no gain: you'll have to use the brute force integration.
This one may actually be easier in Cartesian coordinates or cylindrical-polar.

[edit: strangerep beat me to it]