QM: Commuting the Hamiltonian with position

1. Jan 31, 2009

Niles

1. The problem statement, all variables and given/known data
Hi all.

I am commuting the Hamiltonian (H = p2/(2m) + V(x)) with position. This is what I get:

$$[H,x] = -\frac{i\hbar}{m}p,$$

where p is the momentum operator. But here's my question: The momentum-operator contains d/dx, so does this mean that the commutator is zero, or do I leave it as I have derived it above?

Because my trouble is that I have to take the expectation value of the above commutator, and how does it make sense to take the expectation value of an operator?

I hope you can shed some light on this. Thanks in advance.

Sincerely,
Niles.

Last edited: Jan 31, 2009
2. Jan 31, 2009

Fredrik

Staff Emeritus
Why would it mean that? d/dx doesn't commute with x.

Operators are the only things you ever take expectation values of in QM, so why wouldn't it make sense? The interpretation is that the expectation value tells you the average of a large number of measurements of the operator performed on identically prepared systems.

3. Jan 31, 2009

Niles

First, thanks for replying. What I mean is that we have a d/dx just "hanging" in the air, i.e.:

$$[H,x] = -\frac{i\hbar}{m}\frac{\hbar}{i}\frac{d}{dx}.$$

The d/dx works on a constant ("1", I guess?), so that's why I think that it might be zero. But you say it is not?

Ok, that's comforting to know. That at least takes care of problem #2.

Last edited: Jan 31, 2009
4. Jan 31, 2009

turin

The d/dx does not work on a constant (necessarily). I suppose you could say that it is just hanging in the air. Remember, this is QM, and so observables are operators. What about H? Is H=V(x) because the Laplacian is working on a constant? No; it is also an operator. x is an operator, too, but we treat it like a regular number when we work in the configuration basis (as you are doing), which is the eigenbasis of x, so you can replace the operator x with its eigenvalue. If it makes you feel more comfortable, work with the "matrix elements":

$$\int{}dx\ \psi_1^*(x)[H,x]\psi_2(x)=-\frac{\hbar^2}{m}\int{}dx\ \psi_1^*(x)\frac{d\psi_2(x)}{dx}$$

The only reason it would work on a constant is if a constant could solve the Schroedinger equation.

5. Feb 1, 2009

Niles

Great, thanks to both of you for taking the time to help me.