# Expected value of linear momentum with real wave function

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.

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.

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.

• What happens if $\psi(x)$ is neither even nor odd? Feb 1, 2021 at 10:08
• As long as $\psi(x)$ is infinitely differentiable, we can find a series expansion. This means that the product $\psi(x)\partial\psi/\partial x$ can be expanded into a sum of polynomial terms, which must be either even or odd, and so my argument holds. I can't think of a situation where a wavefunction wouldn't be even or odd, and not differentiable - certainly not within an undergraduate class. Feb 1, 2021 at 10:28