User talk:Jeffrey Wein

This is an statistic model explanation of Jacobi's Triple Identity.
The identity reads


 * $$\prod_{n=1}^\infty (1-q^n) (1+q^{n-\frac{1}{2}}t)(1+q^{n-\frac{1}{2}}/t )= \sum_{n\in\mathbb{Z}}q^{\frac{n^2}{2}}t^n$$.

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
 * $$\psi(z) = \sum_{r\in\mathbb{Z}+1/2}\psi_r z^{r-1/2}r\,\,,\,\, \psi^*(z) = \sum_{r\in\mathbb{Z}+1/2}\psi^*_r z^{r-1/2}$$.

Anticommutation relation


 * $$\{\psi_r,\psi_s\} = \delta_{r+s,0}\,\,,\,\,\{\psi^*_r,\psi^*_s\} = \delta_{r+s,0}$$.

Equipped with a Hamiltonian


 * $$H = E_0\sum_{r\in\mathbb{Z}^+ -1/2} r\left(\psi_{-r}\psi_r +\psi^*_{-r}\psi^*_r\right) = E_0 L_0 $$,

where
 * $$L_0 =\sum_{r\in\mathbb{Z}^+ -1/2} r\left(\psi_{-r}\psi_r +\psi^*_{-r}\psi^*_r \right)$$

is the zero mode of Virasoro algebra.

A fermion number defines as


 * $$N =\sum_{r\in\mathbb{Z}^+ -1/2} \left(\psi_{-r}\psi_r -\psi^*_{-r}\psi^*_r \right)$$.

Partion function of this Grand canonical ensemble

$$Z(q,t) = \sum_{states} \exp (-\beta (E-\mu N)$$, and define the parameter :$$t = e^{\beta\mu}, q = e^{-\beta E_0}$$.

Substitution of the operator expression of N and H

$$Z = Tr q^{\sum_r r(\psi_{-r}\psi_r+\psi_{-r}^*\psi^*_r)} t^{\sum_r\psi_{-r}\psi_r-\psi_{-r}^*\psi^*_r) }$$

$$ =Tr q^{\sum_r r \psi_{-r}\psi_r} t^{\sum_r \psi_{-r}\psi_r} q^{r\sum_r \psi^*_{-r}\psi^*_r} t^{-\sum_r \psi^*_{-r}\psi^*_{r}}$$

$$ = \prod_{r\in\mathbb{N}-\frac{1}{2}}(1+q^r t)(1+q^r/t)$$.

Another way to counting the same system is a Young diagram counting. Classfying the system by the Fermion number N.
 * $$Z(q,t) = \sum_{N\geq 0} t^N Z_N (q)$$.

Consider the N=0 counting of states. At energy level n, the number of states is :$$p[n]$$, the partition number of n. This can see from the following table

The $$N=0$$ counting of states gives the Dedekind eta function

$$Z_0(q,t) = \prod_{m=1}^\infty \frac{1}{1-q^n}$$.

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 $$\psi_{-k-1/2}\psi^*_{-1/2}|k\rangle$$,

where $$|k\rangle = \psi_{-1/2}\psi_{-3/2}\psi_{-5/2}\cdots\psi_{-k+1/2}|0\rangle$$,

is also known as the colored vacuum of Dirac sea. The second excitations are

$$\psi_{-k-1/2}\psi^*_{-3/2}|k\rangle$$,

$$\psi_{-k-3/2}\psi^*_{-1/2}|k\rangle$$.

This argument follows, and the only difference of level k counting and level 0 counting is the energy difference.

$$E_k/E_0 = \sum_{r=1}^k (k-\frac{1}{2}) = \frac{k^2}{2}$$

The level k counting contributs $$Z_k = q^{\frac{k^2}{2}}\prod_{m=1}^{\infty}\frac{1}{1-q^m} $$

this leads the total partion function

$$Z(q,t) = \sum_{n=0}^\infty t^n q^{\frac{n^2}{2}}\prod_{m=1}^{\infty}\frac{1}{1-q^m} $$.

The two ways of counting states should be equivalent. The Jacobi triple identity is proven.

Operator Formalism of Calogero Sutherland Model
Operator Formalism of Calogero Sutherland Model

The CS Hamiltonian

$$H_{CS} = -\frac{1}{2}\partial_{x_i}^2 + \frac{\beta(\beta-1)}{sin^2 (\frac{\pi}{L}x_{ij})}\,\,,\,\,x_{ij}\equiv x_i - x_j$$.

Choose $$z_i = \exp(2i x_i), L=\pi$$. We have

$$d z_i = 2i z_i d x_i\,\,,\,\, \partial_{x_i} = (2i z_i )^{-1} \partial_{x_i}\,\,,\, \partial_{x_i} = (2 i z_i)\partial_{z_i} \sim 2i L_0$$

where $$L_n$$ be the generator of Witt algebra

$$[L_n, L_m]= (n-m)L_{n+m}\,\,,\, L_n = z^{n+1}\partial_z\,\,,\,\,L_0 = z\partial_z $$.

Notice that the conformal map from cylinder to complex plane also maps the translation of $$x_i$$(generated by momentum operator) to a scaling(or inflation) which is generated by $$L_0$$.

It is simple that

$$(\partial_{x_i})^2 = -4(z_i\partial_{z_i} + z_i^2\partial^2_{z_i}) = -4L_0 -4L_{-n}L_n$$.

Another Definition of $$H_{CS}$$ (up to zero point energy)

$$\tilde{H} = \frac{1}{2}\left(\partial_{x_i}+\partial_{x_i}\prod_{l<j}sin^{\beta}x_{lj}\right)\left(\partial_{x_i}-\partial_{x_i}\prod_{k<j}sin^{\beta}x_{kj}\right)$$.

This could be derived from the following calculations.

$$\partial_{x_{i}}\ln \prod_{l<j} sin^\beta (x_{lj}) = \beta \sum_{i\neq j} ctg(x_{ij})$$.

Then the Hamiltonian becomes

$$\tilde{H} = -\frac{1}{2}\partial_{x_i}^2 +\frac{\beta^2}{2}\sum_{j,k\neq i}ctg(x_{ij})ctg(x_{ik}) + \frac{1}{2}\beta\left[\partial_{x_i}, \sum_{k\neq i} ctg(x_{ik})\right]$$.

Since

$$\left[\partial_{x_i}, \sum_{k\neq i} ctg(x_{ik})\right] = \sum_{k\neq i}\frac{-1}{sin^2(x_{ik})}$$ ,

and using the identity

$$\sum_{i,j\neq k}ctg(x_{ij})ctg(x_{ik})+cyclic(i,j,k)=\sum_{i\neq j\neq k}(-1) = -N(N-1)(N-2)$$ ,

we have

$$\frac{\beta^2}{2}\sum_{j,k\neq i} ctg(x_{ij})ctg(x_{ik}) = \frac{-\beta^2}{6}N(N-1)(N+1) +\frac{1}{2}\beta^2 \sum_{j\neq i} \frac{1}{sin^2(x_{ij})}$$.

This leads to the result that $$\tilde{H} = H_{CS}- E_0$$ with

$$E_0 = \frac{\beta}{6} (N-1)N(N+1)$$.

From the Hamiltonian $$\tilde{H}$$, it is clear that

$$\psi_0 = \prod_{i<j} sin^\beta (x_{ij})$$

is the ground state with energy 0. Thus it is also a ground state of $$H_{CS}$$ with ground energy $$E_0$$

Jacobi's Transformation
Moving out the contribution of ground state,

$$H_J \equiv \psi^{-1}_0 \tilde{H} \psi_0 = \psi_0^{-1}(\frac{1}{2}(\partial_i +\psi_0^{-1}\partial_i\psi_0))(\frac{1}{2}(\partial_i-\psi_0^{-1}\partial_i\psi_0))\psi_0$$ ,

noticing $$(\partial_i-\psi_0^{-1}\partial_i \psi_0)\psi_0 = \psi_0\partial_i$$, we have

$$H_J = -\frac{1}{2}(\partial_i + 2\partial_i\psi_0)\partial_i$$.

For $$\partial_{x_i}\psi_0 = \beta\sum_{j\neq i} ctg(x_{ij})$$,

one arrives the complex plane expression of $$H_J$$

$$H_J= 2\sum_i(z_i\partial_{z_i})^2 +2\beta\sum_{i0}\frac{a_{-n}}{n}\sum_{i=1}^N (z_i)^n\right)$$, $$k=\sqrt{\beta}$$

one have

$$ H_J \prod_{i=1}^N V_-^k(z_i)|k_0\rangle = (-i)(\partial_{x_i}+2\beta\sum_{j\neq i}ctg(x_{ij})) k a_{-n}\sum_{i=1}(z_i)^n\prod_{i=1}^N V_-^k(z_i)|k_0\rangle$$

$$ = \left\{(-i)\left[(2i) n k a_{-n} \sum_{i=1}^N (z_i)^n + 2i k^2 a_{-n} a_{-m} \sum_{i=1}^N (z_i)^{n+m}\right] + 2\beta \sum_{i\neq j}\frac{z_i + z_j}{z_i-z_j} k a_{-n} \sum_{i=1}^N (z_i)^n\right\}\prod_{i=1}^N V_-^k(z_i)|k_0\rangle$$

Besides of the last interaction term, all terms can be operatorization

because of the coherent relation $$ a_{n} e^{ k\sum_{m>0} a_{-m}/m \sum_i^N (z_i)^m} = k \sum_i^N(z_i)^ne^{ k\sum_{m>0} a_{-m}/m  \sum_i^N (z_i)^m}$$,

the acting of $$H_J$$ on generating function gives

$$H_J = 2n a_{-n} a_{n} + 2k a_{-n}a_{-m}a_{n+m}+ 2k^3 a_{-n}\sum_{i\neq j}\frac{z_i+z_j}{z_i-z_j}\times  (z_i)^n$$.

Since $$\sum_{i\neq j}\frac{z_i + z_j}{z_i - z_j} (z_i)^n = \frac{1}{2}\sum_{i\neq j}\frac{z_i + z_j}{z_i - z_j} ((z_i)^n-(z_j)^n)$$ $$ = \sum_{i<j}\frac{z_i + z_j}{z_i - z_j}(z_i-z_j)\left(z_i^{n-1} + z_i^{n-2}z_j +\cdots+z_j^{n-1} \right)$$ $$=\sum_{i<j}(z_i^n + 2 (z_i^{n-1}z_j +\cdots z_i z_j^{n-1})+ z_j^n)$$ $$=\sum_{i\neq j} \left\{z_i^n + z_i^{n-1}z_j +\cdots+z_i z_j^{n-1}\right\}$$ $$= N p_n + p_{n-1}p_1 + \cdots + p_1 p_{n-1}- n p_n((i=j) contribution)$$

now all terms can have operator formalism $$\tilde{H}_{CS} = 2n a_{-n}a_n +2k a_{-n}a_{-m}a_{n+m} + 2k a_{-n}(Nk a_{n} - nk a_n + a_{n-m}a_m)$$ $$= 2 \hat{H}_{CS} = 2\left\{k (a_{-n}a_{-m}a_{n+m} + a_{-n-m }a_n a_m) +(1-k^2)n a_{-n} a_n +Nk^2 a_{-n}a_n\right\}$$

Fermionic Representation
Now we forget about the N contribution since when N goes to infinity it will be divergent. Meanwhile we do the substitution $$\frac{a_{-n}}{k}\rightarrow a_{-n}\,,\, k a_{n}\rightarrow a_n$$ the Hamiltonian now reads $$\hat{H}_{CS}= a_{-n-m}a_n a_m + a_{-n}a_{-m}a_{n+m} +(1-k^2)(n a_{-n}a_n - a_{-n}a_{-m} a_{n+m})$$