Common Composite States

This page is a lookup guide to common composite states. It gives the formula, the usual subsystem split, the main interpretation, and the canonical page to read next.

It does not own the derivations. Use it when you need to recognize a state quickly, compare examples, or choose the right page for details.

How to Use This Page

Every state name is incomplete until the relevant split is clear. A Bell state is entangled across two qubits. A two-mode squeezed state is entangled across two modes. A Slater determinant is antisymmetric across formal particle slots, but its physical entanglement depends on the mode, orbital, spin, spatial-region, or observable split being used.

When reading any formula below, ask:

• What is the Hilbert space or Fock space?
• Which factors, modes, regions, or algebras are being treated as subsystems?
• Is the state pure or mixed?
• Is the displayed expression normalized?
• Is the state a physical vector, a density operator, or an idealized generalized state?

The Formula Sheet collects the supporting identities. The Entanglement Diagnostic Table summarizes which tests apply to which state classes. Computational Notebooks gives reproducible checks for these states. The caution that entanglement depends on a specified decomposition is developed in Entanglement Depends on a Decomposition.

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

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

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

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:

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

This is why GHZ entanglement is genuinely multipartite: the phase coherence is not visible in any one qubit or in every small reduction. See GHZ States.

W States

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. Their entanglement pattern is different from GHZ states: losing one qubit leaves the remaining qubits with some entangled component rather than only a classical branch mixture. See W States.

Graph and Stabilizer States

For a graph $G=(V,E)$, a 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 occupation string is usually a product across the chosen mode factors. 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

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 entangled across the two mode factors. 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

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

A single determinant is the standard uncorrelated fermionic reference relative to a chosen orbital basis. Its antisymmetry is required by fermion statistics; it should not automatically be counted as useful particle entanglement. Correlated many-electron states are often superpositions of determinants. 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.

Quick Recognition Guide

• Product pure state: one tensor factor for each subsystem; zero pure-state entanglement.
• Classically correlated separable state: correlated density operator but convex mixture of product states.
• Bell state: two-qubit maximally entangled state with local reductions $I/2$.
• Singlet: rotationally invariant two-spin Bell state with antisymmetric spin factor.
• GHZ state: global branch coherence; small reductions can look classical.
• W state: one excitation coherently delocalized across parties.
• Graph state: stabilizer state built by controlled-$Z$ gates along graph edges.
• Number state: definite occupations of chosen modes.
• Two-mode squeezed state: continuous-variable entangled state with paired occupations.
• Slater determinant: antisymmetric fermionic reference state for chosen spin-orbitals.

Common Mistakes

• Quoting a state name without specifying the subsystem split.
• Treating identical-particle slot labels as observable particle names.
• Calling every correlated state entangled.
• Calling every antisymmetrized fermion state an entanglement resource.
• Treating a coherent state as entangled merely because it is a superposition of number states.
• Confusing a Bell state with the Bell theorem.
• Assuming a GHZ state has pairwise Bell entanglement in every two-qubit reduction.
• Calling a definite occupation string mode-entangled without checking the mode split.
• Forgetting normalization factors in infinite sums, determinants, and symmetric sums.

Cross-Links

• Formula Sheet
• Entanglement Diagnostic Table
• Computational Notebooks
• Product States
• Entangled States
• Separable Mixed States
• Classical Correlation versus Entanglement
• Bell States
• Singlet and Triplet States
• Schmidt Decomposition
• GHZ States
• W States
• Graph States
• Stabilizer States Preview
• Number States
• Occupation-Number Basis
• Mode Decompositions
• Gaussian States Preview
• Squeezed States as Entangled Modes
• EPR State Preview
• Slater Determinants
• Permanents
• Identical-Particle Entanglement Cautions
• Tensor Product Exercises
• Partial Trace Exercises
• Identical Particle Exercises
• Fock Space Exercises

References

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995.
• 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.
• D. M. Greenberger, M. A. Horne, and A. Zeilinger, “Going Beyond Bell’s Theorem,” in Bell’s Theorem, Quantum Theory and Conceptions of the Universe, Kluwer, 1989.
• W. Duer, 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.
• R. Raussendorf and H. J. Briegel, “A One-Way Quantum Computer,” Physical Review Letters 86, 5188-5191, 2001, doi:10.1103/PhysRevLett.86.5188.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.
• D. F. Walls and G. J. Milburn, Quantum Optics, 2nd ed., Springer, 2008.

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 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.