Symmetrizer

In quantum mechanics, a symmetrizer  $$ \mathcal{S}$$ (also known as symmetrizing operator) is a linear operator that makes a wave function of N identical bosons symmetric under the exchange of the coordinates of any pair of bosons. After application of  $$ \mathcal{S}$$ the wave function satisfies the Pauli principle. Since  $$ \mathcal{S}$$ is a projection operator, application of the symmetrizer to a wave function that is already totally symmetric has no effect,  $$ \mathcal{S}$$ is effectively  the identity operator when acting on symmetric wave functions.

Mathematical definition
Consider a wave function depending on the space and spin coordinates of N bosons:

\Psi(1,2, \ldots, N)\quad\hbox{with} \quad i \leftrightarrow (\mathbf{r}_i, \sigma_i), $$ where the position vector ri of particle i is a vector in $$\mathbb{R}^3$$ and &sigma;i  takes on 2I+1  values, where I is the integral intrinsic spin of the boson. For instance, photons have I = 1 and &sigma; can have three values 1, 0, &minus;1. We define a transposition operator $$\hat{P}_{ij}$$ that interchanges the coordinates of particle i and j. In general this operator will not be equal to the identity operator (although in special cases it may be).

The Pauli principle postulates that a wave function of identical bosons  must be an eigenfunction of a transposition operator with unity as eigenvalue

\begin{align} \hat{P}_{ij} \Psi\big(1,2,\ldots,i, \ldots,j,\ldots, N\big) &\equiv \Psi\big(\pi(1),\pi(2),\ldots,\pi(i), \ldots,\pi(j),\ldots, \pi(N)\big) \\ &\equiv \Psi(1,2,\ldots,j, \ldots,i,\ldots, N) \\ &=  \Psi(1,2,\ldots,i, \ldots,j,\ldots, N). \end{align} $$ Here we associated the transposition operator $$\hat{P}_{ij}$$ with the permutation of coordinates &pi; that acts on the set of N coordinates. In this case &pi; = (ij), where (ij) is the cycle notation for the transposition of the coordinates of particle i and j.

Transpositions may be composed (applied in sequence). This defines a product between the transpositions that is associative. Since an arbitrary permutation of N objects can be written as a product of transpositions, it holds for a symmetric function &Psi; that

\hat{P} \Psi\big(1,2,\ldots, N\big) \equiv \Psi\big(\pi(1),\pi(2),\ldots, \pi(N)\big) =  \Psi(1,2,\ldots, N), $$ where we associated the linear operator $$\hat{P}$$ with the permutation &pi;.

The set of all N! permutations with the associative product: "apply one permutation after the other", is a group, known as the permutation group or symmetric group, denoted by SN. After this preamble we are ready to give the definition of the symmetrizer

\mathcal{S} \equiv \frac{1}{N!} \sum_{P \in S_N} \hat{P}. $$

Properties of the symmetrizer
In the representation theory of finite groups the symmetrizer is a well-known object, because the map of all permutations onto unity  forms a one-dimensional (and hence irreducible) representation of the permutation group known as the symmetric representation. The symmetrizer is the character projection operator corresponding to the symmetric representation and is therefore idempotent,

\mathcal{S} = \mathcal{S}^2. $$ This has the consequence that for any N-particle wave function &Psi;(1, ...,N) we have

\mathcal{S}\Psi(1,\ldots, N) = \begin{cases} &0 \\ &\Psi'(1,\dots, N) \ne 0. \end{cases} $$ Either &Psi; does not have a symmetric component, and then the symmetrizer projects onto zero, or it has one and then the symmetrizer projects out this symmetric component &Psi;'. The symmetrizer carries a left and a right representation of the group:

\hat{P} \mathcal{S} = \mathcal{S} \hat{P} = \mathcal{S},\qquad \forall \pi \in S_N, $$ with the operator $$\hat{P}$$ representing the coordinate permutation &pi;. Now it holds, for any N-particle wave function &Psi;(1, ...,N) with a non-vanishing symmetric component, that

\hat{P} \mathcal{S}\Psi(1,\ldots, N) \equiv \hat{P} \Psi'(1,\ldots, N)= \Psi'(1,\ldots, N), $$ showing that the non-vanishing component is indeed symmetric.

Permutations of identical particles are unitary, (the Hermitian adjoint is equal to the inverse of the operator), so that the antisymmetrizer is Hermitian,

\mathcal{S}^\dagger = \mathcal{S}. $$

The symmetrizer commutes with any observable $$\hat{H}$$ (Hermitian operator corresponding to a physical&mdash;observable&mdash;quantity)

[\mathcal{S}, \hat{H}] = 0. $$ If it were otherwise, measurement of $$\hat{H}$$ could distinguish the particles, in contradiction with the assumption that only the coordinates of indistinguishable particles are affected by the symmetrizer.

Connection with the permanent
In the special case that the wave function to be symmetrized is a product of spin-orbitals

\Psi(1,2, \ldots, N) = \psi_{n_1}(1) \psi_{n_2}(2) \cdots \psi_{n_N}(N) $$ the symmetrizer yields a constant times a permanent:



\mathcal{S} \Psi(1,2, \ldots, N) = \frac{1}{N!} \begin{Bmatrix} \psi_{n_1}(1) & \psi_{n_1}(2) & \cdots & \psi_{n_1}(N) \\ \psi_{n_2}(1) & \psi_{n_2}(2) & \cdots & \psi_{n_2}(N) \\ \cdots & \cdots & \cdots & \cdots \\ \psi_{n_N}(1) & \psi_{n_N}(2) & \cdots & \psi_{n_N}(N) \\ \end{Bmatrix} $$ The quantity in curly bracket is a permanent, which is the exact analogue of a determinant, with all N! terms having a plus sign. The correspondence follows immediately from the Leibniz formula for permanents, which reads

\mathrm{perm}(\mathbf{B}) = \sum_{\pi \in S_N}  B_{1,\pi(1)}\cdot B_{2,\pi(2)}\cdot B_{3,\pi(3)}\cdot\,\cdots\,\cdot B_{N,\pi(N)}, $$ where B is the matrix

\mathbf{B} = \begin{pmatrix} B_{1,1} & B_{1,2} & \cdots & B_{1,N} \\ B_{2,1} & B_{2,2} & \cdots & B_{2,N} \\ \cdots & \cdots & \cdots & \cdots \\ B_{N,1} & B_{N,2} & \cdots & B_{N,N} \\ \end{pmatrix}. $$ To see the correspondence we notice that the particle labels, permuted by the terms in the symmetrizer, indicate the different columns (particle labels are second indices). The first indices are orbital indices, n1, ..., nN indicating the rows.

Examples
1. By the definition of the symmetrizer

\mathcal{S} \psi_a(1)\psi_b(2)\psi_c(3) = \frac{1}{6} \Big( \psi_a(1)\psi_b(2)\psi_c(3) + \psi_a(3)\psi_b(1)\psi_c(2) + \psi_a(2)\psi_b(3)\psi_c(1) $$

+\psi_a(2)\psi_b(1)\psi_c(3) + \psi_a(3)\psi_b(2)\psi_c(1)+ \psi_a(1)\psi_b(3)\psi_c(2)\Big). $$ Consider the unnormalized permanent

D\equiv \begin{Bmatrix} \psi_a(1) & \psi_a(2) & \psi_a(3) \\ \psi_b(1) & \psi_b(2) & \psi_b(3) \\ \psi_c(1) & \psi_c(2) & \psi_c(3) \\ \end{Bmatrix}. $$ By the Laplace expansion along the first row of D

D = \psi_a(1) \begin{Bmatrix} \psi_b(2) & \psi_b(3) \\ \psi_c(2) & \psi_c(3) \\ \end{Bmatrix} +\psi_a(2) \begin{Bmatrix} \psi_b(1) & \psi_b(3) \\ \psi_c(1) & \psi_c(3) \\ \end{Bmatrix} +\psi_a(3) \begin{Bmatrix} \psi_b(1) & \psi_b(2) \\ \psi_c(1) & \psi_c(2) \\ \end{Bmatrix}, $$ so that

D= \psi_a(1)\Big( \psi_b(2) \psi_c(3) + \psi_b(3) \psi_c(2)\Big) + \psi_a(2)\Big( \psi_b(1) \psi_c(3) + \psi_b(3) \psi_c(1)\Big) $$

+ \psi_a(3)\Big( \psi_b(1) \psi_c(2) + \psi_b(2) \psi_c(1)\Big). $$ By comparing terms we see that

\frac{1} {6} D = \mathcal{S} \psi_a(1)\psi_b(2)\psi_c(3). $$

2. It is fairly easy to normalize a projected orbital product &Psi;. That is, to compute the normalization constant K from the following:

1 = K^2 \langle \mathcal{S} \Psi | \mathcal{S} \Psi \rangle =K^2 \langle  \Psi | \mathcal{S} \Psi \rangle. $$ The rightmost side follows from the hermiticity and idempotency of the symmetrizer. We introduce the occupation representation

\Psi \rightarrow |\,n_1, n_2, \ldots, n_M \rangle \equiv (\psi_1)^{n_1}\; (\psi_2)^{n_2}\cdots(\psi_M)^{n_M} \quad\hbox{with}\quad N=n_1+n_2+\cdots+n_M= \sum_{k=1}^M n_k, $$ where nk gives the number of times the orbital &psi;k is occupied, i.e., the number of times it appear in the product wave function. Note that nk = 0 implies an unoccupied orbital, one that does not appear in the product. Under the condition that the orbitals are orthonormal, we find

\begin{align} 1 &= K^2 \langle n_1, n_2, \ldots, n_M | \mathcal{S} | n_1, n_2, \ldots, n_M \rangle \\ &= K^2 \frac{1}{N!} (n_1)!(n_2)!\cdots (n_M)! \\ \end{align} $$ so that

K = \left({N \over n_1\,n_2\,\cdots n_M}\right)^{1/2} $$ where the multinomial coefficient is

\left({N \over n_1\,n_2\,\cdots n_M}\right) \equiv \frac{N!}{(n_1)!(n_2)!\cdots (n_M)!}. $$ The following is a normalized and symmetrized boson-orbital product,

\left({N \over n_1\,n_2\,\cdots n_M}\right)^{1/2}\; \mathcal{S}\; |\,n_1, n_2, \ldots, n_M \rangle. $$