# Expectation value of an imaginary operator acting on a real function

In a video (http://youtu.be/r_gBQ_qhg8U?t=9m58s) it's stated that a matrix element of an imaginary operator acting on a real wave function is zero, i.e. $$\langle\text{real}|\text{imaginary}|\text{real}\rangle ~=~ 0,$$ and I don't really understand why.

When we make an actual calculation, won't the $i$ from $l_z$ simply move in front of the integral and have no influence on its actual value?

• Welcome to Physics Stack Exchange. What is an "imaginary operator"? Many users on this site won't click on links to videos. It is actually site policy that you include whatever you need from any link directly into the question. This has several benefits: 1) It avoids link rot, 2) It forces you to think about what material in the link is really relevant, which often helps you solve your own problem, 3) It makes the post much easier to understand, and therefore means you're much more likely to get an answer. – DanielSank Nov 3 '15 at 11:57
• Hint to the question (v1): Can the expectation value of a self-adjoint operator $\hat{L}_z$ be imaginary? Somewhat related: physics.stackexchange.com/q/16678/2451 – Qmechanic Nov 3 '15 at 12:03
• That is kind of the issue... This statement is just thrown out in the video and there not that much justification provided. I thought that maybe I am missing something very basic here and decided to clarify here. At least now I see that I am not the only one confused by this. :) – J.K. Nov 4 '15 at 8:35

I don't know the term "imaginary operator". I take this to be an antihermitian operator, which eigenvalues are purely imaginary. Then the statement is clearly not true. Take as a counter example any hermitian operator $\hat{A}$ and real wavefunction $\psi$ with $\langle \psi | \hat{A} |\psi\rangle = A_\psi \neq 0$. $A_\psi$ is of course real. Take now $\hat{B} = i \hat{A}$, which is antihermitian. It follows $$\langle \psi | \hat{B} |\psi\rangle =i \langle \psi | \hat{A} |\psi\rangle=i A_\psi\neq 0$$
$$\mathbb{R}~\ni~ \langle \psi |\underbrace{\hat{L}_z}_{\text{self-adj.}} |\psi \rangle ~=~ \int\! d^3r ~\underbrace{\overline{\psi({\bf r})}}_{\text{real}} \underbrace{ (-i\hbar) \frac{\partial}{\partial \varphi}}_{\text{imaginary}} \underbrace{\psi({\bf r})}_{\text{real}} ~\in~i\mathbb{R}.$$