The Phase-Flip Channel

Real quantum hardware is never perfectly isolated. The environment continuously disturbs qubit phases, a process called dephasing. The phase-flip channel is the simplest mathematical model of this effect.

Kraus operators and the phase-flip channel

A quantum channel is a completely positive, trace-preserving (CPTP) map. One natural representation uses Kraus operators $\{K_i\}$ , which satisfy $\sum_i K_i^\dagger K_i = I$ . The action on a density matrix $\rho$ is

\mathcal{E}(\rho) = \sum_i K_i \rho K_i^\dagger.

For the phase-flip channel with error probability $p \in [0, \tfrac{1}{2}]$ the two Kraus operators are

K_0 = \sqrt{1-p}\,I, \qquad K_1 = \sqrt{p}\,Z.

Substituting and expanding gives

\mathcal{E}(\rho) = (1-p)\,I\rho I^\dagger + p\,Z\rho Z^\dagger = (1-p)\,\rho + p\,Z\rho Z.

With probability $1-p$ nothing happens; with probability $p$ the qubit suffers a $Z$ error (a phase flip). You can verify the trace-preservation condition directly: $(1-p)I^\dagger I + p Z^\dagger Z = (1-p)I + pI = I$ .

What does a Z error do?

The $Z$ gate is diagonal in the computational basis:

Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.

This is catastrophic for a qubit stored in the $X$ basis. The equal-superposition state

|{+}\rangle = \tfrac{1}{\sqrt{2}}\bigl(|0\rangle + |1\rangle\bigr)

becomes

Z|{+}\rangle = \tfrac{1}{\sqrt{2}}\bigl(|0\rangle - |1\rangle\bigr) = |{-}\rangle.

The two states $|{+}\rangle$ and $|{-}\rangle$ are orthogonal — they are perfectly distinguishable in the $X$ basis, so a single phase flip is detectable and correctable in principle.

Dephasing destroys off-diagonal coherence

Working in the density-matrix picture makes the destruction of coherence explicit. Write the pure state $|{+}\rangle\langle{+}|$ in matrix form:

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

Applying the channel with arbitrary $p$ :

\mathcal{E}\!\left(|{+}\rangle\langle{+}|\right) = \frac{1}{2}\begin{pmatrix} 1 & 1-2p \\ 1-2p & 1 \end{pmatrix}.

The diagonal entries (populations) are unchanged, but the off-diagonal entries (coherences) are multiplied by $(1-2p)$ . At $p = \tfrac{1}{2}$ the coherences vanish entirely and the qubit is maximally mixed, $\tfrac{1}{2}I$ . This is the hallmark of decoherence: the system retains its energy but loses all quantum superposition.

Connection to real hardware

On a physical device dephasing accumulates continuously. The characteristic timescale is $T_2$ (or its Ramsey counterpart $T_2^*$ ). In the discrete model studied here, $p$ increases with the time a qubit spends idle. Keeping gate sequences short — or encoding information in decoherence-free subspaces — is the standard mitigation strategy.

Try it

The exercise below asks you to simulate a single phase-flip event acting on $|{+}\rangle$ . Build the 1-qubit circuit that prepares $|{+}\rangle$ and then applies the $Z$ error, leaving the qubit in $|{-}\rangle$ . The grader compares the full statevector, so the order matters.

Run your code to see the quantum state.

After running your circuit, inspect the statevector: the amplitude of $|0\rangle$ should be $+\tfrac{1}{\sqrt{2}} \approx 0.707$ and the amplitude of $|1\rangle$ should be $-\tfrac{1}{\sqrt{2}} \approx -0.707$ , confirming the phase flip.