B.3.c. States And Operators
Let

α
1
,⋯,
α
N
⟩
(
S/A
)
represent an Nparticle 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 pairwise 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 nrepresentation, 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
β−1
n
α
is the number of pairs of particle
interchanges required to bring b to the leftmost position.