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I was reading a paper titled "Using a Recurrent Neural Network to Reconstruct Quantum Dynamics of a Superconducting Qubit from Physical Observations" and was confused about a stochastic master equation governing the evolution of the single-qubit system. According to the paper, they are denoting the non-unitary evolution of a superconducting qubit. The equation in question is the following: $$ d \rho_{t}=\underbrace{\left(i\left[H_{R}, \rho_{t}\right]+\mathcal{L}\left[\sqrt{\frac{\gamma}{2}} \sigma_{Z}\right] \rho_{t}\right)}_{\text {dissipative evolution }} d t+\underbrace{\sqrt{\eta} \mathcal{H}\left[\sqrt{\frac{\gamma}{2}} \sigma_{Z}\right] \rho_{t}}_{\text {backaction }} d w_{t} $$ In this case, $H_{\mathrm{R}}=\frac{\hbar \Omega_{R}}{2} \sigma_{X}$, "describes a microwave drive at the qubit transition frequency which induces unitary evolution of the qubit state characterized by the Rabi frequency $\Omega_R$." I understand and was able to implement (in MATLAB) the first half of the dissipative evolution, $i[H_R,p_T]$, as this is fairly simple; however, what do the $\mathcal{L}$ and $\mathcal{H}$ mean? The paper describes $\mathcal{L}$ as "the Lindblad superoperator describing the qubit dephasing induced by the measurement of strength $\gamma$"; however I have little idea as to what this means. What exactly does this mean and, if possible, what are methods of encoding this computationally (in something like MATLAB or python)

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  • $\begingroup$ I find the question a little vague? What is "a value" of an operator? Of course, if there is a basis, then you can expand on that basis. It is similarly true for a super operator! $\endgroup$
    – user185597
    Nov 27 '19 at 22:04
  • $\begingroup$ I edited the title. My main problem is I don't know what $L$ is supposed to mean. It appears to be multiplied with the square root of gamma/2 * Z but is $L$ a matrix, a constant, an operation? $\endgroup$ Nov 27 '19 at 22:50
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    $\begingroup$ L is a super-operator, that is, it acts on an operator (the things in the big bracket []) which at the end of acts on the density matrix. In the above example, L is acting on \sigma_z. Like operator, there are many super-operators, in your example, it is the dephasing superoperator which is $\sigma_z\rho\sigma_z - \rho$. There are ways to find a matrix for L in for a given problem. I guess some reading is mandatory here. $\endgroup$
    – user185597
    Dec 2 '19 at 9:50
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    $\begingroup$ Maybe this help more: physics.stackexchange.com/questions/197696/… $\endgroup$
    – user185597
    Dec 2 '19 at 9:52
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    $\begingroup$ A few forms of super-operators have been developed in the last decade! Dephasing, cooling, and heating are a few of those! However, it is no magic and if you expand $\rho$ you get some sense of it. For a two-level system (a qubit) you can expand $\rho= a\sigma_x +b\sigma_y+c\sigma_z+d I$ then knowing some Pauli matrix algebra (such as $\sigma_z\sigma_z = I, \sigma_z\sigma_x\~\sigma_y$) you find out that it leads to x and y mixing. $\endgroup$
    – user185597
    Dec 24 '19 at 11:36
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$\newcommand\dag\dagger$ The Lindblad superoperator is shorthand for a longer operation that generically reads $$\mathcal{L}[\hat{L}]\hat{\rho} = \hat{L} \hat{\rho} \hat{L}^\dag - \tfrac{1}{2}\left(\hat{L}^\dag\hat{L}\hat{\rho} + \hat{\rho}\hat{L}^\dag\hat{L} \right),$$ where the $\dag$ denotes the conjugate transpose (i.e. Hermitian adjoint) of the Lindblad operators $\hat{L}$.
Likewise, the last term (measurement / backaction superoperator) connected to the Wiener noise stands for $$\mathcal{H}[\hat{L}]\hat{\rho} = \hat{L}\hat{\rho} + \hat{\rho}\hat{L}^\dag - \hat{\rho}\,tr(\hat{L}\hat{\rho} + \hat{\rho}\hat{L}^\dag),$$ The stochastic master equation listed in the initial question contains three terms. The commutator (resonant drive) captures unitary evolution of the qubit, wherein it behaves like a closed quantum system. (You would just have Heisenberg's equation of motion, equivalent to the Schrodinger equation, if you truncated there). The two superoperator terms are related to the fact that this qubit is not actually a closed quantum system, but is open to an environmental channel through which it is being continuously and weakly monitored. In particular, $\mathcal{L}$ captures the average / dissipative evolution of the open channel, and the term $\mathcal{H}$ captures conditional evolution due to monitoring with efficiency $\eta$. The particular choice of $\hat{L} = \sqrt{\tfrac{\gamma}{2}}\hat{\sigma}_z$ characterizes the environmental channel through which measurement is being performed, and indicates that the Pauli-z observable is being tracked at a rate $\gamma$ (i.e. a larger $\gamma$ denotes a stronger measurement power, which is more invasive and ``collapses'' the qubit state faster). You can imagine that this is modeling the time-continuum limit of a sequence of very short but weak quantum measurements of $\hat{\sigma}_z$, each of which necessarily returns a random outcome. The SME gives the evolution of the qubit density matrix conditioned on the noisy measurement readout $r \propto \sqrt{\eta}\,tr(\hat{L}\hat{\rho} + \hat{\rho}\hat{L}^\dag) + dw_t/dt$, where the randomness / noise is characterized by $dw_t$, and the resulting individual trajectories undergo something that looks like Brownian motion.

As a further note: The Lindblad form $\mathcal{L}$ generically appears for any open quantum system which interacts with environmental channel(s) such that the resulting dissipation is Markovian. It dates to https://link.springer.com/article/10.1007/BF01608499, and a more recent introduction appears in, for example, https://aip.scitation.org/doi/pdf/10.1063/1.5115323. If you want further context or details about the case where the environment is monitored (i.e. continuous quantum measurement / stochastic quantum trajectories, which is the where $\mathcal{H}$ becomes crucial), you might also consider perusing some of the following...

  1. https://arxiv.org/abs/quant-ph/0108132
  2. https://arxiv.org/abs/quant-ph/0611067
  3. https://arxiv.org/abs/1410.5345
  4. https://arxiv.org/abs/1908.04720
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  • $\begingroup$ Welcome to our community! A superb first answer! $\endgroup$ Jul 23 at 4:04

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