Prove formula for the product of two exponential operators

• astrocytosis
In summary, the problem asks to show that Exp(A+B) = Exp(A)Exp(B)Exp(-1/2[A,B]), given two operators A and B satisfying [A,[A,B]] = 0 and [B,[A,B]] = 0. To solve this, we define T(s) = Exp(As)Exp(Bs) and differentiate with respect to s, giving T'(s) = Exp(As)Exp(Bs)B + Exp(Bs)Exp(As)A. However, this is not allowed if A and B do not commute, so we cannot switch the order of A and B. Instead, we can express T'(s) in terms of T(s) and use the
astrocytosis

Homework Statement

Consider two operators A and B, such that [A,[A, B]] = 0 and [B,[A, B]] = 0 . Show that

Exp(A+B) = Exp(A)Exp(B)Exp(-1/2 [A,B])

Hint: define Exp(As)Exp(Bs) as T(s), where s is a real parameter, differentiate T(s) with respect to s, and express the result in terms of T(s). Then use the Baker-Hausdorff lemma, and finally simply integrate your expression.

Homework Equations

[/B]
Baker-Hausdorff Lemma

e-B A eB = A + [B,A] + 1/2! [B,[B,A]] + 1/3! [B,[B,[B,A]]] +...

The Attempt at a Solution

I did what the hint said and took the derivative of T(s)

T'(s) = Exp(A s)Exp(B s) B + Exp(B s)Exp(A s) A

T'(s) = T(s) * (A + B)

but I am very lost as to how to proceed for here. I don't see how the Baker-Hausdorff lemma can be applied to this. I looked online for a derivation of this formula, but they all seemed more complicated than what the problem is asking for, and none of them defined a function like T(s) (that I saw). I think I must be fundamentally misunderstanding something here but I can't figure out what it is. I tried computing T(s) for the case where A, B depend on s, but that just made things more confusing.

astrocytosis said:

Homework Statement

Consider two operators A and B, such that [A,[A, B]] = 0 and [B,[A, B]] = 0 . Show that

Exp(A+B) = Exp(A)Exp(B)Exp(-1/2 [A,B])

Hint: define Exp(As)Exp(Bs) as T(s), where s is a real parameter, differentiate T(s) with respect to s, and express the result in terms of T(s). Then use the Baker-Hausdorff lemma, and finally simply integrate your expression.

Homework Equations

[/B]
Baker-Hausdorff Lemma

e-B A eB = A + [B,A] + 1/2! [B,[B,A]] + 1/3! [B,[B,[B,A]]] +...

The Attempt at a Solution

I did what the hint said and took the derivative of T(s)

T'(s) = Exp(A s)Exp(B s) B + Exp(B s)Exp(A s) A
You moved the Exp(As) A to the right of Exp(Bs) which is not allowed if A and B don't commute. Be careful to not switch the order of the A and B operators,

I thought an analytic function of an operator returned a function of its eigenvalue so it wouldn't matter... but then how can I write it in terms of T(S)?

1. What is the formula for the product of two exponential operators?

The formula for the product of two exponential operators is eAeB = eA+B, where A and B are operators in a vector space.

2. How do you prove the formula for the product of two exponential operators?

The proof for the formula involves using the properties of the exponential function and the commutative and associative properties of operators. It can also be derived using the Baker-Campbell-Hausdorff formula.

3. What are the applications of the formula for the product of two exponential operators?

The formula is used in quantum mechanics, specifically in the study of time evolution of quantum states. It is also used in differential geometry and Lie algebras.

4. Is the formula for the product of two exponential operators valid for all operators?

No, the formula is only valid for operators that commute with each other. If the operators do not commute, then the product formula becomes more complex and involves the use of the Baker-Campbell-Hausdorff formula.

5. Can the formula for the product of two exponential operators be extended to more than two operators?

Yes, the formula can be extended to any number of operators, not just two. It follows the same pattern: eAeB...eN = eA+B+...+N. However, as mentioned before, the operators must commute with each other for the formula to hold.

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