Expectation value of angular momentum

[fluidistic]

Homework Statement

A particle is under a central potential. Initially its wave function is an eigenfunction $\psi$ such that $\hat {\vec L ^2} \psi = 2 \hbar ^2$ , $\hat L_3 \psi =0$.
Calculate the expectation value of $\hat {\vec L}$ for all times.

Homework Equations

$\frac{d}{dt}<\hat {\vec L}>=\frac{i}{\hbar } [\hat H , \hat {\vec L}]$.
I get that the quantum numbers of the eigenfunction are $l=1$, $m=0$. Not sure if that helps.

The Attempt at a Solution

Using the relevant equation, I get that $\frac{d}{dt}<\hat {\vec L}>=0$ because H commutes with L because it's a central potential problem. This means that the expectation value of the operator L is constant in time.
So I can calculate it at t=0, and use the information that at $t=0$, $\Psi (x,t=0)$ is an eigenfunction (of the Hamiltonian but also of L^2 and L_z apparently).
So here's my problem: $<\hat {\vec L}>(0)=<\psi , \hat {\vec L} \psi>$. I don't know how to calculate this. I've checked in Nouredine Zetilli's book because its said to have many solved problems but I couldn't find a single problem where one have to calculate the expected value of a vector operator such as $\hat {\vec L}$.
Using intuition I guess I should rewrite that expectation value in terms of the expectation values of L^2 and L_z (=L_3) but I don't know how I could do this.

Here's an attempt: $\hat {\vec L^2}=L_x^2+L_y^2+L_z^2$.
$\Rightarrow (L_x^2+L_y^2) \psi =2\hbar ^2 \psi$. Stuck here.

Here's another attempt: I want the expectation value of $<L_x \hat i + L_y \hat j + L_z \hat k>$. I know I can't write this as $<L_x>+<L_y>+<L_z>$. So I'm stuck here.

Any help is appreciated.

 

[Oxvillian]

Here's another attempt: I want the expectation value of $<L_x \hat i + L_y \hat j + L_z \hat k>$. I know I can't write this as $<L_x>+<L_y>+<L_z>$.

Well, obviously not, since it's a vector

But you can write it as $\langle L_x \rangle \hat i+ \langle L_y \rangle \hat j + \langle L_z \rangle \hat k$.

 

[fluidistic]

Well, obviously not, since it's a vector

But you can write it as $\langle L_x \rangle \hat i+ \langle L_y \rangle \hat j + \langle L_z \rangle \hat k$.

I see thanks, with the last term being equal to 0.
So I am left with $<\hat {\vec L}>=<L_x>\hat i + <L_y> \hat j$. I'm not sure on how to proceed.

 

[Oxvillian]

Try expressing $L_x$ and $L_y$ in terms of ladder operators.

 

[fluidistic]

Try expressing $L_x$ and $L_y$ in terms of ladder operators.

Oh right...
$L_x \psi =L_x |1,0>=\frac{\hbar}{\sqrt 2} (|1,1>+|1,-1>)$. Similarly $L_y=\frac{\hbar}{\sqrt 2 i}(|1,1>+|1,-1>)$.
Then if the relation $<L_x>=\frac{\hbar}{\sqrt 2}(<|1,1>>+<|1,-1>>)$ is true, then $<L_x>=0$ and $<L_y>=0$.

This would make that $<\hat { \vec L}>(t)=0 \hat i + 0 \hat j + 0 \hat k$.
So I am learning that the expectation value of a "vector operator" is actually a vector... I would have thought that the expectation value of anything would be a scalar!