# Can average momentum be imaginary?

I am new to quantum physics. We just learnt about wave equations, observables and expectation values today. What really caught my attention was the expectation value of average momentum and energy:

$$\langle p \rangle = \int_{-\infty}^\infty \text{d}x\,\,\, \psi^*(x,t) \frac{\hbar}{i}\frac{\partial}{\partial x}\psi(x,t)$$

$$\langle H \rangle = \int_{-\infty}^\infty \text{d}x\,\,\, \psi^*(x,t) i\hbar\frac{\partial}{\partial t}\psi(x,t)$$

For the first equation, we take the $$\hbar/i$$ outside the integral. Obviously, the value of the integral has to be either real or complex. If it is complex, then it's completely fine as both the $$i$$s get cancelled out. But what if it's real? I read online that since momentum is represented by a Hermitian operator, all of its eigenvalues are real. Does this mean that the integral in this case is always zero?

I have the same question regarding the average energy. If the integral is complex, then nothing to worry about. But if it is real, then does it have to be zero? On the other hand, if the integral can be a real non-zero value, what does it mean for the average momentum to be imaginary?

I'm not sure what exactly I'm missing here. It would be great if someone could help me out with this. Please note, I'm a complete beginner to this whole concept (as mentioned in the beginning).

• You can immediately show that it is real by taking the complex conjugate and showing that the integral is the same. Jan 9 at 10:06
• Thank you, but what if the wave function is a real expression, it's conjugate will be the same thing, right? So then what happens to the i...does the whole thing always turn out to be zero?? Jan 9 at 10:08
• Yes. Real wavefunctions have an average momentum of zero. Jan 9 at 10:09
• Is this independent of the limits we take?? Jan 9 at 10:12
• Yes, because then the integrand is a total derivative. Jan 9 at 10:18

The integral is indeed zero, and it's quite easy to see why, since if $$\psi(x)$$ is real, then $$\psi^*(x) = \psi(x)$$, so:
$$\langle p \rangle = \frac{\hbar}{i}\int_{-\infty}^\infty \text{d}x\,\,\, \psi(x) \frac{\partial \psi}{\partial x} = \frac{\hbar}{2i}\int_{-\infty}^\infty \text{d}x\,\,\,\frac{\partial }{\partial x} \psi^2(x) = \frac{\hbar}{2i} \times \psi^2(x)\Bigg|_{-\infty}^\infty.$$
Since the wavefunction is real, $$\psi^2(x) \equiv |\psi(x)|^2$$, the probability density. And we know this to be zero at $$\pm \infty$$, since that is one of the requirements for a wavefunction.
• My teacher posed this question in class, where I had to find the expectation value for average momentum for the wave function,$$xsin^{2}\frac{2πx}{L}$$ and we were instructed to find this by integrating it within the limits 0 and L. I got L/2 as the answer, but when I tweaked the limits, and integrated it from 0 to L/2, I got a complex number as the answer. Can you please tell me what this means?? Jan 11 at 4:51
• (a) You must have made a mistake with the integral from 0 to $L$: it's clearly zero. (b) When you integrate it between arbitrary limits, this is no longer the expectation value: the expectation value is defined a certain way. You can certainly try to integrate this function from $a$ to $b$, and you will find (from the equation in my answer) that it is $=\frac{\hbar}{2i} \left(\psi^2(b)-\psi^2(a)\right)$. This is a complex number, yes, but it is not the expectation value! It's just an integral. (Curiously, if you do it from 0 to $L/2$ you should still get 0, but that's a coincidence.) Jan 11 at 6:54