Schmidt Decomposition Overview

The Schmidt decomposition finds local bases in which a bipartite pure state is a sum of matched pairs. Its coefficients immediately reveal whether the state factors and how mixed its reduced states are. Schmidt decomposition gives the proof, constructive algorithms, and extensions beyond finite dimensions.

Helpful background. Bipartite systems introduces the coefficient matrix; reduced states explains the local density operators. The singular-value decomposition is useful for computing the coefficients.

Matched Orthonormal Pairs

Every normalized pure state of finite-dimensional factors has the form

$$|\Psi\rangle=\sum_{r=1}^{R}s_r|u_r\rangle_A|v_r\rangle_B, \qquad s_r>0,\qquad\sum_rs_r^2=1,$$

where the $|u_r\rangle$ form an orthonormal set in $A$ and the $|v_r\rangle$ form an orthonormal set in $B$. There are at most $R\leq\min(d_A,d_B)$ nonzero terms. Orthonormality is essential: an arbitrary sum of product vectors is not automatically a Schmidt decomposition.

The positive $s_r$ are the Schmidt coefficients. A state is product exactly when $R=1$; otherwise it is entangled. Tracing over either orthonormal set gives reduced-state eigenvalues $s_r^2$, with additional zeros if a factor has larger dimension.

A State Already in Schmidt Form

For

$$|\Psi\rangle=\frac{\sqrt3}{2}|0\rangle_A|2\rangle_B +\frac12|1\rangle_A|0\rangle_B,$$

the paired local kets are orthonormal, so the coefficients are $\sqrt3/2$ and $1/2$. The qubit reduction has eigenvalues $3/4,1/4$; the qutrit reduction has those same nonzero eigenvalues and one zero. The state is entangled even though the two coefficients differ.

In general, reshape the product-basis amplitudes into $C$ and compute its singular values. The full derivation explains the complex-conjugation convention needed to turn the right singular vectors into the $B$ kets. Degenerate Schmidt coefficients allow freedom in the paired bases; the coefficients remain well defined.

Quick Check

Is $(|0\rangle|0\rangle+|0\rangle|1\rangle)/\sqrt2$ a two-term Schmidt decomposition?

Solution

No: the two $A$ kets are identical, not orthogonal. Factorizing gives $|0\rangle_A|+\rangle_B$, which has one Schmidt coefficient equal to one. Counting displayed product terms is not the same as finding Schmidt rank.

References

• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, sec. 2.5.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, chs. 2 and 6. Author’s manuscript.