Register to reply

Commutator Identity

by kreil
Tags: commutator, identity
Share this thread:
kreil
#1
Oct5-10, 04:11 PM
kreil's Avatar
P: 545
1. The problem statement, all variables and given/known data


Prove the following identity:

[tex]e^{x \hat A} \hat B e^{-x \hat A} = \hat B + [\hat A, \hat B]x + \frac{[\hat A, [\hat A, \hat B]]x^2}{2!}+\frac{[\hat A,[\hat A, [\hat A, \hat B]]]x^3}{3!}+...[/tex]

where A and B are operators and x is some parameter.

2. Relevant equations
[tex] e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...[/tex]
[tex] e^{-x} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...[/tex]

3. The attempt at a solution

[tex] \hat B e^{-x \hat A} = \hat B - [\hat B, \hat A]x + \frac{[[\hat B, \hat A], \hat A] x^2}{2!}+...[/tex]

It seems after I rearrange the commutation orders, the signs all become positive and this is the required result, so I know I must be doing something wrong. I think it has to do with how I'm multiplying out B into the series..

i.e. [tex] \hat B (\hat A x)^2 = [\hat B, \hat A \hat A] x^2....??[/tex]

or [tex] \hat B (\hat A x)^2 = [[\hat B, \hat A], \hat A] x^2...??[/tex]
Phys.Org News Partner Science news on Phys.org
What lit up the universe?
Sheepdogs use just two simple rules to round up large herds of sheep
Animals first flex their muscles
gabbagabbahey
#2
Oct5-10, 06:21 PM
HW Helper
gabbagabbahey's Avatar
P: 5,003
Quote Quote by kreil View Post
It seems after I rearrange the commutation orders, the signs all become positive and this is the required result, so I know I must be doing something wrong. I think it has to do with how I'm multiplying out B into the series..

i.e. [tex] \hat B (\hat A x)^2 = [\hat B, \hat A \hat A] x^2....??[/tex]

or [tex] \hat B (\hat A x)^2 = [[\hat B, \hat A], \hat A] x^2...??[/tex]
Neither is correct.

[tex]\hat{B}(x\hat{A})^2=BA^2x^2[/itex]

You won't have anything involving commutators until you multiply by both exponentials and collect terms in powers of the parameter [itex]x[/itex].
kreil
#3
Oct5-10, 07:23 PM
kreil's Avatar
P: 545
ahh I see it now I think..

[tex]e^{x \hat A} \hat B e^{-x \hat A} = \left ( 1+\hat A x + \frac{1}{2} \hat A^2 x^2 + \frac{1}{6} \hat A^3 x^3+... \right ) \left ( \hat B - \hat B \hat A x + \frac{1}{2} \hat B \hat A^2 x^2 - \frac{1}{6}\hat B \hat A^3 x^3+... \right ) [/tex]

So I multiply this out, collect terms in powers of x, and simplify to the commutator relations

Thanks


Register to reply

Related Discussions
Commutator math problem Advanced Physics Homework 2
Commutator: [E,x] Advanced Physics Homework 20
Differentiating the identity to develop another identity Calculus & Beyond Homework 3
The commutator [L,p] Advanced Physics Homework 7
[Identity relations] Need help at some odd identity relation problem Calculus & Beyond Homework 1