How to differentiate exponentials of Lindblad operators?

Question

Suppose I have a time evolution generate by a Lindbladian $\mathcal{L}(x)$, parameterised by $x\in \mathbb{R}$. Suppose I wish to differentiate $k$ times with respect to $x$, how do I got about this? That is, suppose I am working in the Heisenberg picture and $A$ is an operator, then how should I express the following: $$\frac{\partial^k e^{t\mathcal{L}(x)}}{\partial x^k}(A) \ \ \ \ ?$$

Answer 1

In this question, it is not very relevant that $\mathcal L$ is a superoperator and acts on an operator $A$. For any (super-) operator $L(x)$ that depends on a parameter $x$, the derivative of the matrix exponential is given as follows: $$ \partial_x \mathrm e^{L(x)} = \int_0^1 \mathrm d\lambda\; \mathrm e^{\lambda L(x)}\, L'(x)\, \mathrm e^{(1-\lambda) L(x)} . $$ (source)

To apply this formula in your case, replace $L(x)$ by $t\mathcal L(x)$. To obtain higher derivatives, apply the formula repeatedly using the product rule inside the integral. For $[L'(x), L(x)] = 0$, the result simply becomes $L'(x)\, \mathrm e^{L(x)}$.

Finally I wanted to note that your expression $\mathrm e^{t\mathcal L} A$ strikes me as unusual. I do not know the context, but operators in the Heisenberg picture usually evolve according to the adjoint generator, i.e., $\mathrm e^{t\mathcal L^\dagger} A$.