Probability of one-photon loss from a cavity

In summary, the probability of one-photon loss from a cavity in a time interval is given by the equation ##\kappa \delta t\langle \psi(t)| \hat{a}^{\dagger}\hat{a} |\psi(t)\rangle##, where ##\kappa## is the decay rate. This is similar to Fermi's Golden Rule, but not exactly the same. The equation can be found on page 37 of the textbook "Quantum Optics" by J. P. Berman and V. S. Malinovsky. This assumption is based on the assumption that the probability of losing a photon is proportional to the expectation value of the number of photons in the cavity. ##\k
  • #1
maxverywell
197
2
Why is the probability of one-photon loss from a cavity in the time interval ##[t, t+\delta t]## is:
##\kappa \delta t\langle \psi(t)| \hat{a}^{\dagger}\hat{a} |\psi(t)\rangle##
where ##\kappa## is the decay rate. It looks like the Fermi Golden rule but it's not exactly it.
 
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  • #2
Do you have a source for this equation? I guess, it's indeed an application of Fermi's golden rule, but from which model Lagrangian/Hamiltonian?
 
  • #3
For example here: https://books.google.co.uk/books?id=UnTNBQAAQBAJ&pg=PA37&lpg=PA37&dq=cavity+probability+of+emitting+a+photon&source=bl&ots=ivGtnIlmCz&sig=6qJjKian0MrATDeRY9nt4pDA1xQ&hl=el&sa=X&ved=0ahUKEwiPlq7C9LvPAhVIdR4KHejJD3EQ6AEIODAE#v=onepage&q=cavity probability of emitting a photon&f=false

page 37, equaiton (2.41).

They assume that this probability is proportional to the expectation value of the number of photon in the cavity. Based on what is this assumption?
Also, is ##\kappa## defined as the average number of photons that leak out of cavity per unit of time or as the probability of loosing a photon per unit of time?
 
  • #4
It's the number of photons. It's a reasonable assumption that it is the more probable to loose a photon the more photons are there. It's like the radioactive-decay law: The rate at which a nucleus decays is proportional to the number of nuclei present. This leads to the usual exponential decay. Quantum-theoretically that's a bit more tricky. Strictly speaking the exponential decay law cannot be exactly right (due to unitarity of time evolution). See, e.g., the textbook by Sakurai, Modern Quantum Mechanics, 2nd ed., Addison-Wesley.
 

Related to Probability of one-photon loss from a cavity

1. What is the probability of one-photon loss from a cavity?

The probability of one-photon loss from a cavity depends on several factors, including the quality of the cavity and the materials used. In general, the probability is very low, on the order of 10^-6 to 10^-9, but can vary depending on the specific setup.

2. How is the probability of one-photon loss calculated?

The probability of one-photon loss is calculated using the principles of quantum mechanics, specifically the laws of quantum optics. This involves taking into account the properties of the cavity, such as its reflectivity and transmissivity, as well as the properties of the photon, such as its energy and wavelength.

3. What are the main causes of one-photon loss from a cavity?

There are several possible causes of one-photon loss from a cavity, including imperfect mirrors, absorption by the cavity walls, and scattering by impurities in the cavity material. These factors can all contribute to a decrease in the probability of a photon remaining in the cavity.

4. How can the probability of one-photon loss be reduced?

To reduce the probability of one-photon loss from a cavity, scientists use various techniques such as using high-quality mirrors with low reflectivity, minimizing impurities in the cavity material, and using cryogenic temperatures to decrease absorption. Additionally, optimizing the design of the cavity can also help to reduce losses.

5. Why is the probability of one-photon loss important in cavity quantum electrodynamics?

The probability of one-photon loss is crucial in cavity quantum electrodynamics because it affects the overall efficiency and performance of the system. In experiments involving photons, it is essential to keep the probability of loss as low as possible to ensure accurate and reliable results. Additionally, a low probability of loss is necessary for the successful implementation of quantum information processing and quantum communication protocols.

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