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.

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    $\begingroup$ 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 $\lambda=1$). $\endgroup$ Feb 26, 2021 at 2:23
  • $\begingroup$ 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. $\endgroup$ Feb 26, 2021 at 8:49
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    $\begingroup$ It is just Hadamard's lemma on adjoint action, an elementary stepping stone to all CBH expansions. Taught routinely in Lie theory reviews... $\endgroup$ Feb 26, 2021 at 15:03
  • $\begingroup$ 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. $\endgroup$ Feb 26, 2021 at 15:22

1 Answer 1


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.

An answer to the original question how to find an explicit matrix form for a given superoperator can again be found in Mark Mitchison's answer.


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