Commutators and wave function

1. Aug 3, 2011

atomicpedals

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

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

2. Relevant equations

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

3. 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 ]$\psi$(x)?

2. Aug 3, 2011

George Jones

Staff Emeritus
What is [x,p]?

3. Aug 3, 2011

George Jones

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

4. Aug 3, 2011

atomicpedals

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

5. Aug 3, 2011

George Jones

Staff Emeritus
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.

6. Aug 3, 2011

atomicpedals

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

7. Aug 3, 2011

atomicpedals

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

8. Aug 3, 2011

George Jones

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

9. Aug 4, 2011

atomicpedals

(i $\hbar$)p+p(i $\hbar$) ?

10. Aug 4, 2011

George Jones

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

11. Aug 4, 2011

atomicpedals

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

(i $\hbar$ I)p + p(i $\hbar$ 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 $\psi$(x)? (a complex wave equation in momentum space?)

12. Aug 4, 2011

George Jones

Staff Emeritus
Yes, $i\hbar I$ commutes with all operators, and $IA = A$ for all operators $A$.
$\psi \left( x \right)$ is a complex wave function (not equation) in position space (since $\psi$ is a function of position $x$). What is the position space representation of the momentum operator, i.e., $p = ?$ This also will have been covered earlier in your text/course.

13. Aug 5, 2011

atomicpedals

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