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1. Jun 14, 2016

### Joker93

Hello!
Could somebody please tell me how i can compute the expectation value of the momentum in the case of a free particle(monochromatic wave)? When i take the integral, i get infinity, but i have seen somewhere that we know how much the particle's velocity is, so i thought that we can get it from the expectation value of its momentum.

Also, Wikipedia states(here: https://en.wikipedia.org/wiki/Free_particle#Measurement_and_calculations -- just scroll down a bit) that one CAN do the calculation via the integral and get hbar*k.

Thanks!

2. Jun 14, 2016

### A. Neumaier

The result depends on the state in which you compute the expectation. In a Gaussian state the result is finite.

3. Jun 14, 2016

### Joker93

Hello! No, no, i am talking about a monochromatic wave(single k).

4. Jun 14, 2016

### A. Neumaier

For an unnormalizable wave function such as $|k\rangle$ you cannot talk about expectations, let alone compute them!

5. Jun 14, 2016

### Joker93

So, why does Wikipedia(the link i have provided) state that it is hbar*k?

6. Jun 14, 2016

### A. Neumaier

You must calculate the expectation for wave packets (such as Gaussians), and then take the limit of infinite spread in position, and zero spread in momentum.

7. Jun 14, 2016

### Joker93

Gaussian with infinity variance gives a plane wave or a constant(in space) function?

8. Jun 14, 2016

### A. Neumaier

$\psi(x)=e^{\pm ik\cdot x-x^2/2\sigma^2}$ gives (after normalization, and with the correct sign) a moving Gaussian wave packet with momentum $\langle p\rangle=\hbar k$ and approaches a plane wave in the limit $\sigma\to\infty$

9. Jun 14, 2016

Thanks!