An identity for functions of operators

In summary, there are two identities that need to be proven. The first one is e^{\hat{A}}e^{\hat{B}}=e^{\hat{A}+\hat{B}}e^{[\hat{A},\hat{B}]/2} and the second one is 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{
  • #1
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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  • #2
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]
 
  • #3
Thanks :)
 

1. What is an identity for functions of operators?

An identity for functions of operators is a mathematical expression that allows us to manipulate and solve equations involving operators, which are symbols that represent mathematical operations. It is used to simplify the process of solving equations involving operators.

2. How is an identity for functions of operators derived?

An identity for functions of operators is derived using algebraic manipulation and properties of operators, such as commutativity and associativity. It is also based on the fundamental principles of mathematics, such as the distributive property and the identity property.

3. What is the purpose of using an identity for functions of operators?

The purpose of using an identity for functions of operators is to make solving equations involving operators more efficient and accurate. It allows us to simplify complex equations and express them in a more manageable form, making it easier to find solutions and make further calculations.

4. Can an identity for functions of operators be applied to all types of equations?

Yes, an identity for functions of operators can be applied to a wide range of equations involving operators, such as polynomial equations, logarithmic equations, and trigonometric equations. It can also be used to solve problems in various fields of science and mathematics, including physics, chemistry, and engineering.

5. Are there any limitations to using an identity for functions of operators?

While an identity for functions of operators is a powerful tool for solving equations, it is not always applicable to every equation. Some equations may require more advanced techniques or may not have a simple solution. Additionally, the use of identities may lead to extraneous solutions, so it is important to check the validity of the solution obtained.

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