Consider the following integral:

\begin{equation} L =2 \int_{0}^{\infty} d t \exp\left\{-\hat{\rho}_{\lambda} t\right\} \partial_{\lambda} \hat{\rho}_{\lambda} \exp\left\{-\hat{\rho}_{\lambda} t\right\} \end{equation} where \begin{equation} \partial_\lambda \hat{\rho}_{\lambda}=\mathrm{i}\left[a^{\dagger}+a, \hat{\rho}_{\lambda}\right] \end{equation} with $a$ being the annihilator operator.

I cannot use the truncated Baker-Campbell-Haussdorf formula since the commutator is not a number. Is there anything I could do to evaluate this integral? Any remark/tips are appreciated.

  • 4
    $\begingroup$ Could you check that the signs in the exponents are correct? Otherwise, the integral is too general - I doubt that you can get any help without giving more details on $\rho_\lambda$. $\endgroup$
    – Roger V.
    Feb 15, 2021 at 11:03
  • 2
    $\begingroup$ Exactly. Can you check as @Vadim said, please? Is that a Heisenberg evolution? More details on relation between $\hat{\rho}_{\lambda}$ and $a,a^{\dagger}$ algebra? $\endgroup$
    – joigus
    Feb 15, 2021 at 11:22
  • 1
    $\begingroup$ Do $a$, $a^{\dagger}$ satisfy comm. or anti-comm. relations (bosonic/fermionic)?, etc. $\endgroup$
    – joigus
    Feb 15, 2021 at 11:26

1 Answer 1


It is frankly inconceivable the sign of the second exponential's exponent is not +, instead of -, so I'll switch it, by executive editorial action, \begin{equation} L =2 i\int_{0}^{\infty} d t \exp\left\{-\hat{\rho}_{\lambda} t\right\} \left[a^{\dagger}+a, \hat{\rho}_{\lambda}\right] \exp\left\{\hat{\rho}_{\lambda} t\right\}. \end{equation} To bypass excessive notational clutter, the source of all mistakes, I'll also define $ \hat{\rho}_{\lambda} =A$, and $ a^{\dagger}+a =B$, so that $$ L =-2 i\int_{0}^{\infty}\!\! d t ~~\left ( e^{-t[A,} [A, \right )B ~~~, $$ where I am using the notation $[A, =\operatorname{ad}_A$ $\leadsto \operatorname{ad}_A B= [A,B]$ which many physicists don't know, but is detailed in WP. Many call this the Hadamard formula.

The integrand already involves an antiderivative, so, formally (as a power series expansion), the integral is undone and the boundary terms are $$ L =2 i \left ( e^{-\infty[A,} -I \right )B ~~~. $$ Assuming every nested commutator $[A,...[A,B]]]...]$ is a positive operator, (your issue), the first term vanishes and you get $L=-2iB$. I doubt your problem is that pat, but, anyway, you might glean tricks from here.

Of course, regardless of the integral, your second relation $$ \partial_\lambda A(\lambda) = i[B,A(\lambda)] , $$ has the customary ready solution for $\lambda$-independent B, (check it!), $$ A(\lambda) = e^{i\lambda [B, } A(0)\equiv e^{i\lambda B} A(0) e^{-i\lambda B} . $$ Are you positive you did not reverse symbols someplace?


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