Timeline for Lindblad equation for Heisenberg operators?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Nov 4 at 18:32 | comment | added | NinjaDarth | I worked out the derivation in detail in this reply on an example sufficiently similar to cover the case in the original query. | |
| Jul 29, 2022 at 17:17 | comment | added | Jess Riedel | Yikes, you are right. Sorry about. | |
| Jul 27, 2022 at 8:42 | comment | added | Mark Mitchison | @Wolpertinger thanks, I fixed it :) | |
| Jul 27, 2022 at 8:42 | history | edited | Mark Mitchison | CC BY-SA 4.0 |
reversed incorrect edit
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| Jul 26, 2022 at 23:00 | comment | added | Wolpertinger | @JessRiedel I think your edit is not correct. | |
| Jul 26, 2022 at 6:47 | history | edited | Jess Riedel | CC BY-SA 4.0 |
Fixed math typo: the dagger wasn't swapped on Lindblad eq for operators in Heisenberg picture,
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| Mar 4, 2021 at 12:26 | comment | added | andrix | 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. | |
| Feb 11, 2021 at 13:18 | comment | added | andrix | Ok, I see, great! Thanks for the explanation! | |
| Feb 10, 2021 at 16:55 | comment | added | Mark Mitchison | @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. | |
| Feb 10, 2021 at 13:44 | comment | added | andrix | I have a question - how do you know that equation 4 is true? Based on what did you define adjoint Liouvillian? | |
| Jan 21, 2021 at 23:49 | history | bounty ended | Wolpertinger | ||
| Sep 16, 2016 at 14:42 | comment | added | Mark Mitchison | @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! | |
| Sep 16, 2016 at 14:26 | comment | added | Jagerber48 | 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!) | |
| Sep 16, 2016 at 7:41 | comment | added | yuggib | 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 | |
| Sep 16, 2016 at 7:25 | comment | added | Mark Mitchison | @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! | |
| Sep 16, 2016 at 7:23 | comment | added | yuggib | 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). | |
| Sep 16, 2016 at 6:31 | history | edited | Mark Mitchison | CC BY-SA 3.0 |
added 98 characters in body
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| Sep 16, 2016 at 6:30 | vote | accept | Jagerber48 | ||
| Sep 16, 2016 at 6:22 | history | answered | Mark Mitchison | CC BY-SA 3.0 |