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.

Helpful background. Schmidt decomposition supplies the orthonormal paired modes and their reduced-state eigenvalues.

Schmidt rank and matrix rank

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.

Why one Schmidt term means a product state

For a normalized product $|a\rangle|b\rangle$, the coefficient matrix is $C_{ij}=a_i b_j$ and has rank one. Conversely, a one-term Schmidt decomposition is already a product. An entangled pure state therefore has rank greater than one. This criterion depends on the bipartition, not on the chosen product bases.

Rank can jump under a small perturbation

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.

Local filters cannot increase pure-state Schmidt rank

For a recorded branch of local operations, the unnormalized coefficient matrix transforms as

$$C\longmapsto A C B^{\mathsf T},\qquad \operatorname{rank}(ACB^{\mathsf T})\le\operatorname{rank}C.$$

If the branch probability is nonzero, normalization multiplies this matrix by a scalar and leaves its rank unchanged. Invertible local filters preserve rank; singular filters can reduce it. Each fully recorded branch of an LOCC protocol is of this product form. If outcomes are forgotten, the output is usually mixed and pure-state Schmidt rank no longer applies; use Schmidt number instead.

Rank and entropy behave differently under postselection. For $|\psi\rangle=\sqrt{0.9}|00\rangle+\sqrt{0.1}|11\rangle$, the filter $A=\operatorname{diag}(1/3,1)$ on $A$ produces an unnormalized $\sqrt{0.1}(|00\rangle+|11\rangle)$. Its success probability is $0.2$, and the normalized successful branch is a Bell state. The rank stays two while that branch’s entropy increases. LOCC explains why this does not contradict average entanglement monotonicity.

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.

In finite dimensions the mixed-state extension is the Schmidt number (Terhal and Horodecki):

$$\operatorname{SN}(\rho)=\min_{\rho=\sum_k p_k|\psi_k\rangle\langle\psi_k|} \max_{k:p_k>0}\operatorname{SR}(\psi_k).$$

A particular entangled-state ensemble gives an upper bound, not the minimum. Schmidt number one is exactly separability, because it means a decomposition using only product vectors exists. Thus $\operatorname{SN}(\rho_{\rm cc})=1$ even though the density matrix and its two marginals have rank two.

Exact rank versus a numerical threshold

A coefficient of size $10^{-10}$ is nonzero mathematically but can be indistinguishable from rounding error in numerical data. State the precision and tolerance used by a rank computation. For controlled approximation, report the discarded Schmidt weight $\delta_k=\sum_{r>k}s_r^2$ rather than declaring small coefficients exactly zero; the truncation calculation converts that weight into an error.

References

• R. A. Horn and C. R. Johnson, Matrix Analysis, 2nd ed., Cambridge University Press (2012).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), Sec. 2.5.
• 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), Chs. 1 and 6.

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. Apply the local projective measurement $\{|0\rangle\langle0|,|1\rangle\langle1|\}$ to $A$ of a Bell pair. Find the Schmidt rank in each recorded branch and the Schmidt number if the outcome is forgotten.

Solution

The two branches have probability $1/2$ and states $|00\rangle$ and $|11\rangle$. Each has Schmidt rank one. Forgetting the outcome gives $\rho_{\rm cc}=\tfrac12|00\rangle\langle00|+\tfrac12|11\rangle\langle11|$, whose Schmidt number is one by this product ensemble. Its density-matrix rank is two, a different quantity.

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.