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Use logarithms to base two and set $0{\log }_{2}0=0$. For a density matrix with eigenvalues $({\lambda }_j)$, the Von Neumann entropy isThus it is the Shannon entropy of the spectrum. In dimension $d$, each term is nonnegative, and entropy vanishes exactly when one eigenvalue is one, namely for a pure state. The Jensen inequality applied to the strictly concave function $-x{\log }_{2}x$ gives $S(\rho )\le {\log }_{2}d$, with equality precisely at the maximally mixed state $I/d$. Unitary conjugation preserves eigenvalues, so $S(U\rho U^{†})=S(\rho )$. Product eigenvalues giveThe finite-dimensional entropy is continuous because the eigenvalues depend continuously on the matrix and $-x\log x$ is continuous also at zero.

Define the partial trace over ${\mathcal{K}}_2$ by choosing an orthonormal basis $(e_b)$ and settingEquivalently it is uniquely characterized by $\mathrm{Tr}({\rho }_{1}A)=\mathrm{Tr}(\rho (A\otimes I))$ for every operator $A$ on ${\mathcal{K}}_1$. This identity proves basis independence. Define ${\rho }_2$ similarly, with $\mathrm{Tr}({\rho }_{2}B)=\mathrm{Tr}(\rho (I\otimes B))$. Hermiticity is preserved by the displayed sum. For each $u\in {\mathcal{K}}_1$,Taking the trace in a product orthonormal basis gives $\mathrm{Tr}{\rho }_1=\mathrm{Tr}\rho =1$, and the same proof works for ${\rho }_2$. Thus both reductions are density matrices. This is the partial trace positivity from product vectors argument.

For a bipartite pure state $\rho =|\psi ⟩⟨\psi |$, expand $\psi$ in a product orthonormal basis with coefficient matrix $C$. A singular value decomposition gives the Schmidt decompositionwith orthonormal families on both sides. Taking the two partial traces makes all cross terms vanish:The spectra may have different numbers of zeros but have identical nonzero eigenvalues. Consequently the reduced entropies coincide:This common value is the entanglement entropy. In infinite-dimensional trace-class systems the same partial-trace and Schmidt arguments apply, and the equality remains valid with the value $+\infty$ allowed; the bound ${\log }_{2}d$ is specifically finite-dimensional.

For completeness, the basic mixing and correlation properties also admit short proofs. Define quantum relative entropy by $D(\tau \Vert \sigma )=\mathrm{Tr}\tau ({\log }_2\tau -{\log }_2\sigma )$ when supports are included, and by $+\infty$ otherwise. Here is its nonnegativity proof without merely citing an operator inequality. Diagonalize $\tau ,\sigma$ with eigenvalues $r_i,s_j$, and let $q_{ij}$ be the squared overlaps of their eigenbases. Both row and column sums of $q$ are one. The scalar inequality $r\ln (r/s)\ge r-s$, with zero cases taken by limits, givesEquality forces $r_i=s_j$ on every nonzero overlap, hence $\tau =\sigma$. For $\overline{\rho }={\sum }_i{p}_i{\rho }_i$, expanding the trace definition yieldswhich proves Concavity of Von Neumann entropy.

The complementary entropy bounds for a quantum mixture areTo prove the upper bound, write the spectral pure-state decompositions of the ${\rho }_i$ as one ensemble with weights $w_{ij}=p_i{\lambda }_{ij}$. Form the column matrix with columns $\sqrt{w_{ij}}|{\psi }_{ij}⟩$. Its two Gram operators have the same nonzero spectrum: one is $\overline{\rho }$, and the other, $G$, has diagonal $w$. Relative entropy against $\mathrm{diag}w$ gives $0\le D(G\Vert \mathrm{diag}w)=H(w)-S(G)$. Since $H(w)=H(p)+{\sum }_i{p}_{i}S({\rho }_i)$ and $S(G)=S(\overline{\rho })$, the bound follows. Zero-weight columns are omitted.

Finally the support of $\rho$ is contained in $\mathrm{supp}{\rho }_1\otimes \mathrm{supp}{\rho }_2$. Indeed, a vector in a marginal kernel gives a sum of nonnegative product-vector expectations equal to zero; positivity forces the corresponding product vectors into the kernel of $\rho$. The operator identity $\log ({\rho }_1\otimes {\rho }_2)=\log {\rho }_1\otimes I+I\otimes \log {\rho }_2$ on this support and the partial-trace identities giveThis proves Subadditivity of Von Neumann entropy, with equality exactly for a product state. Purify ${\rho }_{12}$ by adjoining a third system, using its spectral decomposition. Equality of pure-state marginal entropies gives $S({\rho }_{13})=S({\rho }_2)$ and $S({\rho }_3)=S({\rho }_{12})$. Subadditivity on systems $1,3$ then gives $S({\rho }_2)\le S({\rho }_1)+S({\rho }_{12})$; swapping $1,2$ gives the Araki–Lieb inequality $\boxed{|S({\rho }_1)-S({\rho }_2)|\le S({\rho }_{12})}$. These finite-dimensional proofs establish the usual basic entropy bounds, rather than attributing all mixed-state behavior to pure-state Schmidt decompositions.

The four Pauli matrices form an orthogonal basis of $2\times 2$ matrices for the Hilbert-Schmidt inner product, since $\mathrm{Tr}({\sigma }_{\mu }{\sigma }_{\nu })=2{\delta }_{\mu \nu }$. A Hermitian density matrix therefore has real coefficients in this basis, and its trace-one condition fixes the identity coefficient. ThusThe multiplication rule for the Pauli matrices ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$ implies $(n\cdot \sigma {)}^2=\Vert n{\Vert }^{2}I$. Its trace is zero, so for $n\ne 0$ its eigenvalues are $\pm \Vert n\Vert$; for $n=0$ it vanishes. HencePositivity is equivalent to $\Vert n\Vert \le 1$. Conversely every real vector in this unit ball defines a positive trace-one matrix by the boxed formula. Therefore $\Phi (\rho )=n$ is a one-to-one correspondence with the Bloch ball, with the displayed inverse. Each coordinate is linear in $\rho$, proving the affine property under any convex mixture.

A density matrix is pure precisely when its eigenvalues are $1,0$, equivalently when ${\rho }^2=\rho$. The eigenvalue formula makes this equivalent to $\Vert n\Vert =1$. Alternatively $\mathrm{Tr}{\rho }^2=(1+\Vert n{\Vert }^2)/2$. Thus the pure states correspond exactly to the Bloch sphere.

For $n\ne 0$, put $\hat{n}=n/\Vert n\Vert$. The spectral projectors for the larger and smaller eigenvalues areThey are orthogonal rank-one projections, have sum $I$, and give $\rho ={\lambda }_{+}{P}_{+}+{\lambda }_{-}{P}_{-}$. For normalized eigenvectors in this nondegenerate case they equal $|{\varphi }_1⟩⟨{\varphi }_1|$ and $|{\varphi }_2⟩⟨{\varphi }_2|$. Their Bloch vectors are therefore $\boxed{n_1=n/\Vert n\Vert ,n_2=-n/\Vert n\Vert }$, the antipodal eigenprojectors of a qubit density matrix.

At $n=0$, the state is $I/2$ and both eigenvalues coincide. There is no distinguished direction and the printed directional expression is undefined. Any orthonormal eigenbasis gives an antipodal pair of pure-state Bloch vectors, since its two projectors sum to $I$, but that pair can point in any direction. The nonzero-$n$ qualification is essential. The PDF denominators are $\Vert n\Vert$, not the converted TeX denominators involving $n_1,n_2$.

Use binary linear codes $C\subseteq C^{\prime }\subseteq {\mathbb{F}}_2^N$, with dimensions $k,k^{\prime }$. For each coset $u+C$ in $C^{\prime }/C$, defineRepresentatives of one coset give the same vector, and distinct cosets have disjoint computational supports, so these vectors are orthonormal. Their span is the CSS code $\mathcal{X}$, encoding $\boxed{k^{\prime }-k\text{ logical qubits in }N\text{ physical qubits}}$.

For $a,b\in {\mathbb{F}}_2^N$, write $X^a={⨂}_j{X}_j^{a_j}$ and $Z^b={⨂}_j{Z}_j^{b_j}$. The code is stabilized by $X^c$ for $c\in C$, because it permutes the sum, and by $Z^z$ for $z\in (C^{\prime }{)}^{\perp }$, because $z\cdot (u+c)=0$. These Pauli operators commute: $c\cdot z=0$. There are $k+N-k^{\prime }$ independent stabilizer generators, so their common positive eigenspace has dimension $2^{N-(k+N-k^{\prime })}=2^{k^{\prime }-k}$; it is exactly the span above.

The commuting check measurements separate the two types of error. A bit error $X^a$ changes the signs of the $Z$ checks by $(-1{)}^{z\cdot a}$, giving the ordinary syndrome of $a$ for $C^{\prime }$. A phase error $Z^b$ changes the $X$ checks by $(-1{)}^{c\cdot b}$, giving the syndrome of $b$ for the dual code $C^{\perp }$. If $C^{\prime }$ corrects $t_X$ bit errors and $C^{\perp }$ corrects $t_Z$ bit errors, these checks identify the respective patterns, and inverse Pauli operations recover the state without revealing the logical coset. This is the nested-code CSS syndrome recovery procedure.

The sharper distances account for harmless stabilizer errors. A Pauli $X^a{Z}^b$ commutes with every check precisely when $a\in C^{\prime }$ and $b\in C^{\perp }$. Among these, it acts as a stabilizer, up to a phase, precisely when $a\in C$ and $b\in (C^{\prime }{)}^{\perp }$. To justify the latter assertion, $X^a$ with $a\in C^{\prime }\setminus C$ permutes distinct logical cosets, while $Z^b$ with $b\in C^{\perp }\setminus (C^{\prime }{)}^{\perp }$ has unequal signs on two such cosets. Neither can act as a scalar, and a combined operator has the same nontrivial permutation or phase distinction. For positive logical dimension setThe quantum code distance is $\boxed{d=\min (d_X,d_Z)}$: a nontrivial logical Pauli has at least one nonstabilizer component of weight at least the corresponding distance, while a pure $X$ or pure $Z$ logical operator attains the minimum. The weight of $X^a{Z}^b$ counts the union of its supports, so this argument also includes $Y$ errors.

Here is the correction proof for coherent errors, not only a classical decoding argument. Let $P$ be the code projector and let $E,F$ have bit-component weights at most $t_X$ and phase-component weights at most $t_Z$. If $2t_X<d_X$ and $2t_Z<d_Z$, the corresponding components of $E^{†}F$ have weights below $d_X,d_Z$. If this product anticommutes with a stabilizer $S$, then $P(E^{†}F)P=-P(E^{†}F)P=0$, using $SP=P$. If it commutes with all stabilizers, the weight bounds force its two components into $C$ and $(C^{\prime }{)}^{\perp }$, so its code compression is a scalar multiple of $P$. Thus the Knill--Laflamme condition holds for the entire error family.

To see directly why this gives a common recovery, diagonalize its positive Gram matrix $c_{EF}$ and form the corresponding error combinations $F_j$. Their restrictions obey $P{F}_j^{†}{F}_{l}P=c_j{\delta }_{jl}P$. For $c_j>0$, $F_{j}P/\sqrt{c_j}$ is an isometry onto a syndrome subspace, and the different images are orthogonal. Measure these subspaces and apply the inverse isometry; zero-$c_j$ combinations annihilate the code. This recovers every quantum channel whose Kraus operators lie in the error span. Arbitrary operators supported on at most $t$ qubits expand in Pauli operators with the same support. Consequently the CSS code corrects all errors on up to $\boxed{t=\lfloor (d-1)/2\rfloor }$ qubits, and more generally the asymmetric family just described. The sufficient classical bound is obtained by replacing $d_X,d_Z$ by the minimum distances of $C^{\prime }$ and $C^{\perp }$, respectively; degeneracy can improve it.

Use coordinates $1,2,3$ for paper, scissors and stone. Let $X^N$ take values in the lattice simplex $S_N=\{(n_1,n_2,n_3)/N:n_i\ge 0,{\sum }_i{n}_i=N\}$. A uniformly selected unordered pair consists of types $i$ and $j$, $i\ne j$, with probability $2n_i{n}_j/(N(N-1))$. Losing copies the winner, so the possible jumps of the continuous-time Markov chain areTies make no change. For arbitrary event rate $\lambda$, the respective transition rates are $2\lambda N{x}_1{x}_3/(N-1)$, $2\lambda N{x}_2{x}_1/(N-1)$ and $2\lambda N{x}_3{x}_2/(N-1)$. ChooseThen the rates are exactly $N{x}_1{x}_3$, $N{x}_2{x}_1$, $N{x}_3{x}_2$. The infinitesimal generator of this cyclic imitation chain iswhere $\mathcal{J}=\{e_1-e_3,e_2-e_1,e_3-e_2\}$ and the three $\beta$ functions are, in that order, $x_1{x}_3,x_2{x}_1,x_3{x}_2$. Its drift is precisely

Here is a precise fluid limit with its error control. Suppose $X_0^N\to x_0$ in probability, and let $x$ solve $\dot{x}=F(x)$, $x(0)=x_0$. For every finite $T$,Indeed, the martingale decomposition of a density-dependent jump process givesThis follows by compensating each transition count by its integrated rate; the compensated counts are square-integrable martingales, with variance equal to the expected compensator. Since $\Vert \ell {\Vert }^2=2$ and $x_1{x}_2+x_2{x}_3+x_3{x}_1\le 1/3$ on the simplex, the Doob L2 maximal inequality, applied coordinatewise, givesThe polynomial drift is Lipschitz on the compact simplex, with some finite constant $L$. It preserves the sum of the coordinates, and each coordinate solves an equation of the form ${\dot{x}}_i=x_i{h}_i(x)$, so nonnegativity is preserved. Thus the ordinary differential equation stays in the simplex and exists for all time. The Gronwall inequality giveswhich proves the claim by Markov's inequality. This proves the required version directly, rather than invoking an unspecified approximation theorem.

For an interior trajectory, differentiating the logarithm of the product of numbers givesThe same identity for the product itself holds on the boundary, so $x_1(t)x_2(t)x_3(t)$ is conserved. A strictly positive initial product keeps the deterministic trajectory away from the boundary. Except at $(1/3,1/3,1/3)$ these interior trajectories are periodic: the strictly concave function ${\sum }_i\log x_i$ has smooth compact level curves surrounding its unique maximum, and the vector field is nonzero and tangent to each such curve.

The finite Markov chain behaves differently at long times. For $P(x)=x_1{x}_2{x}_3$, the change in $P$ during the first jump isThe other two terms are cyclic permutations. Their first-order contributions cancel, leaving the exact product decay in a cyclic imitation chain identityFor $N\ge 3$, while all three counts are positive, their product is at least $N-2$, so $P(X_t^N)\ge (N-2)/N^3$. A missing type cannot reappear. If $\tau$ is the first loss of a type, thenOnce only two types remain, one always wins, so its count increases until an absorbing monochromatic vertex is reached, almost surely in finite time. The cases of one or two initially present types follow the same argument without the first stage. Every finite chain eventually becomes monochromatic, whereas an interior deterministic solution retains its positive product. The fluid limit holds on each fixed finite interval; it does not justify exchanging the limits $N\to \infty$ and $t\to \infty$.

Separate $L$ into a diffusion generator and the constant killing term $-1/2$. The appropriate stochastic differential equation and its explicit geometric Brownian motion solution areThe second-order term in the Itô formula cancels the drift of $\log |X|$ when $x\ne 0$; if $x=0$, the solution stays zero.

Fix a terminal time $t$. Apply the Itô formula to $e^{-s/2}u(t-s,X_s)$ for $0\le s\le t$. Its drift isAfter localization this is a local martingale, and it is bounded because $u$ is bounded. It is therefore a true martingale. Taking expectations at its endpoints gives the Feynman-Kac formula

For $t>0$ and $x\ne 0$, the change of variable $y=x{e}^z$ in the normal density produces the killed geometric Brownian heat kernel with respect to Lebesgue measure:For $xy>0$, an equivalent form isIn particular,The apparent missing $1/y$ in this last expression is absorbed into the completed square, together with the killing factor; it is not a transcription error. The total mass is $e^{-t/2}$, as expected for killing at rate $1/2$. At $x=0$ the fundamental kernel is instead the measure $e^{-t/2}{\delta }_0(dy)$; it has no Radon-Nikodym derivative with respect to Lebesgue measure. Thus $u(t,0)=e^{-t/2}g(0)$. At $t=0$ the kernel is ${\delta }_x$, and continuity of bounded $g$ gives the initial condition. This also specifies the fundamental solution at the degenerate point omitted by a density-only formula.

First stop the solution on leaving $[-n,n]$, at ${\tau }_n$, and set $F_n(t)=\mathbb{E}{\sup }_{u\le t}|X_{u\wedge {\tau }_n}{|}^2$. The stochastic integral representation, the Doob L2 maximal inequality and the Cauchy-Schwarz inequality give, for $0\le t\le 1$,Here stopping ensures square integrability before the calculation, and $8{\sigma }^2+2t{b}^2\le 8({\sigma }^2+b^2)$ explains the constant. The Gronwall inequality gives $F_n(t)\le e^{8At}-1$. For globally Lipschitz functions as coefficients, the global existence theorem for stochastic differential equations with Lipschitz coefficients supplies a nonexplosive solution. Letting $n$ increase and using Fatou's lemma proves the slightly stronger maximal second-moment bound under linear growth:

For coefficients that are only locally Lipschitz functions, let ${\pi }_n(x)=\max (-n,\min (x,n))$ and replace the coefficients by $\sigma ({\pi }_n(x))$ and $b({\pi }_n(x))$. These are globally Lipschitz functions, agree with the originals on $[-n,n]$, and obey the same linear growth condition for an SDE, since $|{\pi }_n(x)|\le |x|$. The globally defined solutions agree until their common exits from $[-n,n]$ by pathwise uniqueness. The uniform preceding estimate and Markov's inequality implyPatch the solutions before their increasing exit times. Their limiting lifetime exceeds $1$ almost surely by this bound. Thus a pathwise unique strong solution exists throughout $[0,1]$ even for locally Lipschitz coefficients.

For a simple predictable process, define its stochastic integral byCommon refinement of the deterministic partitions makes this definition independent of the representation. Each summand is a martingale increment multiplied by information already available at its left endpoint, so the resulting process is a continuous square-integrable martingale starting at zero. Its terminal value is the same sum with $t$ replaced by $t_n$.

Write ${\Delta }_{k}M=M_{t_{k+1}}-M_{t_k}$. If $j<k$, then $Z_{t_j}{\Delta }_{j}M{Z}_{t_k}$ is ${\mathcal{F}}_{t_k}$-measurable. By the conditional expectation identity for increments of a martingale,The conditional bracket isometry for stopped martingale increments givesFor completeness, this identity follows by applying the optional stopping theorem to $M$ and $M^2-[M]$ and expanding the square. The latter process is a true uniformly integrable martingale here: the Doob L2 maximal inequality bounds $\mathbb{E}{\sup }_t|M_t{|}^2$, stopped square identities and Fatou's lemma give $\mathbb{E}[M{]}_{\infty }<\infty$, and its absolute supremum is bounded by ${\sup }_t|M_t{|}^2+[M{]}_{\infty }$. Consequently all the stopped identities pass to the limit.

Multiplying the conditional identity by $Z_{t_k}^2$, taking expectations, and using the vanished cross terms proves the Itô isometry:

To extend the stochastic integral, use the finite quadratic-variation measure ${\nu }_M(C)=\mathbb{E}\int 1_{C}d[M]$ on the predictable sigma-algebra. The density of simple predictable processes for finite measures gives approximations to every $H\in L^2({\nu }_M)$. The Itô isometry makes the corresponding terminal integrals Cauchy in $L^2$, and the Doob L2 maximal inequality makes the whole processes Cauchy in expected squared supremum. Their limit defines a continuous martingale, and the isometry persists. For a locally bounded predictable process, choose increasing stopping times tending to infinity on which it is bounded, also restrict the time horizon, and perform this construction after stopping. The isometry shows that the stopped definitions agree on overlaps, so they patch into a uniquely defined continuous local martingale $H\cdot M$.