Phy5670/Bethe Ansatz testing: Difference between revisions

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(New page: === Bethe-Ansatz for 1D-Heisenberg Model (Ref[1])=== The Bethe ansatz was originally developed for the one-dimensional Heisenberg model with nearest neighbor interaction and periodic boun...)
 
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=== Bethe-Ansatz for 1D-Heisenberg Model (Ref[1])===
=== 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:
The Bethe ansatz was originally developed for the one-dimensional Heisenberg model with nearest neighbor interaction and periodic boundary conditions:
Line 15: Line 15:
|F\rangle = |\uparrow\uparrow\uparrow...\uparrow\rangle
|F\rangle = |\uparrow\uparrow\uparrow...\uparrow\rangle
</math>
</math>
and when the Hamiltonian act on it, only the last term of the Hamiltonian contribute energy and the ground state energy is:
<math> E_{0}=-\frac{NJ}{4} </math>
==== 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:  
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:  


:<math>
:<math>
|n_1n_2\rangle = |\uparrow\uparrow\underbrace{\downarrow}_{n_1}\uparrow..\uparrow\underbrace{\downarrow}_{n_2} \uparrow...\uparrow\rangle
|n_1,n_2\rangle = |\uparrow\uparrow\underbrace{\downarrow}_{n_1}\uparrow..\uparrow\underbrace{\downarrow}_{n_2} \uparrow...\uparrow\rangle
</math>
</math>


As <math> [S_{z},H]=0 </math>, so the eigenstates of the Hamiltonian are given by the superpositions of states which have same number of flipped spins with the spin being put at different combination of positions.  
As <math> [S_{z},H]=0 </math>, so one may expect <math> |n_1,n_2\rangle </math> 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.


==== case of only one flipped spin ====
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:
Eigenstate is given by superpositions of states with only one flipped spin at the lattice site n:
Line 32: Line 39:
</math>
</math>


The eigenvectors are solutions of the stationary Schrödinger equation. By comparing coefficients, we obtain N difference equations for the coefficients a(n):
:<math>\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}</math>
 
which is a difference equation. By applying periodic boundary condition
<math>a(n+N)=a(n)</math> we obtain plane wave coefficients (we call it spin wave):
 
<math> a(n)=e^{ik_{m}n}, \qquad k_{m}=\frac{2\pi}{N}m \qquad \text{where} \quad m=0,1,... N-1 </math>


<math> 2\left [E-E_{0}a(n)\right ]=J\left [2a(n)-a(n-1)-a(n+1)\right ] </math>
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:


By applying periodic boundary condition <math>a(n+N)=a(n)</math> we obtain plane wave solution:
:<math>
E=E_{0}+J\left [1-cos(k_{m})\right ]
\qquad \text{where} \quad m=0,1,... N-1 </math>


<math> a(n)=e^{ikn}, \qquad k=\frac{2\pi}{N}m \qquad \text{where} \quad m=0,1,... N-1 </math>
==== case of 2 flipped spins ====


Thus, the eigenvectors are given by superpositions of states with only one flipped spin.
Eigenstate is superposition of all states which have same number of flipped spins:
The energy of these states follows from the Schrodinger equation:


:<math>
:<math>
E=E_{0}+J\left [1-cos(k)\right ]
|\Psi\rangle = \sum^N_{n_{1}<n_{2}}a(n_1,n_2)|n_1,n_2\rangle
</math>
</math>
where we have assumed <math> n_1<n_2 </math>.




==== case of 2 flipped spin state ====
First consider n1 and n2 are not neighboring pairs i.e. n1+1<n2. Then by acting the Hamiltonian to the eigenstate, we get


Eigenstate is:
<math>(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 ] </math>
 
adding <math> \langle \Psi| </math> on it we get:
 
<math>\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)  </math>
 
<math>\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 ] </math>
 
as it just like two times of the previous case, so we can easily see <math> e^{ik_1 x_1+ik_2 x_2} </math>
is a solution of the equation. As we can exchange k1 and k2, so  <math> e^{ik_2 x_1+ik_1 x_2} </math>
would also be a solution. So the general solution for this n1+1<n2 case is  


:<math>
:<math>
|\Psi\rangle = \sum^N_{n1<n2}a(n_1,n_2)|n_1,n_2\rangle
a(n_1,n_2)=A_1e^{i(k_1n_1+k_2n_2)}+A_2e^{i(k_1n_2+k_2n_1)}
</math>
</math>
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 <math> a(n_1,n_2)  </math>
as
:<math>
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)}
</math>
Then by periodic boundary condition <math> a(n_1,n_2)=a(n_2,n_1+N) </math> , we get
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 <math> \langle \Psi| </math> on it, we obtain the difference equation:
<math>\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) </math>
we can easy see that
:<math>
a(n_1,n_2)=A_1e^{i(k_1n_1+k_2n_2)}+A_2e^{i(k_1n_2+k_2n_1)}
</math>
is also the solution of equation. Now
:<math>
E=E_{0}+J\left [(1-\cos(k_1))+(1-\cos(k_2))\right ]
</math>
which is the energy


Bethe's approach for the coefficients <math>a(n_1,n_2)</math> are again plane wave but with unknown amplitudes <math>A_1</math> and <math>A_2</math>:
Bethe's approach for the coefficients <math>a(n_1,n_2)</math> are again plane wave but with unknown amplitudes <math>A_1</math> and <math>A_2</math>:
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</math>
</math>


with:
which gives:


:<math>
:<math>
Line 81: Line 143:
</math>
</math>


where the integers <math> \lambda_i = {0,1 ... N} </math> 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:
where the integers <math> \lambda_i = {0,1 ... N-1} </math> 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:


:<math>
E=E_{0}+J\left [(1-\cos(k_1))+(1-\cos(k_2))\right ]
</math>


:<math>
:<math>
E=E_{0}+J\sum_{j=1,2}\left (1-\cos(k_j)\right )
E=E_{0}+J\sum_{j=1,2}\left (1-\cos(k_j,\ lambda{j})\right )
</math>
</math>


 
==== Case of r flipped spins====
==== Case of r flipped spin====


Eigenstates:
Eigenstates:

Latest revision as of 17:51, 9 December 2010

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 ] \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:

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

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_2)=A_1e^{i(k_1n_1+k_2n_2)}+A_2e^{i(k_1n_2+k_2n_1)} }

is also the solution of equation. Now

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_1))+(1-\cos(k_2))\right ] }

which is the energy

Bethe's approach for the coefficients 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)} are again plane wave but with unknown amplitudes 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_1} and 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_2} :

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)} }

The parameters 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_1 } and 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_2 } are determined by the insertion into the Schrodinger equation. This gives the following 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_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:

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+\frac{1}{2}\theta_{12})}+e^{i(k_1n_2+k_2n_1+\frac{1}{2}\theta_{21})} }

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

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 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 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} } 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:

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_1))+(1-\cos(k_2))\right ] }
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\sum_{j=1,2}\left (1-\cos(k_j,\ lambda{j})\right ) }

Case of r flipped spins

Eigenstates:

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_{n1<n2<..<n_r}a(n_1,n_2,..,n_r)|n_1,n_2,..,n_r\rangle }

with coefficients:

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_r)=\sum_{P\in S_r}\exp\left(i\sum^r_{j=1}k_{P_j}n_j+i\sum_{i<j}\theta_{P_iP_j} \right) }

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

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{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 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_1, .. \lambda_r) } satisfying the Bethe equations. The energy of the corresponding state is 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\sum^r_{j=1}\left (1-\cos k_j \right) }
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