QM: Commuting the Hamiltonian with position

• Niles
In summary, Niles is trying to figure out if the momentum-operator is zero because the d/dx does not work on a constant, and Sarah is helping him figure out how to work with the matrix elements.
Niles

Homework Statement

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.Niles.

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Niles said:
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?
Why would it mean that? d/dx doesn't commute with x.

Niles said:
how does it make sense to take the expectation value of an operator?
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.

Niles said:
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?

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?

Fredrik said:
Operators are the only things you ever take expectation values of in QM

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

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

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?
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.

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

1. How does the Hamiltonian commute with position?

The Hamiltonian operator, which represents the total energy of a quantum system, commutes with the position operator because both operators are Hermitian and have continuous eigenvalues. This means that the order in which they are applied does not affect the outcome, and they can be considered independent of each other.

2. What is the significance of commuting the Hamiltonian with position?

Commuting the Hamiltonian with position is significant because it allows us to simultaneously measure both the energy and position of a quantum system. This is known as a compatible set of observables, and it is a fundamental concept in quantum mechanics.

3. Can the Hamiltonian commute with other operators besides position?

Yes, the Hamiltonian can commute with other operators besides position, as long as they are also Hermitian and have continuous eigenvalues. Examples of other operators that commute with the Hamiltonian include momentum, angular momentum, and spin.

4. What is the mathematical expression for the commutator of the Hamiltonian with position?

The commutator of the Hamiltonian with position is given by [H, x] = -iħ(dH/dx), where H is the Hamiltonian operator and x is the position operator. This expression shows how the Hamiltonian and position operators are related and how they commute with each other.

5. How does commuting the Hamiltonian with position relate to the Heisenberg uncertainty principle?

Commuting the Hamiltonian with position is related to the Heisenberg uncertainty principle, which states that it is impossible to simultaneously know the exact position and momentum of a particle. Since the Hamiltonian and position operators commute, this means that the energy and position of a system can be known with arbitrary precision, but this comes at the cost of knowing nothing about the momentum and vice versa.

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