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Determinant of $e^Ae^Be^{-A}e^{-B}$

Step-by-step mathematics solution: Determinant of $e^Ae^Be^{-A}e^{-B}$

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1. Restating the problem in plain language

We are asked to verify the identity

[ \operatorname{Det}!\bigl(e^{A}\,e^{B}\,e^{-A}\,e^{-B}\bigr)=\exp!\bigl(\operatorname{Tr}[A,B]\bigr), \qquad [A,B]=AB-BA, ]

when (A) and (B) are finite‑dimensional matrices (say (n\times n) complex matrices).

The symbol “(\operatorname{Det})” denotes the ordinary matrix determinant, and (\operatorname{Tr}) is the usual trace.
The question is whether the formula holds for any two matrices, or only under extra hypotheses (for instance, when ([A,B]) is a scalar multiple of the identity).

2. Detailed solution

2.1. Two elementary facts

  1. Determinant of an exponential
    For any square matrix (X),

    [ \det(e^{X}) = e^{\operatorname{Tr}X}. ]

    Proof. Write the Jordan (or Schur) decomposition (X = SJS^{-1}) with (J) upper‑triangular. Then (e^{X}=Se^{J}S^{-1}) and (\det(e^{X})=\det(e^{J})).
    Since (e^{J}) is also upper‑triangular and its diagonal entries are (e^{\lambda_i}) where (\lambda_i) are the eigenvalues of (X), we have

    [ \det(e^{J})=\prod_i e^{\lambda_i}=e^{\sum_i\lambda_i}=e^{\operatorname{Tr}X}. ]

  2. Cyclicity of the trace
    For any two matrices (X,Y),

    [ \operatorname{Tr}(XY)=\operatorname{Tr}(YX). ]

    Consequently, (\operatorname{Tr}[A,B]=\operatorname{Tr}(AB)-\operatorname{Tr}(BA)=0).

2.2. Computing the left‑hand side

Using the multiplicativity of the determinant together with fact 1:

[ \begin{aligned} \det!\bigl(e^{A}e^{B}e^{-A}e^{-B}\bigr) &= \det(e^{A})\;\det(e^{B})\;\det(e^{-A})\;\det(e^{-B}) \[4pt] &= e^{\operatorname{Tr}A}\;e^{\operatorname{Tr}B}\;e^{-\operatorname{Tr}A}\;e^{-\operatorname{Tr}B} \[4pt] &= 1. \end{aligned} ]

Thus the determinant of the product is always equal to (1), no matter what (A) and (B) are.

2.3. Computing the right‑hand side

From fact 2 we have

[ \operatorname{Tr}[A,B]=\operatorname{Tr}(AB)-\operatorname{Tr}(BA)=0, ]

hence

[ \exp!\bigl(\operatorname{Tr}[A,B]\bigr)=\exp(0)=1. ]

2.4. Comparing the two sides

Both sides equal (1); therefore the identity

[ \boxed{\displaystyle \det!\bigl(e^{A}e^{B}e^{-A}e^{-B}\bigr)=\exp!\bigl(\operatorname{Tr}[A,B]\bigr) } ]

holds for every pair of finite‑dimensional matrices (A,B).
The statement is true but completely trivial in the finite‑dimensional setting because (\operatorname{Tr}[A,B]=0).

2.5. Why the formula is interesting in infinite dimensions

When (A) and (B) act on an infinite‑dimensional Hilbert space, the ordinary determinant is no longer defined. One replaces it by the Fredholm (or regularized) determinant (\operatorname{Det}), which is multiplicative only under special circumstances.
If the commutator ([A,B]) belongs to the trace‑class ideal, the regularized determinant satisfies

[ \operatorname{Det}!\bigl(e^{A}e^{B}e^{-A}e^{-B}\bigr)=\exp!\bigl(\operatorname{Tr}[A,B]\bigr), ]

a non‑trivial result proved by Pincus, Helton, Howe and others.
For finite matrices the regularized determinant coincides with the ordinary determinant, and the formula collapses to the trivial identity proved above.

3. Final answer

Yes, the formula is true for all finite matrices (A,B).
Both sides equal (1) because

  • (\det(e^{X}) = e^{\operatorname{Tr}X}) gives (\det(e^{A}e^{B}e^{-A}e^{-B}) = 1);
  • (\operatorname{Tr}[A,B]=0) for any pair of matrices, so (\exp(\operatorname{Tr}[A,B]) = 1).

Hence the equality holds, albeit in a completely elementary way.

4. Common mistakes

Mistake Why it’s wrong How to avoid it
Assuming (\operatorname{Tr}[A,B]\neq0) for finite matrices. The trace of any commutator is zero by cyclicity of the trace. Remember the identity (\operatorname{Tr}(AB)=\operatorname{Tr}(BA)).
Using (\det(AB)=\det(A)\det(B)) without checking dimensions. The multiplicative property holds only for square matrices of the same size. Verify that all matrices involved are (n\times n).
Confusing the ordinary determinant with the Fredholm determinant. In infinite dimensions the ordinary determinant is undefined; the quoted formula is about the regularized determinant. Keep the context clear: for finite matrices the two determinants coincide, making the formula trivial.
Omitting the step (\det(e^{X}) = e^{\operatorname{Tr}X}). Without this property the computation cannot be finished. Prove or cite the exponential‑determinant identity (e.g., via Jordan/Schur form).
Trying to expand the product (e^{A}e^{B}e^{-A}e^{-B}) via the Baker–Campbell–Hausdorff formula. That expansion is unnecessary and leads to cumbersome higher‑order commutators. Use the determinant and trace properties directly; they give the result instantly.

By keeping these points in mind, the problem becomes a straightforward application of two basic linear‑algebra identities.

Original question: Determinant of $e^Ae^Be^{-A}e^{-B}$ on Mathematics Stack Exchange, licensed CC BY-SA.

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