Checkpoint: Model a Noisy Qubit

This checkpoint synthesises the density-matrix and noise-channel ideas from module 13. Rather than running a circuit, you will track the probability distribution analytically through a two-step noise process — the same computation a density-matrix simulator performs under the hood.

The bit-flip channel, revisited

The bit-flip channel with error probability $p$ maps a single-qubit density matrix $\rho$ to

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

With probability $1-p$ the qubit is untouched; with probability $p$ the Pauli- $X$ gate fires and flips $|0\rangle \leftrightarrow |1\rangle$ . This is the quantum version of a binary symmetric channel with crossover probability $p$ .

Starting from the pure state $|0\rangle\langle 0|$ :

\mathcal{E}\!\left(|0\rangle\langle 0|\right) = (1-p)\,|0\rangle\langle 0| + p\,|1\rangle\langle 1| = \begin{pmatrix} 1-p & 0 \\ 0 & p \end{pmatrix}.

The qubit is now in a classical mixture: $|0\rangle$ with probability $1-p$ and $|1\rangle$ with probability $p$ .

Applying the channel a second time

What happens when the same channel is applied again to the output $\rho_1$ ?

Because $\rho_1$ is already diagonal, we only need to track how its diagonal entries transform. With probability $1-p$ each entry stays put; with probability $p$ it is swapped with the other diagonal entry (the $X$ gate exchanges the two populations). Writing $q_0 = 1-p$ and $q_1 = p$ for the populations after the first channel:

P_2(0) = (1-p)\,q_0 + p\,q_1 = (1-p)^2 + p^2.

The qubit lands back in $|0\rangle$ either because neither flip fired or because both flips fired — two independent rare events that restore the original state. Conversely

P_2(1) = 2p(1-p),

which is the probability that exactly one of the two flips fired.

Sanity-check with purity

The purity $\gamma = \text{tr}(\rho^2)$ measures how mixed the state is. For a diagonal density matrix $\rho = \text{diag}(a, 1-a)$ the purity is $a^2 + (1-a)^2$ .

After zero channels: $\gamma = 1$ (pure state $|0\rangle$ ).
After one channel with $p = 0.3$ : $\gamma = 0.7^2 + 0.3^2 = 0.58$ .
After two channels: $a = P_2(0) = 0.58$ , so $\gamma = 0.58^2 + 0.42^2 \approx 0.513$ .

Purity decreases monotonically: each application of the channel (with $0 \lt p \lt \tfrac{1}{2}$ ) pushes the state closer to the maximally mixed state $I/2$ , whose purity is $\tfrac{1}{2}$ .

Generalisation: $n$ sequential channels

After $n$ applications of the bit-flip channel with parameter $p$ , the population of $|0\rangle$ satisfies the recurrence

P_n(0) = (1-p)\,P_{n-1}(0) + p\,P_{n-1}(1),

with $P_0(0) = 1$ . The closed form is

P_n(0) = \tfrac{1}{2} + \tfrac{1}{2}(1 - 2p)^n,

which can be verified by induction. As $n \to \infty$ the population converges to $\tfrac{1}{2}$ , the maximally mixed state. This exponential relaxation toward equilibrium is a hallmark of Markovian quantum noise.

For $n = 1$ , $p = 0.3$ : $P_1(0) = \tfrac{1}{2} + \tfrac{1}{2}(0.4) = 0.7$ . For $n = 2$ : $P_2(0) = \tfrac{1}{2} + \tfrac{1}{2}(0.4)^2 = \tfrac{1}{2} + 0.08 = 0.58$ . Both match the direct calculation above.

Try it

Compute $P(\text{measure } 0)$ after two sequential bit-flip channels with $p = 0.3$ , starting from $|0\rangle$ . The grader checks the exact numerical answer.