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.

In 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.

Helpful background. Tensor products defines subsystem factors; local unitary equivalence distinguishes changing local bases from changing the subsystems themselves.

A state together with its subsystems

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. In finite dimensions, a mixed state is separable across the same split if

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

For infinite-dimensional normal states, use the trace-norm closure of finite product mixtures, as in separable mixed states.

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:

Change	Same subsystem split?	Entanglement
Active $U_A\otimes U_B$	Yes	Preserved
Active joint $U_{AB}$	Yes	May change
New factorization	No	A different question

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

A Bell basis used to define different factors

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. For this assignment, the new local $Z$ observables are old joint correlations:

$$\widetilde Z_A=Z_A\otimes Z_B,\qquad \widetilde Z_B=X_A\otimes X_B.$$

The first distinguishes $\Phi$ from $\Psi$ and the second distinguishes $+$ from $-$. These observables act on both 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 in symmetric or antisymmetric wavefunctions label coordinates; they do not assign observable names to individual identical 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 entangled across the tensor product of the two bosonic mode Fock spaces. 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 scaled center-of-mass and relative coordinates

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

The usual center of mass is $X_{\mathrm{cm}}=R/\sqrt2$; the scaling above makes the coordinate transformation orthogonal with absolute Jacobian one. 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.

Take positive finite widths $\sigma_R,\sigma_r$. Consider the normalized Gaussian

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

Its normalization is $N=(\pi\sigma_R\sigma_r)^{-1/2}$. It is product in the $R\vert r$ decomposition. Written in particle coordinates, its exponent is

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

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

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

In finite dimensions, these algebras commute and together generate all of $\mathcal B(\mathcal H)$. Conversely, commuting unital full matrix subalgebras $\mathcal A\cong M_m(\mathbb C)$ and $\mathcal B\cong M_n(\mathbb C)$ that together generate $\mathcal B(\mathcal H)$ determine a factorization $\mathcal H\cong\mathbb C^m\otimes\mathbb C^n$ with the stated local actions. Full generation matters: commutation and scalar centers alone can leave an extra multiplicity space unaccounted for. This finite-dimensional result does not directly settle local algebras in continuum field theory.

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.

Physical meaning of the decomposition

Separated laboratories, individually controlled spins, and selected optical modes supply concrete subsystem definitions. A coordinate transformation can simplify a Hamiltonian without changing which observables the laboratories can actually measure.

For fermionic modes, odd creation and annihilation operators anticommute across modes; the ordinary commuting-subalgebra statement must be applied with the parity restrictions or a suitable graded formulation. For spatial regions in continuum field theory, local algebras require further care. See mode decompositions and entanglement in QFT before transferring finite-dimensional partial-trace formulas unchanged.

Once the subsystem structure and available operations are specified, the entanglement question is precise. Changing that structure asks a different physical question; it does not invalidate the original answer.

Common pitfalls

A joint basis is not automatically a subsystem structure. Declaring Bell vectors to be “product basis vectors” also changes which observables count as local.

Coordinate separation does not imply particle separation. A Gaussian can factor in collective coordinates while correlating the original particle coordinates.

Identical-particle slots are not laboratory labels. State the modes, regions, or accessible observables before interpreting antisymmetric coefficients as usable entanglement.

References

• Howard Barnum, Emanuel Knill, Gerardo Ortiz, Rolando Somma, and Lorenza Viola, “A subsystem-independent generalization of entanglement,” Physical Review Letters 92, 107902 (2004). Preprint.
• Stephen D. Bartlett, Terry Rudolph, and Robert W. Spekkens, “Reference frames, superselection rules, and quantum information,” Reviews of Modern Physics 79, 555–609 (2007). Preprint.
• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, section 2.5.
• Paolo Zanardi, Daniel A. Lidar, and Seth Lloyd, “Quantum tensor product structures are observable induced,” Physical Review Letters 92, 060402 (2004). Preprint.

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 $f(x_1)g(x_2)$. Also compute $\langle x_1x_2\rangle$ and check that it vanishes in the product case.

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

$$\sigma_R=\sigma_r.$$

Since $x_1x_2=(R^2-r^2)/2$ and the probability density has variances $\langle R^2\rangle=\sigma_R^2/2$, $\langle r^2\rangle=\sigma_r^2/2$,

$$\langle x_1x_2\rangle=\frac{\sigma_R^2-\sigma_r^2}{4}.$$

Both means vanish. Thus unequal widths give a nonzero connected correlation, which for this pure state also rules out a product. Equal widths give zero and the explicit factorization above.

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.