Schmidt Decomposition as Linear Algebra

The Schmidt decomposition is the singular value decomposition of a bipartite coefficient matrix, translated back into tensor-product notation.

This page focuses on the finite-dimensional linear algebra. The physics-facing theorem, reduced states, entanglement criterion, and infinite-dimensional cautions live in Schmidt Decomposition.

Helpful background. tensor products for product bases and coefficient arrays; singular value decomposition for orthonormal singular vectors and nonnegative singular values.

Coefficient Matrix of a Bipartite Vector

Let $\mathcal H_A$ and $\mathcal H_B$ be finite-dimensional Hilbert spaces with orthonormal bases

$$\{\lvert i\rangle_A\}_{i=1}^{d_A}, \qquad \{\lvert j\rangle_B\}_{j=1}^{d_B}.$$

Every vector in the tensor product has an expansion

$$\lvert\Psi\rangle = \sum_{i=1}^{d_A} \sum_{j=1}^{d_B} C_{ij}\, \lvert i\rangle_A\otimes\lvert j\rangle_B.$$

The numbers $C_{ij}$ form a $d_A\times d_B$ matrix $C$. Normalization of the vector is the Frobenius-norm condition

$$1 = \langle\Psi\vert\Psi\rangle = \sum_{i,j}\lvert C_{ij}\rvert^2 = \operatorname{Tr}(CC^\dagger).$$

This coefficient matrix depends on the chosen product bases. Its singular values do not depend on local unitary changes of basis.

SVD Derivation

Take the singular value decomposition of $C$:

$$C=U\Sigma V^\dagger.$$

If the positive singular values are $s_1,\ldots,s_R$, then

$$C_{ij} = \sum_{r=1}^{R} U_{ir}s_r V_{jr}^*.$$

Define vectors in the two Hilbert spaces by

$$\lvert u_r\rangle_A = \sum_i U_{ir}\lvert i\rangle_A, \qquad \lvert v_r\rangle_B = \sum_j V_{jr}^*\lvert j\rangle_B.$$

Because the columns of $U$ and $V$ are orthonormal, the vectors $\{\lvert u_r\rangle_A\}$ and $\{\lvert v_r\rangle_B\}$ are orthonormal sets. Substituting the SVD into the product-basis expansion gives

$$\begin{aligned} \lvert\Psi\rangle &= \sum_{i,j} \sum_{r=1}^{R} U_{ir}s_r V_{jr}^* \lvert i\rangle_A\lvert j\rangle_B\\ &= \sum_{r=1}^{R} s_r \left(\sum_i U_{ir}\lvert i\rangle_A\right) \left(\sum_j V_{jr}^*\lvert j\rangle_B\right)\\ &= \sum_{r=1}^{R} s_r \lvert u_r\rangle_A\lvert v_r\rangle_B. \end{aligned}$$

This is the Schmidt decomposition. The Schmidt coefficients are the singular values of $C$.

Why a Complex Conjugate Appears

The factor $V_{jr}^*$ in $\lvert v_r\rangle_B$ comes from the $V^\dagger$ in the matrix SVD. It is a convention consequence of writing a bipartite vector with kets on both sides rather than with one side already dualized.

The important invariant statement is simple: local basis changes multiply $C$ by unitary matrices on the left and right, so the singular values are unchanged.

For example, an active local unitary

$$\lvert\Psi\rangle \mapsto (U_A\otimes U_B)\lvert\Psi\rangle$$

changes the coefficient matrix as

$$C\mapsto U_A C U_B^T.$$

Both $U_A$ and $U_B^T$ are unitary, so the singular values of $C$ are invariant.

Schmidt Rank and Matrix Rank

The Schmidt rank is the number of nonzero Schmidt coefficients:

$$\operatorname{SR}(\Psi) = \#\{r:s_r>0\}.$$

Since the $s_r$ are the positive singular values of $C$,

$$\operatorname{SR}(\Psi) = \operatorname{rank}C.$$

Thus a bipartite pure state is product exactly when its coefficient matrix has rank one. The physics-facing rank invariant is developed in Schmidt Rank.

Reduced States from the Matrix

The reduced state on subsystem $A$ has matrix elements

$$(\rho_A)_{ii'} = \sum_j C_{ij}C_{i'j}^*.$$

In matrix notation,

$$\rho_A=CC^\dagger.$$

With the coefficient convention used here, tracing out $A$ gives

$$(\rho_B)_{jj'}=\sum_i C_{ij}C_{ij'}^*,\qquad \rho_B=C^T C^*=(C^\dagger C)^T.$$

Here $C^*$ means entrywise conjugation. The transpose matters for matrix entries and eigenvectors, although $\rho_B$ and $C^\dagger C$ have the same eigenvalues.

Using $C=U\Sigma V^\dagger$ gives

$$CC^\dagger = U\Sigma\Sigma^\dagger U^\dagger,$$

so the nonzero eigenvalues of $\rho_A$ are

$$s_r^2.$$

The same nonzero eigenvalues occur for $\rho_B$. This is the linear-algebra reason the squared Schmidt coefficients are the probabilities used in Entanglement Entropy.

Product States as Rank-One Matrices

If

$$\lvert\Psi\rangle = \lvert\alpha\rangle_A\otimes\lvert\beta\rangle_B$$

with

$$\lvert\alpha\rangle_A=\sum_i a_i\lvert i\rangle_A, \qquad \lvert\beta\rangle_B=\sum_j b_j\lvert j\rangle_B,$$

then

$$C_{ij}=a_i b_j.$$

The coefficient matrix is an outer product:

$$C=ab^T.$$

It has rank one when both factors are nonzero. Conversely, if $C$ has rank one, it can be written as an outer product, and the bipartite vector factors. This is the linear-algebra core of the pure-state product criterion.

Example: Product State Hidden by the Basis

Consider

$$\lvert\Psi\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert01\rangle \bigr).$$

In the computational product basis,

$$C = \frac{1}{\sqrt2} \begin{pmatrix} 1 & 1\\ 0 & 0 \end{pmatrix}.$$

Then

$$CC^\dagger = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}.$$

The singular values are $1$ and $0$. Therefore the Schmidt rank is one. Indeed,

$$\lvert\Psi\rangle = \lvert0\rangle_A \otimes \frac{1}{\sqrt2} \bigl( \lvert0\rangle_B+\lvert1\rangle_B \bigr).$$

The original expansion had two product-basis terms, but the state was still a product vector.

Example: Bell State

For

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr),$$

the coefficient matrix is

$$C = \frac{1}{\sqrt2} \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}.$$

The singular values are

$$s_1=s_2=\frac{1}{\sqrt2}.$$

Therefore the Schmidt rank is two, and

$$\rho_A = CC^\dagger = \frac12 I.$$

The two equal singular values give one bit of bipartite entanglement entropy.

Degeneracy and Basis Freedom

The Schmidt coefficients are unique up to ordering, because singular values are unique. The Schmidt vectors need not be unique.

If a singular value is nondegenerate, the two corresponding Schmidt kets are fixed up to opposite phase choices:

$$\lvert u_r\rangle_A\mapsto e^{i\theta}\lvert u_r\rangle_A, \qquad \lvert v_r\rangle_B\mapsto e^{-i\theta}\lvert v_r\rangle_B.$$

If several singular values are equal, there is a larger unitary freedom inside the degenerate Schmidt subspace. The state is unchanged, but the displayed Schmidt basis is not unique.

Common Mistakes

• Treating the coefficient matrix as basis-independent. The matrix changes under local basis changes; its singular values do not.
• Forgetting the complex conjugate in the second Schmidt basis when translating $C=U\Sigma V^\dagger$ into kets.
• Counting product-basis terms instead of computing matrix rank or singular values.
• Confusing Schmidt coefficients $s_r$ with reduced-state eigenvalues $s_r^2$.
• Applying the bipartite coefficient-matrix derivation to multipartite states without qualification.
• Applying pure-state Schmidt language directly to mixed density operators.
• Treating degenerate Schmidt vectors as unique.

References

• G. H. Golub and C. F. Van Loan, Matrix Computations, 4th ed., Johns Hopkins University Press, 2013.
• R. A. Horn and C. R. Johnson, Matrix Analysis, 2nd ed., Cambridge University Press, 2013.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• E. Schmidt, “Zur Theorie der linearen und nichtlinearen Integralgleichungen. I. Teil,” Mathematische Annalen 63, 433-476, 1907.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018.

Exercises

1. Write the coefficient matrix for

$$\lvert\Psi\rangle = \frac{1}{\sqrt5} \bigl( 2\lvert00\rangle+\lvert11\rangle \bigr)$$

and find its Schmidt coefficients.

Solution

The coefficient matrix is

$$C = \frac{1}{\sqrt5} \begin{pmatrix} 2 & 0\\ 0 & 1 \end{pmatrix}.$$

It is already diagonal with nonnegative entries, so the Schmidt coefficients are

$$s_1=\frac{2}{\sqrt5}, \qquad s_2=\frac{1}{\sqrt5}.$$

2. Show from the coefficient matrix that

$$\frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert01\rangle \bigr)$$

is a product state.

Solution

The coefficient matrix is

$$C = \frac{1}{\sqrt2} \begin{pmatrix} 1 & 1\\ 0 & 0 \end{pmatrix}.$$

It has rank one, so the Schmidt rank is one. Explicitly,

$$\frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert01\rangle \bigr) = \lvert0\rangle_A \otimes \frac{1}{\sqrt2} \bigl( \lvert0\rangle_B+\lvert1\rangle_B \bigr).$$

3. For

$$C = \frac12 \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix},$$

compute $CC^\dagger$ and the Schmidt coefficients.

Solution

Compute

$$CC^\dagger = \frac14 \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix} = \frac12 I.$$

The eigenvalues of $CC^\dagger$ are $1/2$ and $1/2$, so the Schmidt coefficients are

$$s_1=s_2=\frac{1}{\sqrt2}.$$

4. Let $|\Psi\rangle=|0\rangle_A\otimes(|0\rangle_B+i|1\rangle_B)/\sqrt2$. Find $C$ and $\rho_B$, and check the sign of its upper-right entry.

Solution

The coefficient matrix and reduced state are

$$C=\frac1{\sqrt2}\begin{pmatrix}1&i\\0&0\end{pmatrix},\qquad \rho_B=C^T C^*=\frac12\begin{pmatrix}1&-i\\i&1\end{pmatrix}.$$

This agrees with $|\beta\rangle\langle\beta|$ for $|\beta\rangle=(1,i)^T/\sqrt2$. Using $C^\dagger C$ without the transpose would give the conjugate state. The Schmidt coefficients alone would not detect that mistake because both states have the same spectrum.