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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

Urb
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Jagerber48
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1 Answers1

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$\def\dd{{\rm d}} \def\LL{\mathcal{L}} \def\ii{{\rm i}} \def\ee{{\rm e}}$ 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.

Mark Mitchison
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    Mathematically, there are some serious caveats but the idea is correct. Let's say that as usual there is no problem if $\mathcal{L}$ is a bounded operator (on the Banach space of trace class operators). If else, the evolution $e^{t\mathcal{L}}$ is strongly continuous only on a subspace of the bounded operators (called the adjoint space) where the adjoint generator can be densely defined. On the whole space of bounded operators the evolution is only weakly continuous (so no generator). – yuggib Sep 16 '16 at 07:23
  • @yuggib Everything written here should hold without caveats for the case of a finite Hilbert space dimension, right? In the case of an infinite Hilbert space dimension, I have no idea what happens, but note that in this case, the Lindblad(-Gorini-Kossakowski-Sudarshan) theorem doesn't even hold! – Mark Mitchison Sep 16 '16 at 07:25
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    The Lindblad characterization holds also for bounded generators (norm continuous semigroups) on infinite dimensional spaces. And in general you can do what you say also for unbounded generators, keeping in mind the right domains – yuggib Sep 16 '16 at 07:41
  • You say that there is no problem if $\mathscr{L}$ is bounded. The case I care about is of a photon field $a$ in a cavity which leaks out of a cavity at rate $\kappa$, or a harmonic oscillator $b$ which has some dissipitive damping rate $\gamma$. In this case $L\rightarrow a$, the harmonic oscillator ladder operators. These operators act on the (infinite dimensional) Fock space of number states. In this typical quantum optics situation are the necessary mathematical conditions satisfied for this transformation to work? (sorry if this is too much of a question in the comments!) – Jagerber48 Sep 16 '16 at 14:26
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    @jgerber Here is a non-rigorous answer from a physicist interested in extracting quantitative predictions from theory. Since all physical states have finite energy, only a finite number of states in the Hilbert space will be appreciably occupied in any physical process. Therefore, your harmonic oscillator can always be approximated by a finite-dimensional Hilbert space, with an error that can be made as small as desired by increasing the dimension. The mathematical issues alluded to above thus never arise in practice. Remember that a real cavity cannot hold an infinite number of photons! – Mark Mitchison Sep 16 '16 at 14:42
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    I have a question - how do you know that equation 4 is true? Based on what did you define adjoint Liouvillian? – Andris Erglis Feb 10 '21 at 13:44
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    @AndrisErglis The fourth equation is the definition of the adjoint Liouvillian. It is actually the general definition of the adjoint, i.e. $\langle u,A v\rangle = \langle A^\dagger u, v\rangle$. In this particular case, the abstract vectors $u$ and $v$ are operators and the (Frobenius) inner product between them is $\langle u,v\rangle = {\rm Tr}[u^\dagger v]$. (My answer uses $u = P^\dagger$ and $v=Q$). You can easily check that the explicit form of $\mathcal{L}^\dagger$ follows from this definition using the cyclic invariance of the trace. – Mark Mitchison Feb 10 '21 at 16:55
  • Ok, I see, great! Thanks for the explanation! – Andris Erglis Feb 11 '21 at 13:18
  • I just derived the expression for the $\mathcal{L}^{\dagger}$ by just using cyclic property of the trace as you said. I do not understand though, why the interlude about Heisenberg and Shcrodinger picture seems necessary, since the adjoint Lindbladian comes from pure mathematics, as far as I understand. The definition of Eq. 4 does not even seem necessary as well. – Andris Erglis Mar 04 '21 at 12:26
  • @JessRiedel I think your edit is not correct. – Wolpertinger Jul 26 '22 at 23:00
  • @Wolpertinger thanks, I fixed it :) – Mark Mitchison Jul 27 '22 at 08:42
  • Yikes, you are right. Sorry about. – Jess Riedel Jul 29 '22 at 17:17