Do we need Lindblad operators to describe spontaneous emission

In summary: 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.
  • #1
td21
Gold Member
177
8
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?
 
Physics news on Phys.org
  • #2
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.
 
  • #3
ShayanJ said:
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.
 
  • #4
Demystifier said:
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!
 
  • #5
ShayanJ said:
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.
 
  • Like
Likes ShayanJ

Related to Do we need Lindblad operators to describe spontaneous emission

1. What is spontaneous emission?

Spontaneous emission is the process in which an excited atom or molecule emits a photon without any external influence.

2. Why do we need Lindblad operators to describe spontaneous emission?

Lindblad operators are used in the Lindblad equation, which is a mathematical tool that describes the time evolution of a quantum system. In the case of spontaneous emission, Lindblad operators are necessary to account for the loss of energy from the excited state to the ground state.

3. Can spontaneous emission be described without using Lindblad operators?

No, Lindblad operators are essential for accurately describing spontaneous emission. Without them, the Lindblad equation would not be able to account for the dissipation of energy and the resulting decay of the excited state.

4. Are there other methods for describing spontaneous emission besides using Lindblad operators?

Yes, there are other theoretical approaches that can be used to describe spontaneous emission, such as the Weisskopf-Wigner theory. However, the Lindblad equation is a widely accepted and powerful tool for describing the dynamics of open quantum systems, including spontaneous emission.

5. How do Lindblad operators affect the probability of spontaneous emission?

Lindblad operators contribute to the decay rate of the excited state, which in turn affects the probability of spontaneous emission. The larger the decay rate, the higher the probability of spontaneous emission occurring within a given time interval.

Similar threads

  • Quantum Physics
Replies
15
Views
2K
  • Atomic and Condensed Matter
Replies
6
Views
3K
  • Quantum Physics
Replies
3
Views
1K
  • Quantum Physics
Replies
1
Views
712
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Quantum Physics
Replies
21
Views
2K
  • Quantum Interpretations and Foundations
Replies
21
Views
2K
Replies
4
Views
1K
  • Quantum Physics
Replies
3
Views
2K
Back
Top