Symmetric Group

The symmetric group $S_N$ is the group of all permutations of $N$ labels. It is the finite group behind particle exchange, tensor-slot relabeling, determinants, permanents, and many-body basis symmetries.

This Toolkit page owns the group-theoretic object. The physical rule selecting bosonic or fermionic sectors is the Symmetrization Postulate, and the operator-level treatment of particle exchange is Exchange Operators.

Definition

Let

$$\{1,2,\ldots,N\}$$

be a finite set of labels. The symmetric group $S_N$ is the set of all bijections from this set to itself, with composition as the group operation.

The identity element is the permutation that fixes every label. The inverse of a permutation $\pi$ is the inverse bijection $\pi^{-1}$. The number of elements is

$$\lvert S_N\rvert=N!.$$

For $N=1$, the group is trivial. For $N=2$, there are two elements: the identity and the swap $(12)$. For $N=3$, there are six elements.

Cycle Notation

A cycle

$$(i_1\,i_2\,\cdots\,i_k)$$

sends

$$i_1\mapsto i_2,\quad i_2\mapsto i_3,\quad \ldots,\quad i_k\mapsto i_1,$$

and leaves labels not listed in the cycle unchanged. For example, in $S_4$,

$$(123)(4)$$

is usually written simply as $(123)$.

Every permutation can be written as a product of disjoint cycles, and disjoint cycles commute. The same permutation can also be written as a product of transpositions, where a transposition swaps two labels:

$$(ij).$$

The number of transpositions in such a product is not unique, but its parity is unique. This parity defines the sign representation.

Sign Representation

The sign of a permutation is

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

Equivalently, if $\pi$ is written as a product of transpositions, $\operatorname{sgn}(\pi)=+1$ when the number of transpositions is even and $\operatorname{sgn}(\pi)=-1$ when it is odd.

The sign is a one-dimensional representation:

$$\operatorname{sgn}(\pi\sigma) = \operatorname{sgn}(\pi)\operatorname{sgn}(\sigma).$$

The kernel of this representation is the alternating group $A_N$, the subgroup of even permutations.

Adjacent Transposition Generators

The group $S_N$ is generated by the adjacent transpositions

$$s_i=(i\,\,i+1), \qquad i=1,\ldots,N-1.$$

They obey the Coxeter relations

$$s_i^2=e,$$ $$s_i s_j=s_j s_i \quad \text{if}\quad \lvert i-j\rvert>1,$$

and

$$s_i s_{i+1}s_i = s_{i+1}s_i s_{i+1}.$$

The last relation says that two different ways of interchanging neighboring labels through a three-label block give the same final permutation. It is the finite-permutation cousin of the braid relation; in ordinary permutation groups each generator also squares to the identity.

Action on Tensor Slots

Let $\mathcal h$ be a one-particle vector space or Hilbert space. The $N$-slot tensor product is

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

The symmetric group acts by permuting tensor slots. Define $U(\pi)$ on product vectors by

$$U(\pi) (v_1\otimes v_2\otimes\cdots\otimes v_N) = v_{\pi^{-1}(1)} \otimes v_{\pi^{-1}(2)} \otimes\cdots\otimes v_{\pi^{-1}(N)}.$$

The inverse appears so that

$$U(\pi)U(\sigma)=U(\pi\sigma).$$

If $\mathcal h$ is a Hilbert space, these operators are unitary. They form the permutation representation of $S_N$ on $\mathcal h^{\otimes N}$.

For $N=2$, the nontrivial operator is

$$U(12)(v\otimes w)=w\otimes v.$$

Symmetric and Antisymmetric Projectors

The fully symmetric projector is

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

The fully antisymmetric projector is

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

They satisfy

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

The symmetric sector carries the trivial representation of $S_N$:

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

The antisymmetric sector carries the sign representation:

$$U(\pi)\lvert\Psi\rangle = \operatorname{sgn}(\pi)\lvert\Psi\rangle.$$

For two slots,

$$\mathcal S_2=\frac{1}{2}(I+U(12)), \qquad \mathcal A_2=\frac{1}{2}(I-U(12)).$$

These are the projectors onto symmetric and antisymmetric two-particle states.

Bosons and Fermions Preview

In ordinary nonrelativistic quantum mechanics, identical bosons are represented by states in the fully symmetric sector, and identical fermions are represented by states in the fully antisymmetric sector:

$$\text{bosons:}\quad \mathcal S_N\lvert\Psi\rangle=\lvert\Psi\rangle,$$ $$\text{fermions:}\quad \mathcal A_N\lvert\Psi\rangle=\lvert\Psi\rangle.$$

This page explains the group theory behind those projectors. It does not replace the physical postulate. The postulate and its consequences are developed in Bosons, Fermions, and Symmetric and Antisymmetric Wavefunctions.

Determinants and Permanents

The antisymmetrizer produces determinants. If $\varphi_1,\ldots,\varphi_N$ are one-particle states, then the antisymmetrized tensor

$$\mathcal A_N (\varphi_1\otimes\cdots\otimes\varphi_N)$$

is a signed sum over permutations. In coordinate representation this is the structure behind a Slater determinant.

The symmetrizer produces permanents: the same permutation sum but without signs. This is the corresponding structure for bosonic states with repeated occupation allowed.

The detailed many-body wavefunction constructions belong to Slater Determinants and Permanents.

Irreducible Representations

The trivial and sign representations are only two representations of $S_N$. For $N\geq3$, the symmetric group has additional irreducible representations, organized by partitions of $N$ and Young diagrams.

These mixed-symmetry representations matter in atomic, molecular, nuclear, and many-body theory, especially when spin, flavor, orbital, or internal labels are being organized simultaneously. The full total state of ordinary identical bosons or fermions is still selected by the symmetrization postulate; mixed symmetry is usually a classification tool for parts of the state or for systems with additional structure.

In two spatial dimensions, particle exchange can lead to braid-group representations rather than ordinary symmetric-group representations. Anyons and Braiding develops that quotient boundary, the retained winding data, and its physical representations.

Common Mistakes

• Treating tensor slots as physical particle names for identical particles.
• Forgetting the inverse in the slot action and accidentally making an anti-representation.
• Confusing the sign of a permutation with the sign of a wavefunction under an arbitrary coordinate transformation.
• Assuming every $S_N$ representation is a boson or fermion sector.
• Forgetting that symmetrization or antisymmetrization applies to the total state, including spin and spatial factors.
• Using the antisymmetrizer on more fermions than there are available orthonormal one-particle states and expecting a nonzero result.

Cross-Links

• Groups
• Group Actions
• Representations
• Tensor Products
• Exchange Operators
• Symmetrization Postulate
• Bosons
• Fermions
• Symmetric and Antisymmetric Wavefunctions
• Slater Determinants
• Permanents

References

• B. E. Sagan, The Symmetric Group, 2nd ed., Springer, 2001.
• W. Fulton and J. Harris, Representation Theory: A First Course, Springer, 1991.
• M. Hamermesh, Group Theory and Its Application to Physical Problems, Dover, 1989.
• A. Messiah, Quantum Mechanics, Vol. II, Dover, 1999.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.

Exercises

1. List the elements of $S_3$ in cycle notation.

Solution

They are

$$e,\quad (12),\quad (13),\quad (23),\quad (123),\quad (132).$$

The first is the identity, the next three are transpositions, and the last two are three-cycles.

2. Show that the sign representation is multiplicative.

Solution

Write $\pi$ as a product of $r$ transpositions and $\sigma$ as a product of $s$ transpositions. Then $\pi\sigma$ is written as a product of $r+s$ transpositions, so

$$\operatorname{sgn}(\pi\sigma) = (-1)^{r+s} = (-1)^r(-1)^s = \operatorname{sgn}(\pi)\operatorname{sgn}(\sigma).$$

The parity of the number of transpositions is independent of the chosen decomposition, so the argument is well-defined.

3. Verify that $\mathcal S_2$ is a projector.

Solution

Since $U(12)^2=I$,

$$\mathcal S_2^2 = \frac14(I+U(12))^2 = \frac14(I+2U(12)+I) = \frac12(I+U(12)) = \mathcal S_2.$$

4. Why does the antisymmetrizer kill a product with two identical one-particle factors?

Solution

For two factors,

$$\mathcal A_2(v\otimes v) = \frac12(v\otimes v-v\otimes v) = 0.$$

This is the two-particle form of Pauli exclusion: an antisymmetric state cannot place two fermions in the same one-particle state.