1
$\begingroup$

I am interested in explicitly doing the math for what is decribed in this answer.

Specifically, I am interested in the simplest model for the creation of a coherent state from stimulated emission in a laser.

As @pyramids describes, a photon can either cause stimulated emission in a gain medium or not (the "boring case"). Therefore a photon will be in a superposition of the interesting case and the boring case, and as the photon goes through the medium (or loops in a cavity), this process is repeated over and over, creating a superposition of (near) infinite possible photon numbers.

I could swear that I worked this out by hand a few years ago, but I can't find my notes anywhere and I've been having some trouble reproducing the math.

My attempt:

my photon has a chance of creating a second photon with chance p, and has the "boring case" of no interacting with probability 1-p

$|1\rangle \rightarrow (1-p)|1\rangle + p|2\rangle$

which for now we will say p is small enough that we can replace (1-p) with 1.

$|1\rangle \rightarrow |1\rangle + p|2\rangle$

and using the second quantization notation:

$a^{\dagger^2}|0\rangle = \sqrt{2}|2\rangle \implies |2\rangle = \frac{1}{\sqrt{2}}a^{\dagger^2}|0\rangle $

we get that this:

$a^\dagger|0\rangle \rightarrow a^\dagger|0\rangle + p\frac{1}{\sqrt{2}} a^{\dagger^2}|0\rangle $

which I'm going to interpret as a rule that:

$a^\dagger \rightarrow a^\dagger + p\frac{1}{\sqrt{2}} a^{\dagger^2} $

which means that in the second step, this interaction is going to happen again:

$\begin{align}a^\dagger &\xrightarrow{1} a^\dagger + p\frac{1}{\sqrt{2}} a^{\dagger^2} \\ &\xrightarrow{2} \Big(a^\dagger + p\frac{1}{\sqrt{2}} a^{\dagger^2}\Big) + p\frac{1}{\sqrt{2}} \Big( a^\dagger + p\frac{1}{\sqrt{2}} a^{\dagger^2} \Big) \Big( a^\dagger + p\frac{1}{\sqrt{2}} a^{\dagger^2} \Big) \\ &= \Big(a^\dagger + p\frac{1}{\sqrt{2}} a^{\dagger^2}\Big) + p\frac{1}{\sqrt{2}} \Big( a^{\dagger^2} + 2p\frac{1}{\sqrt{2}} a^{\dagger^3} + 4p^2\frac{1}{2} a^{\dagger^4} \Big) \\ &= a^\dagger + 2 p\frac{1}{\sqrt{2}} a^{\dagger^2} + p^2 a^{\dagger^3} + 4p^3\frac{1}{2\sqrt{2}} a^{\dagger^4} \end{align} $

so I can keep repeating this step over-and-over and write an infinite series, but it doesn't seem like it is approaching the coefficients of a coherent state.

$\endgroup$
2
  • $\begingroup$ I think it's very difficult to start the derivation from a single photon state, like if you were to switch on the laser. It's a fairly large system of differential equations, even if you only consider classical rate equations, which leads to qualitatively different behavior depending on the parameters. $\endgroup$ – A. P. Jan 15 at 12:43
  • $\begingroup$ Instead you might try to show that a coherent state is the steady-state solution, by showing that the gain by a large number of excited atoms and the losses balance each other and result in the very same state. $\endgroup$ – A. P. Jan 15 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.