# Commutator proof

• cahill8

## Homework Statement

Show $$\left[x,f(p)$$$$\right)]$$ = $$i\hbar\frac{d}{dp}(f(p))\right.$$

## Homework Equations

I can use $$\left[x,p^{n}$$$$\right)]$$ = $$i\hbar\\n\right.$$$$p^{n}\right.$$
f(p) = $$\Sigma$$ $$f_{n}$$$$p^{n}$$ (power series expansion)

## The Attempt at a Solution

I started by expanding f(p) to the power series which makes

$$\left[x,\Sigma\\f_{n}\\p^{n}$$$$\right)]$$

and I know I must use the commutator identity [A, BC] = [A,B]C + B[A,C]
but the power series cannot be split up into two products(BC) ? So I'm not sure how to go on

and I know I must use the commutator identity [A, BC] = [A,B]C + B[A,C]
How do you know that?

In a textbook it says it can be shown using that equation

Trying a different method:

[x, f(p)] = [x,$$\sum_{n}\\f_{n}p^{n}$$] = [x,fnpn + $$\sum_{n-1}\\f_{n}p^{n}$$]

using [A, B+C] = [A,B] + [A,C]

= [x, fnpn] + [x, $$\sum_{n-1}\\f_{n}p^{n}$$]

using [A, BC] = C[A,B] + B[A,C]

= fn[x, pn] + pn[x, fn] + [x, $$\sum_{n-1}\\f_{n}p^{n}$$]

using [x, pn] = i$$\hbar$$npn-1

= fni$$\hbar$$npn-1 + pn[x, fn] + [x, $$\sum_{n-1}\\f_{n}p^{n}$$]

[x, fn] = 0 as fn is a const.

= fni$$\hbar$$npn-1 + [x, $$\sum_{n-1}\\f_{n}p^{n}$$]

am I on the right track?

Last edited:
I'm curious why you used
[A, rC] = [A,r]C + r[A,C]​
to pull out a scalar, rather than just using
[A, rC] = r [A,C]​

I'm also curious why you stopped using
[A, B + C] = [A,B] + [A,C]​