An identity for functions of operators

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SUMMARY

The discussion focuses on proving two specific identities involving operators in quantum mechanics: the Baker-Campbell-Hausdorff formula and the transformation of operators under exponentiation. The first identity, e^{\hat{A}}e^{\hat{B}}=e^{\hat{A}+\hat{B}}e^{[\hat{A},\hat{B}]/2}, is identified as incorrect for general operators \hat{A} and \hat{B}. The second identity involves the operator transformation e^{\hat{A}}\hat{B}e^{-\hat{A}}, which can be proven using Taylor series expansion and calculating derivatives of the function f(t)=e^{At}Be^{-At} at t=0.

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  • Understanding of operator algebra in quantum mechanics
  • Familiarity with the Baker-Campbell-Hausdorff formula
  • Knowledge of Taylor series and derivatives
  • Experience with exponential operators and commutators
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  • Study the Baker-Campbell-Hausdorff formula in detail
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  • Explore Taylor series expansions and their applications in physics
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Identity
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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
 
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Your first equation is, for general A,B, wrong.

As for the second one set

[tex]f(t)=e^{At}Be^{-At}[/tex]

calculate derivatives [itex]f^{(n)}(0)[/itex] (calculate first three, and you will see how it goes), and then use Taylor's expansion:

[tex]f(1)=f(0)+f'(0)+\frac{1}{2!}f''(0)+...[/tex]
 
Thanks :)
 

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