Exchange Operators

An exchange operator is a unitary operator that permutes the slot labels in a many-particle tensor product. For identical particles, those slots are bookkeeping devices rather than observable particle names, so exchange operators are the mathematical language for the symmetry constraints on states and observables.

For two particles, the basic operator is $P_{12}$, which swaps slots $1$ and $2$. Its eigenspaces with eigenvalues $+1$ and $-1$ are the symmetric and antisymmetric two-particle sectors. For $N$ particles, the operators $U(\pi)$ representing permutations $\pi\in S_N$ lead to the symmetrizer $\mathcal S_N$ and antisymmetrizer $\mathcal A_N$.

This page is the canonical home for the operator-level treatment. The symmetrization postulate states which sector is used by bosons and fermions; Symmetric and Antisymmetric Wavefunctions works out explicit coordinate-space wavefunctions and normalization.

Slot-Labeled Space

Let $\mathcal h$ be the one-particle Hilbert space for a species of identical particles. The formal $N$-slot tensor product is

$$\mathcal h^{\otimes N} = \underbrace{ \mathcal h\otimes\cdots\otimes\mathcal h }_{N\ \text{factors}}.$$

The factors are slots. A product vector such as

$$\lvert\phi_1\rangle_1 \lvert\phi_2\rangle_2 \cdots \lvert\phi_N\rangle_N$$

is a useful formal object, but the subscripts do not name persistent physical particles when the particles are identical. They label places in the tensor product on which permutation operators act.

This distinction is essential. Exchange operators do not reveal hidden identities. They compare different slot descriptions of the same unlabeled physical situation.

Two-Slot Exchange

For two slots, define the exchange operator $P_{12}$ by its action on product vectors:

$$P_{12} \bigl( \lvert\alpha\rangle_1\lvert\beta\rangle_2 \bigr) = \lvert\beta\rangle_1\lvert\alpha\rangle_2.$$

Linearity extends this definition to every vector in $\mathcal h\otimes\mathcal h$. If a general vector is

$$\lvert\Psi\rangle = \sum_{ij} C_{ij}\, \lvert i\rangle_1\lvert j\rangle_2,$$

then

$$P_{12}\lvert\Psi\rangle = \sum_{ij} C_{ij}\, \lvert j\rangle_1\lvert i\rangle_2.$$

Relabeling dummy summation indices gives

$$P_{12}\lvert\Psi\rangle = \sum_{ij} C_{ji}\, \lvert i\rangle_1\lvert j\rangle_2.$$

Thus, in coefficient language, exchange transposes the coefficient array.

In a coordinate-spin representation with $q=(\mathbf x,s,\ldots)$,

$$(P_{12}\Psi)(q_1,q_2) = \Psi(q_2,q_1).$$

The complete one-particle label is exchanged. If spin, polarization, isospin, band index, or another internal label belongs to the one-particle state, it is part of $q$.

Unitarity and Eigenvalues

The exchange operator preserves inner products. For $u=\lvert\alpha\rangle_1\lvert\beta\rangle_2$ and $v=\lvert\gamma\rangle_1\lvert\delta\rangle_2$,

$$\begin{aligned} \langle P_{12}u\vert P_{12}v\rangle &= \langle\beta\vert\delta\rangle \langle\alpha\vert\gamma\rangle = \langle\alpha\vert\gamma\rangle \langle\beta\vert\delta\rangle. \end{aligned}$$

By linearity and continuity, $P_{12}$ is unitary on the whole two-slot Hilbert space:

$$P_{12}^\dagger P_{12}=I.$$

Applying exchange twice returns the original slot order:

$$P_{12}^2=I.$$

Therefore

$$P_{12}^{-1}=P_{12}, \qquad P_{12}^\dagger=P_{12}.$$

So $P_{12}$ is both unitary and Hermitian. If $\lvert\Psi\rangle$ is an eigenvector,

$$P_{12}\lvert\Psi\rangle = \lambda\lvert\Psi\rangle,$$

then applying $P_{12}$ again gives

$$\lambda^2\lvert\Psi\rangle = \lvert\Psi\rangle.$$

For a nonzero state, $\lambda^2=1$, hence

$$\lambda=\pm1.$$

The $\lambda=+1$ eigenspace is the symmetric two-slot subspace. The $\lambda=-1$ eigenspace is the antisymmetric two-slot subspace.

Two-Slot Projectors

The symmetric and antisymmetric projectors are

$$\Pi_S = \frac12(I+P_{12}), \qquad \Pi_A = \frac12(I-P_{12}).$$

Using $P_{12}^2=I$,

$$\Pi_S^2 = \frac14(I+P_{12})(I+P_{12}) = \frac12(I+P_{12}) = \Pi_S,$$

and similarly

$$\Pi_A^2=\Pi_A.$$

They are orthogonal projectors because

$$\Pi_S\Pi_A = \frac14(I+P_{12})(I-P_{12}) = 0,$$

and

$$\Pi_S+\Pi_A=I.$$

Any two-slot state decomposes uniquely as

$$\lvert\Psi\rangle = \Pi_S\lvert\Psi\rangle + \Pi_A\lvert\Psi\rangle.$$

For identical bosons, the physical two-particle space is the symmetric part. For identical fermions, it is the antisymmetric part.

Projecting Product States

Given two one-particle states $\lvert a\rangle$ and $\lvert b\rangle$, the projectors give

$$\Pi_S \lvert a\rangle_1\lvert b\rangle_2 = \frac12 \bigl( \lvert a\rangle_1\lvert b\rangle_2 + \lvert b\rangle_1\lvert a\rangle_2 \bigr),$$

and

$$\Pi_A \lvert a\rangle_1\lvert b\rangle_2 = \frac12 \bigl( \lvert a\rangle_1\lvert b\rangle_2 - \lvert b\rangle_1\lvert a\rangle_2 \bigr).$$

These projected vectors are not automatically normalized. For distinct orthonormal one-particle states, the normalized versions carry the familiar factor $1/\sqrt2$. If the one-particle states overlap, the normalization changes; if $a=b$, the antisymmetric projection is zero:

$$\Pi_A \lvert a\rangle_1\lvert a\rangle_2 = 0.$$

That last equation is the two-slot operator form of the Pauli exclusion principle.

Permutation Group Preview

For $N$ slots, every permutation $\pi\in S_N$ has a corresponding unitary slot-permutation operator $U(\pi)$. A common coordinate convention is

$$(U(\pi)\Psi)(q_1,\ldots,q_N) = \Psi(q_{\pi^{-1}(1)},\ldots,q_{\pi^{-1}(N)}).$$

The inverse in this formula is a convention tied to how one composes permutations. What matters physically is that the family $U(\pi)$ gives a unitary action of the permutation group on the slot-labeled space, and that transpositions generate all permutations.

For a transposition $(ij)$, $U((ij))$ swaps slots $i$ and $j$. Since any permutation can be written as a product of transpositions, the behavior of a state under pair exchanges determines its behavior under all of $S_N$.

The sign of a permutation is

$$\operatorname{sgn}(\pi) = \begin{cases} +1, & \pi\ \text{even},\\ -1, & \pi\ \text{odd}. \end{cases}$$

An even permutation is a product of an even number of transpositions; an odd permutation is a product of an odd number. The parity is well-defined even though the decomposition into transpositions is not unique.

Thus a fermionic state does not pick up a minus sign under every nonidentity permutation. It picks up a minus sign under odd permutations and no sign change under even permutations. For example, a three-cycle is even because it can be written as two transpositions.

N-Particle Symmetrizers

The $N$-particle bosonic symmetrizer is

$$\mathcal S_N = \frac{1}{N!} \sum_{\pi\in S_N} U(\pi).$$

It projects onto the completely symmetric subspace:

$$\operatorname{Sym}^N\mathcal h = \mathcal S_N \bigl(\mathcal h^{\otimes N}\bigr).$$

The $N$-particle fermionic antisymmetrizer is

$$\mathcal A_N = \frac{1}{N!} \sum_{\pi\in S_N} \operatorname{sgn}(\pi)U(\pi).$$

It projects onto the completely antisymmetric subspace:

$$\wedge^N\mathcal h = \mathcal A_N \bigl(\mathcal h^{\otimes N}\bigr).$$

The group property implies

$$\mathcal S_N^2=\mathcal S_N, \qquad \mathcal A_N^2=\mathcal A_N.$$

The antisymmetrizer automatically kills any product vector with two identical one-particle factors. If slots $r$ and $s$ both carry $\lvert a\rangle$, then the transposition $(rs)$ leaves the product vector unchanged, while antisymmetry demands a minus sign. The only vector compatible with both statements is zero.

This is the operator reason Slater determinants vanish when two columns are identical. The determinant form is developed in Slater Determinants.

Observables Must Commute with Exchange

For identical particles, a physical observable cannot depend on an arbitrary slot name. The operator condition is

$$[O,U(\pi)]=0 \qquad \text{for all }\pi\in S_N.$$

For two slots this becomes

$$[O,P_{12}]=0.$$

Equivalently,

$$U(\pi)OU(\pi)^{-1}=O.$$

This condition says that the observable is invariant under relabeling the formal slots. It does not say that every operator on $\mathcal h^{\otimes N}$ is physically available for identical particles.

For example, on $\mathcal h\otimes\mathcal h$,

$$P_{12}(A\otimes I)P_{12} = I\otimes A.$$

Therefore $A\otimes I$ alone is not exchange-invariant unless it equals $I\otimes A$ on the relevant space. The exchange-invariant one-body observable is the symmetric sum

$$A_1+A_2 = A\otimes I+I\otimes A.$$

For $N$ identical particles, the corresponding one-body observable is

$$\sum_{i=1}^{N} A_i,$$

where the same one-particle operator $A$ acts in each slot. Two-body interactions likewise appear as symmetric sums, such as

$$\sum_{1\le i<j\le N} V_{ij},$$

with the same pair interaction assigned to each pair of identical particles.

Exchange-Invariant Dynamics

If the Hamiltonian is a physical identical-particle Hamiltonian, it commutes with all particle permutations:

$$[H,U(\pi)]=0 \qquad \text{for all }\pi\in S_N.$$

Then exchange symmetry is preserved by time evolution. If

$$U(\pi)\lvert\Psi(0)\rangle = \chi(\pi)\lvert\Psi(0)\rangle,$$

where $\chi(\pi)=1$ for bosons or $\chi(\pi)=\operatorname{sgn}(\pi)$ for fermions, then

$$\begin{aligned} U(\pi)e^{-iHt/\hbar}\lvert\Psi(0)\rangle &= e^{-iHt/\hbar}U(\pi)\lvert\Psi(0)\rangle \\ &= \chi(\pi)e^{-iHt/\hbar}\lvert\Psi(0)\rangle. \end{aligned}$$

Thus the time-evolved state remains in the same exchange sector. A calculation that starts with a properly symmetric or antisymmetric state will not leave that sector if the Hamiltonian treats the identical particles identically.

Exchange Is Not Spatial Reflection

Exchange symmetry is sometimes confused with ordinary spatial symmetries. They are different.

For two particles on a line, exchange sends

$$(x_1,x_2)\longmapsto(x_2,x_1),$$

while parity sends

$$(x_1,x_2)\longmapsto(-x_1,-x_2).$$

These are different transformations on configuration space. A wavefunction can be symmetric under particle exchange and odd under parity, or antisymmetric under exchange and even under parity. The exchange rule concerns particle slots, not reflection through the origin.

Worked Example: One-Body Observable

Let $A$ be a one-particle observable on $\mathcal h$. Define

$$O=A_1+A_2 = A\otimes I+I\otimes A.$$

Using $P_{12}(A\otimes I)P_{12}=I\otimes A$ and $P_{12}^2=I$,

$$\begin{aligned} P_{12}OP_{12} &= P_{12}(A\otimes I)P_{12} + P_{12}(I\otimes A)P_{12}\\ &= I\otimes A+A\otimes I\\ &=O. \end{aligned}$$

Therefore

$$[O,P_{12}]=0.$$

The operator $O$ is a physical observable for two identical particles. It represents the total value of the one-particle quantity $A$, not a measurement of a named particle $1$.

Common Mistakes

• Treating $P_{12}$ as a dynamical process in time rather than a slot-permutation operator.
• Reading the subscripts in $\mathcal h_1\otimes\mathcal h_2$ as observable particle names.
• Forgetting that exchange acts on all one-particle degrees of freedom, including spin and internal labels.
• Saying a fermionic state changes sign under every nonidentity permutation; only odd permutations produce the minus sign.
• Using $A_1$ as a physical observable for identical particles without forming an exchange-invariant sum or density.
• Confusing exchange symmetry with parity, rotation, or mirror symmetry in ordinary space.
• Projecting with $\Pi_S$ or $\Pi_A$ and forgetting to normalize the result when it is nonzero.

Cross-Links

• Indistinguishability
• Symmetrization Postulate
• Symmetric Group
• Anyons and Braiding
• Bosons
• Fermions
• Pauli Exclusion Principle
• Symmetric and Antisymmetric Wavefunctions
• Spin and Spatial Wavefunctions
• Slater Determinants
• Permanents
• Operators on Composite Systems
• Occupation-Number Basis
• Bosonic Fock Space
• Fermionic Fock Space
• Identical Particle Exercises
• Formula Sheet

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• A. Messiah and O. W. Greenberg, “Symmetrization Postulate and Its Experimental Foundation,” Physical Review 136, B248-B267, 1964.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloe, Quantum Mechanics, Wiley, 1977.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover, 2003.

Exercises

1. Exchange algebra. Use $P_{12}^2=I$ and unitarity to show that $P_{12}$ is Hermitian and has only eigenvalues $\pm1$.

Solution

Unitarity gives $P_{12}^{-1}=P_{12}^\dagger$. Since $P_{12}^2=I$, one also has $P_{12}^{-1}=P_{12}$. Therefore $P_{12}^\dagger=P_{12}$.

If $P_{12}\lvert\Psi\rangle=\lambda\lvert\Psi\rangle$ for a nonzero vector, then applying $P_{12}$ again gives

$$\lvert\Psi\rangle = P_{12}^2\lvert\Psi\rangle = \lambda^2\lvert\Psi\rangle.$$

Thus $\lambda^2=1$, so $\lambda=\pm1$.

2. Projectors. Prove that $\Pi_S=(I+P_{12})/2$ and $\Pi_A=(I-P_{12})/2$ are orthogonal projectors.

Solution

Using $P_{12}^2=I$,

$$\Pi_S^2 = \frac14(I+P_{12})^2 = \frac14(I+2P_{12}+I) = \Pi_S.$$

Similarly,

$$\Pi_A^2 = \frac14(I-P_{12})^2 = \frac14(I-2P_{12}+I) = \Pi_A.$$

Their product is

$$\Pi_S\Pi_A = \frac14(I+P_{12})(I-P_{12}) = \frac14(I-P_{12}^2) = 0.$$

Since $P_{12}$ is Hermitian, both projectors are Hermitian.

3. Pauli exclusion from projection. Show that $\Pi_A\lvert a\rangle_1\lvert a\rangle_2=0$.

Solution

By definition,

$$\Pi_A \lvert a\rangle_1\lvert a\rangle_2 = \frac12 \bigl( \lvert a\rangle_1\lvert a\rangle_2 - P_{12}\lvert a\rangle_1\lvert a\rangle_2 \bigr).$$

Since exchanging two identical factors changes nothing,

$$P_{12}\lvert a\rangle_1\lvert a\rangle_2 = \lvert a\rangle_1\lvert a\rangle_2.$$

Therefore the two terms cancel and the antisymmetric projection is zero.

4. Even permutation sign. Write the three-cycle $(123)$ as a product of two transpositions and determine the sign acquired by a fermionic state.

Solution

One decomposition is

$$(123)=(12)(23),$$

up to the chosen convention for composing permutations. It uses two transpositions, so the permutation is even:

$$\operatorname{sgn}(123)=+1.$$

A fermionic state is therefore unchanged by this even permutation:

$$U((123))\lvert\Psi\rangle = \lvert\Psi\rangle.$$

Only odd permutations give a minus sign.

5. Observable invariance. Let $O=A_1+A_2$ with $A_1=A\otimes I$ and $A_2=I\otimes A$. Show that $[O,P_{12}]=0$.

Solution

Exchange swaps the two slots:

$$P_{12}A_1P_{12}=A_2, \qquad P_{12}A_2P_{12}=A_1.$$

Therefore

$$P_{12}OP_{12} = P_{12}(A_1+A_2)P_{12} = A_2+A_1 = O.$$

Multiplying on the right by $P_{12}$ and using $P_{12}^2=I$ gives $P_{12}O=OP_{12}$, hence $[O,P_{12}]=0$.

6. Time evolution. Suppose $[H,P_{12}]=0$ and $P_{12}\lvert\Psi(0)\rangle=\eta\lvert\Psi(0)\rangle$ with $\eta=\pm1$. Show that the exchange eigenvalue is preserved in time.

Solution

Since $[H,P_{12}]=0$, $P_{12}$ also commutes with $e^{-iHt/\hbar}$. Thus

$$\begin{aligned} P_{12}\lvert\Psi(t)\rangle &= P_{12}e^{-iHt/\hbar}\lvert\Psi(0)\rangle\\ &= e^{-iHt/\hbar}P_{12}\lvert\Psi(0)\rangle\\ &= \eta e^{-iHt/\hbar}\lvert\Psi(0)\rangle\\ &= \eta\lvert\Psi(t)\rangle. \end{aligned}$$

The state remains in the same symmetric or antisymmetric sector.