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For a bipartite pure state, entanglement entropy is the Von Neumann entropy of either reduced density matrix. When Alice makes a local measurement with pure conditional branches, Bob's reductions average to his original reduction. Concavity of Von Neumann entropy therefore gives ${\sum }_r{p}_{r}S({\rho }_{B,r})\le S({\rho }_B)$. When Bob measures, apply concavity to Alice's reductions. In each pure branch the two reduced entropies coincide by the Schmidt decomposition. Refining measurement outcomes to individual Kraus operators and iterating proves the inequality for arbitrary finite-round LOCC. Discarded local systems can be measured and their outcomes refined as well, so discarding does not evade the bound.

For any normalized pure two-qubit vector, its squared overlap with a maximally entangled state is at most $(1+C)/2$: in a Schmidt decomposition the overlap is bounded by $(\sqrt{\lambda }+\sqrt{1-\lambda })/\sqrt{2}$. Averaging gives $C(W_F)\ge 2F-1$. Equality follows from the six-state ensemble $\sqrt{F}|{\Psi }_{-}⟩\pm \sqrt{1-F}|t_j⟩$, with equal weights, where $t_1={\Psi }_{+}$, $t_2={\Phi }_{-}$ and $t_3=i{\Phi }_{+}$. The coherences cancel and each coefficient determinant gives pure concurrence $2F-1$.

For a bipartite pure state, entanglement entropy is the Von Neumann entropy of either reduced density matrix. A Schmidt decomposition gives squared coefficients $p_j$ as the nonzero eigenvalues of both reductions, soA product state has one coefficient equal to one and entropy zero. A Bell state has two equal coefficients and entropy one bit when logarithms have base two.

This entropy quantifies how much the subsystems are entangled even though the joint pure state itself has zero Von Neumann entropy. The purity assumption is essential: a classically correlated mixed state can have mixed reductions without entanglement. Reduced-state entropy alone is therefore not a general entanglement measure for arbitrary mixed bipartite states.

Let three parties share the normalized quantum stateFor $n$ copies, group each party's $n$ qubits together. Across any nontrivial division of these three parties, the Schmidt decomposition has one nonzero coefficient for every binary string of length $n$, so its Schmidt rank is $2^n$. Similarly, $m$ copies of the standard GHZ state have Schmidt rank $2^m$ across the same division.

Every recorded LOCC branch acts by a product of local Kraus operators. Schmidt-rank contraction under product operators therefore rules out an exact successful conversion when $m<n$, even if success need only have nonzero probability. This is a rank obstruction for the full target, not merely a bound on its entanglement entropy.

Approximation changes the question. In the expansion of $|{\Psi }_p{⟩}^{\otimes n}$, squared coefficients have the binary-product distribution with parameter $p$. A quantum typical subspace retains weight tending to one while using only $2^{n{h}_2(p)+o(n)}$ strings, where $h_2$ is binary entropy. The normalized truncation approaches the full target in quantum fidelity, although its Schmidt rank can be far smaller than $2^n$. Consequently, the exact finite-copy rank argument does not forbid an asymptotic entanglement dilution protocol with vanishing approximation error.

Under a product unitary operator $U_A\otimes U_B$, the reduced density matrix of $A$ becomes $U_A{\rho }_A{U}_A^{†}$. The unitary on $B$ cancels under the partial trace. Unitary conjugation preserves eigenvalues, so the squared Schmidt coefficients of a bipartite pure state are unchanged. Conversely, two normalized bipartite pure states with the same squared Schmidt coefficients, including multiplicities and zero padding, have Schmidt decompositions carried into each other by local unitaries mapping their Schmidt bases. Thus reduced-state spectra completely classify bipartite pure states up to local unitary operations.

Write a source and target in Schmidt decomposition with squared coefficients $(1-s,s)$ and $(1-t,t)$, where $0<s,t\le 1/2$. If $s\ge t$, Nielsen's pure-state conversion theorem allows deterministic conversion. If $s<t$, the Schmidt-tail entanglement monotone gives $p\le s/t$. Alice attains this bound with a local two-outcome measurement in quantum mechanics whose success Kraus operator is $K_s=\mathrm{diag}(\sqrt{s(1-t)/(t(1-s))},1)$ and whose failure operator is $K_f=\mathrm{diag}(\sqrt{1-s(1-t)/(t(1-s))},0)$. They obey $K_s^{†}{K}_s+K_f^{†}{K}_f=I$. The successful unnormalized state is $\sqrt{s/t}$ times the target, and failure leaves a product state. A classical message tells Bob which branch occurred. Known local changes of Schmidt bases allow the same protocol for arbitrary two-qubit pure states with these coefficients.

The displayed computational basis expansion is already a Schmidt decomposition, with coefficients $\sqrt{1-p}$ and $\sqrt{p}$. For $0<p<1$ both are nonzero, so the Schmidt rank is two. A pure product state has Schmidt rank one. The state is therefore entangled throughout the specified open interval.

Equivalently, the reduced density matrix isThe reduced state is a mixed state, which is impossible for one factor of a pure product state. At the excluded endpoints the Schmidt rank drops to one, giving $|00⟩$ or $|11⟩$.

Grouping by the first qubit givesThe vectors on both sides form orthonormal bases, so this is a Schmidt decomposition with two nonzero Schmidt coefficients $1/\sqrt{2}$. Hence the state is maximally entangled. Independently, its coefficient matrix has determinantwhich excludes a product state by the two-qubit product-state determinant criterion. Each reduced density matrix is $I/2$.

Let $a_{ij}$ be the coefficient of $|ij⟩$. The partial trace contracts the repeated index of the discarded qubit:For the first qubit, each diagonal entry is $1/3+1/6=1/2$ and the upper off-diagonal entry is $1/3-1/6=1/6$. For the second qubit, the diagonal entries are $2/3$ and $1/3$, while the upper off-diagonal entry is $-1/\sqrt{18}+1/\sqrt{18}=0$. Thus the two reduced density matrices, each in computational basis order $0,1$, areBoth have eigenvalues $2/3,1/3$ and trace one, so both are mixed states even though the joint quantum state is pure. Their different eigenvectors are visible in the useful rewriteAbsorbing the minus sign into one local basis vector makes this a Schmidt decomposition with nonnegative coefficients. It also checks both partial traces and exhibits the entanglement responsible for the mixed local descriptions.

The partial trace removes the other orthonormal factor in the Schmidt decomposition:For instance, tracing $|a_k{b}_k⟩⟨a_{\ell }{b}_{\ell }|$ over $B$ gives $⟨b_{\ell }|b_k⟩|a_k⟩⟨a_{\ell }|$, which vanishes unless $k=\ell$. Consequently the nonzero eigenvalues of the two reduced density operators agree, with multiplicity. If the subsystem dimensions differ, the orthonormal Schmidt families can be completed to full bases and the extra eigenvalues are zero.

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 Schmidt decomposition and local unitary operations put any entangled two-qubit pure state intoBoth Schmidt coefficients are nonzero. Alice performs a two-outcome generalized measurement with Kraus operatorsTheir effects sum to the identity, so this is a legitimate local measurement in quantum mechanics. On success,The normalized output is the Bell state ${\Phi }_{+}$. Bob's unitary matrix $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ maps it to the spin singlet state. Alice communicates her success flag; no shared quantum operation is needed. At $\lambda =1/2$ the filter is the identity and success is certain. This is the singlet target case of optimal stochastic conversion of a two-qubit pure state.

For dichotomic local outcomes $A_x,B_y\in \{-1,1\}$, the CHSH inequality isIt assumes a local hidden-variable theory with freely chosen settings independent of the hidden-variable distribution. For the spin singlet state, direct evaluation of its Pauli matrices correlations givesChoose $a_0=e_z$, $a_1=e_x$, $b_0=(e_z+e_x)/\sqrt{2}$, and $b_1=(e_z-e_x)/\sqrt{2}$. The four correlations are $-1/\sqrt{2},-1/\sqrt{2},-1/\sqrt{2},+1/\sqrt{2}$, respectively, giving

The success flag must be recorded and communicated before the subsequent independent choice of Bell-test settings. This conditioning is legitimate because setting-independent heralding preserves Bell locality: a putative local hidden-variable theory for the entire sequential experiment would, after successful filtering, still have local response functions with a setting-independent conditional hidden-variable distribution. More explicitly, if the hidden variable is $\xi$ and $h$ is the already determined success flag, conditioning changes its distribution to $\rho (\xi \mid h=1)$ but leavesClassical messages used in preparation can be included in $\xi$; none are exchanged during the spacelike Bell measurements. Such a conditioned model still obeys the CHSH inequality, contradicting the displayed singlet correlations. Therefore no local hidden-variable theory reproduces all sequential measurement correlations of any entangled two-qubit pure state. This argument does not discard outcomes on the basis of the later measurement settings.

For a classical error-correcting code of length $n$, alphabet size $q$, minimum Hamming distance $d$ and $M$ codewords, the Singleton bound isFor a linear $[n,k,d{]}_q$ code this is $n-k\ge d-1$. Deleting any $d-1$ coordinates is injective on the codewords, since two words agreeing on the remaining coordinates would differ in fewer than $d$ positions; this also explains the classical bound.

For an n-qubit quantum code carrying $k>0$ logical qubits and having quantum code distance $d$, the quantum Singleton bound isMore generally replace $k$ by ${\log }_{2}K$ for code dimension $K>1$. The proof applies to degenerate codes too.

Let $r=d-1$. Purify the maximally mixed encoded input by a reference $R$ of dimension $K$:Distance $d$ makes erasure of any subset of at most $r$ qubits correctable. The allowed quantum erasure correction condition says that a correctable erased set $A$ is decoupled from $R$:This is the matrix-element correction condition in another form: all operators on $A$ have constant diagonal matrix elements and zero off-diagonal matrix elements in the logical basis, so tracing the other physical qubits in $|\Omega ⟩⟨\Omega |$ yields the displayed product.

First justify the sizes needed for the partition. If $2r>n$, partition all physical qubits into $A,B$, both of size at most $r$. Both sets are correctable. Purity of $RAB$ and the Schmidt decomposition giveAdding forces $k=0$, a contradiction. Thus $2r\le n$, without assuming the bound we aim to prove.

Choose disjoint sets $A,B$ of size $r$, and let $C$ contain the remaining $n-2r$ qubits. The joint state $RABC$ is pure, so decoupling and Subadditivity of Von Neumann entropy giveAdding and cancelling gives $k\le S(C)$. The Von Neumann entropy of $D$-dimensional states is at most ${\log }_{2}D$: for positive eigenvalues $p_i$, concavity of the logarithm gives ${\sum }_i{p}_i{\log }_2(1/p_i)\le {\log }_2{\sum }_i{p}_i/p_i\le {\log }_{2}D$. Here $D=2^{n-2r}$, so $k\le n-2r$, proving the quantum bound. The code is assumed to encode information ($K>1$); a one-dimensional code requires a separate distance convention, since simply re-preparing a known state makes its erasure-correction condition vacuous.

Use base-two logarithms for Von Neumann entropy, so the units are bits, and adopt $0{\log }_{2}0=0$. All systems here may be taken finite-dimensional; the entropy arguments also apply when the displayed Von Neumann entropies are finite. The density operator of the joint pure state is ${\rho }_{AB}=|{\Psi }_{AB}⟩⟨{\Psi }_{AB}|$. It is a rank-one orthogonal projector, whose only nonzero eigenvalue is one. ConsequentlyThe reduced density matrices are ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ and ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$. The quantum conditional entropy is defined by the entropy differenceFor the pure state under consideration this becomes $S(B|A)=-S({\rho }_A)$. To determine when it is negative, write the Schmidt decompositionThe eigenvalues of ${\rho }_A$ are the ${\lambda }_j$, together with possible zeros. Each contribution $-{\lambda }_j{\log }_2{\lambda }_j$ is nonnegative, and their sum vanishes exactly when one eigenvalue equals one. Equivalently the Schmidt rank is one and $|{\Psi }_{AB}⟩$ is a product state. If the pure state is entangled, then $r\ge 2$, every nonzero eigenvalue is strictly below one, and the Von Neumann entropy is strictly positive. Thus the pure-state negative conditional entropy criterion givesThe purity assumption matters: for mixed density operators, negative quantum conditional entropy is not necessary for entanglement.

Let a density operator on a Hilbert space ${\mathcal{H}}_S$ have the spectral decompositionIntroduce an auxiliary Hilbert space ${\mathcal{H}}_R$ with an orthonormal basis containing $r$ vectors $|j{⟩}_R$, and setThis vector has norm one. Its partial trace isThis explicitly constructs a purification of a density operator. The same construction works for an arbitrary trace-class density operator: its nonzero eigenvalues form a finite or countable summable family, and the displayed vector converges in the Hilbert space norm. The entropy inequality below is understood with finite Von Neumann entropies, avoiding undefined differences of infinities.

Apply this construction with $S=AB$ to obtain a joint pure state on $ABC$. The Schmidt decomposition across $AB:C$, $A:BC$ and $B:AC$ respectively givesNow apply Subadditivity of Von Neumann entropy to the reduced density matrices on $BC$ and $AC$:These inequalities say $S(A)-S(B)\le S(C)$ and $S(B)-S(A)\le S(C)$. Replacing $S(C)$ by $S(AB)$ proves the Araki–Lieb inequality:The subadditivity used here concerns the actual reduced density operators, rather than statistical independence. In particular it follows from nonnegativity of quantum relative entropy, sinceand similarly for $AC$. Thus no independence assumption on the purification of a density operator is needed.

Diagonalize the reduced density matrix on Alice's Hilbert space:For every positive eigenvalue, define a vector on Bob's Hilbert space byThe definition of the partial trace givesThus the $|b_j⟩$ are orthonormal. Extend the $|a_j⟩$ to an orthonormal basis of Alice's Hilbert space. If $|a⟩$ lies in the zero-eigenvalue subspace of ${\rho }_A$, thenThere is therefore no component of $|{\Psi }_{AB}⟩$ in those additional basis directions. Expansion in Alice's orthonormal basis yields the Schmidt decompositionIts positive Schmidt coefficients are $s_j=\sqrt{{\lambda }_j}$ and its Schmidt rank is $r$. Taking the two partial traces and using the orthonormality of both families givesHence the two reduced density matrices have exactly the same nonzero eigenvalues, including multiplicities. Their numbers of zero eigenvalues may differ when the local Hilbert spaces have different dimensions. In particular their Von Neumann entropies agree.

Factor the coefficients into a tensor product:This is already a Schmidt decomposition with one term. ConsequentlyEach reduced density matrix is the rank-one orthogonal projector onto $(|0⟩-|1⟩)/\sqrt{2}$, so its other eigenvalue is zero. The joint pure state is a product state, with no entanglement.

Expand an initial bipartite pure state in local orthonormal bases asBy the Schmidt decomposition, or equivalently the singular value decomposition of $C$, its Schmidt rank is $\mathrm{rank}C$. In a specified local measurement branch, the product Kraus operator $A\otimes B$ changes the coefficient matrix toIndeed the coefficient of $|k⟩|l⟩$ is ${\sum }_{i,j}{A}_{ki}{C}_{ij}{B}_{lj}$. The elementary matrix rank inequality givesNormalizing a nonzero branch multiplies $C^{\prime }$ by a scalar and therefore leaves its matrix rank unchanged.

Now refine a complete LOCC transcript to record every individual Kraus operator. Although later choices depend on the earlier classical outcomes, fixing a transcript fixes all those choices. Composing the local operations therefore gives a single product Kraus operator $A_{\tau }\otimes B_{\tau }$ in that branch. Each nonzero branch obeys the same Schmidt-rank contraction under product operators. Local auxiliary systems prepared independently are included by local isometries and do not add entanglement. Discarding a local auxiliary system is included by resolving its partial trace in an orthonormal basis; this again yields refined product Kraus operators.

If the protocol produces a specified pure state deterministically, its output density operator is a sum of positive rank-one branch density operators. Every nonzero branch vector must be proportional to that output vector: its squared overlap with any vector orthogonal to the output sums to zero, so each such overlap is separately zero. Thus the output Schmidt rank equals that of each surviving branch, andThis proves monotonicity of Schmidt rank under LOCC, even for a nonzero branch selected by postselection. If the classical transcript is instead discarded and the result is mixed, the refined branch vectors supply an ensemble with Schmidt ranks at most $r_{\mathrm{in}}$; by definition its Schmidt number is at most $r_{\mathrm{in}}$. This is the correct mixed-state extension, rather than a bound on the ranks of its reduced density matrices. In finite dimensions, any limiting exact protocol obeys the same pure-state bound because the set of coefficient matrices of rank at most $r_{\mathrm{in}}$ is closed, being defined by vanishing minors.

Apply the Schmidt decomposition to the bipartite pure state. Local unitary operators put it in the form $|{\psi }_{\theta }⟩=\cos \theta |00⟩+\sin \theta |11⟩$, with $0<\theta \le \pi /4$ after interchanging basis vectors. Both Schmidt coefficients are nonzero because the state is entangled. Put $s=\sin 2\theta >0$.

For this state, direct application of the Pauli operators givesFor example ${\sigma }_x\otimes {\sigma }_x$ interchanges $|00⟩$ and $|11⟩$, giving $2\cos \theta \sin \theta$; the crossed products take these two kets into their orthogonal complement.

Choose the CHSH axes for an entangled pure two-qubit state:Their associated Bloch vectors have unit length, so each is a valid two-outcome projective measurement with projectors $(I\pm A_i)/2$ or $(I\pm B_j)/2$. The correlation combination isA local hidden-variable assignment with outcomes $\pm 1$ gives absolute value at most two: exactly one of $b_0+b_1$ and $b_0-b_1$ vanishes, while the other is $\pm 2$. Averaging these assignments preserves that bound, which is the CHSH inequality.

Therefore $\boxed{\mathcal{S}=2\sqrt{1+{\sin }^{2}2\theta }>2}$. For the original state $(U\otimes V)|{\psi }_{\theta }⟩$, conjugate Alice's observables by $U$ and Bob's by $V$. These conjugations rotate their measurement axes and preserve all the displayed correlations. Thus every entangled pure two-qubit state has the requested violating axes. The maximally entangled case gives $2\sqrt{2}$, while the unentangled limit merely saturates two.

The Schmidt decomposition theorem says that a normalized vector in the tensor product of two finite-dimensional Hilbert spaces can be writtenwhere both families are orthonormal. The numbers $\sqrt{{\lambda }_j}$ are its Schmidt coefficients. The Schmidt rank $r$ and coefficients, including multiplicities, are uniquely determined by the state; the local vectors may be changed within degenerate eigenspaces and by compensating phases. Zero coefficients may be appended to make $n_1$ terms, since $n_2\ge n_1$.

For a proof, form the reduced density matrix ${\rho }_1={\mathrm{Tr}}_2|\psi ⟩⟨\psi |$. It is a Hermitian matrix that is positive semidefinite, with trace one. The spectral theorem supplies an orthonormal basis $u_1,\dots ,u_{n_1}$ of eigenvectors with nonnegative eigenvalues ${\lambda }_j$. Define the unnormalized relative vectors in the second Hilbert space byCompleteness of the first orthonormal basis gives $|\psi ⟩={\sum }_j|u_j⟩|w_j⟩$. In any orthonormal basis $e_b$ of the second factor, the definition of the partial trace givesThus a zero eigenvalue gives $w_j=0$, and for each positive eigenvalue the vector $v_j=w_j/\sqrt{{\lambda }_j}$ has unit norm. Distinct such vectors are orthogonal by the same identity. Substituting them gives the required Schmidt decomposition, with at most $n_1$ nonzero terms.

Taking the other partial trace now givesTherefore the two reduced density matrices have identical nonzero eigenvalues, namely the squared Schmidt coefficients. This also proves uniqueness of the coefficient multiset and shows that the pure state is a product state exactly when $r=1$.