Solve two-magnon scattering

Why can two spin waves not simply pass through the chain with arbitrary amplitudes? When the two down spins become neighbors, the Hamiltonian sees two domain walls instead of four. You will match the separated and adjacent equations and derive the relative amplitude of two plane waves. This is a local scattering calculation; the finite ring will impose an additional condition in the next lesson.

Required background. Derive the hopping equation and dispersion in Solve the one-magnon sector. Use the bond exchange rule from Construct a small spin-chain Hamiltonian.

Helpful background. The Library derivation collects the local and periodic arguments. The convention reference fixes which amplitude ratio is called $S_{12}$.

Two down spins and their contact equation

Keep the ferromagnetic spin-$1/2$ XXX Hamiltonian, with $J\gt0$, $\hbar=1$, lattice spacing one, and vacuum energy zero. In the two-down-spin sector,

$$|\psi\rangle=\sum_{1\leq x\lt y\leq N}\psi(x,y)|x,y\rangle, \qquad |x,y\rangle=S_x^-S_y^-|F\rangle.$$

The ordered labels count each configuration once. There is no physical basis vector with two down spins at the same site, because $(S_x^-)^2=0$ for spin $1/2$.

First inspect sites away from the closing bond, with enough sites to distinguish separated and adjacent pairs. A separated pair has four unlike neighboring bonds, giving

$$\begin{aligned} E\psi(x,y)=2J\psi(x,y)-\frac J2\bigl[& \psi(x-1,y)+\psi(x+1,y)\\ &+\psi(x,y-1)+\psi(x,y+1)\bigr], \qquad 1\lt y-x\lt N-1. \end{aligned}$$

For neighbors, the common bond has two equal down spins and contributes zero. Only the outer two bonds remain:

$$E\psi(x,x+1) =J\psi(x,x+1) -\frac J2\bigl[\psi(x-1,x+1)+\psi(x,x+2)\bigr].$$

These are local equations on the unwrapped site coordinates. On the ring, $(1,N)$ is also an adjacent pair; it must eventually satisfy the same contact rule after the periodic identification. For $N=3$ every pair is adjacent, so there is no separated-pair equation to impose. The examples below use chains long enough that the displayed pair is away from that seam.

Entry check and repair

On six sites, compare $H|2,4\rangle$ and $H|2,3\rangle$. How many unequal bonds surround each configuration, and what are the diagonal coefficients?

Repair. The configuration $|2,4\rangle$ has unequal bonds $(1,2)$, $(2,3)$, $(3,4)$, and $(4,5)$, so its diagonal coefficient is $2J$. In $|2,3\rangle$, the bond $(2,3)$ joins equal spins; only $(1,2)$ and $(3,4)$ are unequal. Its diagonal coefficient is $J$. Thus

$$H|2,3\rangle =J|2,3\rangle-\frac J2\bigl(|1,3\rangle+|2,4\rangle\bigr).$$

The interaction appears through this change in the local equation. Assigning an independent one-magnon diagonal energy to each down spin at contact would miss it.

Match two plane-wave assignments

Set $z_j=e^{ik_j}\ne0$ and try

$$\psi(x,y) =A_{12}z_1^xz_2^y+A_{21}z_2^xz_1^y,\qquad x\lt y.$$

The subscripts record which momentum is assigned to the left and right coordinates. They do not label two distinguishable spin species. Substitution in the separated equation gives

$$E=J\left[2-\frac12 \left(z_1+z_1^{-1}+z_2+z_2^{-1}\right)\right] =J(2-\cos k_1-\cos k_2).$$

For this step, the ratio of the two amplitudes is undetermined.

To impose the contact equation, formally continue the two-exponential expression to coincident coordinates. Subtracting the physical contact equation from the continued separated equation gives

$$\psi(x,x)+\psi(x+1,x+1)=2\psi(x,x+1).$$

The coincident values are an algebraic device, not extra Hilbert-space amplitudes. Inserting the ansatz and cancelling $(z_1z_2)^x$ yields

$$\begin{aligned} 0={}&A_{12}(1+z_1z_2-2z_2)\\ &+A_{21}(1+z_1z_2-2z_1). \end{aligned}$$

When $A_{12}\ne0$ and the denominator below is nonzero, define the exchange amplitude

$$S_{12}\equiv\frac{A_{21}}{A_{12}} =-\frac{1+z_1z_2-2z_2}{1+z_1z_2-2z_1}.$$

The contact equation, rather than a guess about bosonic symmetry, determines this ratio. Our $S_{12}$ is the inverse of Karbach and Müller’s $A/A'=e^{i\theta}$; compare Karbach and Müller 1997, pp. 2–3, equations (9)–(16), arXiv v1 PDF.

Check the phase and one local wave

For real momenta away from singular cases, write $Q=1+z_1z_2-2z_1$. Since $\overline z_j=z_j^{-1}$,

$$1+z_1z_2-2z_2=z_1z_2\,\overline Q.$$

It follows that $|S_{12}|=1$. Exchanging the momenta also gives $S_{21}S_{12}=1$ where both ratios exist. These checks detect many swapped labels and sign errors. Complex momenta need not give a unit-modulus individual exchange amplitude.

For a worked example, choose $k_1=\pi/2$, $k_2=-\pi/2$, and $A_{12}=1$. Then

$$z_1=i,\qquad z_2=-i,\qquad S_{12}=-i,\qquad E=2J.$$

The amplitudes depend on the separation $d=y-x$:

$$\psi(x,y)=e^{-i\pi d/2}-i\,e^{i\pi d/2}.$$

In particular, $\psi(2,3)=1-i$, while $\psi(1,3)=\psi(2,4)=-1+i$. The contact equation becomes

$$J(1-i)-\frac J2[(-1+i)+(-1+i)] =2J(1-i)=E\psi(2,3).$$

This verifies a local interacting equation, not only the free dispersion. It does not yet prove that these momenta are allowed for any chosen ring size.

Rapidities and exceptional choices

For $k\not\equiv0\pmod{2\pi}$ define

$$\lambda=\frac12\cot\frac k2, \qquad z=\frac{\lambda+i/2}{\lambda-i/2}.$$

Algebraic substitution gives

$$S_{12} =\frac{\lambda_1-\lambda_2-i} {\lambda_1-\lambda_2+i}.$$

This form is useful, but it does not remove exceptional cases. A zero-momentum wave has infinite rapidity. A vanishing denominator in the $z$ formula must be examined through the original linear contact relation or a controlled limit. Equal momenta can make the whole wavefunction vanish, even if a formal energy can be written down. Keep these checks separate from solving an algebraic equation.

Exercises

Guided practice: opposite momenta

For $k_1=k$, $k_2=-k$, with real $k\not\equiv0,\pi\pmod{2\pi}$, simplify $S_{12}$ and the energy. Explain the excluded values rather than silently cancelling a zero.

Hint

Use $z_1z_2=1$ and $1-e^{-ik}=-e^{-ik}(1-e^{ik})$.

Solution

For $k\not\equiv0$,

$$S_{12}=-\frac{2-2e^{-ik}}{2-2e^{ik}}=e^{-ik}, \qquad E=2J(1-\cos k).$$

At $k=0$, the displayed ratio is $0/0$; the original contact relation, not this cancelled expression, must decide an admissible state. At $k=\pi$, the two $z$ values coincide at $-1$ and $S_{12}=-1$. The two exponential terms then cancel identically. The formula for energy alone does not rescue that zero vector.

Independent practice: a false eigenstate

Take $z_1=z_2=z\ne0,1$ and $A_{12}=1$. Show that the contact condition gives $S_{12}=-1$ and that the ansatz is identically zero. Why is an eigenvalue residual $H\psi-E\psi=0$ insufficient without a norm check?

Hint

Both coefficients in the linear contact relation equal $(z-1)^2$.

Solution

Since $(z-1)^2\ne0$, the contact relation becomes $1+A_{21}=0$. Consequently

$$\psi(x,y)=z^{x+y}-z^{x+y}=0.$$

The zero vector satisfies $H\psi=E\psi$ for every $E$, but is not an eigenstate. A meaningful numerical check first establishes a nonzero norm and normalizes the candidate, then measures an eigenvector residual. A limiting construction involving coincident parameters would require its own derivation; this substitution alone supplies no such state.

Transfer: change the longitudinal interaction

Replace the Hamiltonian by the explicitly defined anisotropic operator

$$H_\Delta=J\sum_n\left[ \Delta\left(\frac14-S_n^zS_{n+1}^z\right) -\frac12\left(S_n^+S_{n+1}^-+S_n^-S_{n+1}^+\right) \right],$$

with real dimensionless $\Delta$. Keep the same two-exponential ansatz. Find its separated-pair energy and local exchange amplitude. Recover the XXX result at $\Delta=1$.

Hint

An unequal bond now contributes $J\Delta/2$ diagonally, but the exchange coefficient is still $-J/2$. Repeat the subtraction of the separated and contact equations.

Solution

The separated diagonal is $2J\Delta$ and the adjacent diagonal is $J\Delta$. Therefore

$$E_\Delta=J(2\Delta-\cos k_1-\cos k_2),$$

and matching requires

$$\psi(x,x)+\psi(x+1,x+1)=2\Delta\psi(x,x+1).$$

The resulting regular amplitude ratio is

$$S_{12}^{(\Delta)} =-\frac{1+z_1z_2-2\Delta z_2} {1+z_1z_2-2\Delta z_1}.$$

Setting $\Delta=1$ reproduces every XXX formula. This calculation establishes a two-body local matching rule for the stated anisotropic Hamiltonian. It does not supply finite-ring quantization, a many-body commuting family, or spectral completeness.

Add the ring boundary

You have derived the scattering amplitude from a physical contact equation and tested its convention. To turn a nonzero local wave into a finite-ring eigenstate, continue to Quantize magnon momenta on a ring. There the closing bond will constrain the momenta together.

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.