Do we need Lindblad operators to describe spontaneous emission

[td21]

In Griffith and Sakurai QM book, spontaneous emission is treated as a closed system subject to time-dependent perturbation.

Yet in quantum optics sponantanoues emission is treated as in the form master equation of density matrix. Even in two levels system where there is only one spontaneous emission rate, master equation is formulated to described sponantanoues emission. Why do we need Lindblad operators for two level system?

 

[ShayanJ]

A closed quantum system as a whole has a unitary evolution. But sometimes you need to describe the evolution of an open system where you have no access to the degrees of freedom(dof) that the system is interacting with. In those cases you need a quantity that describes the evolution of the dof of the system but not the environment. This can't be done using wave-functions. So you form the density matrix of the system+environment and then trace-out the environmental dof. The resulting quantity is called the reduced density matrix of the system and its evolution is non-unitary and non-deterministic. Such an evolution is an stochastic process and should be described by an stochastic differential equation. Now considering the Heisenberg equation of motion and some simplifying assumptions, you get the Lindblad equation.

 

[Demystifier]

The resulting quantity is called the reduced density matrix of the system and its evolution is non-unitary and non-deterministic. Such an evolution is an stochastic process and should be described by an stochastic differential equation. Now considering the Heisenberg equation of motion and some simplifying assumptions, you get the Lindblad equation.

You are right that Lindblad equation is non-unitary, but it is deterministic and not stochastic.

 

[ShayanJ]

You are right that Lindblad equation is non-unitary, but it is deterministic and not stochastic.

But Lindblad equation is the most general differential equation for a Markovian evolution. Being Markovian is a property of stochastic processes!

 

[Demystifier]

But Lindblad equation is the most general differential equation for a Markovian evolution. Being Markovian is a property of stochastic processes!

One should distinguish Markov process from Markov probability. Markov process is indeed a stochastic process, but such a process is determined by a probability which can have a deterministic dependence on time. For instance, the standard 1-dimensional random walk is a stochastic process with a constant probability (at each step, the probability is always 1/2 for any of the two directions). The Lindblad equation describes only the probability function (more technically, the probability density matrix), not the stochastic process.