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:

$$\left<\hat{L}_x\right> = \frac{1}{2} \left$$

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:

$$= \hbar \left,$$

$$= 0$$

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.

• Hint: what is the result of applying $L_+$ to |l,m> ? May 7 '20 at 20:19

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.