#### B.2.1. Symmetrized Momentum Eigenstates for Bose-Einstein Particles

A symmetrized state can be constructed as

| k 1 ,, k N ( + ) = P P| k 1 ,, k N

= Sum of all N! permutations of the k i  's in | k 1 ,, k N           (B.19)

For example,

P P| k 1 , k 2 , k 3 =| k 1 , k 2 , k 3 +| k 2 , k 3 , k 1 +| k 3 , k 1 , k 2

+| k 2 , k 1 , k 3 +| k 3 , k 2 , k 1 +| k 1 , k 3 , k 2                           (B.20)

Now, if there are n α  particles with momentum k α , there'll be only N! α n α !  distinct terms in (B.19) so that

| k 1 ,, k N ( + ) = P P| k 1 ,, k N   =( α n α ! )× distinct P | k 1 ,, k N

For example, setting k 1 = k 2  in (B.20) gives n α =0  except for n 1 =2  and n 3 =1 , so that only 3! 2!0!1!0! = 3! 2!1! =3  terms are distinct:

P P| k 1 , k 1 , k 3 =| k 1 , k 1 , k 3 +| k 1 , k 3 , k 1 +| k 3 , k 1 , k 1

+| k 1 , k 1 , k 3 +| k 3 , k 1 , k 1 +| k 1 , k 3 , k 1

=2( | k 1 , k 1 , k 3 +| k 3 , k 1 , k 1 +| k 1 , k 3 , k 1 )

The orthonormality of the 1-particle states implies

k a , k b ,, k l | k a' , k b' ,, k l' = δ aa' δ bb' δ ll'                        (B.21)

Thus, for a given permutation P( k 1 ,, k N ) , we have

P( k 1 ,, k N ) | k 1 ,, k N ( + ) = α n α !

so that

k 1 ,, k N | k 1 ,, k N ( + ) =N! α n α !

An orthonormal set of symmetrized states is therefore

| k 1 ,, k N ( S ) = 1 N! α n α ! | k 1 ,, k N ( + )

= 1 N! α n α ! P P| k 1 ,, k N          (B.22)

with

k 1 ,, k N | k 1 ,, k N ( S ) =1                                             (B.23)

Now, in a sum k 1 ,, k N ( k 1 k N ) , only distinct permutations are included.  For example, let k=a,b,c  and N=3 , we have 3 3 =27  terms,

k 1 , k 2 , k 3 =( a,b,c ) ( k 1 k 2 k 3 ) =aaa+aab+aac+baa+bab+bac+caa+cab+cac

+aba+abb+abc+bba+bbb+bbc+cba+cbb+cbc

+aca+acb+acc+bca+bcb+bcc+cca+ccb+ccc

which are all distinct.      On the other hand, for a fixed k 1 ,, k N , all the N! α n α !  distinct permutations P( k 1 ,, k N )  give rise to only 1 distinct symmetrized state, i.e., | k 1 ,, k N ( S ) =P | k 1 ,, k N ( S ) .  Therefore, in a sum k 1 ,, k N | k 1 ,, k N ( S ) , each distinct symmetrized state will appear N! α n α !  times.  Since the completeness relation involves a sum with each distinct orthonormal state counted once, we have

k 1 ,, k N ( 1 N! α n α ! ) | k 1 ,, k N ( S ) k 1 ,, k N |= 1 ˆ ( S )                    (B.24)