Amplitude-Damping Channel

Amplitude damping models energy dissipation: a qubit in its excited state $|1\rangle$ decays to the ground state $|0\rangle$ by emitting a quantum into a cold environment. It is the channel behind spontaneous emission, photon loss, and $T_1$ relaxation, and — unlike the bit-flip or depolarizing channels — it is not a mixture of unitaries and not unital, so it stress-tests every idea in this module.

The physical picture

Let $|0\rangle$ be the ground state and $|1\rangle$ the excited state. In a small time, an excited qubit decays with probability $\gamma \in [0,1]$ , dumping one quantum of energy into the environment; the ground state cannot decay. The two things that can happen — "a quantum was emitted" and "no quantum was emitted" — become the two Kraus operators.

Kraus operators

K_0 = \begin{pmatrix} 1 & 0 \\ 0 & \sqrt{1-\gamma} \end{pmatrix}, \qquad K_1 = \begin{pmatrix} 0 & \sqrt{\gamma} \\ 0 & 0 \end{pmatrix}.

$K_1 = \sqrt{\gamma}\,|0\rangle\langle 1|$ is the emission operator: it sends the excited state to the ground state, weighted by the decay amplitude $\sqrt\gamma$ . $K_0$ is the no-emission branch: it leaves $|0\rangle$ alone but damps the $|1\rangle$ amplitude by $\sqrt{1-\gamma}$ , because the mere fact that no quantum was seen makes the excited state less likely. Trace preservation:

K_0^\dagger K_0 + K_1^\dagger K_1 = \begin{pmatrix} 1 & 0 \\ 0 & 1-\gamma \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & \gamma \end{pmatrix} = \mathbb{1}.

Action on a density matrix

Writing $\rho = \begin{pmatrix} \rho_{00} & \rho_{01} \\ \rho_{10} & \rho_{11}\end{pmatrix}$ , the channel gives

\mathcal{E}(\rho) = \begin{pmatrix} \rho_{00} + \gamma\,\rho_{11} & \sqrt{1-\gamma}\,\rho_{01} \\[4pt] \sqrt{1-\gamma}\,\rho_{10} & (1-\gamma)\,\rho_{11} \end{pmatrix}.

Three effects are visible. The excited population decays: $\rho_{11}\to(1-\gamma)\rho_{11}$ . That lost population flows into the ground state: $\rho_{00}\to\rho_{00}+\gamma\rho_{11}$ , conserving trace. And the coherences shrink by $\sqrt{1-\gamma}$ . Iterating $n$ steps with per-step decay $\gamma$ gives $\rho_{11}\to(1-\gamma)^n\rho_{11}$ ; in continuous time $\rho_{11}(t)=e^{-t/T_1} \rho_{11}(0)$ with $1-\gamma = e^{-\Delta t/T_1}$ .

Not unital, not mixed-unitary

The fixed point is the ground state $|0\rangle\langle 0|$ , not the maximally mixed state: $\mathcal{E}(\mathbb{1}) = \begin{pmatrix} 1+\gamma & 0 \\ 0 & 1-\gamma\end{pmatrix} \ne \mathbb{1}$ , so the channel is not unital. Geometrically it contracts the Bloch ball and translates it toward the north pole $|0\rangle$ — an off-center shrink no mixture of unitaries (which always fixes the center) can reproduce.

Try it

Send the excited state $\rho = |1\rangle\langle 1|$ through the amplitude-damping channel with $\gamma = 0.3$ , using the Kraus operators above. Return the surviving excited-state population $\langle 1|\mathcal{E}(\rho)|1\rangle$ . Physically this is $(1-\gamma) = 0.7$ — compute it from $\sum_k K_k\rho K_k^\dagger$ to confirm.

Run your code to see the quantum state.