# Introduction

The Keldysh path integral can be thought of as a reformulation of the quantum optical master equation, which describes the markovian time evolution of the density operator of an open quantum system in the Schrödinger-picture.

The Keldysh-formalism allows one to calculate Green's functions. But fundamentally the Green's functions should be defined in the Heisenberg-picture, because they contain operators evaluated at different times.

But the formulation of the Heisenberg-picture for open quantum systems requires the introduction of noise terms into the equations of motion. Without the noise terms, the product rule does not hold for the time derivative.

# Question

My question is, that does anyone know any good references for linking these Heisenberg picture calculations based on quantum stochastic calculus to the Green's functions in the Keldysh-formalism?

# Note

If the enviromental variables are explicitly taken into account then the original system + the environment can be regarded as a larger closed system. One can then use the unitary time evolution operator to switch between the Schrödinger and the Heisenberg picture. My question was about linking these different formalisms at the level of the smaller system, where the enviromental degrees of freedom are not present. For example, the Keldysh path integral can be derived from the markovian master equation directly, without going back to the system + environment formulation.

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• Noise terms comes for free from the Master equation, once you encapsulate the external degrees of freedom with some basics hypothesis. You should not worry too much about the representations, what one solves is the Schrödinger equation or the Heisenberg equation written in term of evolution operators $U(t)$, i.e. the wave function reads $\Psi(t)=U(t)\Psi_0$ and any operator reads $A(t)=U(t)A_{0} U^{\dagger}(t)$, then you can switch from the Heisenberg to the Schrödinger equations. Actually what one uses is a mix of both, called the interaction representation. – FraSchelle Mar 7 '18 at 7:03
• Any book containing one of the following many body, quantum field theory, statistical field theory, condensed matter, ... key word should present the interaction representation, and do the derivation towards Green's function (one-body, Matsubara, Keldysh, and beyond). The important hypothesis is the nature of the bath at the switching on of the interaction. – FraSchelle Mar 7 '18 at 7:06
• @FraSchelle Thank you for your answer, but I think my question has been misunderstood. Question is now extended with a note. – DanielTuzes Mar 10 '18 at 13:15
• Then you should put many more details, because as it stands your question is impossible to understand clearly. Usually one says that the environment/external degrees of freedom have been traced out, not that they are absent. The evolution of the complete system follows the Liouville-vonNeuman equation : $\frac{d\rho}{dt}\propto[H,\rho]$, i.e. a Heisenberg-like equation. What do you want to know from that equation ? Please put some math if you do not know the terminology. – FraSchelle Mar 12 '18 at 14:39
• If you haven't found any reference yet, one good reference starting out with master equation (instead of including reservoir explicitly and then integrating out in Schwinger-Keldysh approach) and then converting it to path integral (including source terms generates both correlation functions and response functions) is "Keldysh field theory for driven open quantum systems" – Sunyam Jun 4 '18 at 20:13

$$\partial_t \hat \rho = \mathcal L(\hat \rho) = -i [\hat H, \hat \rho ] + \sum_k \gamma_k (\hat L_k \hat \rho \hat L_k^\dagger - \frac{1}{2} \{ \hat L_k^\dagger \hat L_k, \hat \rho\}),$$ you can obtain the Heisenberg representation $\partial_t \hat O = \mathcal L^*(\hat O)$ for the operator $\hat O$ by requiring that its expectation value is the same in both representations: $$Tr[\hat O \mathcal L(\hat \rho)] = Tr[\mathcal L^*(\hat O) \hat \rho].$$
This gives $$\mathcal L^* (\cdot) = i[\hat H, \cdot] + \sum_k \gamma_k (\hat L_k^\dagger \cdot \hat L_k - \frac{1}{2} \{\hat L_k^\dagger \hat L_k, \cdot \}).$$