What is the solution to [x,p2] and [x,p2]ψ(x) in quantum mechanics?

[atomicpedals]

Homework Statement

Using the results of the previous problem, find [x,p² ] and from that determine [x,p² ][itex]\psi[/itex](x)

Homework Equations

The solution to the previous problem was [A,BC]=[A,B]C+B[A,C]

The Attempt at a Solution

As I'm suppose to use the results of the previous problem I think what I need to do is

[x,pp]=[x,p]p+p[x,p]

But that doesn't really seem right to me... shouldn't this lead me to an identity (it doesn't say but usually in a QM problem p is momentum and this would seem to be leading to a wave equation in momentum space). If I really am approaching this the correct way, then how should I treat [x,p² ][itex]\psi[/itex](x)?

 

[George Jones]

What is [x,p]?

 

[George Jones]

Yes, you are on the right track. I should have started my previous post with this.

 

[atomicpedals]

My first instinct is to answer [x,p]=-[p,x]...but I don't think that was what you were asking?

 

[George Jones]

My first instinct is to answer [x,p]=-[p,x]...but I don't think that was what you were asking?

While this true, you need to evaluate [x,p] explicitly. The answer is something simple, and will have been covered earlier in your text/course.

 

[atomicpedals]

Am I correct in thinking that x is position and p momentum?

 

[atomicpedals]

The canonical commutation? [x,p]=i [itex]\hbar[/itex]

 

[George Jones]

Yes; now, use this in the expression at the end of your first post.

 

[atomicpedals]

(i [itex]\hbar[/itex])p+p(i [itex]\hbar[/itex]) ?

 

[George Jones]

Right, but remember that [itex]\left[ x , p \right] = i\hbar[/itex] really means [itex]\left[ x , p \right] = i\hbar I[/itex], where [itex]I[/itex] is the identity operator.

 

[atomicpedals]

So in the first part I then end up with a result of

(i [itex]\hbar[/itex] I)p + p(i [itex]\hbar[/itex] I)

Do the standard rules of algebra apply here allowing me to combine this?

Then for the second part is my result just the combined result with the [itex]\psi[/itex](x)? (a complex wave equation in momentum space?)

 

[George Jones]

So in the first part I then end up with a result of

(i [itex]\hbar[/itex] I)p + p(i [itex]\hbar[/itex] I)

Do the standard rules of algebra apply here allowing me to combine this?

Yes, [itex]i\hbar I[/itex] commutes with all operators, and [itex]IA = A[/itex] for all operators [itex]A[/itex].

Then for the second part is my result just the combined result with the [itex]\psi[/itex](x)? (a complex wave equation in momentum space?)

[itex]\psi \left( x \right)[/itex] is a complex wave function (not equation) in position space (since [itex]\psi[/itex] is a function of position [itex]x[/itex]). What is the position space representation of the momentum operator, i.e., [itex]p = ?[/itex] This also will have been covered earlier in your text/course.

 

[atomicpedals]

I think I'm starting to grasp this. It seems like the final solution is heading towards something similar to p_x = i [itex]\hbar[/itex] [itex]\partial[/itex]/[itex]\partial[/itex]x .