Schmidt Decomposition

You have met the Bell state |\Phi^+\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle) — two terms, symmetric, beautiful. You have met an arbitrary two-qubit state |\psi\rangle = c_{00}|00\rangle + c_{01}|01\rangle + c_{10}|10\rangle + c_{11}|11\rangle — four amplitudes, no obvious structure, entangled or not depending on whether the amplitudes factor.

Stand those two examples next to each other and a question appears. The Bell state's amplitudes show their structure on sight — you can see the entanglement in the fact that only |00\rangle and |11\rangle appear, and with equal amplitude. The generic state's amplitudes hide whatever structure it has behind four complex numbers. Is there a way to always write a bipartite pure state in a form where the entanglement structure is visible? Where you can read off "this state is a product" or "this state is entangled with such-and-such strength" from the coefficients alone?

Yes. The form is the Schmidt decomposition, and it says something remarkable: for any bipartite pure state, there exist orthonormal bases on each side — not the computational basis in general, but some other basis tailored to the state — in which the amplitudes are real, non-negative, and diagonal. The off-diagonal entries are all zero. The state becomes a sum of "matched" product terms, one per diagonal entry, with the size of each entry measuring how much that term contributes.

This is the entanglement analogue of diagonalising a matrix. It exists for every bipartite pure state. It is unique up to degeneracies. It is the bridge between the amplitude picture of a state and the spectral picture of its reduced density matrix. And it is the first tool you should reach for whenever someone says the word "bipartite entanglement."

What the theorem says

Let Alice hold a d_A-dimensional system and Bob hold a d_B-dimensional system. A joint pure state |\psi\rangle_{AB} lives in the tensor-product Hilbert space \mathcal{H}_A \otimes \mathcal{H}_B of dimension d_A d_B.

Reading the theorem. Three things happen at once. First, the double sum over d_A \cdot d_B amplitudes collapses to a single sum over at most \min(d_A, d_B) terms. Second, the amplitudes are real and non-negative — every complex phase has been absorbed into the basis vectors. Third, the bases on A and B are matched in pairs: the i-th basis vector on A is matched with the i-th basis vector on B, and no other. No cross-terms. The amplitude matrix is diagonal.

That last fact is the full content of the theorem. For a two-qubit state, the generic amplitude matrix is 2\times 2 with four independent complex entries; the Schmidt form is a diagonal matrix with two non-negative real entries. The reduction in complexity is the reduction of entanglement to its bare bones.

The picture: what "diagonal" looks like

Write the joint state's amplitudes as a matrix. For two qubits,

The matrix C is the amplitude matrix — rows indexed by Alice's basis states, columns by Bob's. In general C has four complex entries.

If C were diagonal — say, C = \text{diag}(\lambda_1, \lambda_2) with \lambda_i \geq 0 — then the state is already in Schmidt form in the computational basis:

If C is not diagonal, you can always diagonalise it, not by similarity (which changes eigenvalues) but by the singular-value decomposition (SVD) — which splits C as C = U\,\Sigma\,V^\dagger, where U and V are unitary matrices and \Sigma is diagonal with non-negative entries. The Schmidt bases \{|u_i\rangle\} and \{|v_i\rangle\} are the columns of U and V respectively, and the Schmidt coefficients are the diagonal entries of \Sigma.

This is not an accident of the notation — the SVD is the Schmidt decomposition, rewritten in matrix language instead of ket language. Every linear algebra library you have ever used can compute a Schmidt decomposition for you with a single function call (numpy's svd, scipy's linalg.svd, MATLAB's svd) — you feed it the amplitude matrix, it returns U, \Sigma, V, and the Schmidt decomposition falls out immediately.

The SVD connection — the proof sketch

Here is the full argument in five lines, because the theorem is short enough to deserve a proof you can see in one sitting.

Start with an arbitrary joint state:

Collect the c_{ij} into a matrix C. By the SVD theorem (a theorem of linear algebra, true for any complex matrix), there exist unitary U (of size d_A \times d_A) and V (of size d_B \times d_B), and a diagonal matrix \Sigma with non-negative entries \sigma_i, such that C = U \Sigma V^\dagger. Write the entries: c_{ij} = \sum_k U_{ik}\,\sigma_k\,\overline{V_{jk}}.

Substitute back:

Define |u_k\rangle_A = \sum_i U_{ik}|i\rangle_A and |v_k\rangle_B = \sum_j \overline{V_{jk}}|j\rangle_B. Because U and V are unitary, their columns form orthonormal bases, so \{|u_k\rangle\} and \{|v_k\rangle\} are orthonormal. Setting \lambda_k = \sigma_k:

Why this is everything: the SVD decomposes any matrix into "unitary times diagonal times unitary." The two unitaries become the Schmidt bases, the diagonal becomes the Schmidt coefficients, and the overall sum structure becomes the diagonal pairing. The existence of an SVD for every complex matrix is a theorem of linear algebra; applying it to the amplitude matrix is the existence proof of the Schmidt decomposition. No quantum-mechanical content is needed — the whole theorem is linear algebra in disguise.

Normalisation: the state |\psi\rangle_{AB} is a unit vector, which means \sum_{ij} |c_{ij}|^2 = 1. The Frobenius norm of C equals the sum of squared singular values: \|C\|_F^2 = \sum_i \sigma_i^2. Therefore \sum_i \lambda_i^2 = 1, matching the theorem statement.

Schmidt rank and what it means

The Schmidt rank r is the number of strictly positive \lambda_i in the decomposition. Equivalently, it is the matrix rank of the amplitude matrix C.

Two bounds trap the rank:

• r \geq 1 for any non-zero state (the state has to have some content).
• r \leq \min(d_A, d_B) — the sum can't run longer than the smaller Hilbert space.

For two qubits, d_A = d_B = 2, so r \in \{1, 2\}. Exactly two possibilities.

Rank 1 means product state. If only one \lambda_i is non-zero — say \lambda_1 — the decomposition collapses to |\psi\rangle = \lambda_1 |u_1\rangle_A \otimes |v_1\rangle_B. Since \lambda_1^2 = 1 (normalisation), \lambda_1 = 1. So |\psi\rangle = |u_1\rangle \otimes |v_1\rangle — a single tensor product. No entanglement.

Rank \geq 2 means entangled. Two or more non-zero terms, each a distinct product, means the sum cannot be collapsed to a single product. That is the definition of entanglement (ch.18). So Schmidt rank is a clean binary test: rank 1 ⇔ product; rank > 1 ⇔ entangled.

For two qubits this is as fine a classification as rank gives. For higher dimensions the rank gets more expressive — a qutrit-qutrit system can have Schmidt ranks 1, 2, or 3, with 3 being "more entangled" than 2 in a precise sense the rank alone captures.

Why rank isn't the end of the story: two qutrit-qutrit states can both have Schmidt rank 3 but differ wildly in how evenly the three Schmidt coefficients are distributed. A state with (\lambda_1, \lambda_2, \lambda_3) = (0.99, 0.1, 0.1) has rank 3 but is almost a product. A state with \lambda_i = 1/\sqrt{3} each is maximally entangled at rank 3. The rank counts how many terms appear; the distribution of the \lambda_i says how "spread out" the entanglement is. Together they determine the state up to local unitaries.

Connection to the reduced density matrix

Recall from the partial-trace chapter (ch.9) that \rho_A = \text{tr}_B(|\psi\rangle\langle\psi|_{AB}) is the reduced state of Alice's side when Bob is ignored. The Schmidt decomposition makes the spectrum of \rho_A transparent.

Compute \rho_A using the Schmidt form. Let |\psi\rangle = \sum_i \lambda_i |u_i\rangle_A|v_i\rangle_B. Then

Trace over B:

But \{|v_i\rangle\} is orthonormal, so \langle v_j|v_i\rangle = \delta_{ij}. Only diagonal terms survive:

This is \rho_A's spectral decomposition — written with the Schmidt basis \{|u_i\rangle\} as eigenvectors and the squared Schmidt coefficients \lambda_i^2 as eigenvalues. The Schmidt decomposition is simultaneously the canonical form of the state and the diagonalisation of \rho_A. The eigenvalues of \rho_A are exactly \lambda_1^2, \lambda_2^2, \ldots — the squares of the Schmidt coefficients.

Why this matters operationally: computing the Schmidt decomposition doesn't require actually running SVD every time. If you want the Schmidt coefficients of a known state, you can instead compute \rho_A (take the partial trace over B), diagonalise it to get eigenvalues p_i, and then \lambda_i = \sqrt{p_i}. The two routes — SVD on the amplitude matrix, or diagonalisation of the reduced density matrix — give the same \lambda_i. Whichever is easier for a given problem, use it.

A non-obvious corollary: \rho_A and \rho_B share the same non-zero eigenvalues. From the Schmidt form, \rho_B = \sum_i \lambda_i^2 |v_i\rangle\langle v_i| — same \lambda_i^2 as eigenvalues, just in a different eigenbasis. So tracing out either side gives the same spectrum. This is the Schmidt decomposition's most beautiful consequence: the "entanglement content" is symmetric between the two parties, even though the bases \{|u_i\rangle\} and \{|v_i\rangle\} might be entirely different.

Worked example 1 — Schmidt form of a Bell state

Worked example 2 — a state that requires real computation

Common confusions

• "The Schmidt basis is unique." Only up to degeneracies. If all Schmidt coefficients are distinct, \{|u_i\rangle\} and \{|v_i\rangle\} are uniquely determined (up to overall phases on matched pairs). If two or more coefficients are equal — as in a Bell state where \lambda_1 = \lambda_2 = \tfrac{1}{\sqrt 2} — the basis vectors within the degenerate subspace can be rotated freely. The Schmidt coefficients are always unique; the Schmidt basis is unique only when no coefficient degeneracy exists.

• "Schmidt decomposition generalises to three or more parties." It does not — not as a single sum of matched product terms. The Schmidt decomposition is fundamentally a bipartite statement. For three-qubit states (chapter 39), the analogous "sum of matched product terms" form sometimes exists (states in the GHZ class) and sometimes doesn't (states in the W class). There is no multipartite generalisation that works for every pure state. This is one reason the multipartite entanglement zoo is a zoo.

• "Schmidt rank captures all entanglement." It captures whether the state is entangled or not (rank 1 vs rank > 1), and how many distinct "entangled directions" the state has (rank value). But it is a discrete measure — a rank-2 state can be nearly product (one coefficient tiny) or maximally entangled (coefficients equal). Rank alone does not distinguish. The continuous measure is the entanglement entropy S(\rho_A) = -\sum \lambda_i^2 \log_2 \lambda_i^2, which captures both how many terms appear and how evenly.

• "If I know \rho_A, I know the state." No. Many different pure states |\psi\rangle_{AB} can give the same \rho_A — they differ in the Schmidt basis on B. The pair (\rho_A, \text{Schmidt basis on B}) together determines |\psi\rangle_{AB}; \rho_A alone does not. This is the content of the purification theorem — any mixed state \rho_A on a system can be realised as the reduced state of many different pure states on a larger system, each differing in how the ancilla is structured.

• "Equal Schmidt coefficients ⇒ maximally entangled." Only for a given Schmidt rank. A state with r = 2 and \lambda_i = 1/\sqrt 2 has entropy 1 bit. A state with r = 4 and \lambda_i = 1/\sqrt 4 = 1/2 has entropy 2 bits. "Maximally entangled" always needs a dimensional qualifier: maximally entangled for the given rank. For two qubits the rank is capped at 2, so 1 bit is the ceiling — which Bell states achieve.

• "Schmidt coefficients must be positive." They must be non-negative real numbers, and by convention they are listed in non-increasing order \lambda_1 \geq \lambda_2 \geq \cdots \geq 0. Any complex phase in the original amplitudes is absorbed into the basis vectors |u_i\rangle and |v_i\rangle, not into the \lambda_i.

Going deeper

Why the SVD always exists

The singular-value decomposition C = U \Sigma V^\dagger holds for every complex matrix C, rectangular or square, of any size. The standard proof constructs \Sigma's diagonal entries as the square roots of eigenvalues of the Hermitian matrix C^\dagger C, which is always non-negative definite; hence its eigenvalues are non-negative and their square roots are real. The eigenvectors of C^\dagger C form the columns of V; then U is constructed to make U \Sigma V^\dagger = C work out.

For infinite-dimensional Hilbert spaces (continuous-variable quantum systems — photonic modes, harmonic oscillators), a Schmidt-like decomposition still exists for "reasonable" states, using the singular-value decomposition of compact operators from functional analysis. The sum over i becomes potentially countably infinite, but the \lambda_i still converge to zero fast enough that \sum \lambda_i^2 = 1 holds. The basic theorem survives.

Schmidt rank as an LOCC invariant

Local operations and classical communication (LOCC) — the protocol framework from chapter 39 — cannot increase the Schmidt rank of a bipartite pure state. If Alice and Bob share a state of Schmidt rank r, no sequence of local gates, measurements, and classical communication can produce a state of Schmidt rank r+1 or higher on their qubits. Schmidt rank is LOCC-monotone downward: it can only decrease or stay the same.

This has an operational consequence. A state with Schmidt rank r can simulate any state of rank \leq r via LOCC, but not vice versa. In particular, r = 2 states (Bell states and their cousins) are the LOCC-minimal resource that allows any non-trivial entanglement-assisted protocol; r = 1 (product states) cannot do anything that requires entanglement.

For asymptotic conversions — when Alice and Bob share many copies of one state and want to convert them to many copies of another — the right invariant is not the Schmidt rank but the entanglement entropy. Under LOCC, rate-optimal conversions of pure states are governed by ratios of entropies: n copies of state |\psi\rangle with entropy S_\psi can be converted by LOCC into approximately n S_\psi / S_\phi copies of state |\phi\rangle with entropy S_\phi, in the large-n limit (Bennett, Bernstein, Popescu, Schumacher, 1996). Entropy is the asymptotic currency of bipartite entanglement, and Schmidt decomposition gives you that currency directly.

The purification theorem

Schmidt decomposition enables an elegant structural result. Let \rho_A be any mixed state on a system A — say, \rho_A = \sum_i p_i |a_i\rangle\langle a_i| (spectral decomposition). The purification theorem says: there exists a larger system AB and a pure state |\psi\rangle_{AB} such that \text{tr}_B(|\psi\rangle\langle\psi|) = \rho_A.

The construction is immediate from Schmidt. Set |\psi\rangle_{AB} = \sum_i \sqrt{p_i}\,|a_i\rangle_A \otimes |b_i\rangle_B, where \{|b_i\rangle\} is any orthonormal basis of an ancilla B of dimension at least equal to the rank of \rho_A. Then by direct calculation, tracing out B gives back \rho_A. The Schmidt coefficients of this pure state are \sqrt{p_i} — the "square roots of the probabilities."

Every mixed state "comes from" a pure state on a larger system. In quantum information, this means you can always (mentally, or in theory) promote any mixed state to a pure one at the cost of adding fictitious degrees of freedom — an ancilla, an environment, a reference system. Many proofs use this trick to reduce statements about mixed states to statements about pure states.

Entanglement spectrum in many-body physics

In quantum many-body physics, bipartition a large system into regions A and B — say, the left half and right half of a one-dimensional chain — and compute \rho_A. Its spectrum (eigenvalues) is called the entanglement spectrum of the state across the cut. The entanglement entropy is one number derived from the spectrum; the full spectrum contains much more information.

Li and Haldane (2008) showed that the entanglement spectrum of the fractional quantum Hall state reveals the edge-mode structure of the phase — a ground-state calculation that replaces the traditional need for boundary analysis. This turned the Schmidt spectrum from an entanglement-measurement tool into a diagnostic for topological order. The area-law theorems (Hastings, 2007) for gapped phases establish that the entanglement entropy across a region's boundary scales with the boundary's area, not the region's volume — a quantum analogue of black-hole entropy. Schmidt decomposition sits at the heart of these results.

Matrix-product states and iterated Schmidt

The matrix-product state (MPS) ansatz for one-dimensional quantum systems applies Schmidt decomposition site by site. Take a chain of N qubits in a pure state |\psi\rangle. Bipartition after the first qubit: Schmidt-decompose, keep only the largest \chi Schmidt coefficients (truncation). Bipartition after the second qubit: Schmidt-decompose what's left, keep only \chi coefficients. Continue along the chain.

The result is a compressed representation of |\psi\rangle as a product of \chi \times \chi matrices (with physical indices for the local states). The compression is lossless if the state has small enough entanglement across every cut — which is true for ground states of gapped one-dimensional Hamiltonians, thanks to the area law. Density-matrix renormalization group (DMRG) and tensor-network methods are all, at their core, applications of iterated Schmidt decomposition.

The multipartite gap

Schmidt decomposition works because the SVD works — and the SVD is fundamentally a two-index theorem. The amplitude "matrix" has two indices (one per party). Three parties yield an amplitude "tensor" c_{ijk} with three indices, and no direct analogue of the SVD exists for general tensors. There is no canonical way to diagonalise a three-index tensor into a sum of products of three orthonormal vectors, with real non-negative weights, that works for every tensor.

This is why chapter 39's multipartite classification isn't a Schmidt story. The GHZ and W classes are distinguished by polynomial invariants (the 3-tangle and friends), not by singular values. Partial generalisations exist — the HOSVD (higher-order SVD), the tensor-train decomposition, the PARAFAC decomposition — each captures some of the Schmidt intuition in higher dimensions, but none exactly generalises the clean bipartite story. Schmidt decomposition is not just a tool; it is tied to the fundamental two-party-ness of bipartite entanglement.

Where this leads next

• Entanglement, defined — the definition that Schmidt rank \geq 2 is equivalent to.
• The partial trace — where reduced density matrices come from, and how their spectra connect to Schmidt coefficients.
• Density matrices, preview — the framework in which Schmidt's "eigenvalues of \rho_A" lives.
• Purification theorem — the construction that turns any mixed state into a pure state on a larger system, using Schmidt.
• GHZ and W states — the three-qubit story where bipartite Schmidt stops being enough.
• Entanglement entropy — the continuous measure built from the Schmidt coefficients.

References

1. Wikipedia, Schmidt decomposition — the theorem, the proof sketch, and pointers to applications.
2. Nielsen and Chuang, Quantum Computation and Quantum Information (2010), §2.5 on Schmidt decomposition and purifications — Cambridge University Press.
3. John Preskill, Lecture Notes on Quantum Computation, Ch. 4 on bipartite entanglement — theory.caltech.edu/~preskill/ph229.
4. Wikipedia, Singular value decomposition — the linear-algebra backbone of Schmidt decomposition.
5. H. Li and F. D. M. Haldane, Entanglement Spectrum as a Generalization of Entanglement Entropy (2008) — arXiv:0805.0332. The entanglement spectrum as a phase diagnostic.
6. John Watrous, The Theory of Quantum Information (2018), Ch. 2 on bipartite pure states — cs.uwaterloo.ca/~watrous/TQI.