Common Composite States

These examples make it easier to recognize a state and identify its subsystem split. All displayed vectors are normalized under the assumptions stated beside them. Entropies use base-two logarithms. A named state is only a starting point: its reductions and correlations depend on which qubits, modes, orbitals, or regions are being compared.

Helpful background. Product bases interprets the kets and partial trace gives local states. Use the formula sheet for identities and the diagnostic table to select an entanglement test.

Product States

For two distinguishable subsystems,

$$\lvert\Psi\rangle_{AB} = \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B$$

is a pure product state. Its reduced states are pure:

$$\rho_A = \lvert\psi\rangle\langle\psi\rvert, \qquad \rho_B = \lvert\phi\rangle\langle\phi\rvert.$$

A product density operator has the form

$$\rho_{AB} = \rho_A\otimes\rho_B.$$

It has neither entanglement nor correlation between $A$ and $B$. Product states are the baseline for recognizing entanglement, separability, and correlation. See Product States.

Classically Correlated Separable States

A simple two-qubit classically correlated state is

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

It is separable because it is a convex mixture of product states. It is not a product state, because joint computational-basis outcomes are correlated. Its local reduced states are maximally mixed:

$$\rho_A = \rho_B = \frac12 I.$$

This state is useful because it has the same local marginals as a Bell state but no entanglement. The difference lives in the joint density operator, not in either local density operator alone. See Classical Correlation versus Entanglement and Separable Mixed States.

Bell States

The four Bell states are the standard maximally entangled two-qubit basis:

$$\begin{aligned} \lvert\Phi^+\rangle &= \frac{ \lvert00\rangle+\lvert11\rangle }{\sqrt2}, & \lvert\Phi^-\rangle &= \frac{ \lvert00\rangle-\lvert11\rangle }{\sqrt2}, \\ \lvert\Psi^+\rangle &= \frac{ \lvert01\rangle+\lvert10\rangle }{\sqrt2}, & \lvert\Psi^-\rangle &= \frac{ \lvert01\rangle-\lvert10\rangle }{\sqrt2}. \end{aligned}$$

Each has one-qubit reductions

$$\rho_A = \rho_B = \frac12 I,$$

so each carries one ebit of pure-state entanglement. Bell states are the canonical examples behind teleportation, Bell-basis measurements, dense coding, singlet correlations, and many foundations discussions. The state definitions live at Bell States; Bell inequalities and EPR reasoning belong to Entanglement in Foundations.

Singlet and Triplet States

For two spin-$1/2$ systems, the coupled basis consists of a spin-$1$ triplet and a spin-$0$ singlet:

$$\begin{aligned} \lvert1,1\rangle &= \lvert\uparrow\uparrow\rangle, \\ \lvert1,0\rangle &= \frac{ \lvert\uparrow\downarrow\rangle + \lvert\downarrow\uparrow\rangle }{\sqrt2}, \\ \lvert1,-1\rangle &= \lvert\downarrow\downarrow\rangle, \\ \lvert0,0\rangle &= \frac{ \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle }{\sqrt2}. \end{aligned}$$

The singlet and the $m=0$ triplet are entangled across the first-spin versus second-spin split. The $m=\pm1$ triplet states are product states. Thus “triplet” is not a synonym for “entangled.”

The singlet is also the Bell state $\lvert\Psi^-\rangle$ after identifying $\lvert0\rangle=\lvert\uparrow\rangle$ and $\lvert1\rangle=\lvert\downarrow\rangle$. See Singlet and Triplet States.

Schmidt-Form Template

Any finite-dimensional pure bipartite state can be written in Schmidt form:

$$\lvert\Psi\rangle = \sum_{r=1}^R s_r \lvert r_A\rangle \lvert r_B\rangle, \qquad s_r>0, \qquad \sum_r s_r^2=1.$$

The Schmidt rank $R$ is one for product states and greater than one for entangled pure states. The reduced-state eigenvalues are $p_r=s_r^2$, and the entanglement entropy is

$$S_A = - \sum_r p_r\log_2p_r.$$

This is not a separate named state. It is the normal form used to classify pure bipartite states. See Schmidt Decomposition and Schmidt Rank.

GHZ States

For $n\ge2$, the $n$-qubit GHZ state is

$$\lvert\mathrm{GHZ}_n\rangle = \frac{ \lvert0\rangle^{\otimes n} + \lvert1\rangle^{\otimes n} }{\sqrt2}.$$

It stores coherence in a global branch structure. A single-qubit reduction is maximally mixed, and a two-qubit reduction of $\lvert\mathrm{GHZ}_3\rangle$ is a classically correlated separable state:

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

Across every nontrivial bipartition, the pure GHZ state has two equal Schmidt probabilities. For $n\ge3$, this proves genuine multipartite entanglement. Every proper reduction loses the relative phase, which illustrates why local data can miss global coherence. See GHZ States.

W States

For $n\ge2$, the $n$-qubit W state is the symmetric one-excitation state

$$\lvert W_n\rangle = \frac{1}{\sqrt n} \sum_{j=1}^n \lvert 0\cdots 1_j\cdots 0\rangle.$$

For $n=3$,

$$\lvert W_3\rangle = \frac{ \lvert100\rangle+\lvert010\rangle+\lvert001\rangle }{\sqrt3}.$$

W states distribute one excitation coherently across many parties. For $n\ge3$, losing one qubit leaves an entangled state, not just a mixture containing an entangled component. Indeed, any retained pair has state $(n-2)|00\rangle\langle00|/n+2|\Psi^+\rangle\langle\Psi^+|/n$ with negative partial-transpose eigenvalue $[n-2-\sqrt{(n-2)^2+4}]/(2n)$. This proves the surviving pair entanglement. See W States.

Graph and Stabilizer States

For a simple undirected graph $G=(V,E)$, a qubit graph state is

$$\lvert G\rangle = \prod_{\{u,v\}\in E} \operatorname{CZ}_{uv} \lvert+\rangle^{\otimes \lvert V\rvert}.$$

It is equivalently the simultaneous $+1$ eigenstate of stabilizer generators

$$K_v = X_v \prod_{u\in N(v)} Z_u, \qquad v\in V.$$

Graph states organize entanglement by the edge structure of $G$. They are central in measurement-based quantum computation, stabilizer codes, and many-body toy models. See Graph States and Stabilizer States Preview.

Number States

For bosonic modes, an occupation-number state is

$$\lvert n_1,n_2,\ldots\rangle = \prod_i \frac{(a_i^\dagger)^{n_i}}{\sqrt{n_i!}} \lvert0\rangle.$$

For fermionic modes, each occupation is $0$ or $1$, and a fixed mode ordering is part of the convention:

$$\lvert n_1,\ldots,n_M\rangle = (c_1^\dagger)^{n_1} \cdots (c_M^\dagger)^{n_M} \lvert0\rangle.$$

A definite bosonic occupation string is product across its mode factors. A fermionic occupation string is likewise a product in a fixed ordered occupation representation, with parity restrictions understood for physical operations. Mode entanglement appears in superpositions such as

$$\frac{ \lvert1,0\rangle+\lvert0,1\rangle }{\sqrt2},$$

which is entangled across the two-mode split but is not entanglement between two particles. See Number States, Occupation-Number Basis, and Mode Decompositions.

Coherent and Gaussian States

A one-mode coherent state has number-state expansion

$$\lvert\alpha\rangle = e^{-\lvert\alpha\rvert^2/2} \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} \lvert n\rangle.$$

As a one-mode state, this is not bipartite entanglement. A product of coherent states across modes,

$$\lvert\alpha\rangle_A \lvert\beta\rangle_B,$$

is also unentangled across the $A\vert B$ mode split. Coherent states are nevertheless important reference states for quantum optics, Gaussian states, and number-sector coherence discussions. See Gaussian States Preview and Particle-Number Superselection Preview.

Two-Mode Squeezed States

For finite $r\ge0$, with $\lambda=\tanh r$, the two-mode squeezed vacuum can be written

$$\lvert\mathrm{TMSV}(r)\rangle = \sqrt{1-\lambda^2} \sum_{n=0}^{\infty} \lambda^n \lvert n\rangle_A \lvert n\rangle_B.$$

It is product at $r=0$ and entangled for $r>0$. The correlations become increasingly EPR-like as $r$ grows, but the ideal EPR limit is singular. See Squeezed States as Entangled Modes and EPR State Preview.

Slater Determinants

For $N$ fermions in orthonormal spin-orbitals $\phi_1,\ldots,\phi_N$, the first-quantized Slater determinant is

$$\Phi(x_1,\ldots,x_N) = \frac{1}{\sqrt{N!}} \det \bigl[ \phi_i(x_j) \bigr]_{i,j=1}^N.$$

In occupation-number notation, the same state is

$$\lvert\Phi\rangle = c_1^\dagger c_2^\dagger \cdots c_N^\dagger \lvert0\rangle.$$

Being a single determinant is unchanged by a one-particle orbital rotation, though the same determinant can have orbital entanglement in a different mode split. Its antisymmetry alone does not quantify an operational resource between laboratories. Even a multi-term determinant expansion can simplify to one determinant after rotating orbitals; the chemistry examples show both possibilities. Here the variables $x_j$ include all one-particle labels, including spin when present. See Slater Determinants and Identical-Particle Entanglement Cautions.

Permanents

For $N$ bosons occupying one-particle orbitals, symmetrized wavefunctions use permanents rather than determinants. For two bosons in orbitals $\phi$ and $\chi$,

$$\Psi_B(x_1,x_2) = \frac{ \phi(x_1)\chi(x_2) + \chi(x_1)\phi(x_2) }{\sqrt2}$$

when the orbitals are orthonormal and distinct. The occupation-number version is often cleaner:

$$a_\phi^\dagger a_\chi^\dagger \lvert0\rangle.$$

As with determinants, the symmetrization itself is not the whole entanglement story. One must name the physical split: modes, spatial regions, internal states, species, or an observable algebra. See Permanents and Bosonic Fock Space.

Three states with identical local reductions

The following states all have $\rho_A=\rho_B=I/2$, but their joint properties differ. Entropies and mutual information are in bits.

Joint state	Joint purity and entropy	Correlation across the two qubits
$I_4/4$	$\operatorname{Tr}\rho^2=1/4$, $S_{AB}=2$	Product; $I(A:B)=0$
$\rho_{\rm cc}$	$\operatorname{Tr}\rho^2=1/2$, $S_{AB}=1$	Separable and classically correlated; $I(A:B)=1$
$\lvert\Phi^+\rangle\langle\Phi^+\rvert$	$\operatorname{Tr}\rho^2=1$, $S_{AB}=0$	Entangled; $I(A:B)=2$

Local mixedness alone therefore cannot classify the joint state. The marginal calculations explain what joint measurements add.

Exercises

1. Bell reduction. Compute the reduced density operator of qubit $A$ for $\lvert\Phi^+\rangle$.

Solution

For

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

the density operator is

$$\rho = \frac12 \bigl( \lvert00\rangle\langle00\rvert + \lvert00\rangle\langle11\rvert + \lvert11\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \bigr).$$

Tracing over $B$ removes the off-diagonal terms because $\langle1\vert0\rangle=0$ and $\langle0\vert1\rangle=0$. Therefore

$$\rho_A = \frac12 \lvert0\rangle\langle0\rvert + \frac12 \lvert1\rangle\langle1\rvert = \frac12 I.$$

2. GHZ pair reduction. Trace out qubit $C$ from $\lvert\mathrm{GHZ}_3\rangle$. Is the resulting $AB$ state entangled?

Solution

The state is

$$\lvert\mathrm{GHZ}_3\rangle = \frac{ \lvert000\rangle+\lvert111\rangle }{\sqrt2}.$$

After tracing out $C$,

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

This is a mixture of product states, so it is separable. It is classically correlated but not entangled across $A\vert B$.

3. Normalize the W state. Why does $\lvert W_n\rangle$ have the prefactor $1/\sqrt n$?

Solution

The $n$ basis states

$$\lvert 0\cdots 1_j\cdots 0\rangle, \qquad j=1,\ldots,n,$$

are mutually orthonormal. Therefore the squared norm of the unnormalized sum is $n$. Multiplying by $1/\sqrt n$ gives norm one.

4. Two-mode squeezed normalization. For $\lvert\mathrm{TMSV}\rangle=\sqrt{1-\lambda^2}\sum_{n=0}^{\infty}\lambda^n\lvert n,n\rangle$ with $0\le\lambda<1$, check normalization.

Solution

The states $\lvert n,n\rangle$ are orthonormal, so

$$\langle\mathrm{TMSV}\vert\mathrm{TMSV}\rangle = (1-\lambda^2) \sum_{n=0}^{\infty} \lambda^{2n}.$$

Using the geometric series,

$$\sum_{n=0}^{\infty} \lambda^{2n} = \frac{1}{1-\lambda^2},$$

so the norm is one.

References

• W. Dür, G. Vidal, and J. I. Cirac, “Three qubits can be entangled in two inequivalent ways,” Physical Review A 62, 062314 (2000), doi:10.1103/PhysRevA.62.062314.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill (1971).
• R. Horodecki, P. Horodecki, M. Horodecki, and K. Horodecki, “Quantum entanglement,” Reviews of Modern Physics 81, 865–942 (2009), doi:10.1103/RevModPhys.81.865.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010).
• D. F. Walls and G. J. Milburn, Quantum Optics, 2nd ed., Springer (2008).