Register to reply

Commutator proof

by cahill8
Tags: commutator, proof
Share this thread:
cahill8
#1
Mar27-10, 10:23 PM
P: 31
1. The problem statement, all variables and given/known data
Show [tex]\left[x,f(p)[/tex][tex]\right)][/tex] = [tex]i\hbar\frac{d}{dp}(f(p))\right.[/tex]


2. Relevant equations

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


3. The attempt at a solution
I started by expanding f(p) to the power series which makes

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

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
Phys.Org News Partner Science news on Phys.org
Hoverbike drone project for air transport takes off
Earlier Stone Age artifacts found in Northern Cape of South Africa
Study reveals new characteristics of complex oxide surfaces
Hurkyl
#2
Mar27-10, 10:27 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,099
Quote Quote by cahill8 View Post
and I know I must use the commutator identity [A, BC] = [A,B]C + B[A,C]
How do you know that?
cahill8
#3
Mar28-10, 03:13 AM
P: 31
In a text book it says it can be shown using that equation

Trying a different method:

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

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

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

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

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

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

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

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

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

am I on the right track?

Hurkyl
#4
Mar28-10, 05:49 AM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,099
Commutator proof

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]
after a single addition.

But that aside, everything you wrote looks correct. We won't know if you're on the right track until we see where this path leads, though!
cahill8
#5
Mar28-10, 06:03 AM
P: 31
I see what you mean. [x, fnpn] = fn[x, pn] is fine.

I kept going with the addition and noticed a pattern and managed to solve it. Thanks for the hints :)


Register to reply

Related Discussions
Proof about commutator bracket General Math 2
Commutator: [E,x] Advanced Physics Homework 20
Proof of a commutator algebra exp(A)exp(B)=exp(B)exp(A)exp([A,B]) Quantum Physics 6
Commutator math help Advanced Physics Homework 1
The commutator [L,p] Advanced Physics Homework 7