The Exercise

An atom with ground state $|0\rangle$ and excited state $|1\rangle$ is coupled at resonance to a single mode of the electromagnetic field, which starts in the vacuum. Atom and mode evolve together for a time $t$ under a coupling of strength $g$, and the mode is then discarded. The exercise is to show that the atom has undergone amplitude damping, with damping probability $\gamma = \sin^2(gt)$.

An Atom and a Mode

The atom and the mode are both qubits with state space $H$ and computational basis $|0\rangle, |1\rangle$; for the mode, $|0\rangle$ is the vacuum and $|1\rangle$ holds one photon. The pair has state space $H \otimes H$, the atom’s factor first, and $|a\rangle|c\rangle$ stands for $|a\rangle \otimes |c\rangle$.

For $i, j \in \{0, 1\}$ the matrix unit $|i\rangle\langle j|$ is the rank one operator on $H$ that takes $|j\rangle$ to $|i\rangle$ and annihilates the other basis vector.^✦ For $i = j$ the matrix unit $|i\rangle\langle i|$ is the projector onto $|i\rangle$. The lowering operator $\sigma_-$ of a qubit with basis $|0\rangle, |1\rangle$ is the operator $|0\rangle\langle 1|$ on $H$.

Amplitude Damping

The amplitude damping channel $\mathcal{E}_{AD}$ with damping probability $\gamma$ is the quantum operation on the qubit with operation elements $$E_0 = |0\rangle\langle 0| + \sqrt{1-\gamma}\,|1\rangle\langle 1| , \qquad E_1 = \sqrt{\gamma}\,\sigma_- ,$$(1) which sends an operator $\rho$ on $H$ to $$\mathcal{E}_{AD}(\rho) = E_0 \rho E_0^\dagger + E_1 \rho E_1^\dagger .$$(2)^✦ The element $E_1$ is the loss of one quantum of energy, which befalls an excited atom with probability $\gamma$; $E_0$ is the evolution when nothing is lost.

The Emission Channel

The exchange coupling $S$ of atom and mode is the operator on $H \otimes H$ $$S = |0\rangle\langle 1| \otimes |1\rangle\langle 0| + |1\rangle\langle 0| \otimes |0\rangle\langle 1| ,$$(3)^✦ the interaction of the model of Jaynes and Cummings at zero detuning. The single excitation projector $Q$ on $H \otimes H$ is the projector onto the span of $|1\rangle|0\rangle$ and $|0\rangle|1\rangle$, $$Q = |0\rangle\langle 0| \otimes |1\rangle\langle 1| + |1\rangle\langle 1| \otimes |0\rangle\langle 0| .$$(4)^✦

The resonant evolution $U$ of atom and mode for coupling strength $g$ and time $t$ is the operator on $H \otimes H$ $$U = I + \bigl(\cos(gt) - 1\bigr)\, Q - i \sin(gt)\, S ,$$(5)^✦ with $I$ the identity; it equals $e^{-igtS}$.

The emission dilation $V$ for coupling strength $g$ and time $t$ is the map from $H$ to $H \otimes H$ that takes a vector $|\psi\rangle$ of $H$ to the state of the pair obtained by putting the mode in the vacuum and letting the pair evolve under $U$: $$V|\psi\rangle = U\bigl(|\psi\rangle \otimes |0\rangle\bigr) .$$(6)^✦

The spontaneous emission channel $\mathcal{E}_{SE}$ for coupling strength $g$ and time $t$ is the quantum operation that sends an operator $\rho$ on $H$ to the partial trace $\operatorname{tr}_2$ over the mode of $V\rho V^\dagger$, $$\mathcal{E}_{SE}(\rho) = \operatorname{tr}_2\bigl(V \rho V^\dagger\bigr) .$$(7)^✦

The Claim

Taking the partial trace in the basis of the mode writes $\mathcal{E}_{SE}$ with two operation elements, $(I \otimes \langle 0|)\,V = |0\rangle\langle 0| + \cos(gt)\, |1\rangle\langle 1|$ and $(I \otimes \langle 1|)\,V = -i\sin(gt)\,\sigma_-$. At $\gamma = \sin^2(gt)$ the second is $E_1$ up to a phase, which cancels in $E_1 \rho E_1^\dagger$, and the first is $E_0$ exactly when $\cos(gt) \ge 0$, since $\sqrt{1 - \sin^2(gt)} = |\cos(gt)|$.

What Is Checked

The claim is stated over two channels on the atom: the spontaneous emission channel (7), built from the dilation $V$ of (6) and through it from $U$, $S$ and $Q$, and the amplitude damping channel (2) with its elements $E_0$ and $E_1$.