The Schmidt Decomposition

The Schmidt decomposition is the single most useful structural fact about pure bipartite states. It says that any such state, no matter how it is written, has a canonical form with a single set of non-negative coefficients — and those coefficients answer essentially every question about its entanglement.

The theorem

Schmidt decomposition. For any pure state $|\psi\rangle \in \mathcal{H}_A \otimes \mathcal{H}_B$ there exist orthonormal sets $\{|u_k\rangle\} \subset \mathcal{H}_A$ and $\{|v_k\rangle\} \subset \mathcal{H}_B$ and non-negative reals $s_k$ such that
$|\psi\rangle = \sum_{k=1}^{r} s_k\,|u_k\rangle_A\,|v_k\rangle_B, \qquad \sum_k s_k^2 = 1.$

The numbers $s_k$ are the Schmidt coefficients, and the number $r$ of nonzero ones is the Schmidt rank. Crucially, the same coefficients pair an $A$ -basis vector with a $B$ -basis vector one-to-one — the double sum over independent indices collapses to a single sum.

Where it comes from: the SVD

Write the state in the computational basis as $|\psi\rangle = \sum_{i,j} M_{ij}\,|i\rangle_A |j\rangle_B$ , collecting the amplitudes into a matrix $M$ . The singular value decomposition factors $M = U\Sigma V^\dagger$ with $\Sigma$ diagonal and non-negative. Substituting and absorbing $U$ into the $A$ -basis and $V^*$ into the $B$ -basis turns the double sum into the Schmidt form, with the Schmidt coefficients equal to the singular values of $M$ . The Schmidt decomposition is the SVD in disguise.

The connection to reduced states

Tracing out either subsystem of the Schmidt form gives a diagonal density matrix:

\rho_A = \sum_k s_k^2\,|u_k\rangle\langle u_k|, \qquad \rho_B = \sum_k s_k^2\,|v_k\rangle\langle v_k|.

Both have the same eigenvalues $\{s_k^2\}$ . This immediately proves the symmetry $S(\rho_A) = S(\rho_B)$ used in the previous lesson, and identifies the entanglement entropy as

E = -\sum_k s_k^2 \log_2 s_k^2 = H(\{s_k^2\}),

the Shannon entropy of the squared Schmidt coefficients.

What the Schmidt rank tells you

Schmidt rank $r = 1$ $\iff$ the state is a product $|u_1\rangle|v_1\rangle$ , with $E = 0$ . This is the cleanest test for entanglement: a pure state is entangled iff $r \ge 2$ .
Maximal entanglement $\iff$ all $r = d$ coefficients are equal, $s_k = 1/\sqrt d$ , giving $E = \log_2 d$ .
The rank is invariant under local unitaries on either side — no local operation can change how entangled a pure state is.

A worked example

Take $|\psi\rangle = \tfrac{1}{\sqrt3}(|00\rangle + |01\rangle + |10\rangle)$ . Its coefficient matrix is

M = \frac{1}{\sqrt3}\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}.

Then $\rho_A = MM^\dagger = \tfrac13\begin{pmatrix} 2 & 1 \\ 1 & 1\end{pmatrix}$ has trace $1$ and determinant $\tfrac19$ , so its eigenvalues are $s_\pm^2 = \tfrac{3\pm\sqrt5}{6} \approx \{0.8727,\ 0.1273\}$ . The Schmidt rank is $2$ , so the state is entangled, and

E = -0.8727\log_2 0.8727 - 0.1273\log_2 0.1273 \approx 0.5500\text{ ebits}.

Try it

For $|\psi\rangle = \tfrac{1}{\sqrt3}(|00\rangle + |01\rangle + |10\rangle)$ , build the coefficient matrix, find the squared Schmidt coefficients as the eigenvalues of $\rho_A = MM^\dagger$ , and return the entanglement entropy $E$ in bits.

Run your code to see the quantum state.