# Expectation of momentum for real valued wave function

I have been trying to learn quantum mechanics since 3 hours ago so bear with me if I seem naive. Working through Shankar's book it was easy to do the following exercise: show that for a real wave function, the expectation of the momentum $\langle P \rangle$ is $0$.

I am just wondering: why is this true, intuitively, from a physical standpoint? Thanks.

The eigenstates of the momentum operator are $e^{ikx}$. These are complex.
Let's now try to build a real wave-function out of these eigenstates. It is fairly easy to see that if we add one $e^{ikx}$ state, we can ensure our wavefunction is real by adding the complex conjugate of this state. That is $e^{-ikx}$. Hence in general, our wavefunction will be real, if we add $k$-states in pairs: $$\Psi_k = \frac{1}{\sqrt{2}}\bigg( e^{ikx} + e^{-ikx}\bigg).$$
So to keep the constraint of the wave-function being real, we've had to add up eigenstates, each in pairs moving in opposing directions. That means this wavefunction is symmetrical under $x\rightarrow-x$, and so there is no preferred direction. This means if we look at the momentum, then it must show no preferred direction either, giving $\langle p\rangle=0$.
You can see this explicitly by operating with the momentum operator, $i\hbar \partial_x$ on $\Psi_k$, which gives you zero. From the principle of superposition then, any arbitrary real-valued wave-function built from $\Psi_k$ will also be zero.
Another way to see this in general, is that: $$\langle p \rangle = \int \Psi^* (i \frac{d}{dx}) \Psi dx = i \int \Psi \Psi' dx,$$ where the integral, of the product of two real functions, must be real. Then by equating real and imaginary parts, we see the integral must evaluate to zero.