Spin-chain Hamiltonian conventions

A spin-chain formula is meaningful only after its operator normalization, exchange sign, energy zero and boundary bonds have been fixed. This reference gives explicit conversions for the finite spin-½ XXX chain used in the quantum learning sequence. It also identifies which changes preserve eigenvectors and which change the eigenvalue problem.

Spin, Pauli and permutation operators

Set $\hbar=1$. In the ordered one-site basis $\left(|\uparrow\rangle,|\downarrow\rangle\right)$,

$$S^x=\frac12\begin{pmatrix}0&1\\1&0\end{pmatrix}, \quad S^y=\frac12\begin{pmatrix}0&-i\\i&0\end{pmatrix}, \quad S^z=\frac12\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$

Thus $S^\alpha=\sigma^\alpha/2$, $[S^x,S^y]=iS^z$, and $(S^z)^2=I/4$. On a chain, $S_n^\alpha$ acts at site $n$ and as the identity at every other site. Operators at different sites commute.

For two spin-½ sites, let $P_{12}|a,b\rangle=|b,a\rangle$. Direct multiplication gives

$$P_{12}=2\mathbf S_1\cdot\mathbf S_2+\frac12 I.$$

The normalized singlet has $P_{12}=-1$ and $\mathbf S_1\cdot\mathbf S_2=-3/4$. Each triplet has $P_{12}=1$ and $\mathbf S_1\cdot\mathbf S_2=1/4$. These eigenvalues are quick checks on every conversion below.

Exchange signs and energy shifts

For $N\geq3$ with $N$ periodic nearest-neighbor bonds, our reference Hamiltonian is

$$\begin{aligned} H_{\rm F} &=J\sum_{n=1}^{N}\left(\frac14-\mathbf S_n\cdot\mathbf S_{n+1}\right)\\ &=\frac J4\sum_{n=1}^{N} \left(I-\boldsymbol{\sigma}_n\cdot\boldsymbol{\sigma}_{n+1}\right) =\frac J2\sum_{n=1}^{N}(I-P_{n,n+1}),\qquad J\gt0. \end{aligned}$$

Each bond is $J$ times the singlet projector. Hence $H_{\rm F}$ is positive semidefinite and every fully symmetric spin state has zero energy. This establishes the sign as ferromagnetic without relying on a naming convention.

Written Hamiltonian	Relation to $H_{\rm F}$	What changes?
$H_-=-J\sum_n\mathbf S_n\cdot\mathbf S_{n+1}$	$H_{\rm F}=H_-+JN I/4$	Same eigenvectors; shift every energy by $JN/4$.
$H_+=J\sum_n\mathbf S_n\cdot\mathbf S_{n+1}$	$H_{\rm F}=JN I/4-H_+$	Same eigenvectors; reverse the ordering of energies.
$K\sum_n\boldsymbol{\sigma}_n\cdot\boldsymbol{\sigma}_{n+1}$	$4K\sum_n\mathbf S_n\cdot\mathbf S_{n+1}$	The spin-exchange coefficient is $4K$.

The tutorial by Karbach and Müller 1997, arXiv v1 PDF, pp. 1–2, eqs. (1) and (5) uses $H_-$ and vacuum energy $E_0=-JN/4$. Its excitation energy $E-E_0$ equals our energy. The publication year is 1997; the arXiv version was submitted in 1998. PDF page labels are separate from journal page numbers.

An overall nonzero scale $a$ and shift $bI$ in $aH+bI$ preserve eigenvectors and transform eigenvalues as $E\mapsto aE+b$. A negative $a$ reverses ground- and highest-energy order. It also changes the sign of time evolution at a fixed time coordinate; an energy convention is not a claim that the physical dynamics is unchanged.

Uniform longitudinal field

Let $S^z_{\rm tot}=\sum_nS_n^z$ and use the field convention

$$H_h=H_{\rm F}-hS^z_{\rm tot}.$$

In a sector with $M$ down spins, $S^z_{\rm tot}=N/2-M$. Since $[H_{\rm F},S^z_{\rm tot}]=0$, the same fixed-$M$ eigenvector has

$$E_h=E-h(N/2-M),\qquad E_h-E_{h,\rm vacuum}=E+hM.$$

Here $h$ has units of energy. For a one-magnon state, the energy relative to the all-up vacuum is $J(1-\cos k)+h$. The vacuum remains an eigenstate for either sign of $h$, but need not remain the lowest-energy state. A spatially varying field is not a constant within a fixed-$M$ sector and cannot be handled by this shift.

Boundary bonds and translation

The periodic sum contains the closing bond $(N,1)$. An open chain instead uses $n=1,\ldots,N-1$, so the constant in the conversion from $-J\sum\mathbf S_n\cdot\mathbf S_{n+1}$ is $J(N-1)/4$. Removing the closing bond changes the operator and generally changes its eigenvectors; it is not an additive constant.

For $N=2$, the literal periodic sum contains $(1,2)$ and $(2,1)$, which are the same pair counted twice. A single two-site bond is a different normalization. The learning sequence restricts ring calculations to $N\geq3$ to avoid this ambiguity.

Take the active translation $T|x\rangle=|x+1\rangle$, with site labels modulo $N$. For the coefficient convention

$$|k\rangle=\frac1{\sqrt N}\sum_xe^{ikx}|x\rangle,$$

one finds $T|k\rangle=e^{-ik}|k\rangle$. Choosing $T^{-1}$ instead gives $e^{ik}$. Define the action of translation before attaching a sign to momentum; either choice gives the same dispersion and the condition $e^{ikN}=1$.

Scattering amplitude and rapidity

For ordered positions $x\lt y$, write

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

Define the exchange amplitude as $S_{12}=A_{21}/A_{12}$. For a regular pair with nonzero denominator, the contact equation yields

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

If another source defines its scattering factor as $A_{12}/A_{21}$, that factor is $S_{12}^{-1}$. The physical wavefunction has not changed. In particular the phase $e^{i\theta}$ in Karbach and Müller 1997, arXiv v1 PDF, p. 3, eqs. (14)–(17) is our reciprocal.

With this ordered ansatz, periodicity means $\psi(x,y)=\psi(y,x+N)$, giving

$$z_1^N=S_{12}^{-1},\qquad z_2^N=S_{12}.$$

For finite regular coordinate rapidities, denoted by $\lambda_j$ in this section, define

$$\lambda_j=\frac12\cot\frac{k_j}{2}, \qquad z_j=\frac{\lambda_j+i/2}{\lambda_j-i/2}.$$

Substitution converts the amplitude and equations to

$$\begin{aligned} S_{12}&=\frac{\lambda_1-\lambda_2-i}{\lambda_1-\lambda_2+i},\\ \left(\frac{\lambda_j+i/2}{\lambda_j-i/2}\right)^N &=\frac{\lambda_j-\lambda_\ell+i}{\lambda_j-\lambda_\ell-i}, \quad \{j,\ell\}=\{1,2\},\\ E&=\frac J2\sum_{j=1}^{2}\frac1{\lambda_j^2+1/4}. \end{aligned}$$

The energy follows from $1-\cos k=1/[2(\lambda^2+1/4)]$. For complex momenta the same expressions are algebraic continuations where defined; do not assume each term separately has a real energy.

Zero momentum corresponds to infinite rapidity. Coincident roots, zero denominators and singular limits require returning to the original wavefunction and eigenvalue equation. Multiplying a rational equation by a vanishing denominator may introduce spurious solutions. Consult the Library derivation for the hypotheses and the numerical project for tests of nonzero regular states.

Compare a monodromy creation parameter

The ordered monodromy $T_a(\lambda_B)=L_{aN}(\lambda_B)\cdots L_{a1}(\lambda_B)$ used on this site has an upper-right block $B$. Its one-magnon vector has adjacent coefficient ratio $(\lambda_B-i/2)/(\lambda_B+i/2)$. Matching that vector to the positive-exponent coordinate wave requires

$$\lambda_{\rm coord}=-\lambda_B, \qquad e^{ik_{\rm coord}}=\frac{\lambda_B-i/2}{\lambda_B+i/2}.$$

For $N=4$ and $\lambda_B=1/2$, this is the coordinate state $k=-\pi/2$, $\lambda_{\rm coord}=-1/2$, with active translation eigenvalue $i$ and energy $J$. Using the same numerical rapidity in the coordinate formula would instead give $k=\pi/2$ and translation eigenvalue $-i$, still at energy $J$. The creation-block convention derivation establishes the coefficient ratio and ring condition directly; it does not assume a general many-magnon equivalence.

Negating a regular two-root set reciprocates both sides of its Bethe equations and leaves its energy unchanged. A symmetric pair, or a total translation phase $1$ or $-1$, therefore cannot detect the reversal by these checks alone. The singular-root benchmark checks its limiting vectors explicitly; the formal pair $\pm i/2$ and its energy do not replace that state calculation.

A convention check on six sites

For $k_1=-k_2=k=2\pi/5$, the contact ratio reduces to $S_{12}=e^{-ik}$. The first ring equation becomes $e^{6ik}=e^{ik}$, so $e^{5ik}=1$. Thus this pair has

$$E/J=(5-\sqrt5)/2, \qquad E_-/J=(5-\sqrt5)/2-6/4.$$

The difference is exactly the vacuum shift. At nonzero uniform field the $M=2$ energy gains $-h$ for $N=6$, while its excitation energy above the all-up vacuum gains $2h$. These three comparisons test amplitude orientation, energy zero and field sign independently.

For the algebraic construction, the R-matrix and transfer-matrix reference uses exactly this Hamiltonian and active translation. It derives the spectral normalization and energy offset needed to recover $H_{\rm F}$ from the transfer matrix.

References

• Karbach, Michael, and Gerhard Müller. “Introduction to the Bethe ansatz I.” Computers in Physics 11, 36–43 (1997). DOI. Reading copy: arXiv:cond-mat/9809162v1, submitted 1998. Version record. Open PDF.