Phy5670/Bethe Ansatz for many particle systems

Introduction

Bethe ansatz is a very powerful method for finding the exact solutions of one-dimensional quantum many-body systems. Experts conjecture that each universality class in one dimension contains at least one model solvable by the Bethe ansatz. An ansatz is an educated guess that is verified later by its results. The Bethe ansatz method was invented by Hans Bethe in 1931 to find the exact eigenvalues and eigenvectors of the one-dimensional antiferromagnetic Heisenberg model Hamiltonian. Since then the method has been extended to other models in one dimension: Bose gas, Hubbard model, etc. The exact solutions of the s-d model (by P.B. Wiegmann [2] in 1980 and independently by N. Andrei,[3] also in 1980) and the Anderson model (by P.B. Wiegmann [4] in 1981, and by N. Kawakami an A. Okiji [5] in 1981) are also both based on the Bethe ansatz. The method is also applicable to some of the 2-dimensional classical systems as the transfer matrices of 2D classical system sometimes have common eigenfunctions as some 1-dimensional quantum systems. Some examples of the application of Bethe ansatz in 2D system are stated in Ref[6,7,8].

Models solvable by the Bethe ansatz can be compared to free fermion models. One can say that the dynamics of a free model is one-body reducible: the many-body wave function for fermions (bosons) is the anti-symmetrized (symmetrized) product of one-body wave functions. Models solvable by the Bethe ansatz are not free: the two-body sector has a non-trivial scattering matrix, which in general depends on the momenta. On the other hand the dynamics of the models solvable by the Bethe ansatz is two-body reducible: the many-body scattering matrix is a product of two-body scattering matrices. Many-body collision happen as a sequence of two-body collisions and the many-body wave function can be represented in a form which contains only elements from two-body wave functions. The many-body scattering matrix is equal to the product of pairwise scattering matrices.

In using this method for N-body problem, the N-body wavefunction is represented as a linear combination of N! plane waves with N qusai-momenta. The energy eigenvalue of the lowest energy state in the thermodynamics limit is reduced to a distribution function of the quasi-momenta. The distribution function must satisfy a linear integral equation. The energy per unit length is obtained by solving this integral equation. Elementary excitations from the ground state are expressed by the deviation of the distribution of quasi-momenta from the equilibrium. For general eigenstates, the quasi-momenta are complex numbers, but in many cases they are real number. The total energy is represented by the distribution functions of quasi-momenta.

Bethe Ansatz for Delta-function potential ("quantum billiard ball" problem)

One of the simplest interacting system soluble by Bethe's technique is the "quantum billiard ball" problem. We consider first the classical system of N billiard balls constrained to move along a line, all perfectly elastic and of equal mass, so that in each collision the ingoing and outgoing momenta coincide. As the system develops in time, the original N momenta are preserved, although they keep getting transferred to different balls. The quantum billiard balls can be Boson and Fermions or a mixture of both Bosons and Fermions. The interaction terms in the Hamiltonian is represented by Dirac delta functions. Here we consider different cases separately. The Hamiltonian is in the form:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H=-\sum_{i=1}^{N}\frac{\partial^{2}}{\partial x_{i}^{2}} + 2c\sum_{i<j}\delta(x_{i}-x_{j}) }

Here we will consider periodic boundary condition for all cases.

Bosonic case

2 bosons repulsive potential Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c>0}

If there is only one particle, it will move freely as a plane wave. Now we constrain 2 particle to move along the same line, so that they may collide. And we've assume when they touch each other, they will try to push away each other (c>0), then the Hamiltonian of the system is:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H=-\frac{\partial^{2}}{\partial x_{1}^{2}} -\frac{\partial^{2}}{\partial x_{2}^{2}} + 2c\delta(x_{1}-x_{2}) }

Assume particle 1 originally has quasi-momentum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_{1} } and particle 2 originally has quasi-momentum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_{2} } . As time go on, either the 2 particle do not collide or the particles make elastic collision so after that they exchange their momenta. So we assume the final state function of the system is in the form (ansatz):

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x_{1},x_{2}) = A(12)e^{ik_{1}x_{1}+ik_{2}x_{2}} + A(21)e^{ik_{2}x_{1}+ik_{1}x_{2}} \ \ \ \ for \ x_{1}<x_{2}}

By symmetric condition of the Boson wave function, we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x_{1},x_{2}) = A(12)e^{ik_{1}x_{2}+ik_{2}x_{1}} + A(21)e^{ik_{2}x_{2}+ik_{1}x_{1}} \ \ \ \ for \ x_{2}<x_{1}}

we can easily check that the wave function is continuous at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{1} = x_{2} } but its derivative is discontinuous. Put the Ansatz into the originally Hamiltonian:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} Hf(x_{1},x_{2})&= \left [ -\frac{\partial^{2}}{\partial x_{1}^{2}} -\frac{\partial^{2}}{\partial x_{2}^{2}} + 2c\delta(x_{1}-x_{2}) \right ] f(x_{1},x_{2}) \\ &=(k_{1}^{2} + k_{2}^{2})f(x_{1},x_{2})+2i\delta(x_{1}-x_{2})[A(12)-A(21)](k_{1}-k_{2})e^{i(k_{1}+k_{2})x_{1}} +2c\delta(x_{1}-x_{2})[A(12)+A(21)]e^{i(k_{1}+k_{2})x_{1}} \end{align}}

On the other hand,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Hf(x_{1},x_{2}) = E f(x_{1},x_{2})= \left (k_{1}^{2}+k_{2}^{2}\right )f(x_{1},x_{2})}

So comparing, we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2i(k_{1}-k_{2})\left [A(12)-A(21) \right ] +2c[A(12)+A(21)] =0 }

Thus we obtain the amplitude ratio:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{A(12)}{A(21)}=\frac{k_{1}-k_{2}+ic}{k_{1}-k_{2}-ic} }

3 bosons repulsive potential Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c>0}

Similar to the previous case, we can write the 3-particle ansatz as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} f(x_{1},x_{2},x_{3}) &= A(123)e^{ik_{1}x_{1}+ik_{2}x_{2}+ik_{3}x_{3}} + A(213)e^{ik_{2}x_{1}+ik_{1}x_{2}+ik_{3}x_{3}} +A(132)e^{ik_{1}x_{1}+ik_{3}x_{2}+ik_{2}x_{3}} \\ &+ A(321)e^{ik_{3}x_{1}+ik_{2}x_{2}+ik_{1}x_{3}} +A(231)e^{ik_{2}x_{1}+ik_{3}x_{2}+ik_{1}x_{3}} + A(312)e^{ik_{3}x_{1}+ik_{1}x_{2}+ik_{2}x_{3}} \ \ \ \ for \ x_{1}<x_{2}<x_{3} \end{align}}

Different cases of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{1}<x_{3}<x_{2},.... } can be obtain by switching the order as in previous case. Then we obtain the 6 conditions for the 6 amplitudes A(ijk):

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} &\frac{A(123)}{A(213)}=\frac{k_{1}-k_{2}+ic}{k_{1}-k_{2}-ic} \\ & \frac{A(123)}{A(132)}=\frac{k_{2}-k_{3}+ic}{k_{2}-k_{3}-ic} \\ & \frac{A(213)}{A(231)}=\frac{k_{1}-k_{3}+ic}{k_{1}-k_{3}-ic} \\ & \frac{A(132)}{A(312)}=\frac{k_{1}-k_{3}+ic}{k_{1}-k_{3}-ic} \\ &\frac{A(321)}{A(231)}=\frac{k_{3}-k_{2}+ic}{k_{3}-k_{2}-ic} \\ &\frac{A(321)}{A(312)}=\frac{k_{2}-k_{1}+ic}{k_{2}-k_{1}-ic} \end{align}}

Thus we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} A(123)&=C(k_{1}-k_{2}+ic)(k_{1}-k_{3}+ic)(k_{2}-k_{3}+ic) \\ A(213)&=-C(k_{2}-k_{1}+ic)(k_{1}-k_{3}+ic)(k_{2}-k_{3}+ic) \\ A(132)&=-C(k_{1}-k_{2}+ic)(k_{1}-k_{3}+ic)(k_{3}-k_{2}+ic) \\ A(321)&=-C(k_{2}-k_{1}+ic)(k_{3}-k_{1}+ic)(k_{3}-k_{2}+ic) \\ A(231)&=C(k_{1}-k_{2}+ic)(k_{1}-k_{3}+ic)(k_{2}-k_{3}+ic) \\ A(312)&=C(k_{1}-k_{2}+ic)(k_{3}-k_{1}+ic)(k_{3}-k_{2}+ic) \\ \end{align}}

N bosons, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c=\infty }

We assume the N particle has N quasi-momenta Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_{1}, k_{2}, ...,k_{N} }

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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H=-J\sum^N_{n=1}\vec{S}_n\cdot \vec{S}_{n+1}=-J\sum^N_{n=1}\left[\frac{1}{2}(S_n^+S^-_{n+1} + S_n^-S^+_{n+1})+S^z_nS^z_{n+1} \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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n_1,n_2\rangle = |\uparrow\uparrow\underbrace{\downarrow}_{n_1}\uparrow..\uparrow\underbrace{\downarrow}_{n_2} \uparrow...\uparrow\rangle }

As Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [S_{z},H]=0 } , so one may expect Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle = \sum^N_{n=1}a(n)|n\rangle }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} (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{align}}

which is a difference equation. By applying periodic boundary condition Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a(n+N)=a(n)} we obtain plane wave coefficients (we call it spin wave):

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 (or we call it one magnon) and the energy of these states are given by:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E=E_{0}+J\left [1-cos(k_{m})\right ] }

case of 2 flipped spins

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle = \sum^N_{n_{1}<n_{2}}a(n_1,n_2)|n_1,n_2\rangle }

where we have assumed Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_1<n_2 } .

First consider n1 and n2 are not neighboring pairs i.e. n1+1<n2. Then by acting the Hamiltonian to the eigenstate, we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle \Psi| } on it we get:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 ].....(1) }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{E-E_{0}}{J}a(n_1,n_2) = a(n_{1},n_{2}) -\frac{1}{2}\left [a(n_1+1,n_2)+a(n_1-1,n_2)\right ] + a(n_{1},n_{2}) -\frac{1}{2}\left [ a(n_1,n_2+1) + a(n_1,n_2-1) \right ] }

as it just like two times of the previous case, so we can easily see Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{ik_1 x_1+ik_2 x_2} } is a solution of the equation. As we can exchange k1 and k2, so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{ik_2 x_1+ik_1 x_2} } would also be a solution. So the general solution for this n1+1<n2 case is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a(n_1,n_2)=A_1e^{i(k_1n_1+k_2n_2)}+A_2e^{i(k_1n_2+k_2n_1)} }

where it is important to notice that this solution is valid for equation (1) no matter what n1, n2 are.

As the two terms should contribute the same amplitudes, we can re-write Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a(n_1,n_2) } as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a(n_1,n_2)=e^{i(k_1n_1+k_2n_2+\theta/2)}+A_2e^{i(k_1n_2+k_2n_1-\theta/2)} }

Then by periodic boundary condition Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a(n_1,n_2)=a(n_2,n_1+N) } , we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Nk_{1,\lambda_{1}}=2\pi\lambda_1+\theta\qquad Nk_{2,\lambda_{2}}=2\pi\lambda_2-\theta }

where the integers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_i = {0,1 ... N-1} } named Bethe quantum numbers. To obtain the value of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta } , we need to consider the case whereas 2 spins sit next to each other and then do some matching.

Now consider the 2 spin sit next to each other, i.e. n1+1=n2, again acting the Hamiltonian to the eigenstate and then act Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle \Psi| } on it, we obtain the difference equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{E-E_{0}}{J}a(n_1,n_1+1) = a(n_{1},n_{1}+1) -\frac{1}{2}\left [a(n_1-1,n_1+1) + a(n_1,n_1+2) \right ].....(2) }

we can easy see that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a(n_1,n_1+1)=A_1e^{i(k_1n_1+k_2(n_1+1))}+A_2e^{i(k_1(n_1+1)+k_2n_1)} }

is also the solution of equation. Now as I've mentioned, the solution for eqt (1) is valid for any n1 and n2, so now we choose n2=n1+1 and put it into eqt(1), then comparing with eqt(2), we get a constraint equation that can be used for calculating Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta } :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a(n_1,n_1+1) = \frac{1}{2}[a(n_1,n_1)+a(n_1+1,n_1+1)] }

simplify it, we get

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

And finally we get a equation in $\theta$ which can be solve numerically:

$2\cot {\frac {\theta }{2}}=\cot {\frac {k_{1,\lambda _{1}}}{2}}-\cot {\frac {k_{2,\lambda }}{2}}$

where

$Nk_{1,\lambda _{1}}=2\pi \lambda _{1}+\theta \qquad Nk_{2,\lambda _{2}}=2\pi \lambda _{2}-\theta$

The energy of the system is given by:

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

Case of r flipped spins

Eigenstates is the superposition of all states which have r spins flipped:

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

By the previous works, we consider a ansatz:

$a(n_{1},..n_{r})=\sum _{P}\exp \left(i\sum _{j=1}^{r}k_{P_{j}}n_{j}+i\sum _{i<j}\theta _{P_{i}P_{j}}/2\right)$

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 $(\lambda _{1},....\lambda _{r})$. The energy of the corresponding state is given by

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

Reference

1. http://de.wikipedia.org/wiki/Bethe-Ansatz.

2. P.B. Wiegmann, Soviet Phys. JETP Lett., 31, 392 (1980).

3. N. Andrei, Phys. Rev. Lett., 45, 379 (1980).

4. P.B. Wiegmann, Phys. Lett. A 80, 163 (1981).

5. N. Kawakami, and A. Okiji, Phys. Lett. A 86, 483 (1981).

6. Mehran Kardar, Nuclear Physics B, Volume 290, 1987, Pages 582-602.

7. S. Park and K. Moon, Solid State Communications, Volume 132, Issue 12, December 2004, Pages 851-856.

8. M.J. Martins, Phys. Rev. E 59, 7220–7223 (1999).

9. http://en.wikipedia.org/wiki/Bethe_ansatz.

10. Minoru Takahashi, Thermodynamics of one-dimensional solvable models, 1999.

11. M. Fowler ,J. Appl. Phys. Vol. 53, No.3, March 1982.