# Is the expectation value of momentum of a stationary state zero?

1. Jun 21, 2011

### praharmitra

Given a stationary state

$$H \psi = E \psi \Rightarrow \left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)\right)\psi = E\psi$$

Firstly is it true that

$$\left<p\right> = \frac{\hbar}{i}\int\psi^* \frac{\partial \psi}{\partial x} dx= 0$$ ??

If it is, how do we prove it?

2. Jun 21, 2011

### Matterwave

Yes, it is true. You can prove it by using Ehrenfest's theorem. By the correspondence principle, <p>=m*d<x>/dt. In a stationary state, expectations do not change in time, and so d<x>/dt=0.

3. Jun 21, 2011

### praharmitra

The point I have reached till now, has not yet proved Ehrenfest Theorem (It has been merely pointed out as an amazing coincidence!)

I was wondering if it were possible to show by some brute force calculations and playing around with derivatives?

4. Jun 21, 2011

### Matterwave

Ehrenfest's theorem is only used to prove <p>=m*d<x>/dt. Correspondence principle, though, gives you that result immediately. If you believe in the correspondence principle, then you can get the result immediately.

There probably is a brute force method, but I can't think of one at this moment.

5. Jun 21, 2011

### xlines

Well, I'm not sure this is true for any V(x). Namely, if you take V(x) = 0, then you would not get <p> = 0, obviously. If you are thinking of bound states, it kinda feels one could devise up such V(x,t) to get <p> != 0. But I'll think this trough ...

6. Jun 21, 2011

### Physics Monkey

Some stationary states have non-zero momentum, for example, all the stationary states of H = p^2 + V with V = 0. Of course, these are not bound states.

If you would like to focus on bound states then there are several ways you might proceed.

If the wavefunction is real and normalizable, then it follows from a few manipulations of the definition that <p> = 0. Thus one way to make partial progress would be to investigate when bound states can be made real. You may already know the answer to this question.

You could also just prove <p> = m d<x>/dt for bound states directly. For example, start from the right hand side, bring the time derivatives inside, and use Schrodinger's equation. You'll end up considering the expectation value of a certain commutator.

7. Jun 21, 2011

### Physics Monkey

There is an amusing exception to this rule. A particle on a ring has stationary states with <p> not equal to zero. (A particle on a line also has this feature, but there the momentum "eigenstates" are not normalizable as you know). Ehrenfest's theorem is evaded because there is no single valued operator x, only exp(i x/R) with R the ring radius makes sense.

8. Jun 21, 2011

### praharmitra

I figured it out. Schrodinger's equation is

$$\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)\right)\psi = E\psi$$
Multiply this by $\psi^*$ and its c.c. by $\psi$ and subtract. Then integrate, we get

$$\psi \frac{\partial \psi^*}{\partial x}= c+ \psi^* \frac{\partial \psi}{\partial x}$$
The constant $c$ must be zero because of the boundary conditions of $\psi$ at infinity. (This is simply the statement that the probability current of a stationary state is zero).

Integrating both sides again and using integration by parts we can show

$$\left<p\right> = 0$$

9. Oct 17, 2012

### patricieni

What about the case when the potential is dependent on x squared? In the case of a quantum harmonic oscillator if we apply Ehrenfest's theorem it yields that d<p>/dt=-m*w^2*<x> (<x> is 0) is 0 so it means <p> can be a constant greater than 0.
Am i missing something?

Thanks

10. Oct 18, 2012

### Jazzdude

No, it's not true. The obvious counter example is the free particle hamiltonian and the momentum eigenstates. They're all stationary and can have any momentum expectation you desire.

In general you will also find bound stationary states with non-vanishing angular momentum. Linear momentum will have to vanish because a bound state cannot go anywhere. But only bound states are normalizable, so you should be able to prove that any proper energy eigenstate in fact results in <p> = 0.

Last edited: Oct 18, 2012