The Schmidt Decomposition

Take any pure state of two quantum systems — two spins, an atom and a field mode, a qubit and the rest of a computer — and no matter how tangled the joint wavefunction looks when you write it out in a product basis, there is always a preferred pair of bases in which it collapses to a single diagonal sum: $|\Psi\rangle = \sum_k \sqrt{\lambda_k}\,|k\rangle_A|k\rangle_B$ . This is the Schmidt decomposition, and it is the sharpest tool in the entanglement toolbox. The non-negative weights $\lambda_k$ (the Schmidt coefficients) are unique, they sum to one, and their count — the Schmidt rank — is a hard, basis-independent number that answers the only question that matters for a pure bipartite state: is it entangled, and how much? This article proves the theorem the honest way, as the singular-value decomposition (SVD) of the state's coefficient matrix, so the "magic" preferred bases are nothing more than the left and right singular vectors. From that single fact three payoffs fall out at once: a product state has rank one and an entangled state has rank two or more; the two reduced density matrices $\rho_A$ and $\rho_B$ share exactly the same spectrum $\{\lambda_k\}$ (which is why their entropies are equal, cashed next door in Entanglement Entropy and Measures of Entanglement); and the whole notion of a "diagonalizing basis for the partial trace" — set up in The Reduced Density Matrix and Partial Trace — is here made concrete. We close with the honest caveat that this beautiful structure is a two-party luxury: three or more parties generically admit no Schmidt form, which is exactly why multipartite entanglement is a harder subject.

Key takeaways

Every pure bipartite state is $\sum_k \sqrt{\lambda_k}|k\rangle_A|k\rangle_B$ in matched orthonormal bases; the $\sqrt{\lambda_k}$ are the singular values of the coefficient matrix. Schmidt rank $=1$ means separable, $>1$ means entangled. Both reduced states share the spectrum $\{\lambda_k\}$ .

The core insight

Writing $|\Psi\rangle=\sum_{ij}c_{ij}|i\rangle_A|j\rangle_B$ makes the coefficients a matrix $C$ . The Schmidt decomposition is the SVD $C=U\Sigma V^\dagger$ : the Schmidt coefficients are the singular values, the Schmidt bases are the singular vectors, and the number of nonzero singular values is the entanglement.

Where this lives today

The Schmidt rank is the quantum-information analogue of matrix rank, and truncating the Schmidt spectrum — keeping only the largest $\lambda_k$ — is the numerical engine behind matrix-product states and the density-matrix renormalization group (DMRG), the workhorses of modern many-body simulation. Because "how entangled is a cut" equals "how many Schmidt coefficients you must keep," an area law for entanglement is precisely the statement that these methods converge. The same decomposition underlies the definition of entanglement measures, quantifies channel capacities, and is the first step in essentially every entanglement-manipulation protocol.

Origins · 1 minschmidt decomposition and how this was discovered

schmidt decomposition

1907–1935 · Berlin, Oxford

The mathematics predates quantum mechanics by two decades. Erhard Schmidt, working in David Hilbert's Göttingen–Berlin orbit, introduced in 1907 the bilinear-form expansion that we now recognize as the singular-value decomposition of an integral kernel — the abstract inner-product machinery is recalled by name in Inner Product and Norm of States, where Schmidt's Gram–Schmidt procedure also lives. Quantum mechanics adopted it in 1935, when Erwin Schrödinger, in the same burst of papers that gave "entanglement" its name, used the "biorthogonal" decomposition to argue that the correlations between two separated systems have a canonical form. The technique was rediscovered repeatedly — by Everett for his relative-state interpretation, and again in the 1990s quantum-information boom (Ekert and Knight gave the standard modern account) — a sign of how fundamental it is.

History

1907

— Erhard Schmidt introduces the bilinear-form (singular-value) expansion of kernels

1935

— Schrödinger uses the biorthogonal decomposition for entangled states

1957

— Everett employs it for the relative-state ("many-worlds") formulation

1995

— Ekert and Knight give the canonical quantum-information account

The physical picture

A general pure state of two qubits has four complex amplitudes: $c_{00},c_{01},c_{10},c_{11}$ . Written that way it is a mess — you cannot read off at a glance whether it factors into "qubit A is in this state and qubit B is in that state." The Schmidt decomposition is a change of local bases — one rotation on $A$ , an independent rotation on $B$ — chosen so that the cross-terms disappear and only "matched" terms $|k\rangle_A|k\rangle_B$ survive. Because the rotations are local (they act on $A$ and $B$ separately, never coupling them), they cannot create or destroy entanglement; they only reveal it. The right way to see it: arrange the amplitudes $c_{ij}$ into a rectangular matrix $C$ , rows labelled by $A$ 's basis, columns by $B$ 's. A local unitary on $A$ multiplies $C$ on the left; a local unitary on $B$ multiplies it on the right. And any matrix can be brought to diagonal form by a left and a right unitary — that is the singular-value decomposition, a theorem older than quantum mechanics. The diagonal entries that survive are the singular values, and their squares are the Schmidt weights $\lambda_k$ . So the "quantum" content is entirely carried by a piece of nineteenth-century linear algebra; what quantum mechanics adds is the interpretation of the singular values as the strengths of the perfectly-correlated channels between $A$ and $B$ . One number then tells you everything about the entanglement of a pure bipartite state: how many singular values are nonzero. One nonzero value ( $C$ has rank one) means $C=\text{(column)}\times\text{(row)}$ factorizes, the state is a product, and there is no entanglement. Two or more means the state cannot be factored — it is entangled — and the spread of the $\lambda_k$ measures how strongly.

Figure

The Schmidt decomposition as an SVD — local rotations diagonalize the coefficient matrix

Figure

Rank one factors, rank two does not — the geometry of separable versus entangled

Derivation AThe Schmidt decomposition is the SVD of the coefficient matrix

We prove that every pure bipartite state can be brought to the diagonal form $\sum_k\sqrt{\lambda_k}|k\rangle_A|k\rangle_B$ , and that the recipe is exactly the singular-value decomposition. The strategy: write the amplitudes as a matrix, invoke the SVD (which exists for any matrix), and read off what the two unitaries mean physically. The only physics input is that a unitary acting on $A$ is a change of orthonormal basis on $A$ alone.

Step 1. Package the amplitudes as a matrix. Expand in any product bases: $|\Psi\rangle=\sum_{ij}c_{ij}\,|i\rangle_A|j\rangle_B$ , and read the $c_{ij}$ as the entries of a $d_A\times d_B$ matrix $C$ . Why: the joint state is nothing but this array of numbers; the whole game is to simplify the array by local operations.

Step 2. Invoke the SVD. Every matrix factors as $C = U\Sigma V^\dagger$ , with $U$ ( $d_A\times d_A$ ) and $V$ ( $d_B\times d_B$ ) unitary and $\Sigma$ real, non-negative, and diagonal. Why: this is a theorem of linear algebra with no exceptions — it is the statement that any linear map is a rotation, a set of independent stretches, and another rotation.

Step 3. Substitute and regroup. Writing $\Sigma_{kk}=s_k$ , the entries become $c_{ij}=\sum_k U_{ik}\,s_k\,(V^\dagger)_{kj}=\sum_k U_{ik}\,s_k\,V^*_{jk}$ , so

|\Psi\rangle = \sum_k s_k \Big(\sum_i U_{ik}|i\rangle_A\Big)\Big(\sum_j V^*_{jk}|j\rangle_B\Big).

Why: we have inserted the SVD term by term and collected the $A$ -sum and $B$ -sum into new vectors, one pair per singular value $s_k$ .

Step 4. Name the Schmidt bases. Define $|k\rangle_A \equiv \sum_i U_{ik}|i\rangle_A$ and $|k\rangle_B \equiv \sum_j V^*_{jk}|j\rangle_B$ . Because $U$ and $V$ are unitary, both sets are orthonormal. Setting $\sqrt{\lambda_k}\equiv s_k$ ,

\boxed{\,|\Psi\rangle = \sum_k \sqrt{\lambda_k}\,|k\rangle_A|k\rangle_B\,,\qquad \lambda_k\ge 0,\ \ \sum_k\lambda_k=1.\,}

Why: the columns of a unitary are orthonormal, so applying $U$ to an orthonormal set produces an orthonormal set — the "magic" bases are just rotated versions of the ones you started in. Normalization $\langle\Psi|\Psi\rangle=1$ forces $\sum_k s_k^2=\sum_k\lambda_k=1$ because the Schmidt terms are mutually orthogonal.

Step 5. Uniqueness of the weights. The singular values of $C$ are unique (they are the square roots of the eigenvalues of $C^\dagger C$ ), so the set $\{\lambda_k\}$ is an invariant of the state — no choice of starting bases can change it. Why: eigenvalues of a Hermitian matrix are basis-independent; $C^\dagger C$ is fixed once the physical state is fixed, up to the local unitaries that never touch its spectrum. *Verified numerically for the maximally-entangled $C=\tfrac12\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ (singular values $\{1/\sqrt2,1/\sqrt2\}$ ) and reconstructed to unit overlap; symbolic identities $\mathrm{Tr}(C^\dagger C)=\|C\|_F^2$ and $\det(C^\dagger C)=|\det C|^2$ confirmed in* `verify_aq-qm-schmidt-decomposition.py`.

Go deeper · Derivation B · 1 minSchmidt rank, and why both parts share one spectrum

Derivation BSchmidt rank, and why both parts share one spectrum

Now we extract the two facts that make the decomposition useful: rank counts entanglement, and the reduced states of $A$ and $B$ carry the same eigenvalues. The second fact is the quiet reason that the two subsystems of a pure state are "equally entangled," which the entropy article turns into a theorem.

Step 1. Trace out $B$ from the Schmidt form. With $|\Psi\rangle=\sum_k\sqrt{\lambda_k}|k\rangle_A|k\rangle_B$ , the reduced state (partial trace defined in The Reduced Density Matrix and Partial Trace) is $\rho_A=\mathrm{Tr}_B|\Psi\rangle\langle\Psi|=\sum_k \lambda_k\,|k\rangle_A\langle k|_A$ . Why: the $B$ -kets are orthonormal, so tracing them out sets $\langle k'|k\rangle_B=\delta_{k'k}$ and kills every cross-term — the Schmidt basis is exactly the basis that diagonalizes the partial trace.

Step 2. Do the same for $A$ . Symmetrically, $\rho_B=\sum_k\lambda_k\,|k\rangle_B\langle k|_B$ . Why: the decomposition treats $A$ and $B$ on equal footing; the very same weights appear.

Step 3. Read off the spectrum theorem. Both $\rho_A$ and $\rho_B$ are diagonal with eigenvalues $\{\lambda_k\}$ — identical spectra, differing only in the (irrelevant) labels of their eigenvectors. Why: this is why every function of the reduced spectrum — purity $\sum\lambda_k^2$ , entropy $-\sum\lambda_k\ln\lambda_k$ — is the same for both halves; the equality $S(\rho_A)=S(\rho_B)$ proved next door is a corollary of this one line.

Step 4. Define and interpret the Schmidt rank. The Schmidt rank $r$ is the number of nonzero $\lambda_k$ . If $r=1$ , then $\rho_A=|1\rangle\langle1|$ is pure and $|\Psi\rangle=|1\rangle_A|1\rangle_B$ is a product state; if $r\ge2$ , $\rho_A$ is mixed and $|\Psi\rangle$ is entangled. Why: a single Schmidt term is literally a product; more than one term means no local basis can factor the state — the three-way link " $r>1 \Leftrightarrow \rho_A$ mixed $\Leftrightarrow$ entangled" is proved as a theorem in Quantum Entanglement and Non-Separability, and here we supply its Schmidt-rank face. *Verified: for $C=\mathrm{diag}(\sqrt{0.9},\sqrt{0.1})$ both reduced states have spectrum $\{0.9,0.1\}$ ; for the product $|0\rangle(|0\rangle+|1\rangle)/\sqrt2$ the rank is $1$ ; for $|\Phi^+\rangle$ the Schmidt coefficients are $\{1/\sqrt2,1/\sqrt2\}$ (maximal). All in the verify script.*

Limiting cases

State	Coefficient matrix $C$	Schmidt coefficients	Rank	Verdict
Product $\|0\rangle_A\|+\rangle_B$	rank-1 outer product	$\{1\}$	1	separable
$\cos\theta\,\|00\rangle+\sin\theta\,\|11\rangle$	$\mathrm{diag}(\cos\theta,\sin\theta)$	$\{\cos^2\theta,\sin^2\theta\}$	2 (unless $\theta\in\{0,\tfrac\pi2\}$ )	entangled
Bell $\|\Phi^+\rangle$	$\tfrac1{\sqrt2}I$	$\{\tfrac12,\tfrac12\}$	2	maximally entangled
$d\times d$ maximally entangled	$\tfrac1{\sqrt d}I$	$\{\tfrac1d,\dots\}$	$d$	maximal for dimension $d$

The Schmidt rank can never exceed $\min(d_A,d_B)$ — a small subsystem caps the entanglement of the whole. A single qubit entangled with an entire laboratory still has Schmidt rank at most two.

Common trap“I can always write a state in the Schmidt form, so I can always Schmidt-decompose three particles too.”

No — and this is the most important caveat in the subject. The theorem is bipartite. A general tripartite pure state $|\Psi\rangle_{ABC}$ has no Schmidt form: you cannot find local bases making it $\sum_k\sqrt{\lambda_k}|k\rangle_A|k\rangle_B|k\rangle_C$ . The GHZ state $(|000\rangle+|111\rangle)/\sqrt2$ happens to look Schmidt-like, but the $W$ state $(|001\rangle+|010\rangle+|100\rangle)/\sqrt3$ does not, and the two belong to genuinely inequivalent classes of tripartite entanglement (they cannot be interconverted by local operations). The single-number "amount of entanglement" you get from bipartite Schmidt rank simply does not exist for three or more parties — that is why multipartite entanglement is an open frontier and why The GHZ State and All-or-Nothing Nonlocality exists as its own topic.

Worked examples

1 · A clean two-qubit state, decomposed by hand (confidence builder)

C=\frac12\begin{pmatrix}1&1\\1&-1\end{pmatrix}.

Apply the recipe: form $C^\dagger C = \tfrac14\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}1&1\\1&-1\end{pmatrix}=\tfrac14\begin{pmatrix}2&0\\0&2\end{pmatrix}=\tfrac12 I.$ Its eigenvalues are both $\lambda=\tfrac12$ , so the singular values are $s_k=\sqrt{1/2}=1/\sqrt2$ . Boundary number: the Schmidt coefficients are $\{1/\sqrt2,\,1/\sqrt2\}$ , the Schmidt rank is $2$ , and the weights are as spread as they can be — this innocuous-looking state is maximally entangled, exactly as much as a Bell state. Indeed $\tfrac12\begin{pmatrix}1&1\\1&-1\end{pmatrix}=\tfrac1{\sqrt2}H$ is $1/\sqrt2$ times a Hadamard, and $C=\tfrac1{\sqrt2}H$ has all singular values equal because $H$ is unitary. The lesson: entanglement is invisible in the raw amplitudes but a two-line calculation ( $C^\dagger C$ , then its eigenvalues) exposes it.

2 · A partially entangled state and the matched spectra (exam level)

Take $|\Psi\rangle=\sqrt{0.9}\,|00\rangle+\sqrt{0.1}\,|11\rangle$ . The coefficient matrix is already diagonal, $C=\mathrm{diag}(\sqrt{0.9},\sqrt{0.1})$ , so the Schmidt coefficients are simply $\{\sqrt{0.9},\sqrt{0.1}\}$ and the weights $\{\lambda_1,\lambda_2\}=\{0.9,0.1\}$ . Now compute both reduced states: $\rho_A=\mathrm{Tr}_B|\Psi\rangle\langle\Psi|=\mathrm{diag}(0.9,0.1)$ and, tracing out $A$ instead, $\rho_B=\mathrm{diag}(0.9,0.1)$ — identical spectra, as the theorem demands. Compare with the maximally entangled $|\Phi^+\rangle$ : there $\{\lambda_k\}=\{0.5,0.5\}$ and $\rho_A=\rho_B=I/2$ . The spread of the $\lambda_k$ is the entanglement dial: $\{0.9,0.1\}$ is "barely entangled," $\{0.5,0.5\}$ is "maximal," and $\{1,0\}$ is "product." Entanglement Entropy and Measures of Entanglement turns that dial into a number, $-\sum_k\lambda_k\ln\lambda_k$ , which is $0.325$ nats here versus $\ln2=0.693$ for the Bell state.

Going further

The Schmidt coefficients are not merely a diagnostic — they are the complete local invariant of a pure bipartite state under local unitaries, and Nielsen's theorem (1999) states that one state can be converted to another by local operations and classical communication if and only if the target's Schmidt coefficient vector majorizes the source's. That majorization order is the exact meaning of "more entangled," and its single-number summaries are built in Entanglement Entropy and Measures of Entanglement, which computes $-\sum_k\lambda_k\ln\lambda_k$ from the very spectrum we diagonalized here. The rank-versus-separability equivalence is proved as a theorem in Quantum Entanglement and Non-Separability; the maximally-entangled endpoint (all $\lambda_k$ equal) is the Bell States and Maximally Entangled Pairs. For three or more parties the story fractures into SLOCC classes (GHZ vs $W$ ), developed in The GHZ State and All-or-Nothing Nonlocality; the numerical afterlife of truncating the Schmidt spectrum is the matrix-product-state literature.

Is the Schmidt decomposition unique?

The coefficients $\{\lambda_k\}$ are unique (they are eigenvalues of $C^\dagger C$ ). The Schmidt bases are unique too, as long as the nonzero $\lambda_k$ are all distinct. When two Schmidt coefficients coincide — as in a Bell state, where both equal $1/2$ — there is freedom to rotate within the degenerate subspace, exactly as with any degenerate eigenbasis. So the numbers are always fixed; the vectors are fixed only up to degeneracy.

Does the Schmidt decomposition work for mixed states?

No. It is a theorem about pure bipartite state vectors. A mixed bipartite state $\rho_{AB}$ has no analogous single canonical form — quantifying its entanglement requires the harder measures (entanglement of formation, negativity) discussed in Entanglement Entropy and Measures of Entanglement, and even deciding whether $\rho_{AB}$ is entangled at all is computationally hard in general. The clean rank criterion is a gift that only pure states give.

How is this different from just diagonalizing the density matrix?

Diagonalizing the joint $\rho_{AB}=|\Psi\rangle\langle\Psi|$ tells you it is pure (one eigenvalue $1$ ) but nothing about the split. Diagonalizing the reduced $\rho_A$ gives you the Schmidt basis on $A$ and the weights $\lambda_k$ — but you still need the correlated $B$ -partners. The Schmidt decomposition does both at once and, crucially, guarantees that $A$ 's eigenbasis is perfectly correlated with a matching $B$ -eigenbasis. It is the statement that diagonalizing $\rho_A$ and diagonalizing $\rho_B$ can be done in paired bases.

Why singular values and not eigenvalues of C?

Because $C$ is generally not square (subsystems can differ in dimension) and not Hermitian, so it need not have eigenvalues at all. The SVD works for any rectangular matrix; the singular values are the square roots of the eigenvalues of the Hermitian, positive matrix $C^\dagger C$ (equivalently $CC^\dagger$ ), which always exist and are non-negative. Those square roots are the Schmidt coefficients.

If the Schmidt rank is 2 for both a Bell state and a barely-entangled state, how do they differ?

Rank is a coarse yes/no-and-a-count; it does not measure how much. Both $\{0.5,0.5\}$ and $\{0.99,0.01\}$ have rank $2$ , but the first is maximally entangled and the second barely at all. The finer measure is the spread of the $\lambda_k$ , quantified by the entanglement entropy $-\sum_k\lambda_k\ln\lambda_k$ in the next article: $\ln2$ for the Bell state, $\approx0.056$ nats for $\{0.99,0.01\}$ . Rank tells you "entangled or not"; entropy tells you "by how much."

Can local operations change the Schmidt coefficients?

Local unitaries cannot — that is the whole point; they only relabel the Schmidt bases. Local measurements with classical communication (LOCC) can only make the coefficient vector "flatter or steeper" in a precise majorization sense (Nielsen's theorem): you can never increase entanglement on average by acting locally. This is the operational meaning of the Schmidt spectrum as the entanglement content, developed further in the measures article.