열역학적 베테 가설 풀이(thermodynamic Bethe ansatz)
Pythagoras0 (토론 | 기여)님의 2013년 2월 24일 (일) 11:06 판
개요
- 산란행렬로부터 바닥상태의 에너지를 비섭동적으로 계산할 수 있는 방법
 
basic notions for particle scattering
- infinitely long cylinder of radius \(R\)
 - N species of particles
 - mass of particles \(m_{a}, a=1,\cdots, N\)
 - rapidity \(\theta\) (also called spectral parameter or wave number)
- a notion from relativity
 - http://en.wikipedia.org/wiki/Rapidity
 
 - a notion from relativity
 - energy \(E=m_{a}R\cosh \theta\)
 - momentum \(p=m_{a}R\sinh \theta\)
 - energy-momentum vector \(p^{\mu}=(E,P)\)
 - 산란행렬 S-matrix (factorizable scattering theory)\[S_{ab}(\theta)\]
 - symmetric matrix kernel \[\phi_{ab}(\theta)=-i\frac{d}{d\theta}\log S_{ab}(\theta)\]
 - spectral density of particles \(\epsilon_{a}(\theta)\)
- also called the pseudoenergy
 
 - also called the pseudoenergy
 - Y-system \(Y_{a}(\theta)=e^{-\epsilon_{a}(\theta)}\) i.e. exponential of spectral density
 - ground state energy \(E(R)\)
 - scaling function \(c(R)\) related to the central charge
 - TBA equation
- equation to find the spectral density functions \(\epsilon_{a}(\theta)\)
 
 - equation to find the spectral density functions \(\epsilon_{a}(\theta)\)
 - UV limit
- plateau behaviour
 - \(\epsilon_{a}(\theta)\) becomes constant in a large region for \(\theta\) when r(inverse temperature) is small
 
 - plateau behaviour
 - IR limit
 
 
 
limit
- energy \(E=m_{a}R\cosh \theta\)
 - momentum \(p=m_{a}R\sinh \theta\)
 - in the CFT limit, we regard \theta \to \infty for right movers and -\infty for left movers
 - Thus we get, E=p and E=-p respectively in CFT limit
 
 
 
TBA equation
- a system which interacts dynamically via the scattering matrix and statistically via Fermi statistics\[rm_{a}^{i}\cosh\theta = \epsilon_{a}^{i}(\theta)+\sum_{b=1}^{l}\sum_{j=1}^{\tilde{l}}\int_{-\infty}^{\infty} d\theta' \phi_{ab}^{ij}(\theta-\theta')\ln (1+e^{-\epsilon_{b}^{j}(\theta')})\]
where r is the inverse temperature and \(m_{a}^{i}\) the mass of particle (a,i) 
예 : Yang-Lee 모형
- 1 particle
 - 산란행렬
 
$$ S_{11}(\theta)=\tanh \left(\frac{1}{2} \left(\theta -\frac{2 i \pi }{3}\right)\right) \coth \left(\frac{1}{2} \left(\theta +\frac{2 i \pi }{3}\right)\right) $$
- 커널
 
$$ \phi_{11}(\theta)=-i\frac{d}{d\theta}\log S_{11}(\theta)=\sqrt{3} \left(\frac{1}{2 \cosh (\theta )+1}+\frac{1}{2 \cosh (\theta )-1}\right) $$
$$ N=\frac{1}{2\pi}\int_{-\infty}^{\infty}\phi_{11}(\theta)=1 $$