
#1
Oct510, 04:11 PM

P: 518

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} = 1x+\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] 



#2
Oct510, 06:21 PM

HW Helper
P: 5,004

[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]. 



#3
Oct510, 07:23 PM

P: 518

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 


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