Single Mode Laser
A laser source is a light amplifier sitting in a resonant cavity.
This is a circular cavity which come to the same position after a trip of length L_{\text{cav}}; M_0 and M_1 are consider perfect mirror, and M_S is a semi-transparent mirror with coefficient R and T.
In the steady state the gain of the laser compensate the loss and the beam transmitted outside the cavity; if there is only a single state which is excited, that is a Continuous Wave (CW) laser source.
In this regime, the steady state must be identical to itself after a round trip, so it can be assimilated to a stretched cavity of length L_{\text{cav}} with periodic boundary conditions along that direction:
k_\lambda = \frac{2\pi n c}{L_{\text{cav}}} \quad n \in \mathbf N
The cross section S_{\text{cav}} is much larger than the wavelength so that diffraction can be neglected and it can be considered constant. The quantization volume is:
V_\lambda = S_{\text{cav}}L_{\text{cav}} = V_{\text{cav}}
With good approximation the radiation can be described by a quasi-classical state of the cavity characterized by a complex number | \alpha_\lambda | \gg 1. This corresponds to a high number of photon inside the cavity and it is not complex to calculate it as function of the power of the beam.
It is possible to link the power of the output with the power in the cavity:
P_{\text{out}} = T\,P_{\text{cav}} If there are \langle n_{\text{cav}} \rangle photons in the cavity, the energy is \langle n_{\text{cav}}\rangle \hbar\omega_\lambda and that is equal to the power circulated multiplied by the time it takes to make a loop:
\langle n_{\text{cav}}\rangle \hbar\omega_\lambda = P_{\text{cav}} \frac{L_{\text{cav}}}{c} = \frac{P_{\text{out}}}{T} \frac{L_{\text{cav}}}{c}
It is possible to consider a real example for a Helium-Neon laser with the following inputs:
\begin{aligned} & P_{\text{out}} = 10^{-3} \text{ W} \\ & \lambda_{\lambda} = \frac{2\pi c}{\omega_\lambda} = \frac{2\pi c}{2.98 \times 10^{15}} \text{ nm} = 633 \text{ nm} \\ & T = 10^{-2} = 1\% \\ & L_{\text{cav}} = 0.6 \text{ m} \end{aligned} That gives:
\langle n_{\text{cav}}\rangle = \frac{1}{\hbar\omega_\lambda} \frac{P_{\text{out}}}{T} \frac{L_{\text{cav}}}{c} = \frac{1}{1.0545718 \times 10^{-34} \, \text{m}^2 \text{kg/s} \times 2.98 \times 10^{15} \, \text{rad/s}} \times \frac{10^{-3} \, \text{W}}{10^{-2}} \times \frac{0.6 \, \text{m}}{3 \times 10^8 \, \text{m/s}} = 6.36 \times 10^8
Therefore:
\langle n_{\text{cav}} \rangle \approx 6.4 \times 10^{8} \, \text{photons}
The relative dispersion is therefore small:
\frac{\Delta N_{\text{cav}}}{\langle \mathbf N_{\text{cav}} \rangle} = \frac{1}{\sqrt{\langle \mathbf N_{\text{cav}} \rangle}} = \frac{1}{\sqrt{6.4 \times 10^{8}}} \approx 4 \times 10^{-5}
This dispersion is responsible for the accuracy of measurements made with an ideal laser, whose intensity is stabilized as much as possible. It is what is called the shot noise, which determines the standard quantum limit.