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Suppose $A$ and $B$ are (non-commuting) hermitian $n\times n$ matrices and $k$ is a large positive number. Suppose we write the product of matrix exponentials as

$e^{kA + B} e^{-kA} = e^{C(k)}$

for some matrix $C(k)$. I am interested in how large $C(k)$ is (with respect to some appropriate norm, lets say the operator norm of matrices).

In particular is $\| C(k)\| = O(k)$? Some numerical experiments suggest a bound of the form $||C(k)|| \le O(k) ||[ A,B]||$ but I cannot prove this.

I know the Baker-Campbell-Hausdorff formula gives us an (explicit) equation for $C(k)$. But bounding this expression term-by-term using the triangle inequality gives a terrible bound. I am hoping somebody knows either another argument (or a trick) that gives a more sensible bound.

Thanks a lot!

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1 Answer 1

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$\|C(k)\|$ can be arbitrarily large, since e.g. you can add some large integer multiple of $2\pi i I$ without changing $e^{C(k)}$. If you want to try to avoid this, you might specify that $C(k)$ is the principal branch of the logarithm of $e^{kA+B} e^{-kA}$ (note that although $e^{kA+B} e^{-kA}$ is not hermitian, it has the same eigenvalues as $e^{-kA/2} e^{kA+B} e^{-kA/2}$ which is positive definite). Then the eigenvalues of $C(k)$ will be $O(k)$; unfortunately, since $C(k)$ is not hermitian or normal this does not imply $\|C(k)\| = O(k)$.

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