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 $\mathcal A_N$.

The symmetrization postulate selects the physical sector; here we construct the operators that project onto it. Symmetric and Antisymmetric Wavefunctions supplies coordinate-space normalization examples.

Helpful background. Symmetrization Postulate states the sector rule; Operators on Composite Systems explains tensor-factor actions.

Exchange on a tensor-product space

Let $\mathcal h$ be the complete one-particle Hilbert space. On $\mathcal h\otimes\mathcal h$, 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, use rightmost-first composition: $(\pi\sigma)(i)=\pi(\sigma(i))$. Define the action on product kets by

$$U(\pi)\lvert\phi_1,\ldots,\phi_N\rangle =\lvert\phi_{\pi^{-1}(1)},\ldots,\phi_{\pi^{-1}(N)}\rangle.$$

Then $U(\pi)U(\sigma)=U(\pi\sigma)$. In the coordinate wavefunction, the corresponding argument permutation is

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

The inverse appears in the product-ket ordering, while the coordinate arguments use the forward permutation. Keeping these two actions distinct makes the composition rule consistent. A transposition equals its inverse, which is why this distinction is invisible in two-particle examples.

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:

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

A fixed permutation occurs exactly $N!$ times in the double sum over two group elements. Together with the multiplicative sign, this proves

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

For $N\ge3$, these two projectors generally do not sum to the identity on the unrestricted tensor product: other permutation-symmetry sectors exist. The ordinary boson/fermion postulate excludes those sectors for the species under discussion. It is additional physical input, not a consequence of permutation invariance alone.

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.

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

With the rightmost transposition applied first,

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

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.

References

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