I'm hoping to compute the following integral: $\int_0^T e^{itA}Be^{-itA} dt$ where $iA, iB$ are traceless anti-Hermitian matrices (i.e. $\mathfrak{su}(n)$). I have found the following form for the integral: $\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{(iA)^n(iB)(iA)^m T^{n+m+1}}{n!m!(n+m+1)}$ by simply applying the Taylor series for the matrix exponential, however I can't find a simpler formula for the answer that doesn't involve an infinite series.
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3$\begingroup$ You could diagonalize $A$ (and thus $e^{\pm itA}$ for all $t$) and take it from there. $\endgroup$– Christian RemlingCommented Jun 11, 2014 at 2:18
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8$\begingroup$ From Hadamard's lemma, the expression is also equal to $\int_0^T e^{it \hbox{ad} A} B\ dt = \frac{e^{iT \hbox{ad} A}-1}{i\hbox{ad} A} B$, and from the Duhamel formula it is equal to $\frac{d}{ds} e^{iT(A+sB)}|_{s=0} e^{-iTA}$. (See also the Baker-Campell-Hausdorff formula for some related computations.) But perhaps this is not the kind of "computation" you have in mind... $\endgroup$– Terry TaoCommented Jun 11, 2014 at 5:03
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1$\begingroup$ You mean Dyson? $\endgroup$– Robert IsraelCommented Jun 11, 2014 at 7:45
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$\begingroup$ I haven't assumed that they commute, I'd assumed, incorrectly, that I could copy off my board accurately in to tex! $\endgroup$– BenjaminCommented Jun 11, 2014 at 12:09
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4 Answers
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I hope the following article can help: http://iopscience.iop.org/1063-7869/50/12/A02;jsessionid=EB2515B1F7F6CB626A8CBEF4537BBE05.c1 (Feynman disentangling of noncommuting operators and group representation theory, by V.S. Popov. A Russian version can be found here http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=ufn&paperid=542&option_lang=eng). Namely, it seems, it follows from its equation (10.5) that $$\int_0^T e^{itA}Be^{-itA} dt=\ln{\left [e^{(iA+B)T}e^{-iAT}\right]}.$$
answered Jun 11, 2014 at 6:43
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3$\begingroup$ I find this formula extremely suspicious. The Baker-Campbell-Hausdorff formula applied to the RHS contains terms that are bilinear or higher order in B (e.g. $-\frac{1}{12}[(iA+B)T,[(iA+B)T,-iAT]]$), whereas the LHS only contains terms that are linear in B. Perhaps the Popov paper is implicitly only performing a perturbative expansion and omitting some higher order terms? $\endgroup$ Commented Jun 11, 2014 at 16:35
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1$\begingroup$ I am finding Popov's arguments hard to read (they are written in the language of physics rather than mathematics) but my suspicion is that he has confused a time-ordered exponential with a non-time-ordered exponential somewhere. $\endgroup$ Commented Jun 11, 2014 at 16:49
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8$\begingroup$ OK, I think I found the issue: Popov is interpreting all his formulae using Feyman's operator calculus rather than standard mathematical notation. In this calculus, all operators are implicitly attached to a time parameter, and multiplication is not ordered from left to right, but instead ordered according to the time parameter. So one should be extremely cautious in lifting these formulae out of context, as they are not meant to be interpreted in the way that most mathematicians would read them. $\endgroup$ Commented Jun 11, 2014 at 17:05
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1$\begingroup$ @jjcale: This gives an answer that you can call explicit, but I think the formulae suggested here are elegant and have some extra appeal. $\endgroup$ Commented Jun 11, 2014 at 19:04
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2$\begingroup$ @Benjamin: No, an extra $i$ won't help, as the formula is correct as stated if $A,B$ commute. $\endgroup$ Commented Jun 12, 2014 at 2:46
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This is not an answer; rather, it is an extended comment on Terry's comments on Zurab's answer.
Here's an explicit counterexample to Popov's formula: Let $$ iA = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} . $$ Then the integrand equals $$ \begin{pmatrix} e^t & 0 \\ 0 & e^{-t} \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} e^{-t} & 0 \\ 0 & e^t \end{pmatrix} = \begin{pmatrix} 0 & -e^{2t} \\ e^{-2t} & 0 \end{pmatrix} \equiv F(t) . $$ However, the matrix $C\equiv iA+B$ from Popov's formula equals $$ C = \begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix} , $$ so satisfies $C^2=0$, hence $e^{CT} = 1+CT$, and I think it's already very plausible that we cannot possibly have $e^{CT}e^{-iAT}$ equal to the exponential of the LHS. However, I also computed the LHS explicitly and found $$ \exp \left( \int_0^T F(t)\, dt \right) = \begin{pmatrix} \cos\mu & (a/\mu)\sin\mu\\ -(\mu/a)\sin\mu & \cos\mu\end{pmatrix}, \quad a =-e^T\sinh T, \: \mu =\sqrt{\frac{\cosh 2T - 1}{2}} . $$
Update: I think I've successfully read Popov's mind now: He's really claiming that (notation as above, $F$ is the integrand) $$ Te^{\int_0^T F} = e^{(iA+B)T}e^{-iAT} \quad\quad (1) $$ where (the first) $T$ is the time-ordering "operator" (see my comment above). More succinctly (and precisely), we define the LHS as the fundamental matrix of $Y'=YF$. Then (1) holds trivially (both sides solve the same IVP). While $\int F$ does make an appearance on the LHS, this is a purely notational sleight of hand and the formula really seems useless for the problem at hand.
answered Jun 12, 2014 at 1:11
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1$\begingroup$ Yes, the correct form of Popov's result is (1) and unfortunately it is useless for the problem at hand. I made a mistake not noticing that the exponential is time-ordered in Popov's (10.5). $\endgroup$ Commented Jun 12, 2014 at 12:35
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For numerical purposes can you maybe treat it as a differential equation? $$ u'(t) = e^{itA}Be^{-itA}, \qquad u(0) = 0 $$
answered Jun 12, 2014 at 14:44
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I am not sure if you are ok with diagonalizing $A$ or if the size of the problem makes it unpractical. But if you can calculate eigenvalues and eigenvectors of $A$, and denote them as follows: $ A e_n = \lambda_n e_n$, then simply need to compute $e_m^H B e_n$ (where $e_m^H$ denotes the Hermitian transpose) for all $m$ and $n$: $$ e_m^H \Big( \int_0^T e^{itA} B e^{-itA} dt \Big) e_n = e_m^H B e_n \ \Big(\int_0^T e^{it(\lambda_m - \lambda_n)} dt \Big) $$
