Schmidt Decomposition of Bipartite States

Derivation

Schmidt Decomposition of Bipartite States

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Statement

Let \(\mathcal{H}_A\) and \(\mathcal{H}_B\) be finite-dimensional complex Hilbert spaces with \(\dim\mathcal{H}_A=d_A\) and \(\dim\mathcal{H}_B=d_B\). Every pure state \(|\psi\rangle\in\mathcal{H}_A\otimes\mathcal{H}_B\) can be written as a single sum \(|\psi\rangle=\sum_{k=1}^{r}\sigma_k\,|k\rangle_A\otimes|k\rangle_B\), where \(\{|k\rangle_A\}\) and \(\{|k\rangle_B\}\) are orthonormal sets in \(\mathcal{H}_A\) and \(\mathcal{H}_B\) respectively, the Schmidt coefficients satisfy \(\sigma_k>0\) with \(\sum_k\sigma_k^2=1\), and the number of nonzero terms \(r\le\min(d_A,d_B)\) — the Schmidt rank — is uniquely fixed by \(|\psi\rangle\). This is the singular-value decomposition (SVD) of the coefficient matrix rewritten in vector language, and \(r\) is a basis-independent entanglement measure: \(r=1\) exactly for product states.

Why it matters

The Schmidt decomposition is the structural theorem of bipartite entanglement. It collapses the \(d_Ad_B\) complex amplitudes of a general two-party state into at most \(\min(d_A,d_B)\) non-negative numbers, and does so in local bases — one for Alice, one for Bob — so that the correlation structure is laid bare. Whether a state is entangled, and how much, is read directly from the \(\sigma_k\).

It also unifies the two ways of asking about a subsystem: the Schmidt coefficients squared are simultaneously the eigenvalues of both reduced density matrices \(\rho_A\) and \(\rho_B\). This is why the two subsystems of a pure state always carry the same entropy, and it underlies purification, entanglement entropy, and the Araki–Lieb bound.

Assumptions

The global state is pure.For a mixed \(\rho_{AB}\) no single-sum form exists; one must instead seek separable convex decompositions, and "Schmidt rank" is replaced by the far subtler notion of Schmidt number.

The Hilbert space is a tensor product \(\mathcal{H}_A\otimes\mathcal{H}_B\).Without a fixed bipartition the coefficient array is not a matrix and the SVD has no meaning; the decomposition is defined relative to a chosen cut.

Both factors are finite-dimensional (or the state is Hilbert–Schmidt / trace-class).In infinite dimensions the sum may be countably infinite; the SVD generalises only when the coefficient operator is compact, otherwise no discrete Schmidt spectrum exists.

The inner products are the standard ones inducing the tensor-product inner product.Orthonormality of the Schmidt vectors relies on \(U,V\) being unitary in these inner products; a non-standard metric breaks the isometry step.

Derivation

\[ |\psi\rangle=\sum_{i=1}^{d_A}\sum_{j=1}^{d_B} C_{ij}\,|i\rangle_A\otimes|j\rangle_B \]

Fix any orthonormal product bases \(\{|i\rangle_A\}\), \(\{|j\rangle_B\}\); completeness of the tensor-product basis lets any vector be expanded, defining the coefficient matrix \(C\in\mathbb{C}^{d_A\times d_B}\). A

\[ C=U\,\Sigma\,V^{\dagger},\qquad \Sigma_{k\ell}=\sigma_k\,\delta_{k\ell},\ \ \sigma_k\ge 0 \]

Apply the SVD (prior result) to the rectangular matrix \(C\): \(U\in\mathbb{C}^{d_A\times d_A}\) and \(V\in\mathbb{C}^{d_B\times d_B}\) are unitary and \(\Sigma\) is \(d_A\times d_B\) with non-negative real diagonal entries, the singular values. B

\[ C_{ij}=\sum_{k} U_{ik}\,\sigma_k\,(V^{\dagger})_{kj}=\sum_{k} U_{ik}\,\sigma_k\,V_{jk}^{*} \]

Write the matrix product in components and use \((V^\dagger)_{kj}=V^*_{jk}\); the diagonal \(\Sigma\) collapses the middle index sum to a single \(k\), running to \(\min(d_A,d_B)\). A

\[ |\psi\rangle=\sum_{k}\sigma_k\Big(\sum_i U_{ik}\,|i\rangle_A\Big)\otimes\Big(\sum_j V^*_{jk}\,|j\rangle_B\Big) \]

Substitute step 3 into step 1 and regroup the sums: the bilinearity of \(\otimes\) lets the \(i\)-sum and \(j\)-sum be factored out of the shared coefficient \(\sigma_k\). B

\[ |k\rangle_A\equiv\sum_i U_{ik}\,|i\rangle_A,\qquad |k\rangle_B\equiv\sum_j V^*_{jk}\,|j\rangle_B \]

Define new local vectors as the columns of \(U\) and \(\overline{V}\). This is a change of basis on each factor generated by a unitary. A

\[ {}_A\langle k|k'\rangle_A=\sum_{i,i'}U^*_{ik}U_{i'k'}\,\delta_{ii'}=\sum_i (U^\dagger)_{ki}U_{ik'}=(U^\dagger U)_{kk'}=\delta_{kk'} \]

Orthonormality of \(\{|k\rangle_A\}\) follows from \(U^\dagger U=\mathbb{1}\); the identical computation with \(V^\dagger V=\mathbb{1}\) (the conjugation is undone by the second index) gives \({}_B\langle k|k'\rangle_B=\delta_{kk'}\). Unitaries map orthonormal bases to orthonormal bases. C

\[ |\psi\rangle=\sum_{k:\,\sigma_k>0}\sigma_k\,|k\rangle_A\otimes|k\rangle_B\equiv\sum_{k=1}^{r}\sigma_k\,|k\rangle_A|k\rangle_B \]

Zero singular values contribute nothing, so drop them; \(r\) = number of strictly positive \(\sigma_k\) = rank of \(C\), the Schmidt rank, bounded by \(\min(d_A,d_B)\). A

\[ 1=\langle\psi|\psi\rangle=\sum_{k,k'}\sigma_k\sigma_{k'}\,{}_A\langle k|k'\rangle_A\,{}_B\langle k|k'\rangle_B=\sum_{k=1}^{r}\sigma_k^2 \]

Impose normalisation and use the orthonormality of step 6 to kill cross terms; the surviving diagonal fixes \(\sum_k\sigma_k^2=1\). B

\[ \rho_A=\mathrm{Tr}_B|\psi\rangle\langle\psi|=\sum_{k}\sigma_k^2\,|k\rangle_A\langle k|_A,\qquad \rho_B=\sum_{k}\sigma_k^2\,|k\rangle_B\langle k|_B \]

Trace out \(B\) (prior result) using \({}_B\langle k'|k\rangle_B=\delta_{kk'}\); the Schmidt form is already diagonal, so \(\sigma_k^2\) are the eigenvalues of both reduced states — they share one spectrum. C

Result

\[ \boxed{\ |\psi\rangle=\sum_{k=1}^{r}\sigma_k\,|k\rangle_A\otimes|k\rangle_B,\quad \sigma_k>0,\ \ \sum_{k=1}^{r}\sigma_k^2=1,\ \ \rho_{A,B}=\sum_k\sigma_k^2|k\rangle_{A,B}\langle k|\ } \]

Reading. Any bipartite pure state is a diagonal correlation between two local orthonormal bases: Alice's \(k\)-th mode is perfectly paired with Bob's \(k\)-th mode, weighted by \(\sigma_k\). The Schmidt rank \(r\) counts how many such pairs are active. \(r=1\) means \(|\psi\rangle=|1\rangle_A|1\rangle_B\) — a product state, no entanglement; \(r>1\) means the parties are entangled, and a flat spectrum \(\sigma_k=1/\sqrt r\) is maximal entanglement for that rank. The squared coefficients are the eigenvalues shared by \(\rho_A\) and \(\rho_B\), so the entanglement entropy \(S=-\sum_k\sigma_k^2\log\sigma_k^2\) is symmetric between the parties.

Units check. All quantities are dimensionless. State amplitudes \(C_{ij}\), singular values \(\sigma_k\), and eigenvalues \(\sigma_k^2\) carry no units; \(\sum_k\sigma_k^2=1\) is the (dimensionless) probability normalisation, consistent with \(\mathrm{Tr}\,\rho_A=1\). The entropy \(S\) is measured in nats (with \(\ln\)) or bits (with \(\log_2\)), i.e. dimensionless information units. Both sides of the boxed identity are vectors in \(\mathcal{H}_A\otimes\mathcal{H}_B\) with unit norm.

Limiting cases

Product state (\(r=1\)): a single term, \(|\psi\rangle=|1\rangle_A|1\rangle_B\); \(\rho_A=|1\rangle\langle1|\) is pure, \(S=0\).
Maximal entanglement (\(\sigma_k=1/\sqrt r\)): flat spectrum, \(\rho_A=\mathbb{1}_r/r\), \(S=\log r\) — the largest entropy allowed at rank \(r\).
Qubit–qubit: \(d_A=d_B=2\Rightarrow r\in\{1,2\}\); one Schmidt angle \(\theta\) with \(\sigma_1=\cos\theta,\ \sigma_2=\sin\theta\) parametrises all entanglement up to local unitaries.
Unequal dimensions: \(r\le\min(d_A,d_B)\); the smaller factor caps the entanglement, so a qubit entangled with a qutrit still has \(r\le 2\).
Near-product limit (\(\sigma_2\to 0\)): \(S\to 0\) continuously; entanglement switches on smoothly, but the rank \(r\) jumps discontinuously from 1 to 2.

Breaks when

The state is mixed. For \(\rho_{AB}\) not a rank-one projector there is no single-sum \(\sum_k\sigma_k|k\rangle_A|k\rangle_B\); the eigenvalue spectra of \(\rho_A\) and \(\rho_B\) generically differ, and entanglement must be diagnosed with separability criteria (PPT, entanglement of formation), not a Schmidt rank.
More than two parties. A tripartite \(|\psi\rangle_{ABC}\) has no analogue: the GHZ and W states cannot both be brought to a common single-sum local form. There is no multipartite SVD, and genuine multipartite entanglement is not captured by any one Schmidt spectrum.
Infinite-dimensional factors with non-compact coefficient operator. If \(C\) fails to be Hilbert–Schmidt (e.g. an idealised, unnormalisable EPR state \(\int dx\,|x\rangle_A|x\rangle_B\)), the singular values do not form a discrete summable set and the discrete Schmidt decomposition does not exist.
Ambiguous or dynamical bipartition. If the tensor factorisation \(\mathcal{H}=\mathcal{H}_A\otimes\mathcal{H}_B\) is not fixed (e.g. identical-particle systems where the natural algebra is not a simple tensor product), the Schmidt rank is not well defined until a cut and superselection structure are specified.

Failure modes

Diagonalising the wrong matrix. Taking eigenvalues of the (generally non-Hermitian, rectangular) \(C\) instead of singular values. Schmidt coefficients are singular values of \(C\), equivalently square roots of eigenvalues of \(CC^\dagger\) or \(C^\dagger C\) — never eigenvalues of \(C\) itself.
Forgetting the complex conjugate on Bob's basis. Bob's Schmidt vectors are columns of \(\overline{V}\), i.e. \(|k\rangle_B=\sum_j V^*_{jk}|j\rangle_B\), not of \(V\). Dropping the conjugation gives non-orthogonal or phase-wrong local states.
Confusing \(\sigma_k\) with \(\sigma_k^2\). The reduced-density-matrix eigenvalues (the "Schmidt weights", probabilities) are \(\sigma_k^2\); the amplitudes in the ket are \(\sigma_k\). Plugging \(\sigma_k\) into the entropy \(-\sum p_k\log p_k\) overstates \(S\).
Reading rank off the product basis. Counting nonzero \(C_{ij}\) and calling it the Schmidt rank. \(\tfrac12(|00\rangle+|01\rangle+|10\rangle+|11\rangle)=|+\rangle|+\rangle\) has four nonzero amplitudes but rank 1.
Assuming Schmidt bases are the measurement bases. Treating \(\{|k\rangle_A\}\) as fixed lab bases; they are state-dependent and rotate as \(|\psi\rangle\) changes.
Sign/normalisation slip in maximal entanglement. Claiming \(S=\log r\) for any rank-\(r\) state; it holds only for the flat spectrum \(\sigma_k=1/\sqrt r\).

Discussion

The Schmidt decomposition is the statement that bipartite pure-state entanglement is a one-dimensional problem: up to local unitaries \(U_A\otimes U_B\), a state is completely specified by its ordered list of Schmidt coefficients \((\sigma_1\ge\sigma_2\ge\cdots)\). Everything invariant under local operations — the entanglement entropy, the Rényi entropies, the negativity, the concurrence — is a function of this list alone. Two states are interconvertible by local unitaries iff they share the same Schmidt spectrum, which is why the spectrum, not the state vector, is the physical content of entanglement.

The symmetry of the spectrum between \(A\) and \(B\) has a striking consequence: a small subsystem entangled with a huge environment still has entanglement entropy bounded by \(\log d_A\), the small dimension. This "small factor wins" fact is the seed of area laws in many-body physics and of the Page curve in black-hole information: the entropy of a subregion is set by the boundary that limits its Schmidt rank, not by its volume.

Purification is the decomposition read backwards. Given any mixed \(\rho_A=\sum_k p_k|k\rangle_A\langle k|\), the state \(|\psi\rangle=\sum_k\sqrt{p_k}\,|k\rangle_A|k\rangle_B\) is a pure state on an enlarged space whose reduction is \(\rho_A\); the Schmidt theorem guarantees such a purification always exists and is unique up to a unitary on the ancilla \(B\). This is the formal backbone of the "church of the larger Hilbert space" and of thermofield-double constructions in thermal field theory.

At the operational level the Schmidt coefficients are the currency of entanglement manipulation. Nielsen's theorem states that \(|\psi\rangle\) can be converted to \(|\phi\rangle\) by local operations and classical communication (LOCC) with certainty iff the Schmidt vector of \(|\psi\rangle\) is majorised by that of \(|\phi\rangle\) — a partial order, not a total one, so incomparable states exist and catalysis and probabilistic protocols become necessary. The asymptotic distillable entanglement and entanglement cost both collapse to \(S=-\sum_k\sigma_k^2\log_2\sigma_k^2\) in the many-copy limit, making the Schmidt entropy the unique measure for pure states. The decomposition thus sits at the exact junction where the linear-algebraic SVD, the thermodynamic entropy, and the resource theory of entanglement become the same object.

Common misconceptions. The Schmidt rank is not the number of nonzero amplitudes in some product basis (that is basis-dependent); it is the matrix rank of the coefficient array, an invariant. Nonzero entanglement does not require equal weights — any spectrum with \(r>1\) is entangled, and only the flat spectrum is maximally so. And "same entropy on both sides" is a feature of pure global states only; for mixed \(\rho_{AB}\), \(S(\rho_A)\ne S(\rho_B)\) in general.

Worked examples

Example 1 — A Bell state is already in Schmidt form.

\[ |\Phi^+\rangle=\tfrac{1}{\sqrt2}\big(|0\rangle_A|0\rangle_B+|1\rangle_A|1\rangle_B\big),\qquad C=\frac{1}{\sqrt2}\begin{pmatrix}1&0\\0&1\end{pmatrix} \]

Read off the coefficient matrix in the computational product basis. It is already diagonal. A

\[ CC^\dagger=\tfrac12\begin{pmatrix}1&0\\0&1\end{pmatrix},\qquad \sigma_1=\sigma_2=\tfrac{1}{\sqrt2} \]

Diagonal \(C\) with equal, positive entries: the singular values are the diagonal magnitudes, no rotation needed (\(U=V=\mathbb{1}\)). B

\[ r=2,\qquad \sigma_1^2=\sigma_2^2=\tfrac12,\qquad S=-2\cdot\tfrac12\log_2\tfrac12=1\ \text{bit} \]

Count nonzero singular values for the rank; the flat spectrum gives the maximal entropy \(\log_2 2\). A

\[ r=2,\quad \sigma_1=\sigma_2=\tfrac{1}{\sqrt2},\quad S=1\ \text{bit} \]

Reading. A Bell state is maximally entangled: its reduced state \(\rho_A=\tfrac12\mathbb{1}\) is completely mixed, carrying exactly one bit of entanglement.

Units check. \(\sigma_k\) dimensionless, \(\sum\sigma_k^2=\tfrac12+\tfrac12=1\); \(S\) in bits.

Example 2 — A non-diagonal state requiring the SVD.

\[ |\psi\rangle=\tfrac{1}{\sqrt3}\big(|00\rangle+|01\rangle+|10\rangle\big),\qquad C=\frac{1}{\sqrt3}\begin{pmatrix}1&1\\1&0\end{pmatrix} \]

Amplitudes \(C_{00}=C_{01}=C_{10}=1/\sqrt3,\ C_{11}=0\); three nonzero entries, but rank is not yet read off. A

\[ \rho_A=CC^\dagger=\tfrac13\begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}1&1\\1&0\end{pmatrix}=\tfrac13\begin{pmatrix}2&1\\1&1\end{pmatrix} \]

Form the reduced density matrix; its eigenvalues are the Schmidt weights \(\sigma_k^2\). B

\[ \det(\rho_A-\lambda\mathbb{1})=0:\quad \lambda=\frac{3\pm\sqrt5}{6},\qquad \sigma_k^2=\{0.8727,\ 0.1273\} \]

Solve the characteristic equation of the \(2\times2\) block \(\begin{pmatrix}2&1\\1&1\end{pmatrix}\) (trace 3, det 1), giving \(\lambda=(3\pm\sqrt5)/6\). Symbols first, then numbers with \(\sqrt5\approx2.236\). B

\[ \sigma_1=\sqrt{0.8727}=0.934,\quad \sigma_2=\sqrt{0.1273}=0.357,\qquad r=2 \]

Take positive square roots for the amplitudes; both nonzero, so the Schmidt rank is 2. A

\[ S=-0.8727\log_2 0.8727-0.1273\log_2 0.1273=0.171+0.379=0.550\ \text{bits} \]

Insert the weights into the entanglement entropy. B

\[ r=2,\quad \sigma_1^2=\tfrac{3+\sqrt5}{6}\approx0.873,\ \ \sigma_2^2=\tfrac{3-\sqrt5}{6}\approx0.127,\quad S\approx0.55\ \text{bits} \]

Reading. Despite three nonzero product-basis amplitudes, the state has rank 2 and is only partially entangled — the lopsided spectrum yields well under the maximal 1 bit.

Units check. \(\sigma_1^2+\sigma_2^2=(3+\sqrt5+3-\sqrt5)/6=1\); dimensionless, \(S<\log_2 2\) as required.

Problems

(A) Singlet. Find the Schmidt rank and entanglement entropy of \(|\psi^-\rangle=\tfrac{1}{\sqrt2}(|01\rangle-|10\rangle)\).
Solution
Coefficient matrix \(C=\tfrac{1}{\sqrt2}\begin{pmatrix}0&1\\-1&0\end{pmatrix}\). Then \(CC^\dagger=\tfrac12\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\tfrac12\begin{pmatrix}1&0\\0&1\end{pmatrix}\). Both eigenvalues \(=\tfrac12\), so \(\sigma_1=\sigma_2=1/\sqrt2\), \(r=2\), and \(S=-2\cdot\tfrac12\log_2\tfrac12=1\) bit. Maximally entangled, like every Bell state.
(A) Hidden product state. Show that \(|\psi\rangle=\tfrac12(|00\rangle+|01\rangle+|10\rangle+|11\rangle)\) has Schmidt rank 1, and give its Schmidt form.
Solution
\(C=\tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix}\). Then \(CC^\dagger=\tfrac14\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}1&1\\1&1\end{pmatrix}=\tfrac14\begin{pmatrix}2&2\\2&2\end{pmatrix}=\tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix}\), eigenvalues \(1\) and \(0\). One nonzero singular value \(\sigma_1=1\Rightarrow r=1\), \(S=0\). Explicitly \(|\psi\rangle=|+\rangle_A|+\rangle_B\) with \(|+\rangle=\tfrac{1}{\sqrt2}(|0\rangle+|1\rangle)\): a product state despite four nonzero amplitudes.
(B) Unequal weights. For \(|\psi\rangle=\tfrac{1}{\sqrt5}(2|00\rangle+|11\rangle)\), find \(r\), the Schmidt coefficients, and \(S\) in bits.
Solution
\(C=\tfrac{1}{\sqrt5}\begin{pmatrix}2&0\\0&1\end{pmatrix}\), already diagonal, so \(\sigma_1=2/\sqrt5,\ \sigma_2=1/\sqrt5\), \(r=2\). Weights \(\sigma_1^2=4/5=0.8,\ \sigma_2^2=1/5=0.2\). Entropy \(S=-0.8\log_2 0.8-0.2\log_2 0.2=0.8(0.3219)+0.2(2.3219)=0.2575+0.4644=0.722\) bits. Partially entangled.
(B) Schmidt angle. Write the general two-qubit Schmidt form as \(|\psi\rangle=\cos\theta\,|00\rangle+\sin\theta\,|11\rangle\) with \(0\le\theta\le\pi/4\). Find \(\theta\) that maximises \(S\), and give the maximum.
Solution
Weights \(\sigma_1^2=\cos^2\theta,\ \sigma_2^2=\sin^2\theta\). \(S(\theta)=-\cos^2\theta\log_2\cos^2\theta-\sin^2\theta\log_2\sin^2\theta\). This is the binary entropy of \(p=\cos^2\theta\), maximised at \(p=\tfrac12\), i.e. \(\cos^2\theta=\tfrac12\Rightarrow\theta=\pi/4\). Then \(\sigma_1=\sigma_2=1/\sqrt2\) and \(S_{\max}=1\) bit — the maximally entangled point.
(C) Qubit–qutrit and local-unitary invariance. A qubit \(A\) and qutrit \(B\) share \(|\psi\rangle=\tfrac{1}{\sqrt2}\,|0\rangle_A|0\rangle_B+\tfrac{1}{\sqrt2}\,|1\rangle_A|1\rangle_B\), a state living in the \(2\times3\) space \(\mathcal{H}_A\otimes\mathcal{H}_B\). State the maximum possible Schmidt rank, compute \(S\) here, and argue \(S\) is unchanged by any \(U_A\otimes U_B\).
Solution
Since \(r\le\min(d_A,d_B)=\min(2,3)=2\), the Schmidt rank is at most 2. The given state has \(C=\tfrac{1}{\sqrt2}\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}\); \(CC^\dagger=\tfrac12\mathbb{1}_2\), so \(\sigma_1=\sigma_2=1/\sqrt2\), \(r=2\), \(S=1\) bit — the qutrit's third dimension is unused. Under \(U_A\otimes U_B\), the coefficient matrix transforms as \(C\to U_A\,C\,U_B^{T}\); singular values are invariant under multiplication by unitaries on either side (they are the square roots of eigenvalues of \(U_A CC^\dagger U_A^\dagger\), a similarity transform), so the whole Schmidt spectrum and hence \(S\) are unchanged. This is why \(S\) is a legitimate entanglement measure: it depends only on the state's orbit under local unitaries.