The Bit-Flip Channel

When a qubit travels through a real device it may be accidentally flipped by the environment. The bit-flip channel captures this mathematically: with probability $p$ the Pauli- $X$ gate acts on the qubit, and with probability $1-p$ nothing happens.

Density matrices for mixed states

A pure state $|\psi\rangle$ is described by a ket. A probabilistic mixture requires a density matrix

\rho = \sum_i p_i |\psi_i\rangle\langle\psi_i|, \quad \sum_i p_i = 1, \quad p_i \geq 0.

The $2\times 2$ density matrix of a single qubit is Hermitian, positive semi-definite, and has unit trace. Pure states satisfy $\rho^2 = \rho$ ; mixed states satisfy $\mathrm{tr}(\rho^2) \lt 1$ .

Kraus representation

A quantum channel (a completely positive, trace-preserving map) can always be written in the Kraus form

\mathcal{E}(\rho) = \sum_k K_k \,\rho\, K_k^\dagger, \quad \sum_k K_k^\dagger K_k = I.

The bit-flip channel has two Kraus operators

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

You can verify the completeness relation: $K_0^\dagger K_0 + K_1^\dagger K_1 = (1-p)I + p I = I$ . The channel is therefore trace-preserving for any $p \in [0,1]$ .

Applying the channel to $|0\rangle$

Start with the pure state $\rho_{\mathrm{in}} = |0\rangle\langle 0|$ . Plugging into the Kraus sum:

\mathcal{E}(\rho_{\mathrm{in}}) = (1-p)\,I\,|0\rangle\langle 0|\,I + p\,X\,|0\rangle\langle 0|\,X = (1-p)\,|0\rangle\langle 0| + p\,|1\rangle\langle 1|.

This is a mixture of $|0\rangle$ and $|1\rangle$ with classical weights $(1-p)$ and $p$ . In matrix form:

\rho_{\mathrm{out}} = \begin{pmatrix} 1-p & 0 \\ 0 & p \end{pmatrix}.

The diagonal entries are the probabilities of the two outcomes. Measuring this state in the computational basis gives

P(\text{measure } 0) = \langle 0|\rho_{\mathrm{out}}|0\rangle = 1 - p, \qquad P(\text{measure } 1) = \langle 1|\rho_{\mathrm{out}}|1\rangle = p.

Purity loss

Before the channel the state is pure: $\mathrm{tr}(\rho_{\mathrm{in}}^2) = 1$ . After it

\mathrm{tr}(\rho_{\mathrm{out}}^2) = (1-p)^2 + p^2 = 1 - 2p(1-p).

This is strictly less than 1 for $0 \lt p \lt 1$ , confirming that noise has turned the pure state into a mixed state. The purity is minimised at $p = 1/2$ , where it equals $1/2$ .

Try it

The bit-flip channel with $p = 0.3$ is applied to $|0\rangle$ . Compute the probability of measuring $|1\rangle$ and return it.