The identity reads
.
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
.
Anticommutation relation
.
Equipped with a Hamiltonian
,
where
![{\displaystyle L_{0}=\sum _{r\in \mathbb {Z} ^{+}-1/2}r\left(\psi _{-r}\psi _{r}+\psi _{-r}^{*}\psi _{r}^{*}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e33fc3017495e2dce44d5e9617ea83b75fc67cce)
is the zero mode of Virasoro algebra.
A fermion number defines as
.
Partion function of this Grand canonical ensemble
, and define the parameter :
.
Substitution of the operator expression of N and H
.
Another way to counting the same system is a Young diagram counting.
Classfying the system by the Fermion number N.
.
Consider the N=0 counting of states.
At energy level n, the number of states is :
, the partition number of n. This can see from the following table
Level |
Degeneracy |
States
|
0 |
1 |
|
1 |
1 |
|
2 |
2 |
|
3 |
3 |
|
4 |
5 |
|
The
counting of states gives the Dedekind eta function
.
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
,
where
,
is also known as the colored vacuum of Dirac sea.
The second excitations are
,
.
This argument follows, and the only difference of level k counting and level 0 counting is the energy difference.
The level k counting contributs
this leads the total partion function
.
The two ways of counting states should be equivalent. The Jacobi triple identity is proven.
Operator Formalism of Calogero Sutherland Model
The CS Hamiltonian
.
Choose
. We have
where
be the generator of Witt algebra
.
Notice that the conformal map from cylinder to complex plane also maps the translation of
(generated by momentum operator) to a scaling(or inflation) which is generated by
.
It is simple that
.
Another Definition of
(up to zero point energy)
.
This could be derived from the following calculations.
.
Then the Hamiltonian becomes
.
Since
,
and using the identity
,
we have
.
This leads to the result that
with
.
From the Hamiltonian
, it is clear that
is the ground state with energy 0. Thus it is also a ground state of
with ground energy
Moving out the contribution of ground state,
,
noticing
, we have
.
For
,
one arrives the complex plane expression of
.
Acting on generating function
,
one have
Besides of the last interaction term, all terms can be operatorization
because of the coherent relation
,
the acting of
on generating function gives
.
Since
now all terms can have operator formalism
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
the Hamiltonian now reads