The Photon, the Quantum of Light — Question 6

Question 6

Frequency doubling conserves energy rather than photon number

A lossless idealized frequency doubler converts pairs of input photons at frequency f into single photons at 2f. For a real device, a fraction $\eta=.800$ of incident optical power emerges in the doubled-frequency beam; the rest remains at f.

1. Find doubled-frequency and residual photon rates in terms of input rate $\dot N$.

2. Find the total output photon rate and verify energy conservation.

Question 6 - Solution

Frequency doubling conserves energy rather than photon number

(a) Method and calculation

$$2hf\dot N_{2f}=\eta hf\dot N\quad\Rightarrow\quad\boxed{\dot N_{2f}=.400\dot N}.$$$$hf\dot N_f=(1-\eta)hf\dot N\quad\Rightarrow\quad\boxed{\dot N_f=.200\dot N}.$$

(b) Interpretation and further calculation

$$\boxed{\dot N_{\mathrm{out}}=.600\dot N}.$$$$P_{\mathrm{out}}=hf(.200\dot N)+2hf(.400\dot N)=\boxed{hf\dot N=P_{\mathrm{in}}}.$$

Physical check

Photon number need not be conserved in a nonlinear optical conversion, while the stated energy accounting must hold.