Projection Operators

Permutation Operator: The Permutation Operator is defined by its action on a N-particle wavefunction as follows:
p_ijψ(...,i,...,j,...)= ψ(...,j,...,i,...) ( 1)
In general, p_ij²= e^2iφ
The order of the permutation group of order N is N!.Further, if the hilbert space of each single particle state is d,dimension of the total hilbert space formed is d^N.The particles are assumed to be indistinguishable.

Completely Symmetric & Completely Antisymmetric States and Projection Operators:
Notation: |i1,...,iα,...,iN> is a multiparticle state which means that particle no α is in a state denoted by |iα>.If the |i>'s form a complete orthonormal system in the space of N-particle system,the completely symmetric and completely antisymmetric basis states is defined by:
S_±|i1,...,iα,...,iN>=(1/√ N!) Σ_p(±)^pp|i1,...,iα,...,iN> (2)
By definition,it is clear that S_± are 1-dimensional subspaces.This we can also prove by writing down the projection operators onto these states.The trace of the projection perators is equal to the dimensionality of the hilbert space onto which they project as can be seen from the example of a three dimensional vector.From this elementary consideration, the following properties of the projection operators can be formulated:
1.Tr P < D,where D is the dimensionality of the hilbert space.
2.Tr P is always integral.
3.If Tr P =0 then it projects on to the null space.
4.If P_iP_j=&delta_ij and Tr P_i=1, then i can lie between 1 and D.
5.Iff 1 ≤ i ≤ D, then Σ _i P_i=I

Now, let us give some theorems and their proofs.
Theorem 1: The trace of a projection operator is equal to the dimension of the Hilbert space over which it projects.
Proof:Let the basis of an d dimensional Hilbert space be denoted as {|i 〉 }.The projection operator P_r which projects over a r dimensional subspace can then be written as Σ _k=1 ^r |k 〉 〈 k|.Then Tr P= Σ_l=1 ^D 〈 l| P |l 〉 = Σ_l=1 ^D 〈l Σ_k=1 ^r |k 〉 〈 k|l 〉 =Σ_k=1 ^r Σ_l=1 ^D 〈 l|k 〉〈 k|l 〉=Σ_k=1 ^r1 = r, since
〈 k|l 〉=δ_kl.

Theorem 2:The Projection Operator on to the completely symmetric and the completely antisymmetric states are given by
P_±=(1/√ N!)S_± (2)
Proof:From our definition of the S_± operators we know that there can only be a single symmetric or anti-symmetric state.So let us write
P₊=c₊ S₊.We need to find c₊ such that P₊²=P₊
Now, P₊²|i1,...,iα,...,iN&rang =(1/N!)c₊²&Sigma_qpqp|i1,...,iα,...,iN&rang
=(1/N!)c₊²&Sigma_phh|i1,...,iα,...,iN&rang
=c₊²&Sigma_hh|i1,...,iα,...,iN&rang
(since if pq=h, then p=hq^(-1) and has N! distinct elements.)
=c₊²(√ N!)S₊|i1,...,iα,...,iN&rang
=c₊(√ N!)P₊|i1,...,iα,...,iN&rang
which implies c₊=(1/√ N!)
Thus:
P_±=(1/N!) &Sigma _p(±)^pp|i1,...,iα,...,iN&rang

Theorem 3:Tr P_±=1
Proof: We shall prove this theorem only for the case of P₊.The other one involves more complicated algebra, but gives the same result.
Tr P₊=&Sigma{|i 〉 } &lang i1,...,iα,...,iN|(1/N!)&Sigma_pp|i1,...,iα,...,iN&rang
Only p=1 gives non zero contribution and there are N! such identities are generated due to the sum on p and hence we have the result Tr P₊=1