Amplitude Damping (T₁)

Every physical qubit is weakly coupled to its surroundings. An excited qubit ( $|1\rangle$ ) can spontaneously emit a photon and relax to the ground state ( $|0\rangle$ ). This energy dissipation is called amplitude damping, and its characteristic timescale is $T_1$ (the longitudinal or energy-relaxation time).

Kraus operators of the amplitude-damping channel

For a damping strength $\gamma \in [0, 1]$ , the amplitude-damping channel is defined by two 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}.

You can verify trace-preservation directly:

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} = I. \checkmark

The channel acts on a density matrix $\rho$ as

\mathcal{E}(\rho) = K_0 \rho K_0^\dagger + K_1 \rho K_1^\dagger.

$K_1$ is a lowering operator: it maps $|1\rangle$ to $|0\rangle$ and annihilates $|0\rangle$ . Its presence means that excited-state population drains into the ground state at rate $\gamma$ .

What happens to each element of the density matrix

Let $\rho = \begin{pmatrix}\rho_{00} & \rho_{01} \\ \rho_{10} & \rho_{11}\end{pmatrix}$ . Carrying out the matrix products gives

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

Two effects are visible:

Population transfer. The excited-state population $\rho_{11}$ decreases by the factor $(1-\gamma)$ , and the fraction $\gamma \rho_{11}$ moves to the ground state, raising $\rho_{00}$ .
Coherence decay. The off-diagonal (coherence) terms shrink by the factor $\sqrt{1-\gamma}$ .

At $\gamma = 1$ the channel is maximally damping: the qubit always ends in $|0\rangle$ regardless of the input, and all coherences vanish.

Effect on the excited state

For the pure excited state $\rho = |1\rangle\langle 1| = \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}$ , the formula above simplifies to

\mathcal{E}(|1\rangle\langle 1|) = \begin{pmatrix}\gamma & 0 \\ 0 & 1-\gamma\end{pmatrix}.

The qubit has relaxed to a classical mixture: with probability $\gamma$ it is now in $|0\rangle$ and with probability $1-\gamma$ it remains in $|1\rangle$ . There is no superposition because the off-diagonal entries of $|1\rangle\langle 1|$ were zero to begin with.

Effect on a superposition

A superposition $|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ has the density matrix

|{+}\rangle\langle{+}| = \frac{1}{2}\begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix}.

After the amplitude-damping channel with strength $\gamma$ :

\mathcal{E}(|{+}\rangle\langle{+}|) = \frac{1}{2}\begin{pmatrix}1 + \gamma & \sqrt{1-\gamma} \\ \sqrt{1-\gamma} & 1-\gamma\end{pmatrix}.

The probability of measuring $|0\rangle$ is the $(0,0)$ entry:

P(0) = \frac{1 + \gamma}{2} = \rho_{00} + \gamma \cdot \rho_{11}.

For $\gamma = 0.5$ this gives $P(0) = 0.75$ : the excited-state component has half a chance of falling to the ground state, raising the ground-state probability above the ideal $0.5$ .

Connection to $T_1$ on real hardware

On a physical device the damping strength $\gamma$ grows with idle time $t$ :

\gamma(t) = 1 - e^{-t/T_1}.

A qubit left idle for $t \ll T_1$ suffers little energy loss ( $\gamma \approx t/T_1$ ). For $t = T_1$ , $\gamma \approx 0.63$ ; for $t \gg T_1$ the qubit is almost certain to be found in $|0\rangle$ . Typical superconducting qubits have $T_1$ in the range of tens to hundreds of microseconds (Krantz et al., 2019; cited in the Qiskit Textbook), which constrains the maximum depth of useful circuits.

Try it

The amplitude-damping channel with $\gamma = 0.5$ is applied to $|{+}\rangle$ . Using the formula $P(0) = \rho_{00} + \gamma \cdot \rho_{11}$ , compute and return the probability of measuring $|0\rangle$ after the channel acts.

Run your code to see the quantum state.