Schmidt Rank

Schmidt rank is the number of nonzero Schmidt coefficients of a bipartite pure state. It is the simplest discrete classifier of pure-state entanglement:

$$\operatorname{SR}(\Psi)=1 \quad \Longleftrightarrow \quad \lvert\Psi\rangle \text{ is product},$$

while

$$\operatorname{SR}(\Psi)>1 \quad \Longleftrightarrow \quad \lvert\Psi\rangle \text{ is entangled}$$

for a finite-dimensional bipartite pure state.

Schmidt rank answers a yes-or-no and how-many-modes question. It does not say how evenly the Schmidt weight is distributed. For that, use the Schmidt coefficients themselves, the reduced-state spectrum, or Entanglement Entropy.

Definition

Let

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

be a normalized pure state with Schmidt decomposition

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

The Schmidt rank is

$$\operatorname{SR}(\Psi) = R.$$

Equivalently, it is the number of nonzero eigenvalues of either reduced state:

$$\operatorname{SR}(\Psi) = \operatorname{rank}\rho_A = \operatorname{rank}\rho_B,$$

where

$$\rho_A = \operatorname{Tr}_B \bigl( \lvert\Psi\rangle\langle\Psi\rvert \bigr), \qquad \rho_B = \operatorname{Tr}_A \bigl( \lvert\Psi\rangle\langle\Psi\rvert \bigr).$$

In a product basis,

$$\lvert\Psi\rangle = \sum_{ij} C_{ij} \lvert i\rangle_A\lvert j\rangle_B,$$

the Schmidt rank is also the ordinary matrix rank of the coefficient matrix $C$:

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

This is why Schmidt rank is often the fastest product-versus-entangled test for small finite-dimensional examples.

Product States Have Rank One

If

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

then the state is already in Schmidt form:

$$\lvert\Psi\rangle = 1\cdot \lvert\alpha\rangle_A\lvert\beta\rangle_B.$$

Thus

$$\operatorname{SR}(\Psi)=1.$$

In a product basis, the coefficient matrix factors as

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

so it has matrix rank one.

Conversely, if a normalized pure state has Schmidt rank one, its Schmidt decomposition has one term:

$$\lvert\Psi\rangle = \lvert u_1\rangle_A \lvert v_1\rangle_B.$$

Therefore the state is product. Rank one is exactly the pure-state product condition.

Entangled Pure States Have Rank Greater Than One

For finite-dimensional bipartite pure states, entangled means not product. Since product is equivalent to Schmidt rank one, entanglement is equivalent to Schmidt rank greater than one:

$$\lvert\Psi\rangle \text{ entangled} \quad \Longleftrightarrow \quad \operatorname{SR}(\Psi)>1.$$

For the Bell state

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

the Schmidt form has two nonzero coefficients:

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

Therefore

$$\operatorname{SR}(\Phi^+)=2.$$

For the partially entangled state

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

the Schmidt rank is

$$\operatorname{SR}(\Psi_\theta) = \begin{cases} 1, & \theta=0\text{ or }\theta=\pi/2,\\ 2, & 0<\theta<\pi/2. \end{cases}$$

This illustrates a limitation: Schmidt rank is discontinuous. An arbitrarily small but nonzero second Schmidt coefficient changes the rank from $1$ to $2$, even though the entanglement entropy can be arbitrarily small.

Maximal Schmidt Rank

If

$$d_A=\dim\mathcal H_A, \qquad d_B=\dim\mathcal H_B,$$

then

$$\operatorname{SR}(\Psi) \le \min(d_A,d_B).$$

A state has full Schmidt rank if equality holds:

$$\operatorname{SR}(\Psi) = \min(d_A,d_B).$$

Full Schmidt rank is not the same as maximal entanglement. For two qutrits,

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

has three nonzero Schmidt coefficients, so it has full Schmidt rank:

$$\operatorname{SR}(\Psi)=3.$$

But the coefficients

$$\frac{2}{\sqrt6}, \qquad \frac{1}{\sqrt6}, \qquad \frac{1}{\sqrt6}$$

are not equal. The state is not maximally entangled. A maximally entangled two-qutrit state would have

$$s_1=s_2=s_3=\frac1{\sqrt3}.$$

Schmidt rank counts how many Schmidt directions are occupied; entropy and other measures also care how the weight is distributed among them.

Local Unitary Invariance

Schmidt rank is invariant under local unitaries. If

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

then the reduced state on $A$ transforms as

$$\rho'_A = U_A\rho_A U_A^\dagger.$$

Unitary conjugation does not change matrix rank, so

$$\operatorname{rank}\rho'_A = \operatorname{rank}\rho_A.$$

Therefore

$$\operatorname{SR}(\Psi') = \operatorname{SR}(\Psi).$$

This is physically important. Local basis changes can rotate the Schmidt vectors, and local unitaries can change the product-basis coefficient matrix, but they cannot turn an entangled pure state into a product pure state.

Two-Qubit Determinant Test

For a two-qubit pure state

$$\lvert\Psi\rangle = a\lvert00\rangle + b\lvert01\rangle + c\lvert10\rangle + d\lvert11\rangle,$$

the coefficient matrix is

$$C = \begin{pmatrix} a & b\\ c & d \end{pmatrix}.$$

The state is product exactly when

$$\det C = ad-bc = 0.$$

If $ad-bc\ne0$, then

$$\operatorname{SR}(\Psi)=2,$$

and the pure state is entangled.

This determinant test is special to two qubits. In larger dimensions, use the rank of $C$, equivalently the nonzero eigenvalues of $\rho_A$ or $\rho_B$.

Minimal Product-Term Interpretation

For bipartite pure states, Schmidt rank is also the minimum number of product vectors needed to express the state as a sum:

$$\lvert\Psi\rangle = \sum_{r=1}^{R} \lvert a_r\rangle_A \lvert b_r\rangle_B.$$

The Schmidt decomposition shows that $R=\operatorname{SR}(\Psi)$ product terms are enough. Matrix rank shows that fewer cannot suffice.

This interpretation is useful but should be used with care. The clean equality between matrix rank, reduced-state rank, and minimal product-term number is a bipartite pure-state fact. Multipartite tensor rank is a much harder object, and mixed states require different definitions.

Limitations for Mixed States

Schmidt rank is defined for pure bipartite vectors. It is not the same as:

• the rank of a mixed density operator $\rho_{AB}$;
• the rank of a reduced density operator for a mixed joint state;
• mutual information;
• entanglement entropy for mixed states;
• a general separability criterion for density operators.

For example, the separable mixed state

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

has density-matrix rank $2$ and mixed one-qubit reductions, but it is not entangled. Asking for “the Schmidt rank of $\rho_{\mathrm{cc}}$” is not the right question.

There is a mixed-state extension called Schmidt number: roughly, it asks for the smallest maximum Schmidt rank needed among pure states in an ensemble decomposition of $\rho_{AB}$. That is a more advanced mixed-state entanglement concept. It should not be confused with the Schmidt rank of a pure state.

Common Mistakes

• Confusing Schmidt rank with the Hilbert-space dimension.
• Treating full Schmidt rank as the same thing as maximal entanglement.
• Calling the density-matrix rank of a mixed state its Schmidt rank.
• Forgetting that Schmidt rank depends on the chosen bipartite split.
• Assuming a small nonzero Schmidt coefficient is negligible for rank-based classification.
• Applying the pure-state rank-one criterion directly to mixed states.
• Confusing Schmidt coefficients $s_r$ with probabilities $s_r^2$.

Cross-Links

• Schmidt Decomposition
• Schmidt Decomposition Overview
• Entanglement Entropy
• Renyi Entropies
• Concurrence for Two Qubits
• Subsystem Entropy
• Local Unitary Equivalence
• Product States
• Entangled States
• Separable Mixed States
• Bell States
• Reduced Density Operators
• Multipartite Systems
• Schmidt Decomposition as Linear Algebra
• Spectral Decomposition

References

• E. Schmidt, “Zur Theorie der linearen und nichtlinearen Integralgleichungen. I. Teil,” Mathematische Annalen 63, 433-476, 1907.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• R. A. Horn and C. R. Johnson, Matrix Analysis, 2nd ed., Cambridge University Press, 2012.
• B. M. Terhal and P. Horodecki, “Schmidt Number for Density Matrices,” Physical Review A 61, 040301(R), 2000.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018.

Exercises

1. Find the Schmidt rank of

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

Solution

Factor the state:

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

It is product, so

$$\operatorname{SR}(\Psi)=1.$$

2. Use the determinant test on

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

Solution

The coefficient matrix is

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

Its determinant is

$$\det C = \frac14(-1-1) = -\frac12.$$

Since the determinant is nonzero, $\operatorname{SR}(\Psi)=2$. The state is entangled.

3. Give a full-Schmidt-rank two-qutrit state that is not maximally entangled.

Solution

One example is

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

It has three nonzero Schmidt coefficients, so it has full Schmidt rank for two qutrits. The coefficients are unequal, so it is not maximally entangled.

4. Show that local unitaries preserve Schmidt rank.

Solution

If

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

then

$$\rho'_A = U_A\rho_A U_A^\dagger.$$

Unitary conjugation preserves rank. Since Schmidt rank equals $\operatorname{rank}\rho_A$ for a pure bipartite state,

$$\operatorname{SR}(\Psi') = \operatorname{SR}(\Psi).$$

5. Why is the density-matrix rank of $\rho_{\mathrm{cc}}$ not a Schmidt rank?

Solution

Schmidt rank is defined for pure bipartite vectors. The state

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

is a mixed density operator. Its density-matrix rank is $2$, but it is separable because it is a mixture of product projectors. Calling that matrix rank a Schmidt rank would incorrectly suggest pure-state entanglement language applies directly to this mixed state.