Lindblad and Input-Output Formalism in Quantum Optics

Question

I'm confused about how to apply the Lindblad formalism and the input-output formalism in practice, and how one goes between the two.

Suppose I have a cavity (C) coupled to a reservoir (R), with the Hamiltonian governing the total system (C + R) given by

$$ H = H_C + H_R + H_{CR}, $$ where $$ H_R = \int d\omega\,r^\dagger(\omega) r(\omega),\quad H_{CR} = \int d\omega\,\sqrt{\frac{\kappa(\omega)}{2\pi}} (a^\dagger r(\omega) + r^\dagger(\omega) a). $$ Here $a,a^\dagger$ are bosonic operators for the cavity mode while $r, r^\dagger$ are those for the continuum of bath modes. The cavity Hamiltonian can be left unspecified in what follows. I'll also make the Markov approximation $\kappa(\omega)\approx \kappa$ throughout to make things simpler.

Now, in the Lindblad formalism, we trace over the bath degrees of freedom to find a Master Equation for the density matrix of the cavity: $$ \partial_t\rho = \mathcal{L}[\rho], $$ where the Liouvillian is defined as $$ \mathcal{L}[\rho]\equiv i[\rho,H_C] + 2 L \rho L^\dagger - \{L^\dagger L,\rho\}, $$ and the jump operator here is $L = \sqrt{\frac{\kappa}{4}}a$.

I understand that here we are working in the Schrodinger picture, but we can straightforwardly go to the Heisenberg picture to find that operators evolve under $$ \partial_t A = \mathcal{L}^\dagger[A], $$ with $\mathcal{L}^\dagger$ the adjoint Liouvillian. With this, the equation of motion for $a$ is: $$ \partial_t a = i[H_C,a] - \frac{\kappa}{2}a. $$

Alternatively, I could use the input-output formalism to derive a quantum Langevin equation for $a$, which is given by: $$ \partial_t a = i[H_C,a] - \frac{\kappa}{2}a - \sqrt{\kappa} r_{in}(t) $$ where the input field is defined as the free evolution of the bath modes from some initial time $t_0<t$, $$ r_{in}(t) \equiv \frac{i}{\sqrt{2\pi}} \int d\omega\,r_0(\omega) e^{-i\omega(t-t_0)} $$ with $r_0(\omega)$ defined as $r(\omega)$ evaluated at time $t_0$. The input-output relation then says that the output field $$ r_{out}(t) \equiv \frac{i}{\sqrt{2\pi}} \int d\omega\,r_f(\omega) e^{-i\omega(t-t_f)}, $$ with $t_f>t$ and $r_f(\omega) = r(\omega)|_{t=t_f}$, is simply given as $$ r_{out}(t) = r_{in}(t) + \sqrt{\kappa} a(t). $$

Here are my questions:

1. If I start from the quantum Langevin equation and now try to derive a master equation, I see how the term $-\kappa/2 a$ would appear. However, I'm confused as to what the analogue of the term $r_{in}$ is in the Master equation. Does the master equation need to be modified to account for this somehow?

2. In practice, the input-output formalism suggests that calculating correlation functions of the cavity field alone is sufficient for understanding those of the output fields (assuming $r_{in}$ is coherent etc). But do I calculate correlation functions using the master equation? E.g. do I use $$ \langle a^\dagger a \rangle = Tr[\rho a^\dagger a], $$ with $\rho$ the density matrix evaluated according to the Lindblad master equation? If so, why? And how to account for the "driving" term $r_{in}$ which appears in the quantum Langevin equation but not in the master equation?

EDIT: I can roughly see that if the input is taken as being coherent i.e., $\langle r_{in} \rangle = \beta$, then I could add a "driving" term to the system Hamiltonian $H_C$ of the form $\beta (a + a^\dagger)$, which reproduces the equation of motion for $a$ with $r_{in}$. But this line of reasoning seems too ad-hoc, since the input-output formalism was designed to capture correlations between the bath and the system.

Any help would be appreciated, including references to clear exposition regarding these formalisms and their equivalence (or lack thereof).

Answer 1

There is already a nice answer but I feel that some important aspects deserve additional attention. My answer is simply a list of observations:

1. Master equations involve approximations: It is intuitive that the tracing out procedure that kicks out the bath to give you a Master equation comes at a loss of generality. Typical approximations include the bath being in a stationary state or a semi-classical driving field and the Born-Markov approximation involving the weak system-bath coupling approximation. There are other Master equations where some of these requirements can be relaxed or removed (see e.g. {1,2}), but usually other assumptions appear.
2. Master equations are nice: On the other hand, Master equations are really nice compared to the original coupled system-bath theory. In the Master equation, one is typically left with a hand full of degrees of freedom (some atomic states, some cavity modes, maybe a many-body system if you are doing hard stuff). One can then, for example, simply solve the density matrix equations on a computer.
3. The full system-bath Hamiltonian is hard: On the contrary, the full system-bath Hamiltonian features a continuum of bath modes, which is very difficult to put on a computer and has to be discretized somehow. This challenge can be addressed and there are a number of modern approaches (just giving one example here: matrix product operators {3}), but it is much more demanding than a Master equation.
4. No tracing in the input-output formalism: An important observation on the input-output formalism is that there is not tracing and therefore no assumption on the quantum state of the bath is necessary. While the original presentation by Gardiner-Collett {4} used a Markov approximation, it was later shown that this is not always necessary {5-7}.
5. Input-output is cool: The cool thing about the input-output formalism is that it tells you what the bath does if you know what the system does. You can therefore solve the system-dynamics in any way you like (Master equation or something more fancy) and then simply plug the solution in the input-output relation to get bath correlation functions. As argued in point 4, this has been shown to be extremely general.
6. Input-output is not a solution: That said, the input-output formalism does not actually give you the solution of the system dynamics. You can get Heisenberg-Langevin equations for the operators, but these generally still live on an infinite-dimensional Hilbert space. As mentioned above already, there are various methods to tackle this problem and find the solution, including Master equations but also other methods. These methods therefore complement the input-output formalism.

To give a direct answer to the OP's questions:

1. If I start from the quantum Langevin equation and now try to derive a master equation, I see how the term $−\kappa/2 a$ would appear. However, I'm confused as to what the analogue of the term $r_{in}$ is in the Master equation. Does the master equation need to be modified to account for this somehow?

The input-output relation is more general than the simple Lindblad Master equation in this regard. In the latter, riving terms like $r_{in}$ are usually included only in approximate form, for example as a semi-classical drive.

2. [...] But do I calculate correlation functions using the master equation? [...]

Yes! You can do that if the approximations leading to your Master equation are valid for your situation. Why? Well, you need to solve something to obtain the dynamical solution. As described above, the input-output equations alone do not provide such a solution easily.

If your Master equation approximations are not valid you may have to use someting more fancy (see references given above). However, the input-output relation will still tell you how the bath behaves once you have figured out what the system does.

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**References**
{1} Breuer and Petruccione, *Open quantum systems* ([publisher][1])
{2} de Vega and Alonso *Rev. Mod. Phys.* 89, 015001 (2017) ([journal][2])
{3} Strathearn et al. *Nat. Comm.* 9, 3322 (2018) ([journal][3])
{4} Gardiner and Collett *Phys. Rev. A* 31, 3761 (1985) ([journal][4])
{5} Diósi *Phys. Rev. A* 85, 034101 (2012) ([journal][5]) ([arxiv][6])
{6} Zhang et al. *Phys. Rev. A* 87, 032117 (2013) ([journal][7])
{7} Viviescas and Hackenbroich *Phys. Rev. A* 67, 013805 (2003) ([journal][8])

Answer 2

'I'm confused as to what the analogue of the term $r_{in}$ is in the Master equation'

The analogue of $r_{in}(t)$ in the Master equation is the set of bath modes which, in the master equation, you have traced out. This can be seen by looking at your equation for $r_{in}(t)$ and noticing that $r_{in}(t)$ is written entirely in terms of $r(\omega)$. The input-output formalism is a way of collecting an infinite number of heat bath/environment modes of different frequencies ($r(\omega)$) and writing them as an infinite number of environment modes interacting at different times (r_{in}(t)).

'Any help would be appreciated, including references to clear exposition regarding these formalisms and their equivalence (or lack thereof).'

The link between the input-output formalism and master equation does exist, but the derivation of the Master equation from the input-output formalism is pretty long. If you're interested, it can be found in chapter 6 (particularly section 6.2.1 and Problem 6.1) of 'Quantum Continuous Variables' by Serafini.

'Do I calculate correlation functions using the master equation?' 'Do I use $\langle a^\dagger a \rangle = Tr[\rho a^\dagger a]$ with $\rho$ the density matrix evaluated according to the Lindblad master equation? '

If you have a Quantum Langevin Equation, your operators are in the Heisenberg picture, so the density operator that you use to calculate expectation values is your initial one. In the Heisenberg picture, the density operator stays the same while observables evolve. The wikipedia page for the Heisenberg picture is pretty clear if you want to look this up. The quantum Langevin equation is a stochastic evolution equation for the observables in the Heisenberg picture. You do not need to calculate a 'new' density operator using the master equation.

In some sense, the Lindblad Master equation is for 'noisey' evolution in the Schrodinger picture (since $\rho$ evolves and the operators stay the same), and quantum Langevin equation is for 'noisey' evolution in the Heisenberg picture.

Hope this helps!