The identity reads
∏
n
=
1
∞
(
1
−
q
n
)
(
1
+
q
n
−
1
2
t
)
(
1
+
q
n
−
1
2
/
t
)
=
∑
n
∈
Z
q
n
2
2
t
n
{\displaystyle \prod _{n=1}^{\infty }(1-q^{n})(1+q^{n-{\frac {1}{2}}}t)(1+q^{n-{\frac {1}{2}}}/t)=\sum _{n\in \mathbb {Z} }q^{\frac {n^{2}}{2}}t^{n}}
.
The physical proof of this identity is very interesting. It involves a fermion-antifermion statistic model.
Consider a two-fermion system, known as Neveu-Schwarz fermions with field realization
ψ
(
z
)
=
∑
r
∈
Z
+
1
/
2
ψ
r
z
r
−
1
/
2
r
,
ψ
∗
(
z
)
=
∑
r
∈
Z
+
1
/
2
ψ
r
∗
z
r
−
1
/
2
{\displaystyle \psi (z)=\sum _{r\in \mathbb {Z} +1/2}\psi _{r}z^{r-1/2}r\,\,,\,\,\psi ^{*}(z)=\sum _{r\in \mathbb {Z} +1/2}\psi _{r}^{*}z^{r-1/2}}
.
Anticommutation relation
{
ψ
r
,
ψ
s
}
=
δ
r
+
s
,
0
,
{
ψ
r
∗
,
ψ
s
∗
}
=
δ
r
+
s
,
0
{\displaystyle \{\psi _{r},\psi _{s}\}=\delta _{r+s,0}\,\,,\,\,\{\psi _{r}^{*},\psi _{s}^{*}\}=\delta _{r+s,0}}
.
Equipped with a Hamiltonian
H
=
E
0
∑
r
∈
Z
+
−
1
/
2
r
(
ψ
−
r
ψ
r
+
ψ
−
r
∗
ψ
r
∗
)
=
E
0
L
0
{\displaystyle H=E_{0}\sum _{r\in \mathbb {Z} ^{+}-1/2}r\left(\psi _{-r}\psi _{r}+\psi _{-r}^{*}\psi _{r}^{*}\right)=E_{0}L_{0}}
,
where
L
0
=
∑
r
∈
Z
+
−
1
/
2
r
(
ψ
−
r
ψ
r
+
ψ
−
r
∗
ψ
r
∗
)
{\displaystyle L_{0}=\sum _{r\in \mathbb {Z} ^{+}-1/2}r\left(\psi _{-r}\psi _{r}+\psi _{-r}^{*}\psi _{r}^{*}\right)}
is the zero mode of Virasoro algebra .
A fermion number defines as
N
=
∑
r
∈
Z
+
−
1
/
2
(
ψ
−
r
ψ
r
−
ψ
−
r
∗
ψ
r
∗
)
{\displaystyle N=\sum _{r\in \mathbb {Z} ^{+}-1/2}\left(\psi _{-r}\psi _{r}-\psi _{-r}^{*}\psi _{r}^{*}\right)}
.
Partion function of this Grand canonical ensemble
Z
(
q
,
t
)
=
∑
s
t
a
t
e
s
exp
(
−
β
(
E
−
μ
N
)
{\displaystyle Z(q,t)=\sum _{states}\exp(-\beta (E-\mu N)}
, and define the parameter :
t
=
e
β
μ
,
q
=
e
−
β
E
0
{\displaystyle t=e^{\beta \mu },q=e^{-\beta E_{0}}}
.
Substitution of the operator expression of N and H
Z
=
T
r
q
∑
r
r
(
ψ
−
r
ψ
r
+
ψ
−
r
∗
ψ
r
∗
)
t
∑
r
ψ
−
r
ψ
r
−
ψ
−
r
∗
ψ
r
∗
)
{\displaystyle Z=Trq^{\sum _{r}r(\psi _{-r}\psi _{r}+\psi _{-r}^{*}\psi _{r}^{*})}t^{\sum _{r}\psi _{-r}\psi _{r}-\psi _{-r}^{*}\psi _{r}^{*})}}
=
T
r
q
∑
r
r
ψ
−
r
ψ
r
t
∑
r
ψ
−
r
ψ
r
q
r
∑
r
ψ
−
r
∗
ψ
r
∗
t
−
∑
r
ψ
−
r
∗
ψ
r
∗
{\displaystyle =Trq^{\sum _{r}r\psi _{-r}\psi _{r}}t^{\sum _{r}\psi _{-r}\psi _{r}}q^{r\sum _{r}\psi _{-r}^{*}\psi _{r}^{*}}t^{-\sum _{r}\psi _{-r}^{*}\psi _{r}^{*}}}
=
∏
r
∈
N
−
1
2
(
1
+
q
r
t
)
(
1
+
q
r
/
t
)
{\displaystyle =\prod _{r\in \mathbb {N} -{\frac {1}{2}}}(1+q^{r}t)(1+q^{r}/t)}
.
Another way to counting the same system is a Young diagram counting.
Classfying the system by the Fermion number N.
Z
(
q
,
t
)
=
∑
N
≥
0
t
N
Z
N
(
q
)
{\displaystyle Z(q,t)=\sum _{N\geq 0}t^{N}Z_{N}(q)}
.
Consider the N=0 counting of states.
At energy level n, the number of states is :
p
[
n
]
{\displaystyle p[n]}
, the partition number of n. This can see from the following table
Level
Degeneracy
States
0
1
|
0
⟩
{\displaystyle |0\rangle }
1
1
ψ
−
1
/
2
ψ
−
1
/
2
∗
|
0
⟩
∼
|
Y
o
u
n
g
(
1
,
0
)
⟩
{\displaystyle \psi _{-1/2}\psi _{-1/2}^{*}|0\rangle \sim |Young(1,0)\rangle }
2
2
ψ
−
3
/
2
ψ
−
1
/
2
∗
|
0
⟩
∼
|
Y
o
u
n
g
(
2
,
0
)
⟩
{\displaystyle \psi _{-3/2}\psi _{-1/2}^{*}|0\rangle \sim |Young(2,0)\rangle }
ψ
−
1
/
2
ψ
−
3
/
2
∗
|
0
⟩
∼
|
Y
o
u
n
g
(
1
,
1
)
⟩
{\displaystyle \psi _{-1/2}\psi _{-3/2}^{*}|0\rangle \sim |Young(1,1)\rangle }
3
3
ψ
−
5
/
2
ψ
−
1
/
2
∗
|
0
⟩
∼
|
Y
o
u
n
g
(
3
,
0
)
⟩
{\displaystyle \psi _{-5/2}\psi _{-1/2}^{*}|0\rangle \sim |Young(3,0)\rangle }
ψ
−
3
/
2
ψ
−
3
/
2
∗
|
0
⟩
∼
|
Y
o
u
n
g
(
2
,
1
)
⟩
{\displaystyle \psi _{-3/2}\psi _{-3/2}^{*}|0\rangle \sim |Young(2,1)\rangle }
ψ
−
1
/
2
ψ
−
5
/
2
∗
|
0
⟩
∼
|
Y
o
u
n
g
(
1
,
1
,
1
)
⟩
{\displaystyle \psi _{-1/2}\psi _{-5/2}^{*}|0\rangle \sim |Young(1,1,1)\rangle }
4
5
ψ
−
7
/
2
ψ
−
1
/
2
∗
|
0
⟩
∼
|
Y
o
u
n
g
(
4
,
0
)
⟩
{\displaystyle \psi _{-7/2}\psi _{-1/2}^{*}|0\rangle \sim |Young(4,0)\rangle }
ψ
−
5
/
2
ψ
−
3
/
2
∗
|
0
⟩
∼
|
Y
o
u
n
g
(
3
,
1
)
⟩
{\displaystyle \psi _{-5/2}\psi _{-3/2}^{*}|0\rangle \sim |Young(3,1)\rangle }
ψ
−
3
/
2
ψ
−
3
/
2
∗
ψ
−
1
/
2
ψ
−
1
/
2
∗
|
0
⟩
∼
|
Y
o
u
n
g
(
2
,
2
)
⟩
{\displaystyle \psi _{-3/2}\psi _{-3/2}^{*}\psi _{-1/2}\psi _{-1/2}^{*}|0\rangle \sim |Young(2,2)\rangle }
ψ
−
3
/
2
ψ
−
5
/
2
∗
|
0
⟩
∼
|
Y
o
u
n
g
(
2
,
1
,
1
)
⟩
{\displaystyle \psi _{-3/2}\psi _{-5/2}^{*}|0\rangle \sim |Young(2,1,1)\rangle }
ψ
−
1
/
2
ψ
−
7
/
2
∗
|
0
⟩
∼
|
Y
o
u
n
g
(
1
,
1
,
1
,
1
)
⟩
{\displaystyle \psi _{-1/2}\psi _{-7/2}^{*}|0\rangle \sim |Young(1,1,1,1)\rangle }
The
N
=
0
{\displaystyle N=0}
counting of states gives the Dedekind eta function
Z
0
(
q
,
t
)
=
∏
m
=
1
∞
1
1
−
q
n
{\displaystyle Z_{0}(q,t)=\prod _{m=1}^{\infty }{\frac {1}{1-q^{n}}}}
.
It follows that at arbitrary N, the counting of states is the same as the N=0 case. For example
at level N =k, the first excitation is
ψ
−
k
−
1
/
2
ψ
−
1
/
2
∗
|
k
⟩
{\displaystyle \psi _{-k-1/2}\psi _{-1/2}^{*}|k\rangle }
,
where
|
k
⟩
=
ψ
−
1
/
2
ψ
−
3
/
2
ψ
−
5
/
2
⋯
ψ
−
k
+
1
/
2
|
0
⟩
{\displaystyle |k\rangle =\psi _{-1/2}\psi _{-3/2}\psi _{-5/2}\cdots \psi _{-k+1/2}|0\rangle }
,
is also known as the colored vacuum of Dirac sea .
The second excitations are
ψ
−
k
−
1
/
2
ψ
−
3
/
2
∗
|
k
⟩
{\displaystyle \psi _{-k-1/2}\psi _{-3/2}^{*}|k\rangle }
,
ψ
−
k
−
3
/
2
ψ
−
1
/
2
∗
|
k
⟩
{\displaystyle \psi _{-k-3/2}\psi _{-1/2}^{*}|k\rangle }
.
This argument follows, and the only difference of level k counting and level 0 counting is the energy difference.
E
k
/
E
0
=
∑
r
=
1
k
(
k
−
1
2
)
=
k
2
2
{\displaystyle E_{k}/E_{0}=\sum _{r=1}^{k}(k-{\frac {1}{2}})={\frac {k^{2}}{2}}}
The level k counting contributs
Z
k
=
q
k
2
2
∏
m
=
1
∞
1
1
−
q
m
{\displaystyle Z_{k}=q^{\frac {k^{2}}{2}}\prod _{m=1}^{\infty }{\frac {1}{1-q^{m}}}}
this leads the total partion function
Z
(
q
,
t
)
=
∑
n
=
0
∞
t
n
q
n
2
2
∏
m
=
1
∞
1
1
−
q
m
{\displaystyle Z(q,t)=\sum _{n=0}^{\infty }t^{n}q^{\frac {n^{2}}{2}}\prod _{m=1}^{\infty }{\frac {1}{1-q^{m}}}}
.
The two ways of counting states should be equivalent. The Jacobi triple identity is proven.
Operator Formalism of Calogero Sutherland Model
The CS Hamiltonian
H
C
S
=
−
1
2
∂
x
i
2
+
β
(
β
−
1
)
s
i
n
2
(
π
L
x
i
j
)
,
x
i
j
≡
x
i
−
x
j
{\displaystyle H_{CS}=-{\frac {1}{2}}\partial _{x_{i}}^{2}+{\frac {\beta (\beta -1)}{sin^{2}({\frac {\pi }{L}}x_{ij})}}\,\,,\,\,x_{ij}\equiv x_{i}-x_{j}}
.
Choose
z
i
=
exp
(
2
i
x
i
)
,
L
=
π
{\displaystyle z_{i}=\exp(2ix_{i}),L=\pi }
. We have
d
z
i
=
2
i
z
i
d
x
i
,
∂
x
i
=
(
2
i
z
i
)
−
1
∂
x
i
,
∂
x
i
=
(
2
i
z
i
)
∂
z
i
∼
2
i
L
0
{\displaystyle dz_{i}=2iz_{i}dx_{i}\,\,,\,\,\partial _{x_{i}}=(2iz_{i})^{-1}\partial _{x_{i}}\,\,,\,\partial _{x_{i}}=(2iz_{i})\partial _{z_{i}}\sim 2iL_{0}}
where
L
n
{\displaystyle L_{n}}
be the generator of Witt algebra
[
L
n
,
L
m
]
=
(
n
−
m
)
L
n
+
m
,
L
n
=
z
n
+
1
∂
z
,
L
0
=
z
∂
z
{\displaystyle [L_{n},L_{m}]=(n-m)L_{n+m}\,\,,\,L_{n}=z^{n+1}\partial _{z}\,\,,\,\,L_{0}=z\partial _{z}}
.
Notice that the conformal map from cylinder to complex plane also maps the translation of
x
i
{\displaystyle x_{i}}
(generated by momentum operator) to a scaling(or inflation) which is generated by
L
0
{\displaystyle L_{0}}
.
It is simple that
(
∂
x
i
)
2
=
−
4
(
z
i
∂
z
i
+
z
i
2
∂
z
i
2
)
=
−
4
L
0
−
4
L
−
n
L
n
{\displaystyle (\partial _{x_{i}})^{2}=-4(z_{i}\partial _{z_{i}}+z_{i}^{2}\partial _{z_{i}}^{2})=-4L_{0}-4L_{-n}L_{n}}
.
Another Definition of
H
C
S
{\displaystyle H_{CS}}
(up to zero point energy)
H
~
=
1
2
(
∂
x
i
+
∂
x
i
∏
l
<
j
s
i
n
β
x
l
j
)
(
∂
x
i
−
∂
x
i
∏
k
<
j
s
i
n
β
x
k
j
)
{\displaystyle {\tilde {H}}={\frac {1}{2}}\left(\partial _{x_{i}}+\partial _{x_{i}}\prod _{l<j}sin^{\beta }x_{lj}\right)\left(\partial _{x_{i}}-\partial _{x_{i}}\prod _{k<j}sin^{\beta }x_{kj}\right)}
.
This could be derived from the following calculations.
∂
x
i
ln
∏
l
<
j
s
i
n
β
(
x
l
j
)
=
β
∑
i
≠
j
c
t
g
(
x
i
j
)
{\displaystyle \partial _{x_{i}}\ln \prod _{l<j}sin^{\beta }(x_{lj})=\beta \sum _{i\neq j}ctg(x_{ij})}
.
Then the Hamiltonian becomes
H
~
=
−
1
2
∂
x
i
2
+
β
2
2
∑
j
,
k
≠
i
c
t
g
(
x
i
j
)
c
t
g
(
x
i
k
)
+
1
2
β
[
∂
x
i
,
∑
k
≠
i
c
t
g
(
x
i
k
)
]
{\displaystyle {\tilde {H}}=-{\frac {1}{2}}\partial _{x_{i}}^{2}+{\frac {\beta ^{2}}{2}}\sum _{j,k\neq i}ctg(x_{ij})ctg(x_{ik})+{\frac {1}{2}}\beta \left[\partial _{x_{i}},\sum _{k\neq i}ctg(x_{ik})\right]}
.
Since
[
∂
x
i
,
∑
k
≠
i
c
t
g
(
x
i
k
)
]
=
∑
k
≠
i
−
1
s
i
n
2
(
x
i
k
)
{\displaystyle \left[\partial _{x_{i}},\sum _{k\neq i}ctg(x_{ik})\right]=\sum _{k\neq i}{\frac {-1}{sin^{2}(x_{ik})}}}
,
and using the identity
∑
i
,
j
≠
k
c
t
g
(
x
i
j
)
c
t
g
(
x
i
k
)
+
c
y
c
l
i
c
(
i
,
j
,
k
)
=
∑
i
≠
j
≠
k
(
−
1
)
=
−
N
(
N
−
1
)
(
N
−
2
)
{\displaystyle \sum _{i,j\neq k}ctg(x_{ij})ctg(x_{ik})+cyclic(i,j,k)=\sum _{i\neq j\neq k}(-1)=-N(N-1)(N-2)}
,
we have
β
2
2
∑
j
,
k
≠
i
c
t
g
(
x
i
j
)
c
t
g
(
x
i
k
)
=
−
β
2
6
N
(
N
−
1
)
(
N
+
1
)
+
1
2
β
2
∑
j
≠
i
1
s
i
n
2
(
x
i
j
)
{\displaystyle {\frac {\beta ^{2}}{2}}\sum _{j,k\neq i}ctg(x_{ij})ctg(x_{ik})={\frac {-\beta ^{2}}{6}}N(N-1)(N+1)+{\frac {1}{2}}\beta ^{2}\sum _{j\neq i}{\frac {1}{sin^{2}(x_{ij})}}}
.
This leads to the result that
H
~
=
H
C
S
−
E
0
{\displaystyle {\tilde {H}}=H_{CS}-E_{0}}
with
E
0
=
β
6
(
N
−
1
)
N
(
N
+
1
)
{\displaystyle E_{0}={\frac {\beta }{6}}(N-1)N(N+1)}
.
From the Hamiltonian
H
~
{\displaystyle {\tilde {H}}}
, it is clear that
ψ
0
=
∏
i
<
j
s
i
n
β
(
x
i
j
)
{\displaystyle \psi _{0}=\prod _{i<j}sin^{\beta }(x_{ij})}
is the ground state with energy 0. Thus it is also a ground state of
H
C
S
{\displaystyle H_{CS}}
with ground energy
E
0
{\displaystyle E_{0}}
Moving out the contribution of ground state,
H
J
≡
ψ
0
−
1
H
~
ψ
0
=
ψ
0
−
1
(
1
2
(
∂
i
+
ψ
0
−
1
∂
i
ψ
0
)
)
(
1
2
(
∂
i
−
ψ
0
−
1
∂
i
ψ
0
)
)
ψ
0
{\displaystyle H_{J}\equiv \psi _{0}^{-1}{\tilde {H}}\psi _{0}=\psi _{0}^{-1}({\frac {1}{2}}(\partial _{i}+\psi _{0}^{-1}\partial _{i}\psi _{0}))({\frac {1}{2}}(\partial _{i}-\psi _{0}^{-1}\partial _{i}\psi _{0}))\psi _{0}}
,
noticing
(
∂
i
−
ψ
0
−
1
∂
i
ψ
0
)
ψ
0
=
ψ
0
∂
i
{\displaystyle (\partial _{i}-\psi _{0}^{-1}\partial _{i}\psi _{0})\psi _{0}=\psi _{0}\partial _{i}}
, we have
H
J
=
−
1
2
(
∂
i
+
2
∂
i
ψ
0
)
∂
i
{\displaystyle H_{J}=-{\frac {1}{2}}(\partial _{i}+2\partial _{i}\psi _{0})\partial _{i}}
.
For
∂
x
i
ψ
0
=
β
∑
j
≠
i
c
t
g
(
x
i
j
)
{\displaystyle \partial _{x_{i}}\psi _{0}=\beta \sum _{j\neq i}ctg(x_{ij})}
,
one arrives the complex plane expression of
H
J
{\displaystyle H_{J}}
H
J
=
2
∑
i
(
z
i
∂
z
i
)
2
+
2
β
∑
i
<
j
z
i
+
z
j
z
i
−
z
j
(
z
i
∂
z
i
−
z
j
∂
z
j
)
{\displaystyle H_{J}=2\sum _{i}(z_{i}\partial _{z_{i}})^{2}+2\beta \sum _{i<j}{\frac {z_{i}+z_{j}}{z_{i}-z_{j}}}(z_{i}\partial _{z_{i}}-z_{j}\partial _{z_{j}})}
.
Acting on generating function
∏
i
=
1
N
V
−
k
(
z
i
)
≡
exp
(
k
∑
n
>
0
a
−
n
n
∑
i
=
1
N
(
z
i
)
n
)
{\displaystyle \prod _{i=1}^{N}V_{-}^{k}(z_{i})\equiv \exp \left(k\sum _{n>0}{\frac {a_{-n}}{n}}\sum _{i=1}^{N}(z_{i})^{n}\right)}
,
k
=
β
{\displaystyle k={\sqrt {\beta }}}
one have
H
J
∏
i
=
1
N
V
−
k
(
z
i
)
|
k
0
⟩
=
(
−
i
)
(
∂
x
i
+
2
β
∑
j
≠
i
c
t
g
(
x
i
j
)
)
k
a
−
n
∑
i
=
1
(
z
i
)
n
∏
i
=
1
N
V
−
k
(
z
i
)
|
k
0
⟩
{\displaystyle H_{J}\prod _{i=1}^{N}V_{-}^{k}(z_{i})|k_{0}\rangle =(-i)(\partial _{x_{i}}+2\beta \sum _{j\neq i}ctg(x_{ij}))ka_{-n}\sum _{i=1}(z_{i})^{n}\prod _{i=1}^{N}V_{-}^{k}(z_{i})|k_{0}\rangle }
=
{
(
−
i
)
[
(
2
i
)
n
k
a
−
n
∑
i
=
1
N
(
z
i
)
n
+
2
i
k
2
a
−
n
a
−
m
∑
i
=
1
N
(
z
i
)
n
+
m
]
+
2
β
∑
i
≠
j
z
i
+
z
j
z
i
−
z
j
k
a
−
n
∑
i
=
1
N
(
z
i
)
n
}
∏
i
=
1
N
V
−
k
(
z
i
)
|
k
0
⟩
{\displaystyle =\left\{(-i)\left[(2i)nka_{-n}\sum _{i=1}^{N}(z_{i})^{n}+2ik^{2}a_{-n}a_{-m}\sum _{i=1}^{N}(z_{i})^{n+m}\right]+2\beta \sum _{i\neq j}{\frac {z_{i}+z_{j}}{z_{i}-z_{j}}}ka_{-n}\sum _{i=1}^{N}(z_{i})^{n}\right\}\prod _{i=1}^{N}V_{-}^{k}(z_{i})|k_{0}\rangle }
Besides of the last interaction term, all terms can be operatorization
because of the coherent relation
a
n
e
k
∑
m
>
0
a
−
m
/
m
∑
i
N
(
z
i
)
m
=
k
∑
i
N
(
z
i
)
n
e
k
∑
m
>
0
a
−
m
/
m
∑
i
N
(
z
i
)
m
{\displaystyle a_{n}e^{k\sum _{m>0}a_{-m}/m\sum _{i}^{N}(z_{i})^{m}}=k\sum _{i}^{N}(z_{i})^{n}e^{k\sum _{m>0}a_{-m}/m\sum _{i}^{N}(z_{i})^{m}}}
,
the acting of
H
J
{\displaystyle H_{J}}
on generating function gives
H
J
=
2
n
a
−
n
a
n
+
2
k
a
−
n
a
−
m
a
n
+
m
+
2
k
3
a
−
n
∑
i
≠
j
z
i
+
z
j
z
i
−
z
j
×
(
z
i
)
n
{\displaystyle H_{J}=2na_{-n}a_{n}+2ka_{-n}a_{-m}a_{n+m}+2k^{3}a_{-n}\sum _{i\neq j}{\frac {z_{i}+z_{j}}{z_{i}-z_{j}}}\times (z_{i})^{n}}
.
Since
∑
i
≠
j
z
i
+
z
j
z
i
−
z
j
(
z
i
)
n
=
1
2
∑
i
≠
j
z
i
+
z
j
z
i
−
z
j
(
(
z
i
)
n
−
(
z
j
)
n
)
{\displaystyle \sum _{i\neq j}{\frac {z_{i}+z_{j}}{z_{i}-z_{j}}}(z_{i})^{n}={\frac {1}{2}}\sum _{i\neq j}{\frac {z_{i}+z_{j}}{z_{i}-z_{j}}}((z_{i})^{n}-(z_{j})^{n})}
=
∑
i
<
j
z
i
+
z
j
z
i
−
z
j
(
z
i
−
z
j
)
(
z
i
n
−
1
+
z
i
n
−
2
z
j
+
⋯
+
z
j
n
−
1
)
{\displaystyle =\sum _{i<j}{\frac {z_{i}+z_{j}}{z_{i}-z_{j}}}(z_{i}-z_{j})\left(z_{i}^{n-1}+z_{i}^{n-2}z_{j}+\cdots +z_{j}^{n-1}\right)}
=
∑
i
<
j
(
z
i
n
+
2
(
z
i
n
−
1
z
j
+
⋯
z
i
z
j
n
−
1
)
+
z
j
n
)
{\displaystyle =\sum _{i<j}(z_{i}^{n}+2(z_{i}^{n-1}z_{j}+\cdots z_{i}z_{j}^{n-1})+z_{j}^{n})}
=
∑
i
≠
j
{
z
i
n
+
z
i
n
−
1
z
j
+
⋯
+
z
i
z
j
n
−
1
}
{\displaystyle =\sum _{i\neq j}\left\{z_{i}^{n}+z_{i}^{n-1}z_{j}+\cdots +z_{i}z_{j}^{n-1}\right\}}
=
N
p
n
+
p
n
−
1
p
1
+
⋯
+
p
1
p
n
−
1
−
n
p
n
(
(
i
=
j
)
c
o
n
t
r
i
b
u
t
i
o
n
)
{\displaystyle =Np_{n}+p_{n-1}p_{1}+\cdots +p_{1}p_{n-1}-np_{n}((i=j)contribution)}
now all terms can have operator formalism
H
~
C
S
=
2
n
a
−
n
a
n
+
2
k
a
−
n
a
−
m
a
n
+
m
+
2
k
a
−
n
(
N
k
a
n
−
n
k
a
n
+
a
n
−
m
a
m
)
{\displaystyle {\tilde {H}}_{CS}=2na_{-n}a_{n}+2ka_{-n}a_{-m}a_{n+m}+2ka_{-n}(Nka_{n}-nka_{n}+a_{n-m}a_{m})}
=
2
H
^
C
S
=
2
{
k
(
a
−
n
a
−
m
a
n
+
m
+
a
−
n
−
m
a
n
a
m
)
+
(
1
−
k
2
)
n
a
−
n
a
n
+
N
k
2
a
−
n
a
n
}
{\displaystyle =2{\hat {H}}_{CS}=2\left\{k(a_{-n}a_{-m}a_{n+m}+a_{-n-m}a_{n}a_{m})+(1-k^{2})na_{-n}a_{n}+Nk^{2}a_{-n}a_{n}\right\}}
Fermionic Representation
edit
Now we forget about the N contribution since when N goes to infinity it will be divergent. Meanwhile we do the substitution
a
−
n
k
→
a
−
n
,
k
a
n
→
a
n
{\displaystyle {\frac {a_{-n}}{k}}\rightarrow a_{-n}\,,\,ka_{n}\rightarrow a_{n}}
the Hamiltonian now reads
H
^
C
S
=
a
−
n
−
m
a
n
a
m
+
a
−
n
a
−
m
a
n
+
m
+
(
1
−
k
2
)
(
n
a
−
n
a
n
−
a
−
n
a
−
m
a
n
+
m
)
{\displaystyle {\hat {H}}_{CS}=a_{-n-m}a_{n}a_{m}+a_{-n}a_{-m}a_{n+m}+(1-k^{2})(na_{-n}a_{n}-a_{-n}a_{-m}a_{n+m})}