Schmidt Decomposition

This is the canonical treatment of the Schmidt decomposition, including constructive routes, uniqueness, reduced-state consequences, examples, and finite- versus infinite-dimensional scope. The prerequisite-oriented map is Schmidt Decomposition Overview.

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.

Statement

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 canonical 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. The Reduced States and Partial Trace: First Encounter pages develop the operational meaning and mechanics of these subsystem states.

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 finite-dimensional bipartite pure states 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

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

Its coefficient matrix is

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

The reduced matrices are

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

and

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

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. The eigenvectors of $\rho_A$ and $\rho_B$ supply the corresponding Schmidt bases.

Scope beyond finite dimensions

For normalizable vectors in separable infinite-dimensional Hilbert spaces, the reduced density operators are trace class and hence compact. A countable Schmidt expansion still exists, but the rank may be infinite and the nonzero coefficients may accumulate at zero. Formal expressions built from nonnormalizable position eigenstates require additional care and should not be treated as ordinary finite or countable Schmidt sums without checking the relevant operator domain and topology.

The canonical decomposition page discusses these qualifications further; continuous-variable entanglement belongs in its dedicated advanced treatment.

Practical workflow

For a finite-dimensional bipartite pure state:

1. Fix the bipartition and choose product bases.
2. Put the amplitudes into the coefficient matrix $C$.
3. Check normalization using $\operatorname{Tr}(C C^\dagger)=1$.
4. Find the nonzero singular values of $C$, or diagonalize the smaller of $C C^\dagger$ and $C^{\mathsf T}C^*$.
5. Take positive square roots of the nonzero eigenvalues to obtain $s_r$.
6. Use the eigenvectors and the partner-vector formula to obtain paired Schmidt vectors.
7. Check orthonormality and verify $\sum_r s_r^2=1$.
8. Read off rank, reduced-state spectra, and the product-versus-entangled conclusion.

When only the entanglement criterion is needed, computing $\operatorname{rank}C$ may be enough. When entropies or explicit subsystem modes are needed, the full spectrum and vectors matter.

Example: Partially Entangled Two-Qubit State

For

$$\lvert\Psi_\theta\rangle = \cos\theta\,\lvert00\rangle + \sin\theta\,\lvert11\rangle, \qquad 0\le\theta\le\frac{\pi}{2},$$

the Schmidt coefficients are

$$s_1=\cos\theta, \qquad s_2=\sin\theta.$$

The state is product when $\theta=0$ or $\theta=\pi/2$. It is entangled for $0<\theta<\pi/2$. It is maximally entangled at $\theta=\pi/4$.

The reduced state on $A$ is

$$\rho_A = \cos^2\theta\, \lvert0\rangle\langle0\rvert + \sin^2\theta\, \lvert1\rangle\langle1\rvert.$$

Example: Two Qutrits

The normalized two-qutrit state

$$\lvert\Psi\rangle = \frac{1}{\sqrt6} \bigl( 2\lvert00\rangle+\lvert11\rangle+\lvert22\rangle \bigr)$$

is in Schmidt form with coefficients

$$s_1=\frac{2}{\sqrt6}, \qquad s_2=\frac{1}{\sqrt6}, \qquad s_3=\frac{1}{\sqrt6}.$$

The Schmidt rank is $3$, so the state is entangled across the qutrit-qutrit split. It is not maximally entangled because the three coefficients are not equal.

Common mistakes

• Applying the pure-state Schmidt decomposition directly to a mixed density operator.
• Omitting the bipartition; Schmidt data are always relative to a chosen cut.
• Calling an arbitrary product-basis expansion a Schmidt decomposition even though the local vectors are not orthonormal.
• Reading Schmidt coefficients directly from amplitudes before accounting for phases and basis changes.
• Confusing $s_r$ with reduced-state eigenvalues; the eigenvalues are $s_r^2$.
• Counting appended zero coefficients as part of the Schmidt rank.
• Assuming that equal Schmidt coefficients determine unique Schmidt vectors.
• Treating rank alone as a quantitative measure of entanglement.
• Expecting the larger subsystem of an unequal-dimensional maximally entangled pair to be maximally mixed on its entire Hilbert space.

References

• E. Schmidt, “Zur Theorie der linearen und nichtlinearen Integralgleichungen. I. Teil”, Mathematische Annalen 63, 433–476 (1907).
• 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), Sec. 1.1.
• J. Preskill, Lecture Notes for Physics 229: Quantum Information and Computation, Chapter 2, Sec. 2.4.
• I. Bengtsson and K. Życzkowski, Geometry of Quantum States, 2nd ed., Cambridge University Press (2017), Ch. 15.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer (1995).
• R. A. Horn and C. R. Johnson, Matrix Analysis, 2nd ed., Cambridge University Press (2012).

Exercises

1. Let

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

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. For

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

compute the Schmidt coefficients from $\rho_A$ and verify the normalization condition.

Solution

The coefficient matrix is

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

The reduced state is

$$\rho_A = \frac{1}{3} \begin{pmatrix} 2&1\\ 1&1 \end{pmatrix}.$$

The characteristic equation is

$$\lambda^2-\lambda+\frac19=0,$$

so

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

Hence

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

The normalization check is

$$s_+^2+s_-^2 = \lambda_++\lambda_- =1.$$

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.

Additional exercises retained from the earlier canonical treatment

1. Find the Schmidt coefficients of

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

Solution

The state is already in Schmidt form. The coefficients are

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

Both are nonzero, so the state is entangled.

2. Show that

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

has Schmidt rank one.

Solution

Factor the state:

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

It is product, so the only nonzero Schmidt coefficient is $1$.

3. For

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

compute $\operatorname{Tr}(\rho_A^2)$.

Solution

The reduced state is $\rho_A=I/2$. Therefore

$$\operatorname{Tr}(\rho_A^2) = \operatorname{Tr}\left(\frac14 I\right) = \frac12.$$

The value is less than one, so the pure joint state is entangled.

4. The two-qubit state

$$\lvert\Psi\rangle = \frac12 \bigl( \lvert00\rangle+\lvert01\rangle +\lvert10\rangle-\lvert11\rangle \bigr)$$

has coefficient matrix

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

Find its 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 $1/2$. The Schmidt coefficients are their square roots:

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