Entanglement is easy to point at — |\Phi^+\rangle = \tfrac{1}{\sqrt2}(|00\rangle + |11\rangle) is entangled, |0\rangle\otimes|+\rangle is not — but "point at it" is a poor substitute for a test. Given some tangle of amplitudes across two qubits, is the state secretly a product of two independent pieces, or are the two halves genuinely bound together? Staring at the amplitudes rarely tells you.

The Schmidt decomposition settles it with a single canonical form. It says that any pure state of a two-part (bipartite) system can be rewritten in a strikingly tidy way — a single sum in which each term pairs one basis state of part A with exactly one basis state of part B:

The sets \{|i\rangle_A\} and \{|i\rangle_B\} are each orthonormal, and the Schmidt coefficients \lambda_i are real and non-negative. Count how many of them are non-zero and you have measured, with a single integer, whether — and how much — the state is entangled.

It is just the SVD in disguise

Where does such a clean form come from? Write the state in the plain computational basis, with a matrix of amplitudes C_{jk}:

That grid of amplitudes is an ordinary matrix C, and every matrix has a singular value decomposition C = U\Sigma V^{\mathsf T}. Absorbing the rotation U into a new orthonormal basis for A, and V into a new one for B, collapses the double sum into a single one — and the surviving coefficients are exactly the singular values of C:

So the Schmidt decomposition is not a new theorem at all — it is the SVD of the amplitude matrix, wearing quantum notation. Everything you already know about singular values carries straight over: they are real, non-negative, and unique. The normalisation \sum_i \lambda_i^2 = 1 is just the statement that |\psi\rangle is a unit vector.

The Schmidt rank: a clean entanglement test

The number of non-zero Schmidt coefficients is the Schmidt rank of the state (the ordinary rank of the amplitude matrix C). It is the whole game:

• Rank 1 — a single term \lambda_1 |1\rangle_A|1\rangle_B with \lambda_1 = 1. The state is a plain |a\rangle_A \otimes |b\rangle_B: a product, i.e. separable, not entangled.
• Rank \ge 2 — two or more terms that cannot be factored apart. The state is entangled.

That is the payoff: to decide entanglement you do not need cleverness or intuition, only a rank. One Schmidt coefficient means "not entangled"; two or more means "entangled." A continuous, slippery property has been reduced to counting.

Worked example: the Bell state is maximally entangled

Take the Bell state |\Phi^+\rangle = \tfrac{1}{\sqrt2}(|00\rangle + |11\rangle). Its amplitude matrix is already diagonal:

A diagonal matrix is already in SVD form, so no rotation is needed — the state is already Schmidt-decomposed with |i\rangle_A = |i\rangle_B = |i\rangle. The two Schmidt coefficients are

Schmidt rank 2 — the state is entangled. And because the two coefficients are equal, it is not merely entangled but maximally entangled: no bipartite state of two qubits is more entangled than a Bell state. That is why Bell states are the workhorses of teleportation and quantum key distribution.

Worked example: a product state has rank 1

Now take |0\rangle \otimes |+\rangle = |0\rangle \otimes \tfrac{1}{\sqrt2}(|0\rangle + |1\rangle) = \tfrac{1}{\sqrt2}(|00\rangle + |01\rangle). Its amplitude matrix is

One row is entirely zero, so C has rank 1. Its single non-zero singular value is the length of the surviving row, \sqrt{\tfrac12 + \tfrac12} = 1, giving the Schmidt form

Schmidt rank 1 — a single term with \lambda_1 = 1, so the state is a plain product and not entangled, exactly as we expected from the way we built it. Notice too that the Schmidt basis for B came out as |+\rangle, not |0\rangle or |1\rangle — the bases are chosen by the state, a point we return to in the Watch out! box.

The coefficients are shared, and they live in the reduced state

The Schmidt form makes one more thing fall out for free. Square each coefficient and you get the eigenvalues of the reduced density matrix of either subsystem — the state you are left with after you throw away (partial-trace out) the other half:

Both reduced states carry the same numbers \lambda_i^2 as their eigenvalues — the Schmidt coefficients are shared between the two halves. For the Bell state, \rho_A = \tfrac12|0\rangle\langle 0| + \tfrac12|1\rangle\langle 1| = \tfrac12 I: a completely random single qubit. For the product state, \rho_A = |0\rangle\langle 0|, with eigenvalues 1, 0 — a perfectly definite qubit. This is the deep meaning of maximal entanglement: when all the \lambda_i are equal, each reduced state is maximally mixed — knowing everything about the whole tells you nothing about either part on its own. (The partial trace and mixed states make this precise.)

See it: the spectrum of Schmidt coefficients

Each state has a little bar chart of its Schmidt coefficients. Switch between the three states and read off the entanglement at a glance: one bar means a product (rank 1); two bars mean entangled (rank 2); and two equal bars mean maximally entangled. A partially entangled state sits in between — two bars of different heights.