# Operator integral [closed]

Consider the following integral:

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

• 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 $\rho_\lambda$. Feb 15, 2021 at 11:03
• 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? Feb 15, 2021 at 11:22
• Do $a$, $a^{\dagger}$ satisfy comm. or anti-comm. relations (bosonic/fermionic)?, etc. Feb 15, 2021 at 11:26

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, $$$$L =2 i\int_{0}^{\infty} d t \exp\left\{-\hat{\rho}_{\lambda} t\right\} \left[a^{\dagger}+a, \hat{\rho}_{\lambda}\right] \exp\left\{\hat{\rho}_{\lambda} t\right\}.$$$$ To bypass excessive notational clutter, the source of all mistakes, I'll also define $$\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?