How Does the Commutator [p-hat_x, H-hat] Reveal Quantum Mechanics Fundamentals?

[blueyellow]

Homework Statement

by considering the action of [p-hat (subscript x), H-hat] on a general state, show that

[p-hat (subscript x), H-hat] =-ihbar dV/dx

Homework Equations

H-hat = (((p-hat)^2)/2m) +V(x)

p-hat (subscript x)= -i*h d/dx (partial derivative)

The Attempt at a Solution

tried lookin it up couldn't find anything

 

[Mike Pemulis]

It's right in front of you. Two hints:

1. What is the definition of the commutator of two operators? In other words, how would I write out [A, B]?

2. What is the derivative of a constant?

 

[blueyellow]

so the commutator is AB-BA

so it equals
i*h-bar d/dx(((p-hat)^2)/2m)+V(x)) -(((p-hat)^2)/2m)+V(x)) i*h-bar d/dx
=-ih-bar dV/dx ...
but why does -(((p-hat)^2)/2m)+V(x)) i*h-bar d/dx go to nothing? what do i do please, if the i*h-bar d/dx is put after the hamiltonian?

 

[Mike Pemulis]

what do i do please, if the i*h-bar d/dx is put after the hamiltonian?

In that case, you consider the derivative to be operating on 1 -- in other words d/dx = d/dx(1) = 0. So in the commutator, the only piece that doesn't vanish is the piece that features d/dx operating on a function of x -- which gives you -ih-bar dV/dx.

 

[rwisz]

Based on the given information, it appears that the homework question is asking you to show that the commutator of the x-component of momentum operator (p-hat (subscript x)) and the Hamiltonian operator (H-hat) is equal to -ihbar times the partial derivative of the potential energy function with respect to position (dV/dx). This is a well-known result in quantum mechanics and is related to the Heisenberg uncertainty principle.

To show this, you can start by considering the action of the commutator [p-hat (subscript x), H-hat] on a general state |ψ>. This can be written as:

[p-hat (subscript x), H-hat] |ψ> = (p-hat (subscript x)H-hat - H-hat p-hat (subscript x))|ψ>

Using the definitions of p-hat (subscript x) and H-hat given in the problem statement, we can rewrite this as:

[p-hat (subscript x), H-hat] |ψ> = (-ih d/dx)(((p-hat)^2)/2m)|ψ> + V(x)(-ih d/dx)|ψ> - (-ih d/dx)(V(x)|ψ>)

Expanding the terms and rearranging, we get:

[p-hat (subscript x), H-hat] |ψ> = (-ih/2m)(p-hat)^2d/dx|ψ> + V'(x) (-ih d/dx)|ψ> - (-ih d/dx)(V(x)|ψ>)

where V'(x) represents the derivative of the potential energy function V(x) with respect to position x.

Now, using the commutator relation [p-hat, (p-hat)^2] = 2p-hat, we can simplify the first term as:

(-ih/2m)(p-hat)^2d/dx|ψ> = (-ih/2m)(2p-hat)d/dx|ψ> = (-ih/m)p-hat d/dx|ψ>

Substituting this back into our original equation, we get:

[p-hat (subscript x), H-hat] |ψ> = (-ih/m)p-hat d/dx|ψ> + V'(x