Generalized amplitude damping: entropy and classical capacity

Source and verification

This article explains family 276 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

🏷️ A noisy qubit and its classical messages

A qubit state is a positive $2\times 2$ matrix of trace one. In the basis $|0⟩,|1⟩$, write it as $\rho =\left(\begin{array}{cc}1-q& z\\ \overline{z}& q\end{array}\right)$, where $q\in [0,1]$ is the excited-state population and $|z{|}^2\le q(1-q)$ measures coherence. The state is pure when equality holds.

The generalized amplitude-damping channel $\mathcal{A}={\mathcal{A}}_{\gamma ,\nu }$ acts by

$$\begin{array}{c}\mathcal{A}(\rho )=\left(\begin{array}{cc}1-t& \sqrt{1-\gamma }z\\ \sqrt{1-\gamma }\overline{z}& t\end{array}\right),\\ t=(1-\gamma )q+\gamma \nu ,\\ 0\le \gamma ,\nu \le 1.\end{array}$$

Here $\gamma$ is the damping strength and $\nu$ is the stationary excited-state population. Ordinary amplitude damping has $\nu =0$. At $\gamma =0$ the channel is the identity; at $\gamma =1$ it replaces every input by the fixed state $\mathrm{diag}(1-\nu ,\nu )$.

A quantum channel is completely positive and trace preserving: it remains positive when tensored with the identity on an auxiliary system and preserves total probability. In finite dimensions it has a Kraus representation $\mathcal{N}(P)={\sum }_{\ell }{V}_{\ell }P{V}_{\ell }^{\ast }$ with ${\sum }_{\ell }{V}_{\ell }^{\ast }{V}_{\ell }=I$. For the displayed channel, one can use the two diagonal matrices $\sqrt{1-\nu }\mathrm{diag}(1,\sqrt{1-\gamma })$ and $\sqrt{\nu }\mathrm{diag}(\sqrt{1-\gamma },1)$, together with the transitions $\sqrt{\gamma (1-\nu )}|0⟩⟨1|$ and $\sqrt{\gamma \nu }|1⟩⟨0|$.

To transmit a classical message, a sender chooses an input state according to that message. Across repeated uses, the state may be entangled among all input qubits. The receiver may measure all outputs collectively. The capacity here is unassisted: no preshared entanglement or feedback is available.

🏷️ Entropy and the regularization problem

Use $S(P)=-\mathrm{Tr}P\ln P$, with $0\ln 0=0$, for entropy in nats. The same expression will also be used for unnormalized positive matrices. Write $h(p)=-p\ln p-(1-p)\ln (1-p)$ for binary entropy.

For an ensemble $\{w_a,{\rho }_a\}$, the Holevo information is the entropy of the average output minus the average output entropy. The one-use Holevo capacity $\chi (\mathcal{N})$ is its supremum over finite ensembles, divided by $\ln 2$ to obtain bits. Operationally, the unassisted classical capacity is

$$C(\mathcal{N})=\underset{n\ge 1}{\sup }\frac{1}{n}\chi ({\mathcal{N}}^{\otimes n}).$$

This coding-theorem formula permits arbitrary entangled signals in each block and collective decoding. Product ensembles show that the block Holevo values are superadditive, so the supremum equals the asymptotic normalized limit. A one-use optimizer alone does not remove this regularization.

A qubit state’s eigenvalues are determined by its determinant $u$. Define

$$\begin{array}{c}g(u)=h\left(\frac{1+\sqrt{1-4u}}{2}\right),\\ 0\le u\le \frac{1}{4}.\end{array}$$

Its entropy is $g(u)$. For a pure input of population $p$, direct evaluation of the channel determinant gives

$$\begin{array}{c}\begin{array}{rl}v(p)=& \gamma \nu (1-\nu )\\ & +\gamma (1-\gamma )(p-\nu {)}^2,\end{array}\\ e(p)=g(v(p)),\\ t(p)=(1-\gamma )p+\gamma \nu .\end{array}$$

Thus $e(p)$ is the individual output entropy of any pure signal at latitude $p$, independent of its phase.

Capacity and repeated-use additivity, as stated in the manuscript

For every $\gamma ,\nu \in [0,1]$ and positive integer $n$, write $\mathcal{A}={\mathcal{A}}_{\gamma ,\nu }$. Then

$$\begin{array}{c}\chi ({\mathcal{A}}^{\otimes n})=nC(\mathcal{A}),\\ C(\mathcal{A})=\frac{1}{\ln 2}\\ \times \underset{0\le p\le 1}{\max }\{h(t(p))-e(p)\}.\end{array}$$

The family manuscript claims this formula for all parameters and all block inputs. (OpenAI, 2026) Every maximizing $p$ gives an attaining one-use ensemble of two equiprobable pure signals

$$|{\varphi }_{\pm }⟩=\sqrt{1-p}|0⟩\pm \sqrt{p}|1⟩.$$

Their coherences cancel in the average output, whose entropy is $h(t(p))$, while both individual entropies equal $e(p)$. Uniform independent products of the signals attain every finite-block Holevo optimum. The signals need not be orthogonal: their inner product is $1-2p$. Optimal product encoding therefore does not assert that separate measurements achieve the operational capacity.

🏷️ An entropy inequality for a matrix with a zero block

The proof starts from a matrix inequality. Let $A,B,C$ have the same rectangular shape, let $M=\left(\begin{array}{cc}A& B\\ C& 0\end{array}\right)$, and write $x=\mathrm{Tr}A{A}^{\ast }$, $y=\mathrm{Tr}B{B}^{\ast }$, $z=\mathrm{Tr}C{C}^{\ast }$, with $x+y+z=1$. The zero-block estimate is

$$\begin{array}{rl}S(M{M}^{\ast })\ge & xS(A{A}^{\ast }/x)\\ & +yS(B{B}^{\ast }/y)\\ & +zS(C{C}^{\ast }/z)\\ & +g(yz).\end{array}$$

Terms of zero weight are omitted. The last term is the entropy of the squared singular values of the scalar matrix with entries $\sqrt{x},\sqrt{y},\sqrt{z},0$. It captures entropy caused by the block pattern in addition to the entropy inside the blocks.

The manuscript gives a direct proof of this rectangular case of a cited triangular entropy theorem. (OpenAI, 2026) The proof has three useful ingredients.

For equal-trace positive matrices $U,V$, even of different sizes, the entropy difference is the integral of $\ln \det (I+rU)-\ln \det (I+rV)$ against $dr/r^2$ on $(0,\infty )$. Equal trace cancels the linear term at zero, giving an integrable $O(r^2)$ numerator. At infinity the numerator grows only logarithmically. Integration by parts, followed by the eigenvalue primitive $a\ln (r/(1+ra))$, gives the identity, including zero eigenvalues.

Next, for fixed $Z$, the function

$$\begin{array}{rl}{G}_Z(D,E)=& \ln \det (D+Z{E}^{-1}{Z}^{\ast })\\ & -\ln \det D\end{array}$$

is jointly convex for positive definite $D,E$. An elementary proof introduces the block matrices $\mathcal{D}=\mathrm{diag}(D,E)$ and $\mathcal{C}=\left(\begin{array}{cc}D& Z\\ -Z^{\ast }& E\end{array}\right)$. Although $\mathcal{C}$ need not be Hermitian, its determinant is positive. Along an affine line with derivative $\mathcal{H}$, put $L={\mathcal{D}}^{-1/2}\mathcal{H}{\mathcal{D}}^{-1/2}$ and $K={\mathcal{D}}^{-1/2}(\mathcal{C}-\mathcal{D}){\mathcal{D}}^{-1/2}$. Then $L$ is Hermitian and $K$ is skew-Hermitian, so $R=(I+K{)}^{-1}$ is a contraction. The second derivative is

$$\begin{array}{rl}{G}_Z^{\prime \prime }& =\Vert L{\Vert }_{\mathrm{HS}}^2-\mathrm{Tr}((RL{)}^2)\\ & \ge \Vert L{\Vert }_{\mathrm{HS}}^2-\Vert RL{\Vert }_{\mathrm{HS}}^2\ge 0.\end{array}$$

Here $\Vert B{\Vert }_{\mathrm{HS}}^2=\mathrm{Tr}{B}^{\ast }B$, and $|\mathrm{Tr}{B}^2|\le \Vert B{\Vert }_{\mathrm{HS}}^2$ follows from entrywise Cauchy–Schwarz. No commutativity assumption is used.

Finally, consider the scalar deficit $F(x,y,z)=S(\mathrm{diag}(x,y,z))-S(\mathrm{diag}({\lambda }_{+},{\lambda }_{-}))$, where ${\lambda }_{\pm }$ are the squared singular values of the scalar block matrix. This definition permits any nonnegative masses. The two eigenvalue lists have equal sum, so scaling logarithms cancel and $F$ is homogeneous of degree one. The integral representation and joint convexity of $G$ show that $F(1,y,z)$ is concave. Homogeneity then gives ${\sum }_{i}F(x_i,y_i,z_i)\le F({\sum }_i{x}_i,{\sum }_i{y}_i,{\sum }_i{z}_i)$, with boundary values obtained by continuity.

For the matrix deficit, a Schur complement gives an integrand of the form $\ln \det (I+rA{A}^{\ast })-G_{\sqrt{r}A}(I+rB{B}^{\ast },I+r{C}^{\ast }C)$. Make $A$ diagonal by a singular-value decomposition, then average both arguments of $G$ over the same diagonal sign matrices. Joint convexity bounds $G$ below by its value on the diagonal averages; because it is subtracted, this bounds the entropy deficit above. The resulting scalar deficits combine by the preceding homogeneous inequality. Rectangular blocks are handled by adding zero rows and columns, which leaves their nonzero singular values and entropies unchanged.

🏷️ A thermal mixture with an unchanged scalar term

Generalized damping has upward and downward transitions, so its relevant matrix does not initially have a zero block. Let rectangular $X,Y$ satisfy $\mathrm{Tr}X{X}^{\ast }=\mathrm{Tr}Y{Y}^{\ast }=1$. Take nonnegative $\alpha ,\beta ,\eta ,\delta$ summing to one and satisfying $\alpha \delta -\beta \eta =K^2$, with $K\ge 0$. Put $P_0=\alpha X{X}^{\ast }+\beta Y{Y}^{\ast }$ and $P_1=\eta X{X}^{\ast }+\delta Y{Y}^{\ast }$, and define

$$T=\left(\begin{array}{cc}{P}_0& KX{Y}^{\ast }\\ KY{X}^{\ast }& P_1\end{array}\right).$$

Set $U_0=\alpha +\beta$, $U_1=\eta +\delta$ and $u=U_0{U}_1-K^2=\beta U_1+\eta U_0$. The manuscript’s thermal extension says that $T$ is a state, $0\le u\le 1/4$, and

$$\begin{array}{rl}S(T)\ge & (\alpha +\eta )S(X{X}^{\ast })\\ & +(\beta +\delta )S(Y{Y}^{\ast })+g(u).\end{array}$$

The construction explains why entropy concavity has the needed direction. Hold $X,Y,K,U_0,U_1$ fixed. When $u>0$, the admissible coefficients lie on the segment $U_1\beta +U_0\eta =u$. Its endpoints have $(\beta ,\eta )=(u/U_1,0)$ and $(0,u/U_0)$. They express $T$ as a convex mixture with weights $\beta U_1/u$ and $\eta U_0/u$.

Each endpoint has a Gram factorization with one zero block, after a row swap if necessary. The products of its two corner masses are respectively ${\beta }_1{\delta }_1=u$ and ${\eta }_2{\alpha }_2=u$. Thus both endpoint inequalities contain the same $g(u)$; taking the convex mixture preserves this scalar contribution exactly. If $u=0$, a vanishing row gives an ordinary entropy-concavity statement; otherwise $\beta =\eta =0$ and the zero-block theorem applies directly. When $X=Y$, the matrix is a tensor product of a scalar qubit state and $X{X}^{\ast }$, giving equality. Neither orthogonality of $X,Y$ nor commutativity of their Gram matrices is required.

🏷️ Conditional branches retain their coherence

For any pure input to one damping channel and a finite-dimensional partner $\mathcal{N}$, write

$$|\psi ⟩=\sqrt{1-p}|0⟩|{\psi }_0⟩+\sqrt{p}|1⟩|{\psi }_1⟩.$$

The conditional vectors are normalized but need not be orthogonal. A zero-weight branch may be chosen arbitrarily. If $V_{\ell }$ are Kraus operators for $\mathcal{N}$, give $X,Y$ the respective columns $V_{\ell }{\psi }_0,V_{\ell }{\psi }_1$. Then $X{X}^{\ast }$ and $Y{Y}^{\ast }$ are the branch outputs, while the common column index ensures $X{Y}^{\ast }=\mathcal{N}(|{\psi }_0⟩⟨{\psi }_1|)$. Discarding this cross operator would incorrectly replace the coherent input by a mixture.

For brevity set $a=1-\gamma \nu$, $b=\gamma (1-\nu )$, $c=\gamma \nu$, $d=1-b$ and $k=\sqrt{1-\gamma }$. These satisfy $a+c=b+d=1$ and $ad-bc=k^2$. Substituting $\alpha =a(1-p)$, $\beta =bp$, $\eta =c(1-p)$, $\delta =dp$ and $K=k\sqrt{p(1-p)}$ into the thermal inequality gives the following bound. Write $\Sigma =(\mathcal{A}\otimes \mathcal{N})(|\psi ⟩⟨\psi |)$ for the full output and ${\sigma }_i=\mathcal{N}(|{\psi }_i⟩⟨{\psi }_i|)$ for the branch outputs:

$$\begin{array}{rl}S(\Sigma )& \ge e(p)+(1-p)S({\sigma }_0)\\ & +pS({\sigma }_1).\end{array}$$

The matrices may be rectangular because the partner output dimension and its number of Kraus operators need not coincide.

🏷️ Population convexity and the ensemble bound

The second scalar ingredient is convexity of $e(p)$. The function $s\mapsto g(s^2)$ is nondecreasing and convex for $0\le s\le 1/2$. One way to check it is to put $t=\sqrt{1-4u}$ and $L=\mathrm{atanh}(t)/t$; differentiation gives $g^{\prime }(u)=2L$ and $u{g}^{\prime \prime }(u)+g^{\prime }(u)/2=(L-1)/t^2\ge 0$. The last inequality follows from $\mathrm{atanh}(t)\ge t$, and continuity handles the endpoints. Meanwhile $\sqrt{v(p)}$ is the Euclidean norm of an affine two-component vector. Composing these two convex functions, with the outer one nondecreasing, proves the claim.

Apply the branch inequality repeatedly with the remaining damping uses as partner. Conditional populations average to the original populations because $|0⟩$ and $|1⟩$ are orthogonal; the branch vectors need not be. Convexity of $e$ gives

$$S({\mathcal{A}}^{\otimes n}(|\psi ⟩⟨\psi |))\ge \sum _{j=1}^{n}e(p_j).$$

Entropy concavity and a pure-state decomposition extend this to every mixed block input. For an arbitrary ensemble, let ${\overline{p}}_j$ be its mean input population at site $j$. Another application of convexity bounds the mean individual output entropy below by ${\sum }_{j}e({\overline{p}}_j)$.

The entropy of the average output is at most ${\sum }_{j}h(t({\overline{p}}_j))$: measuring in the computational basis can only increase entropy, and the resulting classical joint entropy is at most the sum of its marginal entropies. Subtract the two bounds and maximize each scalar term to obtain the $n$-use Holevo upper bound. The opposite-phase product ensemble already attains it. This proves the manuscript’s removal of regularization, including entangled input signals. (OpenAI, 2026)

🏷️ Arbitrary partner channels

The same branch estimate yields the stronger manuscript claims

$$\begin{array}{c}\chi (\mathcal{A}\otimes \mathcal{N})=\chi (\mathcal{A})+\chi (\mathcal{N}),\\ S_{\min }(\mathcal{A}\otimes \mathcal{N})\\ =S_{\min }(\mathcal{A})+S_{\min }(\mathcal{N}),\\ C(\mathcal{A}\otimes \mathcal{N})=C(\mathcal{A})+C(\mathcal{N}),\end{array}$$

for every completely positive trace-preserving partner between nonzero finite-dimensional spaces. Here $S_{\min }$ is minimum output entropy in nats. (OpenAI, 2026)

For the Holevo statement, refine the input ensemble to pure states, which cannot decrease Holevo information. Their conditional partner outputs, weighted by each signal’s probability times its branch probability, form a valid ensemble whose mean is exactly the second marginal of the mean full output. The mean full entropy is bounded above by $h(t(\overline{p}))+S({\overline{\sigma }}_{\mathcal{N}})$. Subtracting the branch lower bound leaves one damping scalar objective plus the Holevo information of that partner ensemble. Product ensembles give the reverse bound.

For minimum output entropy, concavity lets one minimize over pure inputs. Since $g$ is nondecreasing and $v(p)$ is minimized at $p=\nu$,

$$S_{\min }(\mathcal{A})=g(\gamma \nu (1-\nu )).$$

The branch bound gives the additive lower bound, and a product of minimizing states attains it. Finally, iterate the Holevo partner identity against ${\mathcal{N}}^{\otimes n}$ before regularizing. This gives the capacity identity while retaining the partner’s full capacity $C(\mathcal{N})$; it does not replace that quantity by $\chi (\mathcal{N})$.

🏷️ Endpoints and a symmetric example

At $\nu =0$, the formula becomes the ordinary amplitude-damping optimization with $v(p)=\gamma (1-\gamma )p^2$. Bit-flip conjugation shows invariance under $\nu \mapsto 1-\nu$.

At the unital midpoint $\nu =1/2$, the scalar objective is concave and symmetric under $p\mapsto 1-p$, so $p=1/2$ maximizes it. The capacity is

$$\begin{array}{c}C({\mathcal{A}}_{\gamma ,1/2})\\ =1-\frac{1}{\ln 2}h\left(\frac{1+\sqrt{1-\gamma }}{2}\right).\end{array}$$

The two signals are $|+⟩$ and $|-⟩$. At $\gamma =0$ capacity is one bit per use; at $\gamma =1$ it is zero for every $\nu$. For $\gamma >0$ and $0<\nu <1$, the determinant formula is strictly positive for every pure input, so this interior thermal family has no pure output. That is why a theorem applying only to channels with pure outputs would not cover it.

🏷️ Key steps and proof outlines

The thermal interpolation has explicit Gram factors

The thermal extension above can be checked directly at the matrix level. Assume first $u>0$. Its two coefficient endpoints are

$$\begin{array}{rl}({\alpha }_1,{\beta }_1,{\eta }_1,{\delta }_1)& =(K^2/U_1,u/U_1,0,U_1),\\ ({\alpha }_2,{\beta }_2,{\eta }_2,{\delta }_2)& =(U_0,0,u/U_0,K^2/U_0),\end{array}$$

with mixture weights $w_1=\beta U_1/u$ and $w_2=\eta U_0/u$. Both have the same row masses and coherence coefficient as the original matrix. Their Gram factors are

$$M_1=\left(\begin{array}{cc}\sqrt{{\alpha }_1}X& \sqrt{{\beta }_1}Y\\ \sqrt{{\delta }_1}Y& 0\end{array}\right),M_2=\left(\begin{array}{cc}\sqrt{{\alpha }_2}X& 0\\ \sqrt{{\delta }_2}Y& \sqrt{{\eta }_2}X\end{array}\right).$$

Multiplication verifies $T_i=M_i{M}_i^{\ast }$, including the cross block $KX{Y}^{\ast }$. Swapping the block rows of $M_2$ puts it into the same zero-corner pattern as $M_1$.

The zero-block theorem contributes $g({\beta }_1{\delta }_1)$ in the first case and $g({\eta }_2{\alpha }_2)$ in the second. Both arguments equal $u$, so concavity yields

$$S(T)\ge w_{1}S(T_1)+w_{2}S(T_2)$$

with exactly one unchanged term $g(u)$. The other entropy coefficients average back to $\alpha +\eta$ and $\beta +\delta$. This is more precise than decomposing the channel into upward and downward damping and averaging their capacities: it decomposes the particular output matrix while keeping its scalar determinant parameter fixed. Vanishing row masses and $u=0$ are handled separately, avoiding division by zero (OpenAI, 2026).

Conditional population bookkeeping survives entangled signals

For two damping uses, decompose a pure input at its first qubit:

$$\psi =\sqrt{1-p_1}|0⟩{\psi }_0+\sqrt{p_1}|1⟩{\psi }_1.$$

Let $p_{2|i}$ be the second-qubit population in branch $i$. The branch entropy inequality gives

$$\begin{array}{rl}S({\mathcal{A}}^{\otimes 2}(\psi {\psi }^{\ast }))\ge & e(p_1)+(1-p_1)e(p_{2|0})\\ & +p_{1}e(p_{2|1})\\ \ge & e(p_1)+e(p_2),\end{array}$$

because $p_2=(1-p_1)p_{2|0}+p_1{p}_{2|1}$ and $e$ is convex. The same identity works for each later qubit, even though its conditional state may remain entangled with all the other later qubits. Orthogonality is needed only between the first-qubit labels $|0⟩,|1⟩$, not between ${\psi }_0,{\psi }_1$.

This illustrates why both inequalities are needed. The thermal matrix theorem retains the coherent cross operator and pays the first-site entropy cost. Population convexity then recombines the conditional later-site costs. Repeating only ordinary entropy concavity would replace the coherent input by a mixture and would not establish the desired lower bound.

Finally, the ensemble argument uses convexity once more, now across different transmitted signals. It compares their average individual entropy with the same populations used to upper-bound the entropy of their average output. The difference becomes a sum of scalar one-site objectives. This matching of population parameters is the step that turns a state entropy bound into additivity of the Holevo optimum and hence the claimed operational capacity (OpenAI, 2026).

🏷️ Connections and verification scope

Photon-number entropy and bosonic communication also derives capacity consequences from a population-sensitive entropy inequality. The current family is finite dimensional and concerns a qubit; its determinant entropy function and stationary population are different from the bosonic photon-number parameters.

The source reading covers the zero-block proof, thermal mixture, branch recursion, ensemble argument, partner identities and endpoint cases. The manuscript’s alternative Hessian treatment and the external coding theorems were not independently audited. Its selected formalization scope covers repeated uses of this channel and the scalar capacity formula, while the arbitrary-partner consequences lie outside that selected scope.

🏷️ Manuscripts and verification scope

• Classical capacity and entropy inequalities for generalized amplitude damping — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The selected scope covers repeated uses of generalized amplitude damping, the attained scalar Holevo maximum, the operational unassisted capacity, and maximizing phase-pair product ensembles. It explicitly excludes arbitrary different partner channels and separate-measurement decoding; the stronger manuscript partner consequences are not assigned that formal coverage here. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026. Classical capacity and entropy inequalities for generalized amplitude damping.