Entanglement Depends on a Decomposition

Entanglement is not a property of a bare vector alone. It is a property of a state together with a specified way of regarding the system as made of parts. In the usual distinguishable bipartite case, that extra structure is a tensor-product decomposition

$$\mathcal H = \mathcal H_A\otimes\mathcal H_B.$$

Only after the factors $A$ and $B$ have been named does it make sense to ask whether a pure state is product, whether a mixed state is separable, or how much entanglement is present across the $A\vert B$ split.

This page is a cautionary page, not a license to be vague. In ordinary laboratory and many-body settings, the subsystem decomposition is usually fixed by spatial localization, species, internal degrees of freedom, modes, control hardware, or accessible observables. Once that structure is fixed, entanglement statements are precise.

The Basic Point

For a chosen bipartite decomposition, a pure state is product if it can be written

$$\lvert\Psi\rangle = \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B.$$

It is entangled across that split if no such factorization exists. A mixed state is separable across the same split if

$$\rho_{AB} = \sum_k p_k\,\rho_A^{(k)}\otimes\rho_B^{(k)}, \qquad p_k\ge 0, \qquad \sum_k p_k=1.$$

The phrases “product,” “separable,” and “entangled” are therefore incomplete unless the relevant factors, modes, regions, or observable algebras are understood.

For example,

$$\mathbb C^4 \cong \mathbb C^2\otimes\mathbb C^2$$

can describe two qubits, but it can also describe a single four-level system. The same four-dimensional abstract vector space does not, by itself, tell us which observables are local, which operations are allowed on one part, or which measurements define correlations between parts.

Local Basis Changes Are Not the Issue

Changing basis inside the same tensor factors does not change entanglement. If

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

then $\lvert\Psi\rangle$ and $\lvert\Psi'\rangle$ are related by a local unitary. They have the same Schmidt coefficients, the same entanglement entropy, and the same status as product or entangled across the same $A\vert B$ split.

The subtlety on this page is different. It concerns changing what counts as subsystem $A$ and subsystem $B$ in the first place. Such a change usually changes the meaning of “local observable” and “local operation.”

The distinction can be summarized as:

$$\begin{array}{c|c|c} \text{operation} & \text{same subsystem split?} & \text{entanglement preserved?}\\ \hline U_A\otimes U_B & \text{yes} & \text{yes}\\ U_{AB}\ \text{generic} & \text{yes} & \text{not necessarily}\\ \text{new factorization} & \text{no} & \text{question changes} \end{array}$$

The last row is not an operation performed on a state. It is a change in the structural question being asked.

A Two-Qubit Warning Example

With respect to the standard qubit split, the Bell state

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \left( \lvert0\rangle_A\lvert0\rangle_B + \lvert1\rangle_A\lvert1\rangle_B \right)$$

is maximally entangled.

Mathematically, one could define a new artificial factorization of the same four-dimensional Hilbert space by declaring

$$\begin{aligned} \lvert\widetilde 0\rangle_{\widetilde A} \lvert\widetilde 0\rangle_{\widetilde B} &= \lvert\Phi^+\rangle,\\ \lvert\widetilde 0\rangle_{\widetilde A} \lvert\widetilde 1\rangle_{\widetilde B} &= \lvert\Phi^-\rangle,\\ \lvert\widetilde 1\rangle_{\widetilde A} \lvert\widetilde 0\rangle_{\widetilde B} &= \lvert\Psi^+\rangle,\\ \lvert\widetilde 1\rangle_{\widetilde A} \lvert\widetilde 1\rangle_{\widetilde B} &= \lvert\Psi^-\rangle. \end{aligned}$$

Relative to the new formal factors $\widetilde A$ and $\widetilde B$, the same vector $\lvert\Phi^+\rangle$ is product. This does not mean Bell entanglement was an illusion. It means that the new “local” observables are highly nonlocal relative to the original qubits. If the physical apparatus addresses the original qubits separately, the original $A\vert B$ split is the relevant one.

This example is useful because it separates two statements:

• As linear algebra, different tensor-product structures can be placed on the same Hilbert space.
• As physics, a tensor-product structure earns its meaning from preparation, control, measurement, locality, and dynamics.

Particle Entanglement and Mode Entanglement

For distinguishable particles, particle labels can correspond to operational subsystems. An electron and a proton in a hydrogen atom have different masses and charges, so a particle split such as

$$\mathcal H = \mathcal H_{\mathrm e}\otimes\mathcal H_{\mathrm p}$$

is physically meaningful.

For identical particles, formal particle labels are not directly observable. The tensor slots used to build symmetric or antisymmetric wavefunctions are bookkeeping devices, not private names attached to individual particles. This is why indistinguishability and the symmetrization postulate require special care.

Mode decompositions often give the physically relevant split. In occupation-number language, a two-mode bosonic Fock space is organized by mode occupations:

$$\lvert n_L,n_R\rangle.$$

The one-particle state

$$\lvert\psi\rangle = \frac{1}{\sqrt2} \left( \lvert1_L,0_R\rangle + \lvert0_L,1_R\rangle \right)$$

is not entanglement between two named particles; there is only one particle. It is a coherent superposition across two modes. Whether it functions as usable entanglement depends on what operations and reference frames are available for the modes, and on any relevant superselection constraints.

For two identical fermions, a single Slater determinant may be nonfactorizable in formal particle slots simply because antisymmetry is required. That nonfactorization should not automatically be counted as operational particle entanglement. The dedicated Identical-Particle Entanglement Cautions page explains how to state the subsystem, mode, region, spin-orbital, or observable split being used.

Center-of-Mass and Relative Coordinates

Even for distinguishable particles, useful variables can reorganize the same Hilbert space. For two one-dimensional particles with equal masses, define

$$R = \frac{x_1+x_2}{\sqrt2}, \qquad r = \frac{x_1-x_2}{\sqrt2}.$$

The Hilbert space can be represented using particle coordinates,

$$L^2(\mathbb R_{x_1})\otimes L^2(\mathbb R_{x_2}),$$

or center-of-mass and relative coordinates,

$$L^2(\mathbb R_R)\otimes L^2(\mathbb R_r).$$

These are unitarily equivalent representations, but they define different tensor factors. A wavefunction can factor in one decomposition and not in the other.

Consider the Gaussian

$$\Psi(x_1,x_2) = N \exp\left( -\frac{R^2}{2\sigma_R^2} -\frac{r^2}{2\sigma_r^2} \right).$$

It is product in the $R\vert r$ decomposition. Written in particle coordinates, its exponent is

$$\begin{aligned} &-\frac14 \left( \frac{1}{\sigma_R^2} + \frac{1}{\sigma_r^2} \right) (x_1^2+x_2^2)\\ &\qquad -\frac12 \left( \frac{1}{\sigma_R^2} - \frac{1}{\sigma_r^2} \right) x_1x_2. \end{aligned}$$

If $\sigma_R\ne\sigma_r$, the cross term $x_1x_2$ prevents factorization into $f(x_1)g(x_2)$. The state is then entangled with respect to the particle split but product with respect to the center-of-mass/relative split. If $\sigma_R=\sigma_r$, the cross term vanishes and the Gaussian factors in both decompositions.

This is a familiar reason that separable Hamiltonians do not automatically imply separable states with respect to every physically interesting split.

Observable-Algebra Viewpoint

A more invariant way to state the issue is to start from observables. A tensor-product split

$$\mathcal H = \mathcal H_A\otimes\mathcal H_B$$

comes with local observable algebras

$$\mathcal A_A = \mathcal B(\mathcal H_A)\otimes I_B, \qquad \mathcal A_B = I_A\otimes\mathcal B(\mathcal H_B).$$

These algebras commute, and together they generate the full algebra of observables on the composite system in the finite-dimensional ideal case. Conversely, under suitable finite-dimensional assumptions, a pair of commuting full matrix algebras with trivial common center can induce a tensor-product structure.

This algebraic language is useful when the natural “parts” are not particles:

• spatial regions in a lattice or field theory,
• modes in quantum optics,
• spin, orbital, or band degrees of freedom,
• encoded subsystems in quantum information,
• superselection sectors and constrained Hilbert spaces.

The algebraic viewpoint also explains why centers and superselection sectors complicate the story. If the accessible observables have a nontrivial center, part of the structure behaves like a classical sector label rather than an ordinary tensor factor. Entanglement questions must then be phrased with the available operations and observables in mind.

Why This Does Not Make Entanglement Arbitrary

The dependence on decomposition is sometimes overstated as “entanglement is arbitrary.” That is not correct.

In practice, physical structure chooses the decomposition:

• separated laboratories define local operations and measurements;
• qubits in a circuit are addressed by different control lines;
• lattice sites define local spin or fermionic algebras;
• optical modes are selected by polarization, frequency, spatial profile, or cavity structure;
• atoms, ions, or dots in distinct traps provide effective subsystems;
• conserved quantities and superselection rules restrict which coherent operations are available.

Once the decomposition and allowed observables are fixed, entanglement is not a matter of taste. Productness, separability, Schmidt coefficients, reduced states, entanglement entropy, and local-unitary equivalence have precise meanings.

The careful statement is:

Entanglement is objective relative to a specified physical decomposition, but the decomposition is part of the physical modeling.

That is the same kind of caution one uses for energy relative to a Hamiltonian, angular momentum relative to a rotation group, or locality relative to a chosen spatial or mode structure.

Common Mistakes

• Asking whether a state is entangled without saying which subsystems, modes, regions, or observables define the split.
• Treating a local basis change as if it could change entanglement across a fixed split.
• Treating every global change of variables as a harmless relabeling of subsystems.
• Calling antisymmetrization itself a useful entanglement resource without specifying an operational mode or observable split.
• Confusing particle entanglement, mode entanglement, spin-position entanglement, and center-of-mass/relative-coordinate factorization.
• Concluding that entanglement is arbitrary merely because several decompositions are mathematically possible.

Cross-Links

• Notation and Subsystem Labels
• Tensor Products of Hilbert Spaces
• Product States
• Separable Mixed States
• Entangled States
• Classical Correlation versus Entanglement
• Local Unitary Equivalence
• Reduced Density Operators
• Schmidt Decomposition
• Entanglement Entropy
• Indistinguishability
• Symmetrization Postulate
• Identical-Particle Entanglement Cautions
• Occupation-Number Basis
• Mode Occupations
• Mode Decompositions
• Particle-Number Superselection Preview
• Entanglement in QFT Preview
• Many-Body Entanglement Overview
• Creation and Annihilation Operators

References

• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• I. Bengtsson and K. Zyczkowski, Geometry of Quantum States: An Introduction to Quantum Entanglement, Cambridge University Press, 2006.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995.
• P. Zanardi, D. A. Lidar, and S. Lloyd, “Quantum Tensor Product Structures Are Observable-Induced,” Physical Review Letters 92, 060402, 2004.
• H. Barnum, E. Knill, G. Ortiz, R. Somma, and L. Viola, “A Subsystem-Independent Generalization of Entanglement,” Physical Review Letters 92, 107902, 2004.
• G. C. Ghirardi and L. Marinatto, “Identical Particles and Entanglement,” Optics and Spectroscopy 99, 386-390, 2005.
• S. D. Bartlett, T. Rudolph, and R. W. Spekkens, “Reference Frames, Superselection Rules, and Quantum Information,” Reviews of Modern Physics 79, 555-609, 2007.

Exercises

1. Bell state under an artificial factorization. Suppose the four Bell states are declared to be the product basis of artificial factors $\widetilde A$ and $\widetilde B$ as in the example above. Is $\lvert\Phi^+\rangle$ entangled relative to $\widetilde A\vert\widetilde B$? Why does this not contradict its ordinary two-qubit entanglement?

Solution

Relative to the artificial factorization,

$$\lvert\Phi^+\rangle = \lvert\widetilde0\rangle_{\widetilde A} \lvert\widetilde0\rangle_{\widetilde B},$$

so it is product. There is no contradiction because the subsystem split has changed. The observables local to $\widetilde A$ or $\widetilde B$ are not generally local to the original qubits $A$ and $B$. Ordinary Bell entanglement is the statement that $\lvert\Phi^+\rangle$ is entangled across the physical two-qubit split.

2. Center-of-mass Gaussian. For

$$\Psi(x_1,x_2) = N \exp\left( -\frac{R^2}{2\sigma_R^2} -\frac{r^2}{2\sigma_r^2} \right), \qquad R=\frac{x_1+x_2}{\sqrt2}, \qquad r=\frac{x_1-x_2}{\sqrt2},$$

find the condition under which the state factors as $f(x_1)g(x_2)$.

Solution

Substituting $R$ and $r$ gives a cross term proportional to

$$\left( \frac{1}{\sigma_R^2} - \frac{1}{\sigma_r^2} \right) x_1x_2.$$

The wavefunction can factor as $f(x_1)g(x_2)$ only if this coefficient vanishes. Thus the Gaussian factors in particle coordinates when

$$\sigma_R=\sigma_r.$$

For unequal widths, it is product in the $R\vert r$ decomposition but nonproduct in the $x_1\vert x_2$ particle decomposition.

3. One particle in two modes. Consider

$$\lvert\psi\rangle = \frac{1}{\sqrt2} \left( \lvert1_L,0_R\rangle + \lvert0_L,1_R\rangle \right).$$

What decomposition is being used if one calls this a mode-entangled state? Why is it not entanglement between two particles?

Solution

The decomposition is a tensor product of mode occupation spaces, one associated with mode $L$ and one associated with mode $R$. The state is a coherent superposition of occupation patterns across those modes. It is not entanglement between two particles because the total particle number is one. Operational use of this coherence also depends on the available mode operations and any particle-number superselection constraints.

4. Single Slater determinant. Let $a_\phi^\dagger a_\chi^\dagger\lvert0\rangle$ be a two-fermion state with orthonormal one-particle modes $\phi$ and $\chi$. Why is it misleading to call this automatically entangled merely because the corresponding first-quantized wavefunction is antisymmetric?

Solution

The antisymmetry is required for identical fermions and reflects exchange structure, not by itself a usable correlation between individually addressable particles. A single Slater determinant represents two occupied orthonormal modes. To make an entanglement claim, one must specify a physical split, such as spatial regions, spin modes, orbitals, or an algebra of accessible observables.