Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 50 / 5 / a / Solution

Label the atomic ground state $|0⟩$ and excited state $|1⟩$, with $0\le p\le 1$. The process is the amplitude damping channel, with Kraus operatorsThey satisfy $A_0^{†}{A}_0+A_1^{†}{A}_1=I$, so the resulting Kraus representation is trace preserving.

The operator $A_1$ represents emission: it maps the excited state to $\sqrt{p}$ times the ground state and annihilates the ground state. Its probability for an initially excited atom is $p$. The operator $A_0$ represents no emission: it leaves the ground state unchanged and retains the excited amplitude with factor $\sqrt{1-p}$. This branch is not generally the identity operation; the absence of a photon itself changes the conditional relative amplitudes. For an arbitrary atomic density operator, the emission probability is $\mathrm{Tr}(A_1\rho A_1^{†})=p{\rho }_{11}$ and the no-emission probability is $1-p{\rho }_{11}$.