0
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Related: physics.stackexchange.com/q/52012/2451 $\endgroup$ – Qmechanic May 26 at 18:32
  • $\begingroup$ You know $\hat{B}=\hat{S}^z$. Is $\hat{A}$ also something specific, or is it arbitrary? $\endgroup$ – J.G. May 26 at 18:40
  • $\begingroup$ @J.G. $\hat{A}$ is basically product of $\hat{S}^+$ and $\hat{S}^-$ operators $\endgroup$ – Roger.Bernstein May 26 at 20:29
  • $\begingroup$ 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. $\endgroup$ – Emilio Pisanty May 27 at 11:59
  • $\begingroup$ @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) $\endgroup$ – Roger.Bernstein May 27 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.