- #1
Identity
- 152
- 0
Is there an easy way to prove the identities:
[tex]e^{\hat{A}}e^{\hat{B}}=e^{\hat{A}+\hat{B}}e^{[\hat{A},\hat{B}]/2}[/tex] and
[tex]e^{\hat{A}}\hat{B}e^{-\hat{A}}=\hat{B}+[\hat{A},\hat{B}]+\frac{1}{2!}[\hat{A},[\hat{A},\hat{B}]]+\frac{1}{3!}[\hat{A},[\hat{A},[\hat{A},\hat{B}]]]+...[/tex]In Zettili they give that: [tex]e^{\hat{A}}=\sum_{n=0}^\infty \frac{1}{n!}\hat{A}^n[/tex]
But I have no idea how I would expand a product of series to prove the identity... the algebra is horrible. Is there an easier way to go about proving it?
Thx
[tex]e^{\hat{A}}e^{\hat{B}}=e^{\hat{A}+\hat{B}}e^{[\hat{A},\hat{B}]/2}[/tex] and
[tex]e^{\hat{A}}\hat{B}e^{-\hat{A}}=\hat{B}+[\hat{A},\hat{B}]+\frac{1}{2!}[\hat{A},[\hat{A},\hat{B}]]+\frac{1}{3!}[\hat{A},[\hat{A},[\hat{A},\hat{B}]]]+...[/tex]In Zettili they give that: [tex]e^{\hat{A}}=\sum_{n=0}^\infty \frac{1}{n!}\hat{A}^n[/tex]
But I have no idea how I would expand a product of series to prove the identity... the algebra is horrible. Is there an easier way to go about proving it?
Thx