# What is the explicit form of exponential superoperators in Liouville space?

In the theory of open quantum systems, operators acting on a density matrix $$\rho$$ are often called superoperators. For example, the time evolution of a closed system may be written as $$\rho(t)=U^\dagger(t)\rho(0)U(t)=\mathcal{U}(t)\rho(0)$$ with $$\mathcal{U}(t)=U^\dagger(t) \bullet U(t)$$ the superoperator. If the Hilbertspace is denoted by $$\mathcal{V}$$ (we need the $$\mathcal{H}$$ symbol later), then $$\rho$$ is living in the vector space $$\mathcal{L} = \mathcal{V} \otimes \mathcal{V}$$ called the Liouville space. Now every density matrix $$\rho$$ and every operator $$A$$ may be thought of as a vector in $$\mathcal{L}$$ with linear operators such as $$\mathcal{U}(t)$$ acting on them like a matrix acts on a vector.

Original Question: What is the explicit form of a superoperator like $$\mathcal{U}(t)[\bullet]=e^{-i[H,\bullet]t}$$?

Updated Question: How to prove that $$e^{-i[H,\bullet]t}=e^{-iHt}\bullet e^{iHt}$$?

The background to my question is that the dynamics of a density matrix is given by the von-Neumann equation $$\dot{\rho} = -i [H, \rho] = \mathcal{H}[\rho]$$ with the solution $$\rho(t)=\mathcal{U}(t)[\rho(0)]=e^{\mathcal{H}[\bullet]t}\rho(0)$$. We can identify the exponential with the time evolution superoperator, i.e. $$\mathcal{U}(t)=e^{\mathcal{H}[\bullet]t}$$. On the other hand, we know that the solution to the von-Neumann equation can be expressed in Hilbertspace using the time evolution matrix as $$\rho(t)=U^\dagger (t)\rho(0)U(t)$$ with $$U(t)=e^{-iHt}$$. Clearly both ways of looking at the problem should coincide so that $$\mathcal{U}(t)[\bullet] = U^\dagger(t)\bullet U(t)$$ or explicity $$e^{-i[H,\bullet]t}=e^{-iHt}\bullet e^{iHt}$$. It is this last equation where I do not see its validity.

• The explicit matrix form of a super operator can be found using the answer(s) given here. Does this answer your question? Or are you just asking for a proof of the last equation, which can basically be proved just by differentiating both sides (or more formally: set $H \to \lambda H$ on the RHS, differentiate with respect to $\lambda$, solve the resulting linear 1st order ODE with obvious boundary conditions at $\lambda=0$, and then find the solution at $\lambda=1$). Feb 26, 2021 at 2:23
• I think your answer to the linked question answers my question. I did not fully understand what you were aiming at with the ODE approach (What should I solve for? What is the obvious boundary condition?). I think splitting $e^{-i[H,\bullet]t}$ as $e^{-i\mathcal{L}(H)[\bullet]t}e^{-i\mathcal{R}(H)[\bullet]t}$ is a more elegant way to prove the equation. I think the only thing left to show would be that indeed $e^{-i\mathcal{L}(H)[\bullet]t}\rho = e^{-i H t}\rho$, respectively with $\mathcal{R}$. If you want to put this into an answer I will accept it, otherwise I can answer the question myself. Feb 26, 2021 at 8:49
• It is just Hadamard's lemma on adjoint action, an elementary stepping stone to all CBH expansions. Taught routinely in Lie theory reviews... Feb 26, 2021 at 15:03
• Indeed, I was not aware of that. Now I also see what Mark Mitchison was referring to. I still prefer the decomposition of the superoperator into left and right acting part because it does not involve the trick of introducing a derivative. Feb 26, 2021 at 15:22

This is an answer to the updated question. As was first pointed out by Mark Mitchison, one can express the action $$e^{-i[H,\bullet]t}$$ on an operator as a combination of left and right acting matrices. To that end define $$\mathcal{L}(A)[\rho] = A\rho$$ and $$\mathcal{R}(A)[\rho] = \rho A$$ as the left (right) acting superoperator. We need three properties:
1. $$\mathcal{L}(A)$$ and $$\mathcal{R}(A)$$ commute because it does not matter if we multiply some $$\rho$$ first from the left or first from the right.
2. The commutator superoperator $$[H,\bullet] = \mathcal{L}(H) - \mathcal{R}(H)$$
3. The action of an exponential of the decomposition into $$\mathcal{L,R}$$ is $$e^{\mathcal{L}(A)[\bullet]t}\rho = e^{At}\rho$$ and $$e^{\mathcal{R}(A)[\bullet]t}\rho = \rho e^{At}$$. This can be checked e.g. by choosing an explicit vectorizing procedure of $$\rho$$.
We can express the l.h.s. as $$e^{-i[H,\bullet]t} = e^{-i\mathcal{L}(H)[\bullet]t}e^{i\mathcal{R}(H)[\bullet]t}=e^{-iHt}\bullet e^{iHt}$$ proving the equation.
Alternatively, the equation may be proven (thanks Cosmas Zachos for the link) by differentiation. Personally, I find the $$\mathcal{L,R}$$ decomposition more elegant.