Does [p,x^n] commute with f(x)=x^n?

[oswaler]

Homework Statement

Prove:
[σ⋅(p-qA)]²=(p-qA)²-q$$\bar{h}$$σ⋅B

where B=∇×A , p=i$$\bar{h}$$∇ and q is constant

Homework Equations

The Attempt at a Solution

if the x component is: px-qAx
and the y component is py-qAy

Then the x and y components shouldn't commute [px-qAx,py-qAy] <>0. This is where the extra term q$$\bar{h}$$σ⋅B comes from.

However, it seems that these values do commute:
(px-qAx)(py-qAy)-(py-qAy)(px-qAx)=0

Am I missing something here that would cause these values not to commute?

 

[Avodyne]

Ay can be a function of x, so it does not commute with px (and vice versa).

 

[oswaler]

so what would [px-qAx,py-qAy] evaluate as?

 

[Avodyne]

In one dimension,
$$[p,f(x)]=-i\hbar f'(x)$$

 

[oswaler]

oops, I asked the question wrong. I was already given what it evaluates as, I mean to ask why.
[p,f(x)]=pf(x)-f(x)p... I guess I'm not seeing how the f' and ih terms are coming.

 

[Avodyne]

Consider f(x)=x^n. Using [A,BCD...]=[A,B]CD... +B[A,C]D... + AC[B,D]... + ...,
$$[p,x^n] = [p,x]x^{n-1}+x[p,x]x^{n-2}+\ldots+x^{n-1}[p,x]x+[p,x]x^n$$
In each term, $[p,x]=-i\hbar$, and this is a number that can be pulled out front. There are n terms, and each has n-1 powers of x, so we get
$$[p,x^n] = -i\hbar nx^{n-1}$$
This is $-i\hbar$ times the derivative of x^n with respect to x. So to compute [p,f(x)], we do a Taylor expansion, and get the derivative term by term.