# Prove Commutator Identity: e^xA B e-xA = B + [A,B]x + ...

• kreil
In summary, the conversation is about proving an identity involving operators and a parameter. The attempted solution involves rearranging commutation orders, but the correct approach is to multiply both exponentials and then collect terms in powers of the parameter x.
kreil
Gold Member

## Homework Statement

Prove the following identity:

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

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

## Homework Equations

$$e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$
$$e^{-x} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...$$

## The Attempt at a Solution

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

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. $$\hat B (\hat A x)^2 = [\hat B, \hat A \hat A] x^2...??$$

or $$\hat B (\hat A x)^2 = [[\hat B, \hat A], \hat A] x^2...??$$

kreil said:
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. $$\hat B (\hat A x)^2 = [\hat B, \hat A \hat A] x^2...??$$

or $$\hat B (\hat A x)^2 = [[\hat B, \hat A], \hat A] x^2...??$$

Neither is correct.

$$\hat{B}(x\hat{A})^2=BA^2x^2[/itex] You won't have anything involving commutators until you multiply by both exponentials and collect terms in powers of the parameter $x$. ahh I see it now I think.. [tex]e^{x \hat A} \hat B e^{-x \hat A} = \left ( 1+\hat A x + \frac{1}{2} \hat A^2 x^2 + \frac{1}{6} \hat A^3 x^3+... \right ) \left ( \hat B - \hat B \hat A x + \frac{1}{2} \hat B \hat A^2 x^2 - \frac{1}{6}\hat B \hat A^3 x^3+... \right )$$

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

Thanks

## 1. What is the commutator identity?

The commutator identity is a mathematical expression that shows the relationship between the exponentiation of a matrix or operator and its commutator. It is commonly used in quantum mechanics and linear algebra.

## 2. Why is the commutator identity important?

The commutator identity is important because it allows us to simplify complex calculations involving exponentiation and commutators. It also helps us understand the behavior of operators in quantum mechanics and other areas of mathematics and physics.

## 3. How do you prove the commutator identity?

The commutator identity can be proven using basic algebraic manipulations and the properties of exponentiation and commutators. It is a straightforward but lengthy process that requires careful attention to detail.

## 4. What are the applications of the commutator identity?

The commutator identity has many applications in mathematics, physics, and engineering. It is commonly used in quantum mechanics, where it helps in the analysis of systems with time-dependent Hamiltonians. It is also used in control theory, signal processing, and other fields.

## 5. Are there any variations of the commutator identity?

Yes, there are several variations of the commutator identity, each with its own specific form and applications. Some common variations include the Jacobi identity, the Baker-Campbell-Hausdorff formula, and the BCH expansion. These variations are useful in different contexts and can be derived from the basic commutator identity.

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