Expected value of linear momentum with real wave function

Question

Let's suppose that a quantum system is characterized by a real wave function $\psi(x)$. How can I prove that in this case the expected value of the linear momentum $\mathbf{p}$ is zero, using the position representation?

I tried using the expression of this operator in the position representation, but I fail to see why the resulting integral has to equal zero.

Answer 1

This is quite similar to this question

The expectation value is:

$$ \langle p\rangle = \frac{\hbar}{i}\int_{-\infty}^{\infty}dx\ \psi^*(x)\frac{\partial \psi(x)}{\partial x} \underbrace{=}_{\psi \ \mathrm{real}} \frac{\hbar}{i} \int_{-\infty}^{\infty}dx\ \psi(x)\frac{\partial \psi(x)}{\partial x}\ $$

The result then follows directly by integrating by parts (assuming $\psi$ vanishes at spatial infinity, as it must to be normalizable): $$ \langle p \rangle = -\frac{\hbar}{i}\int_{-\infty}^{\infty}dx\ \frac{\partial \psi(x)}{\partial x}\psi(x) $$ So that $\langle p\rangle = -\langle p \rangle$, which implies that $\langle p\rangle = 0$.

There are many other ways of seeing this. The first is that $p$ is a Hermitian operator, so the expectation value is necessarily real. As the integral is real, but there's a factor $\hbar/i$ in front, the integral must be zero.

An alternative way is to consider the Fourier decomposition of $\psi(x)$, which can only contain the $k = 0$ term if $\psi(x)$ is real.

Answer 2

According to this answer, the derivative of an even function is odd, and the derivative of an odd function is even. The momentum operator in 1-D is $$ \hat{p} = -i\hbar\frac{\partial}{\partial x}, $$ so the expectation of the momentum is $$ \langle p \rangle = -i\hbar\int_{-\infty}^\infty \psi(x) \frac{\partial \psi}{\partial x}dx. $$ The integrand must be odd, and the integral of an odd function over a domain $(-L,L)$ for any $L$ is zero.