Phase-Damping Channel

Phase damping — pure dephasing — is the most quantum kind of noise: it destroys superposition without dissipating any energy. Populations are untouched; only the off-diagonal coherences decay. It is the channel responsible for $T_2$ decoherence and is the reason a qubit can lose its "quantumness" long before it loses its energy.

Definition and Kraus operators

Phase damping with parameter $\lambda \in [0,1]$ has Kraus operators

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

with $K_0^\dagger K_0 + K_1^\dagger K_1 = \operatorname{diag}(1,1-\lambda) + \operatorname{diag}(0,\lambda) = \mathbb{1}$ . Acting on a general $\rho$ ,

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

The diagonal — the populations — is exactly preserved, so no energy moves. Only the coherences $\rho_{01},\rho_{10}$ are damped, by $\sqrt{1-\lambda}$ .

Equivalent forms

Phase damping is a dephasing channel: defining $p=\tfrac12\big(1-\sqrt{1-\lambda}\big)$ it can be rewritten as a random- $Z$ channel,

\mathcal{E}(\rho) = (1-p)\,\rho + p\,Z\rho Z,

since $Z\rho Z$ flips the sign of the coherences while fixing the diagonal, and the convex mixture multiplies the coherences by $(1-2p) = \sqrt{1-\lambda}$ . So phase damping is a mixed-unitary channel (unlike amplitude damping) and it is unital: $\mathcal{E}(\mathbb{1})=\mathbb{1}$ .

Geometry: the Bloch ball collapses to an axis

In Bloch coordinates the map is

(r_x, r_y, r_z) \;\longmapsto\; \big(\sqrt{1-\lambda}\,r_x,\; \sqrt{1-\lambda}\,r_y,\; r_z\big).

The equatorial ( $x,y$ ) components shrink while the polar ( $z$ ) component is untouched. As $\lambda\to1$ the Bloch ball is squashed onto the $z$ -axis: the only surviving states are the classical mixtures of $|0\rangle$ and $|1\rangle$ . The poles are fixed; the equator — the maximally coherent superpositions $|\pm\rangle$ , $|\pm i\rangle$ — decays fastest.

$T_2$ , $T_1$ , and the $2T_1 \ge T_2$ bound

In continuous time, coherences decay as $e^{-t/T_2}$ . Amplitude damping ( $T_1$ relaxation) already shrinks coherences by $\sqrt{1-\gamma}$ , contributing $1/(2T_1)$ to the decoherence rate; pure dephasing adds an independent rate $1/T_\phi$ . Hence

\frac{1}{T_2} = \frac{1}{2T_1} + \frac{1}{T_\phi} \;\ge\; \frac{1}{2T_1} \quad\Longrightarrow\quad T_2 \le 2T_1 .

Pure phase damping is the limit $T_1\to\infty$ (no energy loss), where coherence still decays at rate $1/T_\phi$ — decoherence with no relaxation.

Try it

Send $\rho = |+\rangle\langle+| = \tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix}$ through the phase-damping channel with $\lambda = 0.5$ , using $\mathcal{E}(\rho)=\sum_k K_k\rho K_k^\dagger$ . Return the overlap $\langle +|\mathcal{E}(\rho)|+\rangle$ , the probability the state is still found in $|+\rangle$ . It equals $\tfrac12\big(1+\sqrt{1-\lambda}\big)$ — compute it and check.

Run your code to see the quantum state.