Quantum Information, Explained

Chapter 07

Entanglement and the Schmidt decomposition

Given the four amplitudes of a two-qubit pure state, how can we tell whether it is a product of two one-qubit states? And if it is not, how close can any product state come to it?

Before you start: Tensor products (Chapter 1); reduced states and the partial trace (Chapter 3); the Bell pair (Chapter 4). Eigenvalues of a $2\times2$ matrix. Notation guide.

A direct test for two qubits

Write a normalised state as $|\Psi\rangle=a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$, and arrange its coefficients in a matrix $C=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$. For a product state with local vectors $u$ and $v$, the coefficients satisfy $C_{ij}=u_iv_j$. This is an outer product $uv^T$ and has rank one.

Conversely, a nonzero rank-one matrix factors in that way; the factors can be normalised while preserving the normalised joint state. Thus for a pure two-qubit state,

$$\text{product state}\quad\Longleftrightarrow\quad ad-bc=0.$$

For $|\Phi^+\rangle$, $C=I/\sqrt2$ and $\det C=1/2$, so no factorisation exists. For $|+\rangle|1\rangle$, $C=\left(\begin{smallmatrix}0&1/\sqrt2\\0&1/\sqrt2\end{smallmatrix}\right)$ and the determinant vanishes. This determinant test is specific to pure two-qubit states; it is not a criterion for arbitrary mixed states.

Derive the Schmidt form

For arbitrary finite subsystem dimensions, collect the amplitudes into a rectangular matrix $C$. A singular value decomposition (SVD) is a factorisation $C=UDV^\dagger$: $U$ and $V$ are unitary changes of basis, and the rectangular diagonal matrix $D$ has nonnegative entries $s_k$, called singular values. We can construct the nonzero entries using an ordinary eigenvalue problem, so no prior SVD technique is required here.

Constructing the SVD from an eigenvalue problem

Here is the construction behind that factorisation. Diagonalise the Hermitian positive matrix $C^\dagger C$. Let its orthonormal eigenvectors be $v_k$, with eigenvalues $s_k^2\ge0$. For every $s_k>0$, define $u_k=Cv_k/s_k$. The resulting vectors are orthonormal because

$$u_\ell^\dagger u_k=\frac{v_\ell^\dagger C^\dagger Cv_k}{s_\ell s_k} =\frac{s_k^2}{s_\ell s_k}\,v_\ell^\dagger v_k=\delta_{\ell k}.$$

The symbol $\delta_{\ell k}$ equals one if the two indices agree and zero otherwise. Vectors with eigenvalue zero lie in the kernel of $C$, meaning $C$ sends them to zero, and contribute nothing. Inserting the complete eigenbasis gives $C=\sum_{s_k>0}s_ku_kv_k^\dagger$. Completing the $u_k$ to a full orthonormal basis yields $U$, while the $v_k$ are the columns of $V$. This proves the factorisation needed for the state expansion.

Its physical purpose is to choose local bases in which the joint coefficients pair matching indices only. In the original basis every $A$ label could accompany every $B$ label; the diagonal form reorganises the same state into a sum of orthogonal pairs. A single surviving pair is a product state. Several surviving pairs cannot be absorbed into one pair of local state vectors.

Substituting into $|\Psi\rangle=\sum_{ij}C_{ij}|i\rangle_A|j\rangle_B$ gives

$$|\Psi\rangle=\sum_k s_k|u_k\rangle_A|v_k^*\rangle_B,\qquad \sum_k s_k^2=1.$$

Here $|u_k\rangle=\sum_iU_{ik}|i\rangle$ and $|v_k^*\rangle=\sum_jV_{jk}^*|j\rangle$. The complex conjugation follows from the dagger in the SVD. Renaming the second orthonormal family as $|w_k\rangle$ gives the usual Schmidt form $\sum_ks_k|u_k\rangle|w_k\rangle$. Each family is orthonormal. The number of nonzero coefficients is the Schmidt rank.

Tracing one subsystem now takes one line:

$$\rho_A=\sum_ks_k^2|u_k\rangle\langle u_k|,\qquad \rho_B=\sum_ks_k^2|w_k\rangle\langle w_k|.$$

Cross terms vanish by orthogonality. The reduced states have the same nonzero eigenvalues, even if their dimensions differ. Schmidt rank one gives pure reduced states and a product joint state. Higher rank gives mixed reduced states and entanglement. Degenerate coefficients allow different Schmidt bases, but the coefficient list is fixed.

A state whose Schmidt basis is not obvious

Consider $|\Psi\rangle=(|00\rangle+|01\rangle+|10\rangle)/\sqrt3$. Its coefficient matrix and first reduced state are

$$C=\frac1{\sqrt3}\begin{pmatrix}1&1\\1&0\end{pmatrix},\qquad\rho_A=CC^\dagger=\frac13\begin{pmatrix}2&1\\1&1\end{pmatrix}.$$

The identity $\rho_A=CC^\dagger$ is the partial trace written in coefficients: $(\rho_A)_{ii'}=\sum_j C_{ij}C^*_{i'j}$, because tracing over $B$ pairs the second index of $C$ with the second index of $C^*$. Likewise $\rho_B=(C^\dagger C)^T$ has the same nonzero eigenvalues, which is why the construction above could start from $C^\dagger C$.

The reduced matrix has trace one and determinant $1/9$. Its characteristic equation $\lambda^2-\lambda+1/9=0$ gives

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

Both coefficients are nonzero, so the state is entangled. Its purity is $\lambda_+^2+\lambda_-^2=1-2/9=7/9$. It is less entangled than a Bell pair according to the entanglement entropy defined below (interpreted in Chapter 15): the Bell pair's reduced spectrum $(1/2,1/2)$ is more uniform than $(\lambda_+,\lambda_-)\approx(0.87,0.13)$. We did not need to find the local eigenvectors to establish entanglement or compute purity.

The squared Schmidt coefficients also answer the second half of the opening question directly. The largest squared overlap $|\langle u\otimes v|\Psi\rangle|^2$ with any normalised product state is $s_{\max}^2$, the largest squared Schmidt coefficient. The reason is that $\langle u\otimes v|\Psi\rangle=u^\dagger C\,v^*$, where $v^*$ is the complex conjugate of $v$, and a matrix cannot lengthen a unit vector by more than its largest singular value; the top Schmidt pair attains that value. For the worked state the best product state reaches $s_{\max}^2=\lambda_+\approx0.873$. For a Bell pair no product state does better than $0.5$.

From coefficient matrix to Schmidt coefficients Three rows. The product state has coefficient matrix with entries zero and one over root two in the right-hand column only, determinant zero, and Schmidt squares one and zero. The worked state has entries one over root three except a zero in the bottom right, determinant minus one third, and Schmidt squares 0.873 and 0.127. The Bell pair has one over root two on the diagonal, determinant one half, and Schmidt squares one half and one half. coefficients C det C Schmidt s² |+⟩|1⟩, a product 0 1/√2 0 1/√2 0 1 0 (|00⟩+|01⟩+|10⟩)/√3 1/√3 1/√3 1/√3 0 −⅓ 0.873 0.127 (|00⟩+|11⟩)/√2, a Bell pair 1/√2 0 0 1/√2 ½ ½ ½
Three states, from coefficient matrix to squared Schmidt coefficients $s^2$. Rows of $C$ are qubit $A$'s label, columns qubit $B$'s. The product state has a single nonzero $s^2$, equal to 1, and $\det C=0$. The worked state has two unequal values, $s^2\approx0.873$ and $0.127$; the Bell pair has two equal values, $\tfrac12$ each. For two qubits $|\det C|=s_1s_2$, so the determinant is zero exactly when one coefficient is missing, and largest, $\tfrac12$, when the two are equal. The largest $s^2$ in each row is the best squared overlap any product state can reach.

What local unitaries can change

Under local unitaries $U_A\otimes U_B$, the coefficient matrix becomes $C'=U_ACU_B^T$. Since $U_A$ and $U_B$ are unitary, the singular values remain unchanged. Local unitaries can rotate the Schmidt bases but cannot change the Schmidt coefficients. In particular, no independent rotations on the two halves can turn a product state into a Bell pair.

For pure bipartite states the coefficients also specify entanglement entropy, $-\sum_ks_k^2\log_2s_k^2$. This is zero for a product and one for a two-qubit Bell state. We postpone its information-theoretic interpretation until entropy has been defined.

Mixed-state entanglement needs a different definition

A mixed state is separable if it admits a convex combination of product states, $\rho_{AB}=\sum_jp_j\rho_A^{(j)}\otimes\rho_B^{(j)}$. Otherwise it is entangled. The classically correlated mixture of $|00\rangle$ and $|11\rangle$ is separable by inspection, although both reduced states are mixed. A spectral decomposition of a mixed matrix is not a Schmidt decomposition of a pure vector, and the pure-state tests above cannot be transferred to it.

The subsystem split also belongs in the problem statement. Entanglement here refers to independently addressable tensor factors $A$ and $B$. Antisymmetrising a state of identical particles does not by itself establish that labelled particles supply two independently controllable qubits.

Where this leaves us

Arrange the amplitudes in the matrix $C$. For two qubits the state is a product exactly when $\det C=0$. In general, the singular values of $C$ are the Schmidt coefficients, and their squares $s^2$ are the eigenvalues of either reduced state. One nonzero $s^2$ means a product; several mean entanglement. How close a product state can come is measured by the largest $s^2$: about $0.873$ for the worked state, $\tfrac12$ for a Bell pair. Local unitaries cannot change those coefficients. All of this is for pure states; mixed states need the separate definition above.

So far entanglement is a property of a state vector. What does it predict that classically correlated systems could not? The next chapter answers with a measurement that two distant labs can perform, and a number that no local classical model can exceed.

Problems

1. Find the Schmidt coefficients of $(|00\rangle+|01\rangle+|10\rangle-|11\rangle)/2$.

Worked answer

Arrange the amplitudes as $C=\frac12\left(\begin{smallmatrix}1&1\\1&-1\end{smallmatrix}\right)$. Compute the reduced matrix explicitly:

$$CC^\dagger=\frac14\begin{pmatrix}1+1&1-1\\1-1&1+1\end{pmatrix}=I/2.$$

Both reduced eigenvalues are $1/2$, so both singular values, which are the Schmidt coefficients, are $1/\sqrt2$. One Schmidt expression is $(|0\rangle|+\rangle+|1\rangle|-\rangle)/\sqrt2$. Applying $H$ to the second factor sends it to $(|00\rangle+|11\rangle)/\sqrt2$, confirming equivalence to a Bell state under a local unitary.

The tempting wrong turn is to count nonzero amplitudes. This state has four, and the worked state in the text has three, yet this one is as entangled as a Bell pair, while a product such as $|+\rangle|+\rangle$ also has four. The number of terms depends on the basis; the Schmidt coefficients do not.

2. Show that $\sqrt p|00\rangle+e^{i\phi}\sqrt{1-p}|11\rangle$ has reduced purity $p^2+(1-p)^2$. Can $\phi$ change its entanglement?

Worked answer

The diagonal terms of the joint projector have weights $p$ and $1-p$. Its cross term is $e^{-i\phi}\sqrt{p(1-p)}|00\rangle\langle11|$ and its conjugate; both vanish under either partial trace. Thus

$$\rho_A=\operatorname{diag}(p,1-p),\qquad\operatorname{Tr}\rho_A^2=p^2+(1-p)^2.$$

To see directly why the phase does not affect entanglement, apply $V=\operatorname{diag}(1,e^{-i\phi})$ to the first qubit. The $|00\rangle$ term is unchanged, while the phase on $|11\rangle$ becomes $e^{i\phi}e^{-i\phi}=1$. A local basis change removes it without changing either Schmidt coefficient.

References and further reading

Nielsen & Chuang, §2.1.10 (singular values), §2.5 (Schmidt decomposition), and §12.5 (entanglement).