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

$Z=Trq^{\sum _{r}r(\psi _{-r}\psi _{r}+\psi _{-r}^{*}\psi _{r}^{*})}t^{\sum _{r}\psi _{-r}\psi _{r}-\psi _{-r}^{*}\psi _{r}^{*})}$

$=Trq^{\sum _{r}r\psi _{-r}\psi _{r}}t^{\sum _{r}\psi _{-r}\psi _{r}}q^{r\sum _{r}\psi _{-r}^{*}\psi _{r}^{*}}t^{-\sum _{r}\psi _{-r}^{*}\psi _{r}^{*}}$

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

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

Level	Degeneracy	States
0	1	$|0\rangle$
1	1	$\psi _{-1/2}\psi _{-1/2}^{*}|0\rangle \sim |Young(1,0)\rangle$
2	2	$\psi _{-3/2}\psi _{-1/2}^{*}|0\rangle \sim |Young(2,0)\rangle$ $\psi _{-1/2}\psi _{-3/2}^{*}|0\rangle \sim |Young(1,1)\rangle$
3	3	$\psi _{-5/2}\psi _{-1/2}^{*}|0\rangle \sim |Young(3,0)\rangle$ $\psi _{-3/2}\psi _{-3/2}^{*}|0\rangle \sim |Young(2,1)\rangle$ $\psi _{-1/2}\psi _{-5/2}^{*}|0\rangle \sim |Young(1,1,1)\rangle$
4	5	$\psi _{-7/2}\psi _{-1/2}^{*}|0\rangle \sim |Young(4,0)\rangle$ $\psi _{-5/2}\psi _{-3/2}^{*}|0\rangle \sim |Young(3,1)\rangle$ $\psi _{-3/2}\psi _{-3/2}^{*}\psi _{-1/2}\psi _{-1/2}^{*}|0\rangle \sim |Young(2,2)\rangle$ $\psi _{-3/2}\psi _{-5/2}^{*}|0\rangle \sim |Young(2,1,1)\rangle$ $\psi _{-1/2}\psi _{-7/2}^{*}|0\rangle \sim |Young(1,1,1,1)\rangle$

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.

Jeffrey Wein, you are invited to the Teahouse

Hi Jeffrey Wein! Thanks for contributing to Wikipedia. Be our guest at the Teahouse! The Teahouse is a friendly space where new editors can ask questions about contributing to Wikipedia and get help from peers and experienced editors. I hope to see you there! Technical 13 (I'm a Teahouse host) Visit the Teahouse This message was delivered automatically by your robot friend, HostBot (talk) 20:41, 10 January 2014 (UTC)

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 $z_{i}=\exp(2ix_{i}),L=\pi$ . We have

$dz_{i}=2iz_{i}dx_{i}\,\,,\,\,\partial _{x_{i}}=(2iz_{i})^{-1}\partial _{x_{i}}\,\,,\,\partial _{x_{i}}=(2iz_{i})\partial _{z_{i}}\sim 2iL_{0}$

where $L_{n}$ 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

$(\partial _{x_{i}})^{2}=-4(z_{i}\partial _{z_{i}}+z_{i}^{2}\partial _{z_{i}}^{2})=-4L_{0}-4L_{-n}L_{n}$.

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,

$H_{J}\equiv \psi _{0}^{-1}{\tilde {H}}\psi _{0}=\psi _{0}^{-1}({\frac {1}{2}}(\partial _{i}+\psi _{0}^{-1}\partial _{i}\psi _{0}))({\frac {1}{2}}(\partial _{i}-\psi _{0}^{-1}\partial _{i}\psi _{0}))\psi _{0}$ ,

noticing $(\partial _{i}-\psi _{0}^{-1}\partial _{i}\psi _{0})\psi _{0}=\psi _{0}\partial _{i}$ , 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 $H_{J}$

$H_{J}=2\sum _{i}(z_{i}\partial _{z_{i}})^{2}+2\beta \sum _{i<j}{\frac {z_{i}+z_{j}}{z_{i}-z_{j}}}(z_{i}\partial _{z_{i}}-z_{j}\partial _{z_{j}})$ .

Acting on generating function

$\prod _{i=1}^{N}V_{-}^{k}(z_{i})\equiv \exp \left(k\sum _{n>0}{\frac {a_{-n}}{n}}\sum _{i=1}^{N}(z_{i})^{n}\right)$, $k={\sqrt {\beta }}$

one have

$H_{J}\prod _{i=1}^{N}V_{-}^{k}(z_{i})|k_{0}\rangle =(-i)(\partial _{x_{i}}+2\beta \sum _{j\neq i}ctg(x_{ij}))ka_{-n}\sum _{i=1}(z_{i})^{n}\prod _{i=1}^{N}V_{-}^{k}(z_{i})|k_{0}\rangle$

$=\left\{(-i)\left[(2i)nka_{-n}\sum _{i=1}^{N}(z_{i})^{n}+2ik^{2}a_{-n}a_{-m}\sum _{i=1}^{N}(z_{i})^{n+m}\right]+2\beta \sum _{i\neq j}{\frac {z_{i}+z_{j}}{z_{i}-z_{j}}}ka_{-n}\sum _{i=1}^{N}(z_{i})^{n}\right\}\prod _{i=1}^{N}V_{-}^{k}(z_{i})|k_{0}\rangle$

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

because of the coherent relation $a_{n}e^{k\sum _{m>0}a_{-m}/m\sum _{i}^{N}(z_{i})^{m}}=k\sum _{i}^{N}(z_{i})^{n}e^{k\sum _{m>0}a_{-m}/m\sum _{i}^{N}(z_{i})^{m}}$,

the acting of $H_{J}$ on generating function gives

$H_{J}=2na_{-n}a_{n}+2ka_{-n}a_{-m}a_{n+m}+2k^{3}a_{-n}\sum _{i\neq j}{\frac {z_{i}+z_{j}}{z_{i}-z_{j}}}\times (z_{i})^{n}$ .

Since $\sum _{i\neq j}{\frac {z_{i}+z_{j}}{z_{i}-z_{j}}}(z_{i})^{n}={\frac {1}{2}}\sum _{i\neq j}{\frac {z_{i}+z_{j}}{z_{i}-z_{j}}}((z_{i})^{n}-(z_{j})^{n})$ $=\sum _{i<j}{\frac {z_{i}+z_{j}}{z_{i}-z_{j}}}(z_{i}-z_{j})\left(z_{i}^{n-1}+z_{i}^{n-2}z_{j}+\cdots +z_{j}^{n-1}\right)$ $=\sum _{i<j}(z_{i}^{n}+2(z_{i}^{n-1}z_{j}+\cdots z_{i}z_{j}^{n-1})+z_{j}^{n})$ $=\sum _{i\neq j}\left\{z_{i}^{n}+z_{i}^{n-1}z_{j}+\cdots +z_{i}z_{j}^{n-1}\right\}$ $=Np_{n}+p_{n-1}p_{1}+\cdots +p_{1}p_{n-1}-np_{n}((i=j)contribution)$

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

Fermionic Representation

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