Quantum mechanics momentum operator

[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:

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

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

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

 

[alxm]

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

 

[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.

 

[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.

 

[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?

 

[xepma]

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?

Yes, exactly. In non-relativistic quantum mechanics time is not considered a separate 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!

 

[mohammed.omar]

Thanks a lot xepma. Your reply was very useful.