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.


1 Answer 1


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.

As far as physical intuition, I'd say it's more showing that the constraint of asking for a real wave-function is very limiting.


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