Schmidt Decomposition

The Schmidt decomposition is the canonical normal form of a bipartite pure state. It replaces a general double sum of product-basis amplitudes by a single sum of orthonormal, paired subsystem modes:

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

The nonnegative numbers $s_r$ are the Schmidt coefficients, and $R$ is the Schmidt rank. From these data one can immediately decide whether the state is product or entangled, read off both reduced-state spectra, and identify the subsystem degrees of freedom that are correlated with one another.

Helpful background. Product bases organize the amplitudes into a matrix; partial trace explains the reduced states used below.

Schmidt decomposition for a bipartite vector

Let $\mathcal H_A$ and $\mathcal H_B$ be finite-dimensional Hilbert spaces, and let

$$\lvert\Psi\rangle \in \mathcal H_A\otimes\mathcal H_B$$

be normalized. There are orthonormal sets

$$\left\lbrace \lvert u_r\rangle_A \right\rbrace_{r=1}^{R}, \qquad \left\lbrace \lvert v_r\rangle_B \right\rbrace_{r=1}^{R},$$

and positive real numbers $s_1,\ldots,s_R$ such that

$$\lvert\Psi\rangle = \sum_{r=1}^{R} s_r\, \lvert u_r\rangle_A\otimes\lvert v_r\rangle_B.$$

Normalization gives

$$\sum_{r=1}^{R}s_r^2=1,$$

and the number of terms obeys

$$1\le R\le \min\!\left\lbrace \dim\mathcal H_A,\dim\mathcal H_B \right\rbrace.$$

It is conventional to order the coefficients as

$$s_1\ge s_2\ge\cdots\ge s_R>0.$$

Only nonzero coefficients are counted. One may instead extend the sum to the smaller subsystem dimension by appending zeros, but then the number of written terms is no longer the Schmidt rank.

The theorem requires two structural choices:

• a pure state vector, rather than a general mixed density operator;
• a specified bipartition $A\mathbin{:}B$.

A multipartite state can be Schmidt-decomposed across a chosen cut, such as $A\mathbin{:}BC$, but there is generally no single three-party analogue with all three families of vectors orthonormal.

From a coefficient matrix to paired modes

Choose arbitrary orthonormal product bases and expand

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

The amplitudes form a $d_A\times d_B$ matrix $C$. State normalization is the Frobenius-norm condition

$$\operatorname{Tr}(C C^\dagger) = \sum_{i,j}\lvert C_{ij}\rvert^2 =1.$$

Write a singular value decomposition in the form

$$C=U S W^{\mathsf T},$$

where $U$ and $W$ are unitary and $S$ is rectangular diagonal with nonnegative entries. This is the usual SVD $C=U S V^\dagger$ with $W=V^*$. Defining

$$\begin{aligned} \lvert u_r\rangle_A &= \sum_i U_{ir}\lvert i\rangle_A, \\ \lvert v_r\rangle_B &= \sum_j W_{jr}\lvert j\rangle_B, \end{aligned}$$

turns the matrix factorization into the Schmidt decomposition. The nonzero diagonal entries of $S$ are precisely the $s_r$.

This gives a useful dictionary:

• the singular values of $C$ are the Schmidt coefficients;
• $\operatorname{rank}C$ is the Schmidt rank;
• the paired singular-vector data from $U$ and $W$ give the Schmidt vectors.

The detailed linear-algebra derivation is at Schmidt Decomposition as Linear Algebra; the underlying matrix theorem is reviewed in Singular Value Decomposition.

A reduced-state route

Often one does not need a full SVD. In the chosen product bases, the reduced density matrices have coefficient matrices

$$\rho_A=C C^\dagger, \qquad \rho_B=C^{\mathsf T}C^*.$$

Diagonalize either one. If

$$\rho_A\lvert u_r\rangle_A = \lambda_r\lvert u_r\rangle_A, \qquad \lambda_r>0,$$

then

$$s_r=\sqrt{\lambda_r}.$$

The matching vector on $B$ can be recovered directly from the state:

$$\lvert v_r\rangle_B = \frac{1}{s_r} \left( {}_A\langle u_r\rvert\otimes I_B \right) \lvert\Psi\rangle.$$

These partner vectors are normalized and mutually orthogonal. Inserting them back into the state reconstructs

$$\lvert\Psi\rangle = \sum_r s_r\, \lvert u_r\rangle_A\lvert v_r\rangle_B.$$

This route is especially efficient when one reduced state is smaller or easier to diagonalize than the original coefficient matrix. Eigenvectors with zero eigenvalue lie outside the support of the state and contribute no Schmidt term.

Schmidt rank

The Schmidt rank has several equivalent descriptions:

$$R = \operatorname{rank}C = \operatorname{rank}\rho_A = \operatorname{rank}\rho_B.$$

It is also the smallest number of product vectors needed to express $\lvert\Psi\rangle$ as a sum. A representation with many product terms need not be minimal; the Schmidt form finds the irreducible count after local basis changes.

For two qubits, $C$ is a $2\times2$ matrix, so

$$R= \begin{cases} 1, & \det C=0,\\ 2, & \det C\ne0. \end{cases}$$

In larger dimensions, the relevant test is the matrix rank rather than a single determinant. The Schmidt Rank page develops rank bounds, local-operation behavior, and the distinct mixed-state notion called Schmidt number.

Entanglement criterion

For a bipartite pure state,

$$R=1 \quad\Longleftrightarrow\quad \lvert\Psi\rangle \text{ is a product state},$$

whereas

$$\begin{aligned} R>1 &\Longleftrightarrow \lvert\Psi\rangle \text{ is entangled} \\ &\hphantom{\Longleftrightarrow} \text{ across }A\mathbin{:}B. \end{aligned}$$

The forward product implication is immediate: a one-term Schmidt decomposition is a tensor product. Conversely, if

$$\lvert\Psi\rangle = \lvert a\rangle_A\lvert b\rangle_B,$$

then its coefficient matrix is an outer product and has rank one.

Rank is a sharp yes-or-no criterion for pure-state entanglement, but it is not a complete measure of its amount. For example, the coefficient lists

$$\left( \sqrt{0.999},\sqrt{0.001} \right) \quad\text{and}\quad \left( \frac{1}{\sqrt2},\frac{1}{\sqrt2} \right)$$

both have rank two, although the second state has much more evenly distributed Schmidt weight. Quantitative measures use the full set $\{s_r^2\}$; see Entanglement Entropy.

Reduced density matrices

Starting from Schmidt form,

$$\lvert\Psi\rangle = \sum_r s_r\, \lvert u_r\rangle_A\lvert v_r\rangle_B,$$

the joint projector is

$$\lvert\Psi\rangle\langle\Psi\rvert = \sum_{r,t} s_r s_t\, \lvert u_r\rangle\langle u_t\rvert_A \otimes \lvert v_r\rangle\langle v_t\rvert_B.$$

Taking the partial trace over $B$ contracts the second factor:

$$\begin{aligned} \rho_A &= \sum_{r,t} s_r s_t\, \lvert u_r\rangle\langle u_t\rvert_A \langle v_t\vert v_r\rangle_B \\ &= \sum_r s_r^2\, \lvert u_r\rangle\langle u_r\rvert_A. \end{aligned}$$

Similarly,

$$\rho_B = \sum_r s_r^2\, \lvert v_r\rangle\langle v_r\rvert_B.$$

The two reduced states therefore have the same nonzero eigenvalues, including multiplicities:

$$\operatorname{spec}_{\ne0}(\rho_A) = \operatorname{spec}_{\ne0}(\rho_B) = \left\lbrace s_1^2,\ldots,s_R^2\right\rbrace.$$

They can have different numbers of zero eigenvalues when $d_A\ne d_B$. Their supports are exactly the spans of their Schmidt vectors:

$$\begin{aligned} \operatorname{supp}\rho_A &= \operatorname{span} \left\lbrace\lvert u_r\rangle_A\right\rbrace, \\ \operatorname{supp}\rho_B &= \operatorname{span} \left\lbrace\lvert v_r\rangle_B\right\rbrace. \end{aligned}$$

This yields another pure-state entanglement test:

$$\begin{aligned} \operatorname{Tr}(\rho_A^2)=1 &\quad\Longleftrightarrow\quad R=1, \\ \operatorname{Tr}(\rho_A^2)<1 &\quad\Longleftrightarrow\quad R>1. \end{aligned}$$

For fixed Schmidt rank $R$,

$$\frac{1}{R} \le \operatorname{Tr}(\rho_A^2) \le1,$$

with the lower value attained when all nonzero Schmidt coefficients are equal. The upper value occurs only for $R=1$; for every state with fixed $R>1$, the upper inequality is strict, although the purity can approach one as all but one coefficient approach zero. Reduced density operators explain why these matrices contain all local measurement predictions.

Maximal entanglement and unequal dimensions

Let

$$d=\min(d_A,d_B).$$

A pure state is maximally entangled across the full smaller subsystem when

$$R=d, \qquad s_r=\frac{1}{\sqrt d} \quad \text{for all }r.$$

The smaller reduced state is then maximally mixed:

$$\rho_{\mathrm{small}}=\frac{I_d}{d}.$$

If $d_A<d_B$, the larger reduced state is not $I_{d_B}/d_B$. Instead it is maximally mixed only on the $d_A$-dimensional Schmidt support:

$$\rho_B = \frac{1}{d_A}P_{\operatorname{supp}\rho_B}.$$

This distinction matters whenever the subsystem dimensions differ.

Uniqueness and basis freedom

The ordered list of Schmidt coefficients is fixed by the state and the chosen bipartition. The Schmidt vectors require more care:

• If a nonzero coefficient is nondegenerate, its pair of Schmidt vectors is unique up to opposite phase choices.
• Terms may be permuted together without changing the state.
• If several coefficients are equal, the corresponding Schmidt vectors can be unitarily mixed within that degenerate subspace.
• Basis vectors in the zero-eigenvalue complements of the reduced states are arbitrary and are not part of the nonzero Schmidt sum.

For a nondegenerate term, the phase freedom is

$$\begin{aligned} \lvert u_r\rangle_A &\longmapsto e^{i\theta_r}\lvert u_r\rangle_A, \\ \lvert v_r\rangle_B &\longmapsto e^{-i\theta_r}\lvert v_r\rangle_B. \end{aligned}$$

If a block has a repeated coefficient, a unitary rotation $M$ of the $A$-side vectors must be accompanied by the complex-conjugate rotation of the paired $B$-side vectors. The coefficient list stays fixed even though the displayed basis changes.

Local unitaries also preserve the Schmidt coefficients. If

$$\lvert\Psi'\rangle = (U_A\otimes U_B)\lvert\Psi\rangle,$$

then a Schmidt form for the new state is obtained by replacing each pair with

$$\begin{aligned} \lvert u_r\rangle_A &\longmapsto U_A\lvert u_r\rangle_A, \\ \lvert v_r\rangle_B &\longmapsto U_B\lvert v_r\rangle_B. \end{aligned}$$

Thus two bipartite pure states in the same finite-dimensional subsystem spaces are related by local unitaries if and only if they have the same Schmidt coefficients, including multiplicities.

Examples

Product state

Consider

$$\lvert\Psi\rangle = \lvert0\rangle_A \otimes \frac{\lvert0\rangle_B+\lvert1\rangle_B}{\sqrt2}.$$

It already has one Schmidt term:

$$s_1=1, \qquad R=1.$$

The reduced states are pure projectors, so the state is not entangled.

Bell state

The Bell state

$$\lvert\Phi^+\rangle = \frac{\lvert00\rangle+\lvert11\rangle}{\sqrt2}$$

is already in Schmidt form with

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

Both reduced states are $I_2/2$. Equal coefficients make the Schmidt bases nonunique: for example,

$$\lvert\Phi^+\rangle = \frac{ \lvert+\rangle_A\lvert+\rangle_B + \lvert-\rangle_A\lvert-\rangle_B }{\sqrt2}.$$

A state diagonal in different local bases

Let

$$\lvert\Psi\rangle = \frac{ \lvert00\rangle+\lvert01\rangle +\lvert10\rangle-\lvert11\rangle }{2}.$$

In the computational product basis this is a four-term expansion. Grouping the $B$-side vectors gives

$$\lvert\Psi\rangle = \frac{ \lvert0\rangle_A\lvert+\rangle_B + \lvert1\rangle_A\lvert-\rangle_B }{\sqrt2}.$$

Because $\{\lvert0\rangle,\lvert1\rangle\}$ and $\{\lvert+\rangle,\lvert-\rangle\}$ are orthonormal sets, this is Schmidt form. The state is maximally entangled even though the same basis is not used on both sides.

A generic two-qubit coefficient matrix

Consider

$$\lvert\Psi\rangle = \frac{ \lvert00\rangle+i\lvert01\rangle+\lvert11\rangle }{\sqrt3}.$$

Its coefficient matrix is

$$C= \frac{1}{\sqrt3} \begin{pmatrix} 1&i\\ 0&1 \end{pmatrix}.$$

The reduced matrices are

$$\rho_A = C C^\dagger = \frac{1}{3} \begin{pmatrix} 2&i\\ -i&1 \end{pmatrix},$$

and

$$\rho_B = C^{\mathsf T}C^* = \frac{1}{3} \begin{pmatrix} 1&-i\\ i&2 \end{pmatrix}.$$

Their eigenvalues are

$$\lambda_\pm = \frac{3\pm\sqrt5}{6},$$

so the Schmidt coefficients are

$$s_\pm = \sqrt{\frac{3\pm\sqrt5}{6}}.$$

Both are nonzero, hence $R=2$ and the state is entangled. The unequal coefficients show that it is not maximally entangled. Choose eigenvectors of $\rho_A$ and use the partner-vector formula to get the corresponding vectors on $B$. Diagonalizing both matrices independently leaves arbitrary phases that must be matched before reconstructing the state.

Normalizable infinite-dimensional states

For a normalized vector in a tensor product of separable Hilbert spaces, the reduced states are positive trace-class operators. Their nonzero spectrum is countable, and a Schmidt expansion exists with $\sum_r s_r^2=1$, possibly infinitely many terms. The partial sums converge in the Hilbert-space norm. For infinite rank, replace the finite span of Schmidt vectors by its closure when describing the support.

An ideal EPR state written with position eigenkets is not a normalized Hilbert-space vector. Its formal integral is not an ordinary Schmidt sum; the EPR state explains the distinction from finite squeezing.

Keeping only the largest Schmidt terms

Suppose $s_1\ge s_2\ge\cdots$ and keep the first $k$ terms. Write

$$|\Psi_k\rangle=\sum_{r=1}^k s_r|u_rv_r\rangle, \qquad \delta_k=\sum_{r>k}s_r^2.$$

Orthonormality gives $\|\Psi-\Psi_k\|^2=\delta_k$: the discarded squared coefficients measure the error, whereas merely counting discarded terms does not. This is the optimal unnormalized rank-at-most-$k$ approximation in Hilbert norm, by the truncated-SVD theorem.

After normalization, $|\widehat\Psi_k\rangle=|\Psi_k\rangle/\sqrt{1-\delta_k}$ has squared overlap $|\langle\Psi|\widehat\Psi_k\rangle|^2=1-\delta_k$. For probabilities $(0.8,0.19,0.01)$, keeping two terms discards weight $0.01$ and gives squared overlap $0.99$, although the exact Schmidt rank remains three. This distinction is central to tensor-network approximations.

Common pitfalls

Amplitudes are not automatically Schmidt coefficients. The two local vector families must be orthonormal, and phases belong in vectors so the coefficients are positive. A general four-term product-basis expansion need not have Schmidt rank four.

Independent eigenvectors need phase matching. Recover the second vector from the first with the partner-vector formula. This avoids arbitrary complex phases from two separate eigensolvers.

Mixed marginals alone do not prove entanglement. The reduced-spectrum criterion here assumes a pure joint state. Mixed joint states need different tests, such as the PPT criterion.

References

• R. A. Horn and C. R. Johnson, Matrix Analysis, 2nd ed., Cambridge University Press (2012), singular values and low-rank approximation.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), Sec. 2.5.
• 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), Ch. 1.

Exercises

1. Let

$$\lvert\Psi\rangle = \sqrt p\, \lvert0\rangle_A\lvert+\rangle_B + e^{i\phi}\sqrt{1-p}\, \lvert1\rangle_A\lvert-\rangle_B.$$

Here $0\le p\le1$.

Find its Schmidt coefficients and rank. For which values of $p$ is the state product, entangled, or maximally entangled?

Solution

The local families $\{\lvert0\rangle,\lvert1\rangle\}$ and $\{\lvert+\rangle,\lvert-\rangle\}$ are orthonormal. The phase $e^{i\phi}$ can be absorbed into either vector of the second pair. The nonzero Schmidt coefficients are therefore the nonzero members of

$$\left\lbrace \sqrt p,\sqrt{1-p} \right\rbrace.$$

For $p=0$ or $p=1$, only one coefficient is nonzero, so $R=1$ and the state is product. For $0<p<1$, both are nonzero, so $R=2$ and the state is entangled. It is maximally entangled when

$$p=1-p=\frac12.$$

2. Show directly that

$$\lvert\chi\rangle = \frac{ \lvert00\rangle+\lvert01\rangle +\lvert10\rangle+\lvert11\rangle }{2}$$

has Schmidt rank one, even though four computational-basis amplitudes are nonzero.

Solution

Factor the state:

$$\begin{aligned} \lvert\chi\rangle &= \frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2} \otimes \frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2} \\ &= \lvert+\rangle_A\lvert+\rangle_B. \end{aligned}$$

This is a one-term Schmidt decomposition with $s_1=1$. The number of nonzero amplitudes in one chosen product basis is not the Schmidt rank.

3. A normalized pure state has Schmidt probabilities $(0.8,0.19,0.01)$. Find the squared norm error of the unnormalized rank-two truncation and the squared overlap of its normalized version with the original state.

Solution

The discarded weight is $\delta_2=0.01$. Thus $\|\Psi-\Psi_2\|^2=0.01$, while $|\langle\Psi|\widehat\Psi_2\rangle|^2=0.99$. The normalized approximation has coefficients $\sqrt{0.8/0.99}$ and $\sqrt{0.19/0.99}$. Approximation does not change the original state’s exact rank, which is three.

4. Suppose $\rho_A\lvert u_r\rangle=s_r^2\lvert u_r\rangle$ with $s_r>0$. Prove that

$$\lvert v_r\rangle_B = \frac{1}{s_r} \left( {}_A\langle u_r\rvert\otimes I_B \right) \lvert\Psi\rangle$$

defines an orthonormal family.

Solution

For two such vectors,

$$\begin{aligned} \langle v_t\vert v_r\rangle &= \frac{1}{s_t s_r} \langle\Psi\rvert \left( \lvert u_t\rangle\langle u_r\rvert\otimes I_B \right) \lvert\Psi\rangle \\ &= \frac{1}{s_t s_r} \langle u_r\rvert\rho_A\lvert u_t\rangle \\ &= \frac{s_t^2}{s_t s_r}\delta_{rt} = \delta_{rt}. \end{aligned}$$

In the final step, the Kronecker delta forces $r=t$ whenever the expression is nonzero, so the ratio is one.

5. Starting from Schmidt form, show that $\rho_A$ and $\rho_B$ have the same purity.

Solution

The two reduced states are

$$\begin{aligned} \rho_A &= \sum_r s_r^2 \lvert u_r\rangle\langle u_r\rvert, \\ \rho_B &= \sum_r s_r^2 \lvert v_r\rangle\langle v_r\rvert. \end{aligned}$$

Orthonormality makes each expression a spectral decomposition. Therefore

$$\operatorname{Tr}(\rho_A^2) = \sum_r s_r^4 = \operatorname{Tr}(\rho_B^2).$$

The same reasoning shows that every spectral function of the nonzero eigenvalues agrees on the two sides.

6. Verify the degenerate-basis identity

$$\frac{\lvert00\rangle+\lvert11\rangle}{\sqrt2} = \frac{\lvert++\rangle+\lvert--\rangle}{\sqrt2}.$$

What does it show about Schmidt vectors?

Solution

Using

$$\lvert+\rangle = \frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2}, \qquad \lvert-\rangle = \frac{\lvert0\rangle-\lvert1\rangle}{\sqrt2},$$

one finds

$$\lvert++\rangle+\lvert--\rangle = \lvert00\rangle+\lvert11\rangle.$$

Both expansions are Schmidt decompositions with coefficients $1/\sqrt2,1/\sqrt2$. Because the coefficients are degenerate, the individual Schmidt vectors are not unique.

7. Let $\dim\mathcal H_A=2$, $\dim\mathcal H_B=3$, and

$$\lvert\Psi\rangle = \frac{ \lvert0\rangle_A\lvert a\rangle_B + \lvert1\rangle_A\lvert b\rangle_B }{\sqrt2},$$

where $\lvert a\rangle_B$ and $\lvert b\rangle_B$ are orthonormal. Find the spectra of both reduced states. Why is $\rho_B$ not $I_3/3$?

Solution

The Schmidt coefficients are both $1/\sqrt2$. Therefore

$$\operatorname{spec}(\rho_A) = \left\lbrace\frac12,\frac12\right\rbrace,$$

while

$$\operatorname{spec}(\rho_B) = \left\lbrace\frac12,\frac12,0\right\rbrace.$$

The state is maximally entangled across the two-dimensional smaller subsystem. The support of $\rho_B$ is only $\operatorname{span}\{\lvert a\rangle,\lvert b\rangle\}$, so $\rho_B$ is maximally mixed on that support, not on all of the three-dimensional Hilbert space.

8. Compare the Bell projector

$$\rho_{\mathrm{Bell}} = \lvert\Phi^+\rangle\langle\Phi^+\rvert$$

with the mixed state

$$\rho_{\mathrm{cc}} = \frac12\lvert00\rangle\langle00\rvert + \frac12\lvert11\rangle\langle11\rvert.$$

Both have reduced states $I_2/2$. Why does the Schmidt criterion apply to the first state but not the second?

Solution

$\rho_{\mathrm{Bell}}$ is a rank-one projector onto a bipartite state vector. That vector has Schmidt coefficients $1/\sqrt2,1/\sqrt2$, so it is entangled.

By contrast, $\rho_{\mathrm{cc}}$ is mixed and is explicitly a convex mixture of product-state projectors. It is separable. Its maximally mixed marginals do not imply entanglement, and it has no pure-state Schmidt decomposition. Equal reduced states can arise from coherent quantum correlations or from classical mixing; the joint state distinguishes them.