Symmetric and Antisymmetric Tensor Products

Symmetric and antisymmetric Hilbert powers are closed subspaces of an ordinary tensor power, obtained by averaging the permutation representation. Their projectors carry exact normalization factors, which become especially important when the one-particle vectors are not orthogonal. The canonical Symmetric and Antisymmetric Wavefunctions article supplies exchange physics; the construction here supplies its operator foundation.

Helpful background. Hilbert Space Tensor Products supplies completed tensor powers. Symmetrization Postulate explains why the bosonic or fermionic sector is selected for the physical model.

Permutations act by bounded unitaries

Let $\mathcal h$ be a one-particle Hilbert space and fix a positive integer $n$. For a permutation $\pi\in\mathfrak S_n$, define

$$U_\pi(x_1\otimes\cdots\otimes x_n) =x_{\pi^{-1}(1)}\otimes\cdots\otimes x_{\pi^{-1}(n)}.$$

The inverse in this convention ensures $U_\pi U_\sigma=U_{\pi\sigma}$. Product inner products are preserved, so this algebraic action extends uniquely to a unitary on $\mathcal h^{\otimes n}$, with $U_\pi^*=U_{\pi^{-1}}$.

Define

$$S_n=\frac1{n!}\sum_{\pi\in\mathfrak S_n}U_\pi, \qquad A_n=\frac1{n!}\sum_{\pi\in\mathfrak S_n}\operatorname{sgn}(\pi)U_\pi.$$

Here $\mathfrak S_n$ denotes the permutation group and $S_n$ its averaging operator.

Group averaging produces orthogonal projections

Both sums are self-adjoint: inversion permutes the group and preserves the sign character. For either character $\chi(\pi)=1$ or $\operatorname{sgn}(\pi)$, write $P_\chi=(n!)^{-1}\sum_\pi\chi(\pi)U_\pi$. In its square, every fixed product $\tau=\pi\sigma$ appears exactly $n!$ times, and $\chi(\pi)\chi(\sigma)=\chi(\tau)$. Therefore

$$P_\chi^2 =\frac1{(n!)^2}\sum_\tau n!\,\chi(\tau)U_\tau =P_\chi.$$

Thus $S_n,A_n$ are orthogonal projections. Their ranges are closed Hilbert subspaces,

$$\operatorname{Sym}^n\mathcal h=S_n\mathcal h^{\otimes n}, \qquad \bigwedge\nolimits^n\mathcal h=A_n\mathcal h^{\otimes n}.$$

Multiplying the average by $U_\sigma$ shows that range vectors transform respectively by $1$ and $\operatorname{sgn}(\sigma)$. Conversely a vector with that transformation rule is fixed by the corresponding average, proving the range characterization.

For $n\geq2$, the two ranges are orthogonal: a transposition has eigenvalues $+1$ and $-1$ on them. For $n\geq3$, their sum generally does not exhaust the whole tensor power; other permutation symmetry types can occur. Selecting bosonic or fermionic states is extra physical input, not a claim that only these two mathematical subspaces exist.

Gram matrices determine normalization

Given one-particle vectors $x_1,\ldots,x_n$, let $G_{ij}=\langle x_i,x_j\rangle$ and $X=x_1\otimes\cdots\otimes x_n$. Projection idempotence gives

$$\|A_nX\|^2=\langle X,A_nX\rangle =\frac{\det G}{n!}.$$

The last step expands the permutation sum into the determinant. Replacing signs by plus signs gives

$$\|S_nX\|^2=\frac{\operatorname{per}G}{n!},$$

where the permanent has the determinant’s permutation products without alternating signs.

With the convention $x_1\wedge\cdots\wedge x_n=\sqrt{n!}\,A_nX$, the wedge norm squared is $\det G$. An orthonormal list therefore gives a unit wedge. A linearly dependent list has zero Gram determinant and produces the zero vector. This is the precise linear-algebra mechanism behind exclusion from a repeated complete one-particle state.

For symmetric products drawn from an orthonormal mode basis with occupation counts $n_j$, the Gram permanent is $\prod_jn_j!$. Accordingly $\sqrt{n!/\prod_jn_j!}\,S_nX$ is normalized. This factor handles repeated bosonic modes without treating them as distinct orthogonal vectors.

Three nonorthogonal fermion orbitals

Choose orthonormal $e_1,e_2,e_3$ and

$$x_1=e_1,\qquad x_2=\frac{e_1+e_2}{\sqrt2},\qquad x_3=\frac{e_2+e_3}{\sqrt2}.$$

Their Gram matrix is

$$G= \begin{pmatrix} 1&1/\sqrt2&0\\ 1/\sqrt2&1&1/2\\ 0&1/2&1 \end{pmatrix}, \qquad \det G=\frac14.$$

Thus $x_1\wedge x_2\wedge x_3$ has norm $1/2$, although all three orbitals are individually normalized. Multiplying the wedge by two normalizes the three-particle state. The naive orthonormal-orbital normalization would miss a factor.

Particle labels are not automatically subsystem labels

The construction starts with labeled tensor slots, but the physical symmetric or antisymmetric subspace need not inherit a tensor factorization into independently addressable particles. A one-particle reduced operator can still be defined through the ambient tensor space; its interpretation and normalization must distinguish particle counting from a bipartite subsystem density operator.

Mode or spatial-region entanglement requires a declared observable split. Symmetrization alone should not be identified with an operational entanglement resource. The tensor projectors also do not prove the relativistic spin–statistics theorem; they implement a sector choice made elsewhere.

Exercises

1. For finite $\dim\mathcal h=d$, determine the dimensions of the symmetric and antisymmetric $n$-particle spaces.

Solution

An antisymmetric basis chooses $n$ distinct modes, giving $\binom dn$, zero for $n>d$. A symmetric basis chooses nonnegative occupation counts summing to $n$, giving $\binom{d+n-1}{n}$. These counts are independent of a particular one-particle basis choice.

2. For $d=2,n=3$, show that the symmetric and antisymmetric sectors do not exhaust the tensor power.

Solution

The full tensor power has dimension $2^3=8$. The symmetric sector has dimension $\binom43=4$, while the antisymmetric sector is zero because three distinct modes cannot be chosen from two. The remaining four dimensions have other permutation symmetry type.

References

• Dereziński, Jan, “Introduction to Representations of the Canonical Commutation and Anticommutation Relations,” in Large Coulomb Systems, Lecture Notes in Physics 695, Springer, 2006, pp. 63–143, Section 5 — permutation projections and Fock sectors. Paper; preprint.
• Dereziński, Jan, and Christian Gérard, Mathematics of Quantization and Quantum Fields, Cambridge University Press, 2013, Chapter 3 — symmetric and exterior tensors. Publisher.