Exact Unassisted Quantum Capacity of the Qubit Depolarizing Channel

Exact Unassisted Quantum Capacity of the Qubit Depolarizing Channel

Definitions and mathematical statement

For a qubit state $\rho$, the depolarizing channel is $\mathcal D_p(\rho)=(1-p)\rho+pI_2/2$, $0\le p\le1$. A block code uses a completely positive trace-preserving encoder into $n$ qubits and a decoder after $\mathcal D_p^{\otimes n}$. A rate $R$ is achievable if $k_n/n\to R$ and the entanglement-transmission error for $k_n$ logical qubits tends to zero. The unassisted quantum capacity is the supremum of achievable rates. Determine \[ Q(\mathcal D_p)\quad\text{for every }p\in[0,1]. \] No preshared entanglement or auxiliary classical messages are available. The parameter $p$ is the replacement-by-maximally-mixed-state probability in this formula, not another convention's total Pauli-error probability.

Formulation and concept references: [S1].

Short English statement

What is the exact rate for reliably transmitting quantum information through a qubit depolarizing channel using arbitrary block codes but no extra shared resources?

Research attempt

Conventions and source audit.

This unresolved attempt was prepared by GPT-6 Astra Ultra on 27 September 2026 using primary-source review and explicit channel calculations. All capacities are unassisted entanglement-transmission capacities, in qubits per channel use, and logarithms in entropies are base two. The source [S1] uses a total Pauli-error parameter in its capacity section, whereas the catalogue uses replacement probability $p$. Their conversion is made explicitly below. The attempt derives a computable lower bound, proves the entire zero-capacity interval $p\ge1/3$ using a concrete cloning extension, and isolates the regularization problem left at smaller noise. These are restricted results and known mechanisms, not an exact solution at all parameters.

The Pauli probabilities and a coherent-information lower bound.

For every qubit density matrix, \[ X\rho X+Y\rho Y+Z\rho Z=2I-\rho. \] Consequently the channel is a Pauli mixture with probabilities \[ q_0=1-\frac{3p}{4},\qquad q_X=q_Y=q_Z=\frac p4. \] The total nonidentity Pauli probability is $q=3p/4$. In particular a threshold quoted in terms of $q$ must be multiplied by $4/3$ before being used for the parameter in the question.

Input half of the Bell state, or equivalently use input density matrix $I/2$ with a reference purification. The output qubit remains maximally mixed, with entropy one. The joint reference-output state is diagonal in the four Bell states with eigenvalues $q_0,q_X,q_Y,q_Z$. A Stinespring purification gives the environment the same nonzero spectrum. Thus the coherent information for this input is \begin{equation}\label{wsref-0-61-32-37-33-3a-68-61-73-68-69-6e-67} I_c(I/2,\mathcal D_p)= 1-H(q_0,q_X,q_Y,q_Z) =1-h_2(3p/4)-(3p/4)\log_2 3. \end{equation} The quantum coding theorem, used here in the form discussed in [S1], makes positive coherent information an achievable asymptotic rate. Together with the trivial zero rate this yields the rigorous bound \[ Q(\mathcal D_p)\ge \max\{0,1-h_2(3p/4)-(3p/4)\log_2 3\}. \] The formula is a lower bound, not a converse. No claim is made that the maximally mixed input optimizes every block length.

At $p=0$ it gives one, which is also the maximum possible rate for a noiseless qubit channel. Its positive root occurs near replacement probability $0.2524$. This numerical value refers only to the displayed bound's zero, not to the unknown positivity threshold of the actual capacity. The numerical check uses the entropy expression itself and keeps the Pauli and replacement parameters separate.

A symmetric extension at replacement probability one third.

Construct an isometry from one input qubit to two output qubits $B,B'$ and an auxiliary qubit $E$ by \[ \begin{split} U|0\rangle&=\sqrt{2/3}|00\rangle_{BB'}|0\rangle_E +\sqrt{1/6}(|01\rangle+|10\rangle)_{BB'}|1\rangle_E,\\ U|1\rangle&=\sqrt{2/3}|11\rangle_{BB'}|1\rangle_E +\sqrt{1/6}(|01\rangle+|10\rangle)_{BB'}|0\rangle_E. \end{split} \] The two images are orthogonal unit vectors, so $U$ is an isometry. The reduced state on $B$ for input $|0\rangle$ has diagonal entries $5/6,1/6$, and the reversed entries occur for input $|1\rangle$. Tracing the off-diagonal operator $U|0\rangle\langle1|U^*$ over $B'E$ gives $(2/3)|0\rangle\langle1|$. Hence the channel to $B$ is exactly $\mathcal D_{1/3}$. Symmetry of the two output qubits gives that same channel to $B'$.

For the channel to $B$, the environment is $B'E$. Discarding $E$ recovers a copy of the channel output on $B'$. This is an explicit antidegrading map: the environment can reproduce everything delivered to the receiver. The construction is checked on a matrix basis, so it holds for arbitrary inputs and their entanglement with a reference, not merely for computational-basis states.

Why antidegradability forces zero unassisted capacity.

For completeness, the needed obstruction can be stated in entropic terms. Let $RBE$ be a purification after transmitting an input with reference $R$. If $B$ can be obtained from $E$ by a channel, coherent information cannot increase under that processing of $E$: \[ I(R\rangle B)\le I(R\rangle E). \] Purity gives $I(R\rangle E)=-I(R\rangle B)$, so $I(R\rangle B)\le0$. Tensor powers remain antidegradable by applying the same degrading map to each environment. Thus every input to every block length has nonpositive coherent information. The regularized capacity theorem in [S1] implies zero capacity. Equivalently, a reliable unassisted code would let both receiver and environment recover the same arbitrary unknown quantum message, contradicting asymptotic no-cloning. The entropic argument specifies the blockwise reason; it does not assume coherent information is additive for this channel.

The depolarizing channels compose according to \[ \mathcal D_s\circ\mathcal D_t=\mathcal D_{s+t-st}. \] For $1/3\le p\le1$, choose $s=(3p-1)/2\in[0,1]$. Then $\mathcal D_p=\mathcal D_s\circ\mathcal D_{1/3}$. A receiver can simulate extra output noise locally, so postprocessing cannot increase unassisted capacity. Therefore \begin{equation}\label{wsref-0-61-32-37-33-3a-7a-65-72-6f} Q(\mathcal D_p)=0\qquad(1/3\le p\le1). \end{equation} The endpoint matches total Pauli error $q=1/4$, the convention used in [S1]. Mistaking that $1/4$ for replacement probability would assert a strictly larger zero-capacity interval without proof.

A separate upper bound and monotonicity check.

Let $\mathcal E_p$ be the qubit erasure channel with an orthogonal classical erasure flag. Replacing that flag by $I/2$ and leaving the nonerased qubit unchanged produces $\mathcal D_p$. Hence \[ Q(\mathcal D_p)\le Q(\mathcal E_p)=\max\{0,1-2p\}, \] using the established unassisted erasure-capacity formula discussed in [S1]. The comparison uses a channel with a larger output alphabet and then discards information; it does not give the receiver an erasure flag in the original task. Combined with (a273:zero), it provides a valid bracket, although it is not intended as the best available upper bound at each noise value.

Likewise, if $0\le p\le p'<1$, then $\mathcal D_{p'}=\mathcal D_s\circ\mathcal D_p$ with $s=(p'-p)/(1-p)\in[0,1]$. Thus capacity is nonincreasing in $p$. The case $p=1$ is handled directly as a constant-output channel. Monotonicity and continuity-type intuitions do not determine where inside the gap between the proved lower and upper regions capacity first becomes zero.

Covariance does not remove the optimization.

Depolarizing noise is invariant under unitary conjugation, but coherent information is a difference of two entropies. It need not be concave in the input for this channel. Every pure input has coherent information zero, because the joint receiver-environment output is pure and its two marginal entropies coincide. At $p=0.3$, the value in (a273:hashing) is negative. Twirling a pure input to $I/2$ therefore decreases coherent information in this explicit example. A covariance argument claiming that twirling can only improve the objective is false. Replacing a negative achievable expression by zero is necessary, but does not establish the true optimizer.

There is a second, independent issue: even exact maximization of the single-use expression would not prove its additivity across uses. The capacity theorem has the form \[ Q(\mathcal D_p)=\sup_{n\ge1}\frac1n \max_{\rho_n} I_c(\rho_n,\mathcal D_p^{\otimes n}), \] where $\rho_n$ may correlate all $n$ input qubits. Product inputs give an additive lower bound, not an upper bound on that maximization. The source discusses block-code improvements, so it would be especially inappropriate to declare the elementary hashing curve exact merely because it has a simple closed expression.

What would close the remaining interval.

One possible route is a matching family of achievable block rates and a converse applying to every encoder and decoder with the prescribed vanishing entanglement error. A different route would prove a valid single-letter or otherwise computable characterization of the regularized optimization. A finite search over stabilizer codes only supplies lower bounds, and omits nonstabilizer codes; a finite search over input states at several block lengths supplies no all-length converse. Assisted capacities and classical communication protocols also change the resource model and cannot replace that missing bound.

Verification and outcome.

Finite matrix checks verify the isometry, both cloning marginals, the Pauli probabilities, the entropy values, and the composition law. The coding and erasure-capacity theorems are explicit external inputs, not claimed to have been reproved here. The strongest exact conclusions are $Q(\mathcal D_0)=1$ and (a273:zero), together with the displayed bounds at smaller noise. The first unresolved step is a sharp converse or exact regularized computation for the remaining parameters $0<p<1/3$. The requested capacity function remains UNRESOLVED.

Sources

• S1 Quantum Information chapter10, 2025 snapshot. https://www.preskill.caltech.edu/ph219/chap10_6A_2025.pdf. Formulation source.