Schmidt decomposition

Remarks

Fix orthonormal bases and write $\psi = \sum_{j,k} c_{jk} \, u_j \otimes v_k$. The singular value decomposition $C = U \Sigma V^*$ of the $n \times m$ matrix $C = (c_{jk})$ gives the result. The $\lambda_i$ are the nonzero singular values of $C$, the $e_i$ come from the columns of $U$, and the $f_i$ from the columns of $\overline{V}$. Uniqueness of $r$ and of the $\lambda_i$ follows because they are the rank and the singular values of $C$.

The $\lambda_i$ are the Schmidt coefficients and $r$ is the Schmidt rank. If $\psi$ is a unit vector, then $\sum_i \lambda_i^2 = 1$. The partial traces of $|\psi\rangle\langle\psi|$ are $\rho_A = \sum_i \lambda_i^2 |e_i\rangle\langle e_i|$ and $\rho_B = \sum_i \lambda_i^2 |f_i\rangle\langle f_i|$. So the two reduced states have the same nonzero eigenvalues. A unit vector $\psi$ is entangled if and only if its Schmidt rank is at least $2$.

Example: $\psi = \tfrac12(|00\rangle + |01\rangle + |10\rangle + |11\rangle) = |{+}\rangle \otimes |{+}\rangle$ with $|{+}\rangle = (|0\rangle + |1\rangle)/\sqrt{2}$ has Schmidt rank $1$, although it has four terms in the standard basis.

The decomposition is special to two factors. For three or more factors, a vector in general has no expansion $\sum_i \lambda_i \, e_i \otimes f_i \otimes g_i$ with orthonormal families in each factor.

Going deeper

For a student

The Schmidt decomposition is the singular value decomposition in the language of tensor products. It shows at once that the two reduced density matrices of a pure state share their nonzero spectrum, so they have the same von Neumann entropy. That common entropy is the standard measure of entanglement of a pure bipartite state. The decomposition is also the main step in the proof that purifications exist and are unique up to a unitary on the purifying system.