Lindblad equation for Heisenberg operators?

Question

Very related to this question:

Is it possible to go from the Master Equation formalism to Heisenberg-Langevin equations

I don't yet have enough reputation to comment so I'm asking the new question here.

I'm coming from a physics background on this question here. My question boils down to the following. On the Wikipedia page for the Lindblad Equation page for the Lindblad equation there are two equations:

$$\dot{\rho} = -\frac{i}{\hbar}[H,\rho] + h \left(L \rho L^{\dagger}-\frac{1}{2}\left(\rho L^{\dagger}L + L^{\dagger}L\rho\right)\right)$$ $$\dot{A} = +\frac{i}{\hbar}[H,A] + \frac{1}{\hbar}\left(L^{\dagger}AL - \frac{1}{2}\left(AL^{\dagger}L + L^{\dagger}LA\right)\right)$$

I've set $h_{1,1}=h$ and $h_{n,m}=0$ otherwise in comparison to the Wikipedia formulas to simplify things. I'm also curious if a factor of $h$ has been lost somewhere in the second equation but that's not my main question.

I want to know how to go from the first equation to the second. From user yuggib's response on the linked question it is clear that it is related to some duality between the space of density operators and the space of observables, but I'm not familiar at all with Von Neumann algebras or semigroups etc.

In their response in the linked question yuggib also writes down similar form for the superoperators to what is found on Wikipedia:

$$(\mathscr{L})_*\rho= -i[H,\rho]+\frac{1}{2}\sum_j\Bigl([U_j\rho,U_j^{\dagger}]+[U_j,\rho U_j^{\dagger}]\Bigr)\; .$$

$$\mathscr{L}X= i[H,X]+\sum_j\Bigl(U_j^\dagger X U_j -\tfrac{1}{2}\{U^\dagger_jU_j,X\}\Bigr)\; .$$

(I've directly quoted yuggib's equations so $L$'s have turned into $U$'s and there is now a summation.)

Similarly it is not clear to me how to go from the first line to the second. It seems a bit more explicit in this case that the two operations should be related (since they're dual in some way I don't quite understand..) however I don't see (computationally speaking) how the jump was made from one to the other. It looks like you just sort of take a hermitian conjugate in terms of how the super operator acts but why is that sufficient and/or necessary to ensure the dual behavior?

Thank you for any help with this.

In this reference on page 70 they make the same jump from the time evolution of the density matrix to the time evolution of the operators

Answer 1

$\def\dd{{\rm d}} \def\LL{\mathcal{L}} \def\ii{{\rm i}} \def\ee{{\rm e}}$ The trick here is very simple and physically motivated. You simply demand that the expectation value of an operator $A$ is the same in the Schroedinger picture (density matrix evolves) and the Heisenberg picture (operator evolves). That is, $$\langle A\rangle = \mathrm{Tr}\left\lbrace A(0) \rho(t)\right\rbrace = \mathrm{Tr}\left\lbrace A(t)\rho(0) \right\rbrace,$$ where $A(t)$ [$\rho(0)$] is the operator [density matrix] in the Heisenberg picture, while $A(0)$ [$\rho(t)$] is the operator [density matrix] in the Schroedinger picture.

The Lindblad equation can be written in the Schroedinger picture as $$ \frac{\dd\rho}{\dd t} = \LL[\rho],$$ where the Liouvillian is defined by ($\hbar = 1$) $$\LL[\rho] = -\ii[H,\rho] + \sum_k \left ( L_k \rho L_k^\dagger - \frac{1}{2} \lbrace L_k^\dagger L_k,\rho\rbrace \right).$$ The formal solution of the Lindblad equation can then be written as $\rho(t) = \ee^{\LL t} [\rho(0)]$, where the exponential of the superoperator $\LL$ is defined as usual by its Taylor series expansion. Now, define the adjoint Liouvillian $\LL^\dagger$ by $$ \mathrm{Tr}\left\lbrace P \LL[Q] \right\rbrace = \mathrm{Tr}\left\lbrace \LL^\dagger[P] Q \right\rbrace,$$ where $P$ and $Q$ are arbitrary operators. It follows from the definitions that $$ \mathrm{Tr}\left\lbrace A(0) \rho(t)\right\rbrace = \mathrm{Tr}\left\lbrace A(0) \ee^{\LL t}[\rho(0)]\right\rbrace = \mathrm{Tr}\left\lbrace \ee^{\LL^\dagger t}[A(0)] \rho(0)\right\rbrace,$$ from which we identify $A(t) = \ee^{\LL^\dagger t}[A(0)]$ as the operator $A$ in the Heisenberg picture, which obviously satisfies the differential equation $$ \frac{\dd A}{\dd t} = \LL^\dagger[A].$$ All that remains is to check that indeed $$ \LL^\dagger[A] = \ii [H,A] + \sum_k \left ( L^\dagger_k A L_k - \frac{1}{2} \lbrace L_k^\dagger L_k ,A\rbrace \right),$$ which can be shown using the definition of $\LL^\dagger$ and the cyclicity of the trace.