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

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Homework Help Overview

The discussion revolves around the commutation relations in quantum mechanics, specifically focusing on the commutator [x, p2] and its application to the wave function ψ(x). The original poster attempts to derive these relations using prior results and seeks clarification on the implications of their findings.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of the commutation relation [x, p] and its implications for calculating [x, p2]. Questions arise regarding the treatment of operators and the nature of the wave function in relation to momentum space.

Discussion Status

Some participants provide guidance on the canonical commutation relation and its application, while others question the original poster's reasoning and suggest evaluating expressions explicitly. Multiple interpretations of the results are being explored, particularly regarding the representation of the momentum operator.

Contextual Notes

There is an emphasis on using previous knowledge from the course material, and participants note the importance of understanding the position space representation of operators. The discussion reflects a collaborative effort to clarify concepts without reaching a definitive conclusion.

atomicpedals
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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)?
 
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What is [x,p]?
 
Yes, you are on the right track. I should have started my previous post with this.
 
My first instinct is to answer [x,p]=-[p,x]...but I don't think that was what you were asking?
 
atomicpedals said:
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.
 
Am I correct in thinking that x is position and p momentum?
 
The canonical commutation? [x,p]=i [itex]\hbar[/itex]
 
Yes; now, use this in the expression at the end of your first post.
 
(i [itex]\hbar[/itex])p+p(i [itex]\hbar[/itex]) ?
 
  • #10
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.
 
  • #11
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?)
 
  • #12
atomicpedals said:
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].
atomicpedals said:
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.
 
  • #13
I think I'm starting to grasp this. It seems like the final solution is heading towards something similar to px = i [itex]\hbar[/itex] [itex]\partial[/itex]/[itex]\partial[/itex]x .
 

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