Depolarizing Channel

The depolarizing channel contracts every traceless component of a state by the same factor. For a finite-dimensional system with $d\ge2$,

$$\mathcal D_\lambda(\rho) = \lambda\rho + (1-\lambda)\frac{I}{d}.$$

The parameter $\lambda$ is the shrink factor. It multiplies every traceless part of the state by the same amount, so no direction in state space is singled out.

This symmetry is why the depolarizing channel is useful as a benchmark model. It is also why it is often unrealistic as a microscopic noise model. Dephasing, relaxation, leakage, coherent miscalibration, and correlated noise usually have structure that depolarizing noise deliberately erases.

Helpful background. The Kraus representation describes a channel by matrix operations. The Choi matrix tests it on one half of an entangled pair.

Depolarizing maps and parameter conventions

To define the channel as a linear map on arbitrary operators $X$, write

$$\mathcal D_\lambda(X) = \lambda X + (1-\lambda)\operatorname{Tr}(X)\frac{I}{d}.$$

For density operators this reduces to the state formula above. The channel is trace preserving because

$$\operatorname{Tr}\mathcal D_\lambda(X) = \lambda\operatorname{Tr}X + (1-\lambda)\operatorname{Tr}X = \operatorname{Tr}X.$$

It is also unital:

$$\mathcal D_\lambda(I)=I.$$

Complete positivity imposes the finite-dimensional bound

$$-\frac{1}{d^2-1} \le \lambda \le 1.$$

If $\lambda=1$, the channel is the identity. If $\lambda=0$, it is the completely depolarizing channel:

$$\mathcal D_0(\rho)=\frac{I}{d}.$$

Many sources instead use a replacement-probability parameter $q$:

$$\mathcal D_q^{\mathrm{rep}}(\rho) = (1-q)\rho + q\frac{I}{d}, \qquad 0\le q\le1.$$

This is the same formula with

$$\lambda=1-q.$$

In this replacement convention, $q$ is literally the probability of discarding the input state and replacing it by $I/d$. The full completely positive family allows slightly negative $\lambda$ as well; those channels are valid but no longer have this literal replacement-probability interpretation.

Traceless-Part Picture

Every state can be decomposed as

$$\rho = \frac{I}{d}+\tau, \qquad \operatorname{Tr}\tau=0.$$

The depolarizing channel acts as

$$\mathcal D_\lambda(\rho) = \frac{I}{d} + \lambda\tau.$$

Thus $I/d$ is the unique fixed state for $\lambda\ne1$ in the completely positive range. When $\lambda<0$, the traceless part also changes sign: the output passes to the opposite side of $I/d$, while its distance from $I/d$ shrinks.

Under repeated application, the shrink factors multiply. The fixed-point and composition viewpoint is summarized in Channel Composition and Fixed Points.

The purity transforms as

$$\operatorname{Tr} \left[ \mathcal D_\lambda(\rho)^2 \right] = \frac{1}{d} + \lambda^2 \left( \operatorname{Tr}\rho^2-\frac{1}{d} \right).$$

Pure states are mixed unless $\lambda=1$. In the usual Markovian decay convention, $0\le\lambda\le1$ and the purity decreases monotonically toward $1/d$.

Qubit Bloch-Sphere Action

For a qubit,

$$\rho = \frac{1}{2} \left( I+\mathbf r\cdot\boldsymbol\sigma \right),$$

where $\mathbf r\in\mathbb R^3$ is the Bloch vector. The depolarizing channel gives

$$\mathcal D_\lambda(\rho) = \frac{1}{2} \left( I+\lambda\mathbf r\cdot\boldsymbol\sigma \right).$$

Hence

$$\mathbf r\longmapsto \lambda\mathbf r.$$

Every Bloch-sphere direction contracts by the same factor. This is the key distinction from dephasing, which contracts only directions transverse to a preferred axis.

For a literal replacement probability $q$,

$$\mathbf r\longmapsto (1-q)\mathbf r.$$

For the full qubit depolarizing family, complete positivity allows

$$-\frac{1}{3}\le\lambda\le1.$$

The negative part of this interval is mathematically allowed, but it is not the ordinary picture of random replacement by $I/2$.

Qubit Pauli Representation

A common qubit convention writes the channel as equally likely Pauli errors:

$$\mathcal P_p(\rho) = (1-p)\rho + \frac{p}{3} \left( X\rho X + Y\rho Y + Z\rho Z \right), \qquad 0\le p\le1.$$

This is the isotropic special case of a Pauli channel.

This is a depolarizing channel with shrink factor

$$\lambda = 1-\frac{4p}{3}.$$

The conversion to replacement probability is therefore

$$q = 1-\lambda = \frac{4p}{3}.$$

This is the most common convention trap. If $p$ means “probability that a nonidentity Pauli error occurred,” then $p=1$ gives $\lambda=-1/3$, not the completely mixed output. If $q$ means “probability of replacing the state by $I/2$,” then complete depolarization is $q=1$, which corresponds to $p=3/4$ in the Pauli-error convention.

The Pauli representation is a Kraus representation with

$$K_0=\sqrt{1-p}\,I, \qquad K_1=\sqrt{\frac{p}{3}}\,X, \qquad K_2=\sqrt{\frac{p}{3}}\,Y, \qquad K_3=\sqrt{\frac{p}{3}}\,Z.$$

The completeness condition is

$$\sum_{\mu=0}^3K_\mu^\dagger K_\mu=I.$$

Generalized Pauli Form

Let $\{W_a\}_{a=0}^{d^2-1}$ be a unitary error basis with $W_0=I$ and

$$\operatorname{Tr}(W_a^\dagger W_b) = d\,\delta_{ab}.$$

The channel with equal probability over all nonidentity errors is

$$\Phi_p(\rho) = (1-p)\rho + \frac{p}{d^2-1} \sum_{a=1}^{d^2-1} W_a\rho W_a^\dagger.$$

Using

$$\sum_{a=0}^{d^2-1} W_a\rho W_a^\dagger = d\,\operatorname{Tr}(\rho)I,$$

one obtains

$$\Phi_p(\rho) = \lambda\rho + (1-\lambda)\frac{I}{d}, \qquad \lambda = 1-\frac{d^2p}{d^2-1}.$$

This is the $d$-dimensional analogue of the qubit Pauli-error convention.

Choi Matrix and Complete Positivity

Using the output-input convention of Choi Matrix, the Choi matrix of $\mathcal D_\lambda$ is

$$J_{\mathcal D_\lambda} = \lambda \lvert\Omega\rangle\langle\Omega\rvert + \frac{1-\lambda}{d} I\otimes I,$$

where

$$\lvert\Omega\rangle = \sum_{i=1}^d \lvert i\rangle\otimes\lvert i\rangle$$

is unnormalized.

The vector

$$\lvert\Omega_N\rangle = \frac{1}{\sqrt d} \lvert\Omega\rangle$$

is an eigenvector with eigenvalue

$$\frac{1+(d^2-1)\lambda}{d}.$$

Every vector orthogonal to $\lvert\Omega_N\rangle$ has eigenvalue

$$\frac{1-\lambda}{d}.$$

Therefore $J_{\mathcal D_\lambda}\ge0$ exactly when

$$-\frac{1}{d^2-1} \le \lambda \le 1.$$

Testing only pure inputs gives the weaker condition $-1/(d-1)\le\lambda\le1$, obtained from their output eigenvalues. Complete positivity must also accommodate entangled inputs.

For a qubit with $\lambda=-1/2$, every isolated state still maps to a valid state: $|0\rangle\langle0|$ becomes $\operatorname{diag}(1/4,3/4)$. But the normalized Choi output has an eigenvalue $(1+3\lambda)/4=-1/8$. The map therefore fails on half of a Bell pair and is not a physical channel acting on arbitrary inputs.

The Choi rank is $1$ at $\lambda=1$, full rank $d^2$ in the interior, and $d^2-1$ at the lower complete-positivity boundary.

Markovian Depolarizing Semigroup

In a frame with no additional Hamiltonian rotation, a time-homogeneous depolarizing semigroup with rate $\Gamma\ge0$ has

$$\mathcal D_t(\rho) = e^{-\Gamma t}\rho + \left( 1-e^{-\Gamma t} \right) \frac{I}{d}.$$

Equivalently,

$$\lambda(t)=e^{-\Gamma t}.$$

For normalized states this solves

$$\frac{d\rho}{dt} = \Gamma \left( \frac{I}{d}-\rho \right).$$

For a qubit, one Lindblad form is

$$\frac{d\rho}{dt} = \frac{\Gamma}{4} \sum_{j=x,y,z} \left( \sigma_j\rho\sigma_j-\rho \right).$$

In Bloch-vector language, this gives

$$\frac{d\mathbf r}{dt} = -\Gamma\mathbf r, \qquad \mathbf r(t)=e^{-\Gamma t}\mathbf r(0).$$

This semigroup explores only the $0\le\lambda\le1$ part of the depolarizing family. Negative shrink factors are valid channels but not produced by this simple continuous-time relaxation toward $I/d$.

Physical uses and limits

Depolarization is useful for benchmark calculations with deliberate isotropy. A general device may instead have a preferred axis, nonunital relaxation, leakage, coherent errors, or correlations between uses. In particular, a unital channel cannot describe cooling to a unique pure ground state, and a fixed-dimensional channel cannot represent an orthogonal loss flag without enlarging the output space.

Unitary twirling and average fidelity

Uniformly averaging $U^\dagger\Phi(U\rho U^\dagger)U$ over all unitaries, with the invariant Haar measure, turns a CPTP map on a fixed $d$-dimensional space into a depolarizing channel. An exact unitary 2-design gives the same average. The average pure-state fidelity with the identity operation is preserved:

$$F_{\mathrm{avg}}=\int d\psi\,\langle\psi|\Phi(|\psi\rangle\langle\psi|)|\psi\rangle =\frac{1+(d-1)\lambda}{d}.$$

Thus $\lambda=(dF_{\mathrm{avg}}-1)/(d-1)$. Pauli twirling alone generally leaves different contractions along different axes; it need not produce depolarization.

Matching this one fidelity does not preserve relaxation direction, coherent accumulation, or behavior under repeated correlated uses. Leakage outside the chosen space must first be modeled explicitly. Pauli Noise and Depolarizing Channels develops parameter conversion and applications to quantum-information protocols.

Simple Examples

Pure input state

For a pure state $\rho=\lvert\psi\rangle\langle\psi\rvert$,

$$\mathcal D_\lambda(\rho) = \lambda \lvert\psi\rangle\langle\psi\rvert + (1-\lambda)\frac{I}{d}.$$

The eigenvalue along $\lvert\psi\rangle$ is

$$\lambda+\frac{1-\lambda}{d},$$

and the $d-1$ orthogonal eigenvalues are

$$\frac{1-\lambda}{d}.$$

For $0\le\lambda\le1$, this is a convex mixture of the input and an isotropic background. For negative $\lambda$, that interpretation fails even though the displayed eigenvalues can all be nonnegative.

Computational basis state of a qubit

For $\rho=\lvert0\rangle\langle0\rvert$,

$$\mathcal D_\lambda(\rho) = \frac{1+\lambda}{2} \lvert0\rangle\langle0\rvert + \frac{1-\lambda}{2} \lvert1\rangle\langle1\rvert.$$

Unlike dephasing in the $Z$ basis, depolarizing noise changes the populations of $\lvert0\rangle$ and $\lvert1\rangle$.

Completely depolarizing channel

At $\lambda=0$,

$$\mathcal D_0(\rho)=\frac{I}{d}$$

for every input state. All input information is erased from the output state, although the output system itself remains present. This differs from an erasure channel, where a flagged loss state is introduced.

Exercises

Bloch shrink from Pauli errors

Starting from

$$\mathcal P_p(\rho) = (1-p)\rho + \frac{p}{3} \left( X\rho X+Y\rho Y+Z\rho Z \right),$$

show that the Bloch vector transforms as $\mathbf r\mapsto(1-4p/3)\mathbf r$.

Solution

Write

$$\rho = \frac12 \left( I+r_xX+r_yY+r_zZ \right).$$

Conjugation by $X$ preserves $r_x$ and flips $r_y,r_z$. Conjugation by $Y$ preserves $r_y$ and flips $r_x,r_z$. Conjugation by $Z$ preserves $r_z$ and flips $r_x,r_y$. Therefore the average of the three nonidentity Pauli conjugations sends

$$\mathbf r\longmapsto -\frac{1}{3}\mathbf r.$$

Combining this with the identity branch gives

$$\mathbf r' = (1-p)\mathbf r + p \left( -\frac{1}{3}\mathbf r \right) = \left( 1-\frac{4p}{3} \right)\mathbf r.$$

Choi bound

Use the Choi eigenvalues to derive the complete-positivity interval for $\mathcal D_\lambda$.

Solution

The Choi matrix is

$$J_{\mathcal D_\lambda} = \lambda \lvert\Omega\rangle\langle\Omega\rvert + \frac{1-\lambda}{d} I\otimes I.$$

On the normalized maximally entangled direction, the eigenvalue is

$$\frac{1+(d^2-1)\lambda}{d}.$$

On the orthogonal subspace, the eigenvalue is

$$\frac{1-\lambda}{d}.$$

Both must be nonnegative. Hence

$$1+(d^2-1)\lambda\ge0, \qquad 1-\lambda\ge0.$$

Therefore

$$-\frac{1}{d^2-1} \le \lambda \le 1.$$

Purity transformation

Let $\rho=I/d+\tau$ with $\operatorname{Tr}\tau=0$. Show that

$$\operatorname{Tr} \left[ \mathcal D_\lambda(\rho)^2 \right] = \frac{1}{d} + \lambda^2 \left( \operatorname{Tr}\rho^2-\frac{1}{d} \right).$$

Solution

Since

$$\mathcal D_\lambda(\rho) = \frac{I}{d}+\lambda\tau,$$

we have

$$\operatorname{Tr} \left[ \mathcal D_\lambda(\rho)^2 \right] = \operatorname{Tr} \left[ \frac{I}{d^2} + \frac{2\lambda}{d}\tau + \lambda^2\tau^2 \right].$$

The cross term vanishes because $\operatorname{Tr}\tau=0$, and $\operatorname{Tr}(I/d^2)=1/d$. Thus

$$\operatorname{Tr} \left[ \mathcal D_\lambda(\rho)^2 \right] = \frac{1}{d} + \lambda^2\operatorname{Tr}\tau^2.$$

Also

$$\operatorname{Tr}\rho^2 = \frac{1}{d} + \operatorname{Tr}\tau^2.$$

Substituting gives the claimed formula.

Dephasing versus depolarizing

Compare the action of complete dephasing in the $Z$ basis and complete depolarization on $\lvert0\rangle\langle0\rvert$.

Solution

Complete dephasing in the $Z$ basis is

$$\Delta_Z(\rho) = P_0\rho P_0+P_1\rho P_1.$$

Since $\lvert0\rangle$ is already a $Z$-basis state,

$$\Delta_Z(\lvert0\rangle\langle0\rvert) = \lvert0\rangle\langle0\rvert.$$

Complete depolarization gives

$$\mathcal D_0(\lvert0\rangle\langle0\rvert) = \frac{I}{2}.$$

Thus dephasing leaves this basis population state unchanged, while depolarization randomizes it.

References

• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002). Publisher.
• M.-D. Choi, “Completely positive linear maps on complex matrices,” Linear Algebra and its Applications 10, 285–290 (1975). DOI.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary edition, Cambridge University Press (2010).
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018). Author’s text.