# I Probability of one-photon loss from a cavity

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1. Oct 8, 2016

### maxverywell

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.

2. Oct 9, 2016

### vanhees71

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. Oct 10, 2016

### maxverywell

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. Oct 10, 2016

### vanhees71

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.