The Depolarizing Channel

The depolarizing channel is the most symmetric model of noise: with some probability the qubit is replaced by the maximally mixed state, otherwise it survives untouched. It is isotropic — it has no preferred direction — which makes its action on the Bloch ball especially clean and makes it the workhorse of average-case error analysis.

Definition

For a single qubit, the depolarizing channel with parameter $p \in [0,1]$ is

With probability $1-p$ nothing happens; with probability $p$ the qubit is discarded and reset to the maximally mixed state $\mathbb{1}/2$. At $p=0$ it is the identity; at $p=1$ it is the completely depolarizing channel that erases everything.

Pauli (Kraus) form

Because $\frac{\mathbb{1}}{2} = \frac14(\rho + X\rho X + Y\rho Y + Z\rho Z)$ for any single-qubit $\rho$ (the Pauli "twirl" identity), the channel is a mixture of Pauli operations:

The Kraus operators are therefore $K_0 = \sqrt{1-\tfrac{3p}{4}}\,\mathbb{1}$ and $K_{X,Y,Z} = \sqrt{\tfrac{p}{4}}\,\{X,Y,Z\}$. Check completeness: $\big(1-\tfrac{3p}{4}\big)\mathbb{1} + 3\cdot\tfrac{p}{4}\mathbb{1} = \mathbb{1}$, since each Pauli squares to $\mathbb{1}$. The channel is CPTP, and it is unital ($\mathcal{E}_p(\mathbb{1})=\mathbb{1}$) because each Pauli term is.

Action on the Bloch ball

Write $\rho = \tfrac12(\mathbb{1} + \vec r\cdot\vec\sigma)$. Each Pauli conjugation flips the sign of the two Bloch components orthogonal to it (e.g. $X\rho X$ sends $(r_x,r_y,r_z)\to(r_x,-r_y,-r_z)$). Averaging over $X,Y,Z$ kills all three components equally, so the channel shrinks the Bloch vector isotropically toward the center:

The Bloch ball contracts uniformly by the factor $1-p$; at $p=1$ every state collapses to the origin $\mathbb{1}/2$. This isotropy is the defining geometric signature of depolarizing noise — contrast it with amplitude damping, whose Bloch map is an asymmetric contraction plus a shift toward the north pole.

Purity and information loss

The shrink factor controls how much information survives. The purity of the output starting from a pure state ($|\vec r\,|=1$) is

falling from $1$ (pure) at $p=0$ to $\tfrac12$ (maximally mixed) at $p=1$. The same $(1-p)$ factor appears in the channel's average fidelity and in its entanglement-breaking threshold (the qubit depolarizing channel becomes entanglement-breaking at $p = 2/3$).

Try it

For the qubit depolarizing channel with $p = 0.4$ acting on the pure input $|0\rangle$ (Bloch vector $\vec r = (0,0,1)$), compute the output purity $\operatorname{tr}\big(\mathcal{E}_p(\rho)^2\big)$. Use the Bloch picture: the output Bloch vector has length $(1-p)$, and for a single qubit $\operatorname{tr}(\sigma^2) = \tfrac12(1 + |\vec s\,|^2)$. Return that number.

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