A gentle intro to the Bethe ansatz

Published by Jeff on October 31, 2021October 31, 2021
Jeff

A gentle intro to the Bethe ansatz

Many modern lattice problems (six-vertex, eight-vertex, Ising, Heisenberg-XYZ) can be solved by obtaining eigenvalues of the transfer matrix or Hamiltonian by the Bethe ansatz. This is a very general method for solving certain classes of very large eigenvalue problems.

We use as an example the quantum -XX model: a spin chain with nearest neighbor dipole-dipole interactions

    \[H=-J\Big({\hbar\over 2}\Big)^2\sum_{i=1}^{N, \, PBC} \Big(\sigma_i^- \sigma_{i+1}^+ + \sigma_i^+ \sigma_{i+1}^-\Big), \qquad \{\sigma_i^+, \sigma_i^-\}=1\]

    \[ (\sigma_i^-)^2=(\sigma_i^+)^2=0.\]

and [\sigma_i^{\pm}, \sigma_j^{\pm}]=0 if i\ne j. Bethe developed the technique to solve the XXX model, add a term \sigma_j^z\sigma_{j+1}^z onto the Hamiltonian above. His method has become a mainstay of many-body and statistical physics.

Let me remind you that what we mean by \sigma_j^z is

    \[\sigma_j^z=1\otimes1\otimes 1\otimes\cdots\otimes 1\otimes \sigma^z\otimes 1\otimes\cdots\otimes 1\]

with the Pauli spin matrix in the j^{th} slot. Denote eigenstates of \sigma_z by

    \[|\tfrac{1}{2},\tfrac{1}{2}\rangle=\uparrow, \quad |\tfrac{1}{2},-\tfrac{1}{2}\rangle=\downarrow\]

    \[\sigma^z \uparrow=\uparrow, \quad \sigma^z \downarrow=-\downarrow\]

    \[\sigma^+ \uparrow=0, \quad \sigma^+\downarrow=\uparrow,\quad \sigma^- \downarrow=0, \quad \sigma^-\uparrow=\downarrow\]

First we work out a few useful identities. Acting on two-particle spin tensors

(1)   \begin{eqnarray*} \Big(\sigma_i^- \sigma_{i+1}^+ + \sigma_i^+ \sigma_{i+1}^-\Big) \uparrow\otimes \uparrow & = & 0, \nonumber\\ \Big(\sigma_i^- \sigma_{i+1}^+ + \sigma_i^+ \sigma_{i+1}^-\Big) \downarrow\otimes \downarrow & = & 0, \nonumber\\ \Big(\sigma_i^- \sigma_{i+1}^+ + \sigma_i^+ \sigma_{i+1}^-\Big) \uparrow\otimes \downarrow & = & \downarrow\otimes \uparrow, \nonumber\\ \Big(\sigma_i^- \sigma_{i+1}^+ + \sigma_i^+ \sigma_{i+1}^-\Big)\downarrow\otimes \uparrow & = & \uparrow\otimes \downarrow.\nonumber\end{eqnarray*}

The ansatz (the goal of which is to produce all eigenvectors and eigenvalues of H) has several stages.
\bullet 1. Establish a “vacuum” and “first excited” state spectrum.
Beginning with a lowest \mathcal{\sigma}_z eigenvalue state (we refer to this as our Bethe vacuum) \Phi_0=\downarrow\otimes\downarrow\otimes \cdots \downarrow\otimes\downarrow, create a “singly excited state” by flipping one spin, for example the n^{th}

    \[\Phi_{n}=\sigma_n^+ \, \Phi_0\]

and prove that

    \[ H \Phi_n=-J\Big(\Phi_{n-1}+\Phi_{n+1}\Big)\]

Let’s call s_i={\hbar\over 2}\sigma_i and rewrite in terms of actual spin operators. Every term in H kills \Phi_0 so H\Phi_0=0, but

    \[\Big(s_{n-1}^- s_{n}^+ + s_{n-1}^+ s_{n}^-\Big)\Phi_n=0+\Big(\tfrac{\hbar}{2}\Big)^2\Phi_{n-1}\]

    \[\Big(s_{n}^- s_{n+1}^+ + s_{n}^+ s_{n+1}^-\Big)\Phi_n=\Big(\tfrac{\hbar}{2}\Big)^2\Phi_{n+1}+0\]

    \[\implies\quad H \Phi_n=-J\tfrac{1}{4} \hbar^2\Big(\Phi_{n-1}+\Phi_{n+1}\Big)\]

Apply H to

    \[ \Psi=\sum_n a_n \Phi_n,\quad H\Psi=\sum_n a_n E_n\Phi_n=-J\tfrac{1}{4} \hbar^2\sum_{n}a_n\Big(\Phi_{n-1}+\Phi_{n+1}\Big)\]

take inner products \langle \Phi_m|\Phi_n\rangle=\delta_{m,n} with both sides

    \[\sum_n a_n E_n\delta_{m,n}=-J\tfrac{1}{4} \hbar^2\sum_{n}a_n\Big(\delta_{m,n-1}+\delta_{m,n+1}\Big)\]

    \[E a_m=-J\tfrac{1}{4} \hbar^2\Big(a_{m-1}+a_{m+1}\Big).\]

We have seen this recursion relation in 711 in our studies of normal modes: the ansatz a_n = c_1 \xi^n gives E=-J\tfrac{1}{4} \hbar^2(\xi^{-1}+\xi)
for which periodic boundary conditions a_n=a_{n+N} forces \xi^N=1, so \xi(a)=e^{2\pi i {a\over N}} with wavenumbers labeled by a=0,1,2,\cdots, N-1. This is why I present the periodic simple atomic chain in 711, to prepare you for learning the Bethe ansatz first step.
Our first excited state spectrum is E_a=-J\tfrac{1}{2} \hbar^2\cos({2\pi a\over N}), an “optical branch” dispersion relation.

\bullet 2. Progress to the doubly excited states. This is a critical step in which the “ansatz equation” is obtained. Formulate doubly excited states (two flipped spins)

    \[ \Phi_{n,m}=\sigma_n^+ \, \sigma_m^+ \, \Phi_0\]

and propose Hamiltonian eigenfunctions

    \[\Psi=\sum_{n,m} a_{n,m} \, \Phi_{n,m}\]

and prove that the secular equations are

(2)   \begin{eqnarray*}E a_{n,m}&=& -J\tfrac{1}{4} \hbar^2\Big(a_{n-1,m}+a_{n+1, m}+a_{n,m-1}+a_{n,m+1}\Big),\quad m>n+1\nonumber\\ E a_{n,n+1}&=& -J\tfrac{1}{4} \hbar^2\Big(a_{n-1,n+1}+a_{n+1, n+1}+a_{n,n}+a_{n,n+2}\Big)\nonumber\end{eqnarray*}

in which the coefficients a_{n,n}, a_{n+1,n+1} are for absurd/forbidden terms \Phi_{n,n}, \Phi_{n+1,n+1} which we retain since they are a critical part of the ansatz. These are called “unwanted” terms. The condition that eliminates the unwanted terms is where the magic lies.

Here it goes:

    \[\Psi=\sum_{n,m} a_{n,m} \, \Phi_{n,m} \quad\mbox{with}\quad \Phi_{n,m}=\sigma_n^+ \, \sigma_m^+ \, \Phi_0\]

first let m>n+1 so the two flipped spins are not nearest neighbors, there will be four terms in H that do not annihilate \Psi; P_{n-1,n}\Phi_{n,m}=\Phi_{n-1,n}, P_{n,n+1}\Phi_{n,m}=\Phi_{n+1,m} ,P_{m-1,m}\Phi_{n,m-1}=\Phi_{n,m-1}, P_{m,m+1}\Phi_{n,m}=\Phi_{n,m+1}, so application oh H to \Psi followed by inner products \langle \Phi_{j,k}|\Phi_{n,m}\rangle=\delta_{j,n}\delta_{k,m} gives the first of

(3)   \begin{eqnarray*}E a_{n,m}&=& -J\tfrac{1}{4} \hbar^2\Big(a_{n-1,m}+a_{n+1, m}+a_{n,m-1}+a_{n,m+1}\Big),\quad m>n+1\nonumber\\ E a_{n,n+1}&=& -J\tfrac{1}{4} \hbar^2\Big(a_{n-1,n+1}+a_{n+1, n+1}+a_{n,n}+a_{n,n+2}\Big)\nonumber\end{eqnarray*}

We can simply set m=n+1 in the first equation to get the second, but the second is not the correct Hamiltonian action on a_{n,n+1}, that would be E a_{n,n+1}= -J\tfrac{1}{4} \hbar^2\Big(a_{n-1,n+1}+a_{n,n+2}\Big) since only P_{n-1,n} and P_{n+1,n+2} do not annihilate \Phi_{n,n+1}.

The secret of the Bethe ansatz is to apply P_{i,i+1} operators to every \Phi_{n,m} abstractly, without regard to physical meaning of \Phi_{n,n}. Think of \Phi_{n,m} as an abstract tensor and P_{i,i+1} as a label permutation and nothing more.

Then we demand that the true a_{n,n+1} equation agree with the naive version gotten from the general a_{n,m} m>n+1 which results in an identity (our ansatz equation)

    \[ a_{n,n}+a_{n+1,n+1}=0\quad \mbox{on the "unwanted" terms $a_{j,j}$}\]

(this identity is usually much more complex than in this example, which is probably the simplest in its category).

\bullet 3. Don’t just take a_{n,n}=a_{n+1,n+1}=0. Instead propose a form

    \[a_{n,m}=c_1 \xi_1^n \xi_2^m+c_2 \xi_2^n \xi_1^m\]

( ie a sum over permutations of the products of the wavenumbers). Take

    \[a_{n,m}=c_1 \xi_1^n \xi_2^m+c_2 \xi_2^n \xi_1^m=c_e \xi_1^n \xi_2^m+c_{\wp(e)} \xi_{\wp(1)}^n \xi_{\wp(2)}^m\]

    \[\mbox{and apply}\quad a_{n,n}+a_{n+1,n+1}=0\]

where e is the identity permutation and here \wp=(12) the transposition. This leads to

    \[ (c_1+c_2)(1+\xi_1\xi_2)=0, \qquad \Big(c_1-c_2 \xi_1^N\Big) \xi_1^n \xi_2^m =\Big(c_1\xi_2^N-c_2\Big) \xi_2^n \xi_1^m\]

    \[ (\xi_1 \xi_2)^N=1, \quad 1+\xi_1\xi_2=0\]

the latter two are incompatible, forcing c_1+c_2=0 to be the solution to the first equation (again, this is usually quite complex), and the solutions to the second must be (\xi_1^n\xi_2^m and \xi_1^m\xi_2^n are independent exponentials)

    \[ c_1(1+\xi_1^N)=0, \quad \xi_1^N=-1, \qquad c_2(\xi_2^N+1)=0, \quad \xi_2^N=-1\]

    \[\implies\quad \xi_1(a_1)=e^{{2\pi \over N} i(a_1+\tfrac{1}{2})}, \quad \xi_2(a_2)=e^{{2\pi \over N} i(a_2+\tfrac{1}{2})}\]

independent of one another. Usually there is a complicated relation between the wavenumbers a_1,a_2. Note that this tells us that coefficients of terms in the proposed wavefunction have opposite signs if the particle labels in the two terms are switched (a wavefunction antisymmetric under particle label transpositions).

\bullet 4. The final stage of the Bethe ansatz is to assume that higher-order excitations Bethe ansatz wave-functions are

    \[ \Psi=\sum_{n_1, \cdots, n_r} a_{n_1, n_2, \cdots, n_r} \, \Phi_{n_1, n_2, \cdots, n_r}\]

    \[ a_{n_1, n_2, \cdots, n_r}=\sum_\wp c_\wp \, \xi_{\wp(1)}^{n_1} \xi_{\wp(2)}^{n_2}\cdots \xi_{\wp(r)}^{n_r}\]

and the ansatz conditions lead to

    \[ c_{(ij)\wp}=-c_\wp, \qquad \xi_k(a_k)=e^{{2\pi \over N} i(a_k+\tfrac{1}{2})}\]

in which \wp is a permutation of the r labels, and (ij)\wp is the permutation gotten by switching labels i and j after permuting by \wp. Can you see where this comes from? This is the crux of the technique, and you might see it more readily if you explicitly write out the details for a triple excitation, but this is rarely done.

This wavefunction describes a system of free fermions ( because the wave-function is antisymmetric under the interchange of any pair of up-spin location labels). If \wp decomposes into n_\wp successive relative transpositions, c_\wp=sgn(\wp)=(-1)^{n_\wp}
and the Bethe wave-function is the Slater determinant a=\det M, M_{i,j}=\xi_i^{n_j}

    \[a_{n_1, n_2, \cdots, n_r}=\sum_\wp sgn(\wp) \, \xi_{\wp(1)}^{n_1} \xi_{\wp(2)}^{n_2}\cdots \xi_{\wp(r)}^{n_r}\]

In the spirit of the ansatz we would show that for 2-fold excitations

    \[E=-J\tfrac{1}{4} \hbar^2\sum_{k=1}^2 \Big(\xi_{k}+\xi_{k}^{-1}\Big)\]

and assume that the result carries over to r-fold excitations. Note that

(4)   \begin{eqnarray*}a_{n-1,m}+a_{n,m-1}&=&c_1\Big(\xi_1^{n-1}\xi_2^m+\xi_1^{m}\xi_2^{n-1}\Big)+c_1\Big(\xi_1^{n}\xi_2^{m-1}+\xi_1^{m-1}\xi_2^{n}\Big)\nonumber\\ &=&c_1(\xi_1^{-1}+\xi_2^{-1})\Big(\xi_1^{n}\xi_2^m+\xi_1^{m}\xi_2^{n}\Big)=(\xi_1^{-1}+\xi_2^{-1})a_{n,m}\nonumber\\ a_{n+1,m}+a_{n,m+1}&=&c_1\Big(\xi_1^{n+1}\xi_2^m+\xi_1^{m}\xi_2^{n+1}\Big)+c_1\Big(\xi_1^{n}\xi_2^{m+1}+\xi_1^{m+1}\xi_2^{n}\Big)\nonumber\\ &=&c_1(\xi_1^{1}+\xi_2^{1})\Big(\xi_1^{n}\xi_2^m+\xi_1^{m}\xi_2^{n}\Big)=(\xi_1^{1}+\xi_2^{1})a_{n,m}\nonumber\end{eqnarray*}

Adding these and substituting into the secular equation for Ea_{n,m} lets you cancel a_{n,m} to get

    \[E=-J\Big(\tfrac{\hbar}{2}\Big)^2\Big(\xi_1^{-1}+\xi_2^{-1}+\xi_1^{1}+\xi_2^{1}\Big)\]

which is a sum of two single excitations but with different wavenumbers. We suspect then that

    \[E=-J\tfrac{1}{4} \hbar^2\sum_{k=1}^r \Big(\xi_{k}+\xi_{k}^{-1}\Big)\]

An English translation of Bethe’s original Zeits f. Physik 71, 205 (1931) paper appears in Daniel C. Mattis {\it The Many-Body Problem}, World Scientific, (1993), p. 689.

Related Posts

Protected: Physics 715: Transfer matrix-II. Classical in D+1 dimensions dual to quantum in D.
Pauli exclusion principle
Physics 715: The hyperscaling hypothesis.
Home 2.0
error: Content is protected !!