Solve the one-magnon sector

Why does a single flipped spin behave like a wave? In the periodic XXX chain, the Hamiltonian moves that down spin to neighboring sites. You will diagonalize this discrete hopping equation and obtain both its dispersion $E(k)=J(1-\cos k)$ and its allowed wave numbers $k=2\pi m/N$. In this one-magnon sector, a finite Fourier basis supplies all $N$ states.

Required background. Use the bond exchange rule and fixed-magnetization basis from Construct a small spin-chain Hamiltonian. The entry repair reviews the complex exponentials used below.

Helpful background. The spin-chain convention reference explains why a translation eigenvalue depends on the direction chosen for the translation operator.

A down spin hopping on a ring

Use $N\geq3$, $\hbar=1$, lattice spacing one, and the model Hamiltonian

$$H=J\sum_{n=1}^N\left(\frac14-\mathbf S_n\cdot\mathbf S_{n+1}\right), \qquad J\gt0,\qquad \mathbf S_{N+1}=\mathbf S_1.$$

The all-up state $|F\rangle$ has energy zero. The orthonormal basis of the $M=1$ sector is

$$|n\rangle=S_n^-|F\rangle,\qquad n=1,\ldots,N.$$

Only two bonds touch the down spin. Each contributes $J/2$ on the diagonal and $-J/2$ for moving it one site:

$$H|n\rangle =J|n\rangle-\frac J2\bigl(|n-1\rangle+|n+1\rangle\bigr),$$

with site labels read modulo $N$. For $|\psi\rangle=\sum_n\psi(n)|n\rangle$, the coefficient of $|n\rangle$ in $H|\psi\rangle=E|\psi\rangle$ gives

$$E\psi(n)=J\psi(n)-\frac J2\bigl(\psi(n-1)+\psi(n+1)\bigr).$$

This is a discrete second-difference operator. It conserves the number of down spins; a magnon is a delocalized excitation within this sector, not an extra site or an extra spin added to the chain.

Entry check and repair

For $\psi(n)=e^{ikn}$, find $\psi(n+1)/\psi(n)$ and $\psi(n-1)/\psi(n)$. When is $\psi(n+N)=\psi(n)$?

Repair. Exponential addition gives the ratios $e^{ik}$ and $e^{-ik}$. Euler’s identity implies $e^{ik}+e^{-ik}=2\cos k$. Since the amplitude is never zero, periodicity requires $e^{ikN}=1$, or $kN=2\pi m$ for an integer $m$. Wave numbers differing by $2\pi$ give identical amplitudes at integer sites.

Derive the dispersion and quantization

Insert the plane wave in the lattice equation and divide by $e^{ikn}$:

$$E=J-\frac J2(e^{-ik}+e^{ik})=J(1-\cos k).$$

The bulk equation determines energy as a function of $k$. The boundary condition then selects

$$k_m=\frac{2\pi m}{N},\qquad m=0,1,\ldots,N-1.$$

These are different steps. A plane wave with arbitrary real $k$ satisfies the bulk difference equation, but it generally fails at the closing bond of a finite ring.

The normalized eigenstates are

$$|k_m\rangle=\frac1{\sqrt N}\sum_{n=1}^N e^{ik_mn}|n\rangle.$$

For $m\ne m'$ modulo $N$, the finite geometric sum gives

$$\langle k_{m'}|k_m\rangle =\frac1N\sum_{n=1}^N e^{2\pi i(m-m')n/N}=0.$$

Each vector has norm one, and there are $N$ orthogonal vectors in an $N$-dimensional sector. This proves completeness here. It does not prove completeness of a two-magnon or many-magnon Bethe construction.

The lattice equation and dispersion agree with Karbach and Müller 1997, p. 2, equations (3)–(8), arXiv v1 PDF, after replacing their $E-E_0$ by our vacuum-shifted $E$.

Work through six sites

For $N=6$, the spectrum is

$m$	$k_m$	$E/J$
$0$	$0$	$0$
$1,5$	$\pi/3,\ 5\pi/3$	$1/2$
$2,4$	$2\pi/3,\ 4\pi/3$	$3/2$
$3$	$\pi$	$2$

At $k=\pi$, neighboring amplitudes have opposite signs. Consequently each off-diagonal term adds to the positive diagonal contribution:

$$J(-1)^n-\frac J2\bigl((-1)^{n-1}+(-1)^{n+1}\bigr) =2J(-1)^n.$$

At $k=0$, all amplitudes are equal and the terms cancel. This zero mode is

$$|0\rangle_{\rm wave}=\frac1{\sqrt N}S_{\rm tot}^-|F\rangle, \qquad S_{\rm tot}^-=\sum_nS_n^-.$$

The subscript distinguishes this wave state from a site label. It belongs to the same total-spin multiplet as the vacuum. Zero excitation energy does not make it the zero vector: its norm is one, and it lies in a different magnetization sector from $|F\rangle$.

Fix the translation direction

Define the active one-site translation by $T|n\rangle=|n+1\rangle$. Relabeling the sum gives

$$T|k\rangle =\frac1{\sqrt N}\sum_n e^{ik(n-1)}|n\rangle =e^{-ik}|k\rangle.$$

Our coefficient convention is $e^{+ikn}$, so this active translation has eigenvalue $e^{-ik}$. The inverse translation has eigenvalue $e^{+ik}$. An energy calculation cannot detect a mistaken choice here because $E(k)=E(-k)$; comparing translation eigenvalues can.

The wave numbers $k$ and $-k$ are generally distinct states with equal energies. They coincide modulo $2\pi$ only for $k=0$ and, when allowed, $k=\pi$.

Exercises

Guided practice: a five-site eigenstate

On $N=5$ sites, take $k=4\pi/5$. Write the normalized amplitudes, find the exact energy, and check the equation at site $1$, including the bond to site $5$.

Hint

Use $z=e^{4\pi i/5}$, so $z^5=1$. In the site-$1$ equation, the amplitude at site $5$ equals the continued amplitude at site $0$.

Solution

The amplitudes are $\psi(n)=z^n/\sqrt5$. Their norm is one because all five squared magnitudes equal $1/5$. Since $\cos(4\pi/5)=-(1+\sqrt5)/4$,

$$E=\frac{5+\sqrt5}{4}\,J.$$

At site $1$,

$$J\psi(1)-\frac J2\bigl(\psi(5)+\psi(2)\bigr) =\frac J{\sqrt5}\left(z-\frac{1+z^2}{2}\right) =J(1-\cos k)\frac z{\sqrt5}.$$

The equality $\psi(5)=1/\sqrt5$ is precisely where periodic quantization enters the endpoint check.

Independent practice: a degenerate pair

For $N=6$, combine the $k=\pi/3$ and $-k$ states into normalized real cosine and sine states. Show that both have energy $J/2$. Are they individually eigenstates of the active translation $T$?

Hint

Take the sum and difference of the two orthonormal Fourier states. Translate amplitudes by replacing $n$ with $n-1$.

Solution

Define

$$\begin{aligned} |c\rangle&=\frac{|k\rangle+|-k\rangle}{\sqrt2} =\sqrt{\frac26}\sum_n\cos(kn)|n\rangle,\\ |s\rangle&=\frac{|k\rangle-|-k\rangle}{i\sqrt2} =\sqrt{\frac26}\sum_n\sin(kn)|n\rangle. \end{aligned}$$

The two states are normalized and orthogonal because this is a unitary change of basis within the Fourier pair. Both retain the common energy $J/2$. Translation mixes them:

$$\begin{aligned} T|c\rangle&=\cos k\,|c\rangle+\sin k\,|s\rangle,\\ T|s\rangle&=-\sin k\,|c\rangle+\cos k\,|s\rangle. \end{aligned}$$

Here $\sin k\ne0$, so neither real state is a translation eigenstate. An arbitrary eigensolver can return this real basis of a degenerate energy eigenspace. To assign a definite wave number, diagonalize translation within that eigenspace.

Transfer: add a longitudinal field

Change the Hamiltonian to

$$H_h=H+h\sum_n\left(\frac12-S_n^z\right),$$

where $h$ is a real parameter with units of energy. Determine the one-magnon eigenvectors, energies, and allowed wave numbers. For $h\gt0$, does the $k=0$ state still have zero energy?

Hint

The added operator counts down spins. Evaluate it throughout the fixed-$M$ sector before attempting a new Fourier calculation.

Solution

The added term is $hM$. In the $M=1$ sector it is simply $h$ times the identity, so the Fourier eigenvectors and the condition $k=2\pi m/N$ are unchanged:

$$E_h(k)=J(1-\cos k)+h.$$

The vacuum still has energy zero, while the $k=0$ one-magnon state has energy $h$. For positive $h$, its degeneracy with the vacuum is lifted. More generally, every fixed-$M$ block is shifted by $hM$; no claim about which sector is the ground state for arbitrary negative $h$ follows from the one-magnon calculation alone.

From one wave to two interacting waves

You have an exact and complete finite Fourier solution in the one-magnon sector. The quadratic small-$k$ behavior $E(k)=Jk^2/2+O(k^4)$ follows by Taylor expansion; on a finite ring, $k$ still takes discrete values. With two down spins, their adjacency changes the local equation. Continue to two-magnon scattering to find the extra matching condition.

References

• Karbach, Michael, and Gerhard Müller. “Introduction to the Bethe Ansatz I.” Computers in Physics 11, 36–43 (1997). DOI. Author version arXiv:cond-mat/9809162v1 (1998); Open PDF.