# Quantum mechanics momentum operator

1. May 22, 2010

### mohammed.omar

Hi All,

I've seen many derivations for the momentum operator, but I've a rather naive problem that I cannot figure out in the derivation done by Griffiths in "Introduction to Quantum Mechanics" book. In chapter 1, when he derives the momentum operator he states:

$$\frac{d <x> }{dt} = \frac{d}{dt}\int x |\psi (x,t)|^2 dx = \int x \frac{\partial}{\partial t} |\psi (x,t)|^2 dx$$

i.e. He assumed $$\frac{\partial}{\partial t} x = 0$$

Why did he do that? Is there any justification for it?

2. May 22, 2010

### alxm

X is just a coordinate. Why would it depend on t?

3. May 24, 2010

### mohammed.omar

Thanks alxm.

I got it, but I mixed up the wave nature with the particle nature. I thought x would represent the position of the particle with time. This is very embarrassing.

Thanks a lot alxm.

4. May 25, 2010

### alxm

No need for embarrassment. If you figured that out now, you're earlier than a lot of students I've seen who don't think about it enough, and then suddenly get surprised when they move to many-particle systems and can't figure out why the wave function is suddenly 3N-dimensional.

5. May 25, 2010

### mohammed.omar

Thanks alxm.

This is rather interesting. For an N-particle system, there will be 3N dimensional wave function other than time? 3 generalized for each particle?

6. May 27, 2010

### xepma

Yes, exactly. In non-relativistic quantum mechanics time is not considered a seperate degree of freedom; only the spatial coordinates are.

PS In Griffiths the coordinate x is always an operator, and therefore time independent. But plenty of other books use x(t) (or sometimes q(t)) to denote the position of the particle, so it's good to be weary of that!

Last edited: May 27, 2010
7. May 27, 2010