# Homework Help: Show that: Commutator relations (QM)

1. Sep 4, 2011

### Diomarte

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

Show that: [p,x] = -iħ,
Show that: [p,x^n] = -niħ x^(n-1), n>1
Show that: [p, A] = -iħ dA/dx

Where p = -iħ d/dx, and A = A(x) is a differentiable function of x.

2. Relevant equations

[p,x] = px - xp;

3. The attempt at a solution

So far I understand part of each of these, to begin with the first part:
[p,x] = [(-iħ d/dx) (x)] - [(x) (-iħ d/dx)]
- From this, I simply get -iħ on the left hand side, which is already my answer that I'm looking for, but I do not understand how the right hand side simply cancels out or goes away, maybe I'm mistaken in my understanding of commutators?

For the 2nd part, I as well have -niħ x^(n-1) on the left, yet again what my goal is, and thus same problem with right hand side... Any assistance in this would be greatly appreciated, thank you.

2. Sep 4, 2011

### vela

Staff Emeritus
It's useful to have the commutator act on a function to see what's going on.
$$[\hat{p},\hat{x}]\psi = \left(-i\hbar\frac{d}{dx}\right) x\psi - x\left(-i\hbar\frac{d}{dx}\right)\psi$$
Use the product rule to expand the first term.

To prove the second relation, use induction and the fact that [A,BC] = [A,B]C + B[A,C].

3. Sep 4, 2011

### Diomarte

Thank you very much Vela, but honestly I'm still not sure how, with the first part, I'm supposed to come up with a solution of -iħ? I'm not trying to get you to solve the problem for me obviously, I'm just having a very difficult time understanding if I'm taking the right approach to this problem. Could you possibly explain to me what having the commutator operate on a function does for this problem, or I suppose, how it helps? I really do appreciate your help here, and I'm fairly certain I'm missing something rather fundamental and simple in my approach.

4. Sep 4, 2011

### vela

Staff Emeritus
What's d/dx of xΨ(x)?

5. Sep 4, 2011

### Diomarte

I figured it was something along the lines of a distribution like that, so if I simply follow product rule with d/dx of xΨ(x) = Ψ(x) + x d/dx Ψ(x) and plug that in, I end up with

[-iħ (Ψ(x) + x (d/dx Ψ(x))] on the left hand side? And then by the order of how commutators work and the fact that you can't interchange the order of multiplications, the right hand side ends up being [x(-iħ d/dx Ψ(x))]? If this is the case, I'm still left with an extra .... oh fantastic, thank you. This should all cancel out then to just leave -iħ from the left hand side, if I am correct.

Last edited: Sep 4, 2011
6. Sep 4, 2011

### vela

Staff Emeritus
No, use that in the right-hand side. It's essentially the first term without the constant factor.

7. Sep 4, 2011

### Diomarte

I suppose then I'm misunderstanding this whole commutator thing... You made the suggestion to use the product rule to expand the first term, which I took to mean the left hand side, and thus I got what I had mentioned above, I ended up getting

-iħ [(Ψ(x) + x (d/dx Ψ(x))] - [x(-iħ d/dx Ψ(x))] which then after distributing the -iħ, I am left with:
-iħ Ψ(x) + -iħ[x (d/dx Ψ(x))] - -iħ[x(d/dx Ψ(x))] which then the second and third terms cancel out, leaving me with simply -iħ Ψ(x), which I thought is what you were leading me towards. Did I make a mistake here?

8. Sep 4, 2011

### vela

Staff Emeritus
Sorry for the confusion. Typically when you say lefthand or righthand side, it refers to sides of the equation.

So you have [p,x]Ψ = -ihΨ, so the operator [p,x] simply has the effect of multiplying by -ih, so you would write [p,x] = -ih.

9. Sep 4, 2011

### Diomarte

Thank you very much Vela, your assistance is greatly appreciated in this.