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I have quite a problem handling the following commutator involving the exponential of the integral of an operator $$\Bigg[\hat{A},\exp\!\Bigg(\int_0^td\tau\,\hat{B}(\tau)\Bigg)\Bigg]$$ especially as in the original problem the operator $\hat{B}(\tau)$ equals the spin-$z$ operator $\hat{S^z}(\tau)$ and I absolutely don't have any idea how to handle such expression. I read about the Magnus expansion, but it's only an approximation like the Dyson series. Are there better or more "elegant" solutions or methods solving the problem?

Edit: The problem I'm dealing with is about the commutator $$\Bigg[S^\pm_{i,j} S^\pm_{k,\ell}\dotsm S^\pm_{m,n},\exp\!\Bigg(-\varepsilon\int_0^td\tau\,\big(S^z_{x,y}(0)-S^z_{x,y}(\tau)\big)^2\Bigg)\Bigg]$$ if this helps.

Caesar.tcl
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  • Related: https://physics.stackexchange.com/q/52012/2451 – Qmechanic May 26 '21 at 18:32
  • You know $\hat{B}=\hat{S}^z$. Is $\hat{A}$ also something specific, or is it arbitrary? – J.G. May 26 '21 at 18:40
  • @J.G. $\hat{A}$ is basically product of $\hat{S}^+$ and $\hat{S}^-$ operators – Caesar.tcl May 26 '21 at 20:29
  • Are you certain that your problem involves an operator exponential and not a time-ordered exponential? For situations where $[\hat B(t),\hat B(t')]\neq0$ for $t\neq t'$, the non-time-ordered exponential is rather uncommon in real-world scenarios. – Emilio Pisanty May 27 '21 at 11:59
  • @EmilioPisanty I'm in the sense certain that this is the correct form as I constructed this operator to have particular properties. It basically measures the "distance" between two states and equals one if nothing happened and zero if the state has changed (and remains zero afterwards) – Caesar.tcl May 27 '21 at 12:55
  • @Roger.Bernstein If you're sure, go for it, but you should double-check the assumptions that underlie your construction. In particular, if you're relying on a derivative of the form $$\frac{\mathrm d}{\mathrm dt}e^{\int_0^td\tau,\hat{B}(\tau)} \stackrel{?}{=} ​\hat B(t) e^{\int_0^td\tau,\hat{B}(\tau)}$$ at any point, then that only works for the time-ordered exponential (and it is false for the normal operator exponential). (For more details, see this thread.) – Emilio Pisanty May 27 '21 at 13:21

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