Phy5670/Bethe Ansatz testing

Bethe-Ansatz for 1D-Heisenberg Model

The Bethe ansatz was originally developed for the one-dimensional Heisenberg model with nearest neighbor interaction and periodic boundary conditions:

$H=-J\sum _{n=1}^{N}{\vec {S}}_{n}\cdot {\vec {S}}_{n+1}=-J\sum _{n=1}^{N}\left[{\frac {1}{2}}(S_{n}^{+}S_{n+1}^{-}+S_{n}^{-}S_{n+1}^{+})+S_{n}^{z}S_{n+1}^{z}\right]$

where J>0 for Ferromagnet and J<0 for Anti-Ferromagnet.

In the ferromagnetic ground state, all spins are aligned in one direction along. We take it as z-direction. Thus, the ground state can be described as:

$|F\rangle =|\uparrow \uparrow \uparrow ...\uparrow \rangle$

and when the Hamiltonian act on it, only the last term of the Hamiltonian contribute energy and the ground state energy is:

$E_{0}=-{\frac {NJ}{4}}$

case of only one flipped spin

If now two of the up spins are flipped to down spins, and if these 2 spins are in position n1 and n2, we can specify the state as:

$|n_{1},n_{2}\rangle =|\uparrow \uparrow \underbrace {\downarrow } _{n_{1}}\uparrow ..\uparrow \underbrace {\downarrow } _{n_{2}}\uparrow ...\uparrow \rangle$

As $[S_{z},H]=0$, so one may expect $|n_{1},n_{2}\rangle$ to be a eigenstate of the Hamiltonian, but it is not as the flipping operators change the position of the spin and so change the state.

However the superpositions of states which have same number of flipped spins with the spin being put at different combination of positions give the eigenstate of the Hamiltonian.

Eigenstate is given by superpositions of states with only one flipped spin at the lattice site n:

$|\Psi \rangle =\sum _{n=1}^{N}a(n)|n\rangle$

${\begin{aligned}(H-E_{0})|n\rangle &=J|n\rangle -{\frac {J}{2}}(|n+1\rangle |+|n-1\rangle )\\\langle \Psi |(H-E_{0})|n\rangle &=J\langle \Psi |n\rangle -{\frac {J}{2}}\langle \Psi |(|n+1\rangle |+|n-1\rangle )\\2\left[E-E_{0}\right]a(n)&=J\left[2a(n)-a(n-1)-a(n+1)\right]\end{aligned}}$

which is a difference equation. By applying periodic boundary condition $a(n+N)=a(n)$ we obtain plane wave coefficients (we call it spin wave):

$a(n)=e^{ik_{m}n},\qquad k_{m}={\frac {2\pi }{N}}m\qquad {\text{where}}\quad m=0,1,...N-1$

Thus, the eigenvectors are given by superpositions of states with only one flipped spin with spin wave coefficient and the energy of these states are given by:

$E=E_{0}+J\left[1-cos(k_{m})\right]\qquad {\text{where}}\quad m=0,1,...N-1$

case of 2 flipped spins

Eigenstate is superposition of all states which have same number of flipped spins:

$|\Psi \rangle =\sum _{n_{1}<n_{2}}^{N}a(n_{1},n_{2})|n_{1},n_{2}\rangle$

where we have assumed $n_{1}<n_{2}<math>.Firstconsidern1andn2arenotneighboringpairsie<math>n_{1}+1<n_{2}$. Then by acting the Hamiltonian to the eigenstate, we get

$(H-E_{0})|n_{1},n_{2}\rangle =2J|n_{1},n_{2}\rangle -{\frac {J}{2}}\left[|n_{1}+1,n_{2}\rangle +|n_{1}-1,n_{2}\rangle +|n_{1},n_{2}+1\rangle +|n_{1},n_{2}-1\rangle \right]$

adding $\langle \Psi$ on it we get:

${\frac {E-E_{0}}{J}}a(n_{1},n_{2})=2a(n_{1},n_{2})-{\frac {1}{2}}\left[a(n_{1}+1,n_{2})+a(n_{1}-1,n_{2})+a(n_{1},n_{2}+1)+a(n_{1},n_{2}-1)\right]$

Bethe's approach for the coefficients $a(n_{1},n_{2})$ are again plane wave but with unknown amplitudes $A_{1}$ and $A_{2}$:

$a(n_{1},n_{2})=A_{1}e^{i(k_{1}n_{1}+k_{2}n_{2})}+A_{2}e^{i(k_{1}n_{2}+k_{2}n_{1})}$

The parameters $A_{1}$ and $A_{2}$ are determined by the insertion into the Schrodinger equation. This gives the following amplitude ratio:

${\frac {A_{1}}{A_{2}}}=e^{i\theta }=-{\frac {e^{i(k_{1}+k_{2})}+1-2e^{ik_{1}}}{e^{i(k_{1}+k_{2})}+1-2e^{ik_{2}}}}$

which gives:

$a(n_{1},n_{2})=e^{i(k_{1}n_{1}+k_{2}n_{2}+{\frac {1}{2}}\theta _{12})}+e^{i(k_{1}n_{2}+k_{2}n_{1}+{\frac {1}{2}}\theta _{21})}$

Using periodic boundary conditions we obtain the wave numbers $k_{1},k_{2}$ and the angle $\theta =\theta _{12}=-\theta _{2,1}$ satisfying the following equations:

$2\cot {\frac {\theta }{2}}=\cot {\frac {k_{1}}{2}}-\cot {\frac {k_{2}}{2}}\qquad Nk_{1}=2\pi \lambda _{1}+\theta \qquad Nk_{2}=2\pi \lambda _{2}-\theta$

where the integers $\lambda _{i}={0,1...N}$ are called Bethe quantum numbers. Thus all eigenvectors for 2 flipped spin case is determined by all possible pairs that satisfy the equations. The energy is then given by:

$E=E_{0}+J\sum _{j=1,2}\left(1-\cos(k_{j})\right)$

Case of r flipped spins

Eigenstates:

$|\Psi \rangle =\sum _{n1<n2<..<n_{r}}^{N}a(n_{1},n_{2},..,n_{r})|n_{1},n_{2},..,n_{r}\rangle$

with coefficients:

$a(n_{1},..n_{r})=\sum _{P\in S_{r}}\exp \left(i\sum _{j=1}^{r}k_{P_{j}}n_{j}+i\sum _{i<j}\theta _{P_{i}P_{j}}\right)$

The sum runs over all possible $r!$ permutation of the numbers ${1,..,r}$. Inserting into the Schrödinger equation and applying the periodic boundary conditions lead to:

Failed to parse (unknown function "\begin{alignat}"): {\displaystyle \begin{alignat} \cdot 2 \cot \frac{\theta_{ij}}{2}&=\cot\frac{k_i}{2}-\cot\frac{k_j}{2} &\qquad \text{where}\quad& i,j=1..r \\ Nk_i&=2\pi\lambda_i+\sum_{j \neq i}\theta_{ij}&&\lambda_i={1,..,N-1} \end{alignat} }

The eigenvectors are given with all combinations of Bethe quantum numbers $(\lambda _{1},..\lambda _{r})$ satisfying the Bethe equations. The energy of the corresponding state is given by

$E=E_{0}+J\sum _{j=1}^{r}\left(1-\cos k_{j}\right)$