Confusion over ladder operators

Question

A system is in the state $|l,m\rangle$, an eigenstate of the angular momentum operators $\hat{L}^2$ and $\hat{L}_z$. I'm using ladder operators, $\hat{L}_+$ and $\hat{L}_-$, to calculate $\left<\hat{L}_x\right>$.

I have gotten as far as:

$$ \left<\hat{L}_x\right> = \frac{1}{2} \left<l, m|(\hat{L}_- + \hat{L}_+)|l,m\right> $$

I know this should equal zero. I was under the impression that the raising and lowering operators cancel each other out, but I'm unsure how to show this mathematically. I have tried using the fact that $\hat{L}_+$ and $\hat{L}_-$ are a Hermitian conjugate pair and found:

$$ = \hbar \left<l, m+1| Re(c_{l,m}^{+})|l, m+1\right>, $$

$$ = 0 $$

where $c_{l,m}^{+} = \sqrt{(l-m)(l+m+1)} $.

Is this correct or if not, how should I go about showing $\left<\hat{L}_x\right>=0$?

Edit: The $c_{l,m}^{+}$, is supposed to be + not $\dagger$, as it corresponds to the raising operator.

Answer 1

You're correct that you can write $\langle L_x\rangle$ as $$ \langle\hat{L}_x\rangle = \frac{1}{2} \langle l, m|\hat{L}_- + \hat{L}_+|l,m\rangle $$But your calculation of this bracket seems confused. You should use the idea that $\langle\psi|\hat{A}+\hat{B}|\psi\rangle=\langle\psi|\hat{A}|\psi\rangle+\langle\psi|\hat{B}|\psi\rangle$.

Thus, you'd get two terms $$\langle l, m|\hat{L}_- |l,m\rangle\text{ and }\langle l, m| \hat{L}_+|l,m\rangle$$ I will discuss the first term first. Since the lowering operator would lower the ket, its inner product with the bra of the original vector would vanish, i.e., $\langle l, m|\hat{L}_- |l,m\rangle\propto\langle l,m|l,m-1\rangle=0$. And you can make the exact same argument for the second term.