What is the restriction on the norm of the operator $L_k$ in the Lindblad master equation $\dot{\rho} = \sum_k L_k \rho L_k^\dagger + \frac{1}{2}\left\{ L_k^\dagger L_k, \rho\right\}$? Although there doesn't seem to be a specific reference that addresses this question, $L_k$ must be a bounded operator. Is it typical to set the operator norm to be bounded by $1$? For example, should $L = \sigma_x + \sigma_z$ be normalized? Additionally, what are the common experimental/physical Lindblad operators that people use? It'd be great if anyone can link me to papers/references that deal with physically motivated Lindbladians.
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$\begingroup$ See for example this tutorial arxiv.org/pdf/1110.2122.pdf the choice of jump operator comes from which system operators interact with the bath. In practice these are often something like Pauli matrices for spin dephasing, transition operators for spontaneous decay, or creation and annihkiation operators for photonic cavity loss and gain $\endgroup$– AwkwardWhaleCommented Apr 13, 2023 at 8:47
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$\begingroup$ The requirement of the $L_k$ being bounded as linear operators (which is trivially satisfied in finite dimensions) is equivalent to the resulting dynamics being uniformly continuous (in $t$), while unbounded operators correspond to "just" strong continuouity. But even in the uniformly continuous case there is no universal bound on the norms (i.e. there is no such thing as "it must hold that $\|L_k\|\leq 1$ for all $k$") as the answer below rightly shows. $\endgroup$– Frederik vom EndeCommented May 7 at 9:41
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For dissipation in harmonic oscillators one often takes the $L_k$ to be annihilation operators for modes $k$. These are not bounded operators!
Now, do you think $\dot{\rho}$ should be bounded? What if I write a state $$\rho(t)=\begin{pmatrix}e^{-t/\tau}&0\\0&1-e^{-t/\tau}\end{pmatrix}?$$ Then $$\dot{\rho}=-\frac{1}{\tau}\begin{pmatrix}e^{-t/\tau}&0\\0&-e^{-t/\tau}\end{pmatrix}.$$ I can choose any $\tau$ that I want and this will be a valid state, so I can make $\tau$ infinitesimally small, so I can make $\dot{\rho}$ arbitrariliy large (I might need to set something like $t=0$ or $t=0.1\tau$ too but that's no problem; there is some time for which the derivative can be arbitrarily large). This seems to imply that one need not bound the operators $L_k$.
answered Apr 13, 2023 at 1:01
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1$\begingroup$ To complete this answer: this $\rho(t)$ comes about, e.g., via Lindblad dynamics induced by the single jump operator $$ L_1(\tau)=\begin{pmatrix}0&0\\\frac1{\sqrt\tau}&0\end{pmatrix}\,. $$ So for a single $\tau$, $L_1(\tau)$ is of course bounded (with operator norm $\frac1{\sqrt\tau}$) because every matrix is bounded, but $\|L_1(\tau)\|\to\infty$ as $\tau\to\infty$ so it is not uniformly bounded. In other words the faster the exponential decay the larger the norm of the jump operators has to be. $\endgroup$ Commented May 7 at 9:42
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$\begingroup$ @FrederikvomEnde that's a useful update - just a quick typo because I think you probably mean $\tau\to 0$ $\endgroup$ Commented May 7 at 15:26
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1$\begingroup$ You are of course right, thank you for pointing that out! $\endgroup$ Commented May 7 at 15:27