Identical-Particle Entanglement Cautions

Entanglement is always relative to a specified decomposition of a physical system into subsystems, modes, regions, or observable algebras. Identical particles make that statement unavoidable: the formal particle slots in $\mathcal h^{\otimes N}$ are not directly observable subsystems.

This page is a cautionary guide. It does not assert that identical particles cannot be entangled, and it does not count every symmetrized or antisymmetrized wavefunction as useful entanglement. It separates three different ideas:

• nonfactorization caused by required exchange symmetry;
• entanglement between physical modes, regions, internal degrees of freedom, or laboratories;
• operationally accessible entanglement under stated local operations and superselection constraints.

Terminology in this subject is not perfectly uniform. The safest habit is to state the subsystem or observable split before making an entanglement claim.

Helpful background. Entanglement Depends on a Decomposition defines the subsystem question; Occupation-Number Basis explains the mode notation.

Particle slots and physical subsystems

For distinguishable systems $A$ and $B$, the standard pure-state question is posed on a tensor product:

$$\mathcal H_A\otimes\mathcal H_B.$$

A pure state is product if it can be written as

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

If it cannot be written that way, it is entangled relative to the $A\vert B$ split.

For $N$ identical bosons or fermions built from one one-particle Hilbert space $\mathcal h$, the physical fixed-particle-number spaces are instead

$$\operatorname{Sym}^N\mathcal h, \qquad \wedge^N\mathcal h.$$

These are subspaces of the formal slot-labeled space $\mathcal h^{\otimes N}$. The slots are not automatically physical subsystems. Asking whether a vector is entangled between “particle 1” and “particle 2” can therefore be a question about mathematical representation rather than an operational question about what can be prepared, measured, or shared.

Slot Nonfactorization Is Not Enough

Consider two identical fermions in orthonormal one-particle states $\lvert a\rangle$ and $\lvert b\rangle$. The antisymmetric state is

$$\lvert a,b\rangle_A = \frac{1}{\sqrt2} \bigl( \lvert a\rangle_1\lvert b\rangle_2 - \lvert b\rangle_1\lvert a\rangle_2 \bigr).$$

As a vector in the formal slot tensor product, this is not a product of slot $1$ and slot $2$ states. But that nonfactorization is forced by antisymmetry. It is a single Slater determinant, and in occupation notation it is simply

$$\lvert1_a,1_b\rangle_F.$$

For the mode split into mode $a$ and mode $b$, this is a definite occupation product:

$$\lvert1\rangle_a\otimes\lvert1\rangle_b,$$

up to the sign conventions needed for fermionic mode ordering. There is no extra mode entanglement between $a$ and $b$ in this state.

The same point holds for two bosons in two distinct modes:

$$\frac{1}{\sqrt2} \bigl( \lvert a\rangle_1\lvert b\rangle_2 + \lvert b\rangle_1\lvert a\rangle_2 \bigr) \quad \longleftrightarrow \quad \lvert1_a,1_b\rangle_B.$$

The slot expression is symmetrized, but the occupation state has one boson in each mode. Exchange symmetry alone should not be advertised as a useful bipartite resource.

What Counts as a Subsystem?

For identical particles, physically meaningful subsystem choices often come from:

• spatial regions, such as left and right wells;
• modes, such as optical paths, lattice sites, frequency modes, or spin-orbitals;
• internal degrees of freedom, such as spin or hyperfine states, when they are operationally accessible;
• laboratories or parties with specified local operations;
• subalgebras of observables assigned to regions or modes.

The entanglement statement changes when the chosen split changes. A state can be unentangled in one mode basis and entangled in another. A state can be mathematically entangled across occupation modes but not directly useful as a shared resource if the allowed local operations cannot access the relevant coherences.

A change of mode basis can mix the two laboratories. Such a change is generally a nonlocal operation relative to the original split; it is not a contradiction if the corresponding entanglement changes.

Mode Entanglement

Mode entanglement treats occupation states of modes as the subsystem degrees of freedom. For two bosonic modes $L$ and $R$, the state

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

is not entangled between two particles; it contains one particle. It is, however, nonfactorizable across the two-mode occupation basis:

$$\lvert\psi\rangle \ne \lvert\chi\rangle_L \otimes \lvert\eta\rangle_R.$$

Suppose each laboratory is restricted to operations that conserve its local particle number and there is no shared phase reference. Averaging over the unknown relative phase gives

$$\mathcal G(\rho)=\sum_n \Pi_n^L\rho\Pi_n^L =\tfrac12\bigl(\lvert1,0\rangle\langle1,0\rvert +\lvert0,1\rangle\langle0,1\rvert\bigr),$$

where $\Pi_n^L$ projects onto $n$ particles on the left. For a single copy and those allowed local operations, the state has no accessible bipartite entanglement even though its unrestricted mode entropy is one bit. This restriction depends on available operations and reference resources; it is not a ban on single-particle interference. Additional copies or a shared reference can change what is accessible, as discussed by Wiseman and Vaccaro and by Bartlett, Rudolph, and Spekkens.

A two-boson NOON-like state,

$$\lvert\psi\rangle = \frac{ \lvert2_L,0_R\rangle + \lvert0_L,2_R\rangle }{\sqrt2},$$

is also mode-entangled across $L\vert R$. Tracing over the right mode gives

$$\rho_L = \frac12 \bigl( \lvert2\rangle\langle2\rvert + \lvert0\rangle\langle0\rvert \bigr),$$

so the mode entanglement entropy is one bit if logarithms are base $2$. The interpretation still depends on what local mode operations and phase references are available.

Spin and Region Entanglement

Identical particles can carry operationally meaningful entanglement when physical regions or modes define the parties. For fermions, order the modes as $L\uparrow,L\downarrow,R\uparrow,R\downarrow$. Suppose there is exactly one particle in each region and the state is

$$\lvert\Psi\rangle = \frac{ c_{L\uparrow}^\dagger c_{R\downarrow}^\dagger - c_{L\downarrow}^\dagger c_{R\uparrow}^\dagger }{\sqrt2} \lvert0\rangle.$$

The symbols $L$ and $R$ are physical regions or modes, not hidden particle names. If the regions are controlled by two laboratories and each laboratory can measure its local spin, this state carries the same operational spin correlations as a singlet shared between the regions.

The identical-particle nature still matters. The state must be written with the correct bosonic or fermionic operator algebra, and exchange symmetry is already built into the Fock-space expression. But the entanglement claim is about the left and right local degrees of freedom, not about a private identity tag on particle $1$ or particle $2$.

Occupation-Number Entanglement

Occupation-number notation is often the cleanest language. Choose a mode split

$$\mathcal h = \mathcal h_A\oplus\mathcal h_B.$$

For bosons, the Fock spaces factor as

$$\mathcal F_B(\mathcal h_A\oplus\mathcal h_B) \cong \mathcal F_B(\mathcal h_A) \otimes \mathcal F_B(\mathcal h_B).$$

Then standard tensor-product entanglement questions can be asked between the mode collections $A$ and $B$.

For fermions, the corresponding factorization carries sign and parity conventions. One can represent it with an ordered tensor product or a graded tensor product. The physical content is usually expressed through local fermionic modes or local even observables rather than through observable particle labels.

In either case, the mode split must be stated. The same vector can look different under a different one-particle basis, because occupation is occupation of chosen modes.

Algebraic Viewpoint Preview

A more invariant way to phrase the issue is to define subsystems by subalgebras of observables. Instead of starting with particle labels, one specifies which observables belong to region $A$ and which belong to region $B$.

For finite-dimensional tensor factors, let $\mathcal A$ and $\mathcal B$ be their commuting local observable algebras. A state is separable across this split if its expectation values admit a finite convex mixture of products:

$$\omega(AB) = \sum_k p_k\, \omega_k^{(A)}(A)\, \omega_k^{(B)}(B),$$

for $A\in\mathcal A$, $B\in\mathcal B$, $p_k\ge0$, and $\sum_k p_k=1$. If no such representation exists, the state is entangled with respect to that split.

For normal states on an infinite-dimensional Hilbert-space tensor product, separability instead uses the trace-norm closed convex hull of product states. An exact countable sum need not exist, as Holevo, Shirokov, and Werner show. The rigorous extension explains this distinction. More general observable algebras require their own state-space and closure choices; the displayed finite-dimensional criterion is only a preview of that setting.

Exercises

1. Slot nonfactorization. Explain why the antisymmetric state

$$\frac{1}{\sqrt2} \bigl( \lvert a\rangle_1\lvert b\rangle_2 - \lvert b\rangle_1\lvert a\rangle_2 \bigr)$$

should not automatically be advertised as useful entanglement between two particles.

Solution

The labels $1$ and $2$ are formal slots, not observable particle identities. The nonfactorization is required by fermionic antisymmetry. In occupation notation the same single Slater determinant is $\lvert1_a,1_b\rangle_F$, which is a definite occupation state of modes $a$ and $b$. Whether there is entanglement depends on a physical split into modes, regions, or observables, not on the formal slot labels alone.

2. Two copies and local number. One copy of $(|1_L,0_R\rangle+|0_L,1_R\rangle)/\sqrt2$ has no accessible entanglement under local number conservation. Now prepare two copies in four orthogonal bosonic modes $L_1,R_1,L_2,R_2$, with the same relative phase. Alice controls $L_1,L_2$ and Bob controls $R_1,R_2$. Each may mix their two local modes while conserving their total local number. Without a shared phase reference between laboratories, project onto Alice’s total number $N_A=0,1,2$. Find the probabilities and the average entanglement accessible within these sectors.

Solution

Regrouping by laboratory, with each two-digit ket listing that laboratory’s two modes, gives

$$|\Psi\rangle=\frac12\bigl( |11\rangle_A|00\rangle_B+|10\rangle_A|01\rangle_B +|01\rangle_A|10\rangle_B+|00\rangle_A|11\rangle_B\bigr).$$

The sector probabilities are $1/4,1/2,1/4$. The outer sectors are product states. Conditioned on $N_A=1$, the normalized state is

$$|\Psi_1\rangle=\frac{|10\rangle_A|01\rangle_B+|01\rangle_A|10\rangle_B}{\sqrt2}.$$

Each local logical qubit has exactly one particle; its two basis states differ in the occupied mode. Local number conservation therefore permits the qubit operations, and the reduced entropy is one bit. The sector-weighted average is $(1/4)0+(1/2)1+(1/4)0=1/2$ ebit per pair of copies. A successful projection yields one ebit with probability $1/2$, not one ebit deterministically.

The copy labels distinguish physical modes, not particle identities. A joint local phase average preserves the middle-sector coherence. Independently randomizing the two copies’ relative phases would destroy it. This is the two-copy example of Wiseman and Vaccaro; the allowed operations and phase relation are essential.

3. NOON reduced state. For

$$\lvert\psi\rangle = \frac{ \lvert2_L,0_R\rangle + \lvert0_L,2_R\rangle }{\sqrt2},$$

compute the reduced density operator of mode $L$.

Solution

The density operator is

$$\rho = \frac12 \bigl( \lvert2,0\rangle\langle2,0\rvert + \lvert2,0\rangle\langle0,2\rvert + \lvert0,2\rangle\langle2,0\rvert + \lvert0,2\rangle\langle0,2\rvert \bigr).$$

Tracing over mode $R$ kills the cross terms because $\langle0_R\vert2_R\rangle=0$. Therefore

$$\rho_L = \frac12 \bigl( \lvert2_L\rangle\langle2_L\rvert + \lvert0_L\rangle\langle0_L\rvert \bigr).$$

4. Region spin entanglement. Why is

$$\frac{ c_{L\uparrow}^\dagger c_{R\downarrow}^\dagger - c_{L\downarrow}^\dagger c_{R\uparrow}^\dagger }{\sqrt2} \lvert0\rangle$$

better described as entanglement between left and right regions than as entanglement between particle $1$ and particle $2$?

Solution

The labels $L$ and $R$ refer to physical regions or modes that can be measured locally. Particle labels $1$ and $2$ are not observable identities for identical particles. If each region contains one particle and each laboratory can measure local spin, the state has operational singlet-like correlations between the regions.

References

• S. D. Bartlett, T. Rudolph, and R. W. Spekkens, “Reference Frames, Superselection Rules, and Quantum Information”, Reviews of Modern Physics 79, 555–609 (2007).
• F. Benatti, R. Floreanini, and U. Marzolino, “Entanglement in systems of identical particles: a unified approach,” Annals of Physics 325, 924-935, 2010.
• K. Eckert, J. Schliemann, D. Bruß, and M. Lewenstein, “Quantum correlations in systems of indistinguishable particles,” Annals of Physics 299, 88-127, 2002.
• A. S. Holevo, M. E. Shirokov, and R. F. Werner, “Separability and Entanglement-Breaking in Infinite Dimensions”, arXiv:quant-ph/0504204 (2005), Definition 1 and Theorem 3.
• R. Lo Franco and G. Compagno, “Indistinguishability of elementary systems as a resource for quantum information processing,” Physical Review Letters 120, 240403, 2018.
• A. M. L. Messiah and O. W. Greenberg, “Symmetrization Postulate and Its Experimental Foundation”, Physical Review 136, B248–B267 (1964).
• J. Schliemann, J. I. Cirac, M. Kuś, M. Lewenstein, and D. Loss, “Quantum correlations in two-fermion systems,” Physical Review A 64, 022303, 2001.
• H. M. Wiseman and J. A. Vaccaro, “Entanglement of Indistinguishable Particles Shared between Two Parties”, Physical Review Letters 91, 097902 (2003).