#### B.3.c. States And Operators

Let | α 1 ,, α N ( S/A )  represent an N-particle state obtained by properly symmetrizing the state where particle j is in the α j  state.  We shall assume the quantum numbers { α 1 ,, α N }  are always arranged in ascending order so that [see B.3.b]

α 1 < α 2 << α N                                                              (a)

Let { α 1 ,, α m }  be the subset of { α 1 ,, α N }  that includes only distinct states.  We define

| α 1 ,, α N ( S/A ) =| 0,0, n α 1 ,, n α m ,0,                                (b)

where n α  is the number of particles in state a with N= j=1 m n α j .

Consider now the action of the creation and annihilation operators on (b).  In order to keep track of the sign changes required to maintain the prescribed order for all possible states, we shall adopt the convention that these operators only affect the state at the leftmost position of the ket.  Thus, the creation of a state b is given by

a β + | α 1 ,, α N ( S/A ) c β +  ( S/A )    | β, α 1 ,, α N ( S/A )                  (c)

where, using the results of section B.3.a., we have

c β +  ( S ) = n β +1        and         c β +  ( A ) =1 n β = 1 n β                    (d)

Permutating b to its proper position gives

a β + | α 1 ,, α N ( S/A ) = ( ± ) P c β +  ( S/A ) | α 1 ,,β, α N ( S )               (e)

where P is the number of pair-wise interchanges required.

On applying the annihilation operator a β  on (B.39a), the result is zero if b is not in the set { α 1 ,, α m } .  Otherwise, we have

a β | α 1 ,,β,, α N ( S/A ) = ( ± ) P a β | β, α 1 ,,,, α N ( S/A )

= ( ± ) P c β ( S/A )    | α 1 ,,,, α N ( S/A )                  (f)

where P is the permutation required to bring b to the leftmost position.  Using the results of section B.3.a., we have

c β ( S ) = n β       and         c β ( A ) = n β = n β                                                (g)

Thus, (b) can be written as

| α 1 ,, α N ( S/A ) =| 0,0, n α 1 ,, n α m ,0,

= α=0 ( a α + ) n α n α ! |Φ   = 1 α=0 n α ! α=0 ( a α + ) n α |Φ

= 1 α=0 n α ! ( a α 1 + ) n α 1 ( a α 2 + ) n α 2 ( a α m + ) n α m |Φ                 (h)

where |Φ=| 0,,0  is the zero occupation state.  In this last form, it is obvious that the symmetries of | α 1 ,, α N ( S/A )  can be satisfied if we set

[ a α + ,   a α' + ] =0                 for all a and a'                                       (i)

Taking the adjoint of (i) gives

[ a α ,   a α' ] =0                    for all a and a'                                       (j)

An obvious generalization to eqs(11,14) is therefore

[ a α ,   a α' + ] = δ αα'                               (k)

Since eqs(11,14) are designed to keep the states |n  orthonormal, we see that (h) implies

n 0 n α n | n 0 n α n = α=0 δ n α n α                            (B.51)

In terms of the n-representation, eqs(e,f) become

For bosons

a β | , n β , = n β | , n β 1,                                   (B.48)

a β + | , n β , = n β +1 | , n β +1,                              (B.49)

For fermions

a β | , n β , = ( ) Σ β n β | , n β 1,                          (B.69)

a β + | , n β , = ( ) Σ β 1 n β | , n β +1,                             (B.70)

where

Σ β = α=0 β1 n α

is the number of pairs of particle interchanges required to bring b to the leftmost position.