# Translation invariant for p

1. Oct 30, 2014

### MickaelPC

Hi everyone,

Let $\psi (x)$ be a one dimension wave function. We suppose $\langle p \rangle =q$ How can I compute the new $\langle p \rangle$ when we set $\psi _1 (x) := e^{\frac{ip_0x}{\hbar}}\psi (x)$ I want to comute it with the formula $$\langle p \rangle=-i \hbar \int \psi _1^* \frac{\partial \psi _1}{\partial x}\mathrm{d}x=q + p_0 \int \psi ^* (x) \psi (x) e^{\frac{ip_0}{\hbar}} \mathrm{d}x$$ but I'm not able to compute the last term, which I want to be equal to zero.

2. Oct 30, 2014

### Staff: Mentor

Well first of all you made a mistake with the last term - you left out the phase factor associated with the conjugate which would cancel the other phase factor, and since the wave function is normalised the integral is one.

But I think you need to explain why you think it should be zero? In multiplying it by what you did you have shifted its momentum by p0 and that's exactly what you got - p0 added to the momentum. To see this imagine your wave-function is in an eigenstate of momentum p. Multiply it by what you did and you get a wave-function in an eigenstate of p + p0.

Thanks
Bill

Last edited: Oct 30, 2014
3. Oct 30, 2014

### MickaelPC

You are right, I made a mistake !

I do not know why I was sure that the result would be the same ... and your comment explain exactly how to have the intuition for answering right !