Spin Chain Basis Construction

A spin-chain basis converts local spin operators into transitions between integer-labeled configurations. Explicit bit conventions eliminate factor-of-two and site-order errors before diagonalization. The numerical basis construction here supports sparse Hamiltonian assembly and exact diagonalization.

Bits and dimensionless spin operators

For $L$ spin-$1/2$ sites, let bit $b_j=1$ mean spin up and $b_j=0$ mean spin down, with integer label $s=\sum_j2^jb_j$. Site 0 is the least significant bit. Use dimensionless operators $S^\alpha=\sigma^\alpha/2$; physical angular momentum is $\hbar S^\alpha$.

Then

$$S_j^z|s\rangle=(b_j-\tfrac12)|s\rangle, \qquad S_j^+|s\rangle= \begin{cases}|s+2^j\rangle,&b_j=0,\\0,&b_j=1.\end{cases}$$

The lowering operator reverses that transition. No fermionic sign accompanies a spin flip on a different site. This distinction matters when comparing to Jordan–Wigner transformed fermions.

In ascending local bit order $(|\downarrow\rangle,|\uparrow\rangle)$, the actual stored matrices are

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

They obey $[S^x,S^y]=iS^z$ with $S^x=(S^++S^-)/2$. Importing the usual displayed Pauli matrices in the opposite $(|\uparrow\rangle,|\downarrow\rangle)$ order without changing basis reverses the stored $y$ and $z$ components. To compare bit transitions with Kronecker matrices, order tensor factors from site $L-1$ down to site 0 so the last factor is the least significant bit. Check a complex $S^y$ action as well as a real exchange spectrum.

The full dimension is $2^L$. If total magnetization is conserved, enumerate only states with a fixed popcount $N_\uparrow$, giving dimension $\binom{L}{N_\uparrow}$ and $S^z_{\rm tot}=N_\uparrow-L/2$.

Heisenberg exchange by bit operations

For one bond,

$$H_{ij}=J\left[ S_i^zS_j^z+\frac12(S_i^+S_j^-+S_i^-S_j^+) \right].$$

Its diagonal contribution is $J(b_i-1/2)(b_j-1/2)$. If the two bits differ, flip both with exclusive-or and add matrix element $J/2$. If they agree, there is no off-diagonal transition. This rule gives a sparse matrix directly without global Kronecker products.

Use a dictionary or combinatorial rank map from allowed bitstrings to row indices. The integer label $s$ is not generally the row number in a symmetry-restricted basis.

Two-spin singlet benchmark

For a single bond and $S^z_{\rm tot}=0$, in basis $(|\uparrow\downarrow\rangle,|\downarrow\uparrow\rangle)$,

$$H=J\begin{pmatrix}-1/4&1/2\\1/2&-1/4\end{pmatrix}.$$

Its eigenvalues are $-3J/4$ for the singlet and $J/4$ for the symmetric triplet state. The fully polarized sectors also have energy $J/4$. Recombining the sectors reproduces the triplet degeneracy.

This test diagnoses use of Pauli matrices where spin matrices were intended: that error multiplies all exchange energies by four. A spectrum alone is not enough if the same wrong convention was used to compute the reference.

Boundary conditions and completeness

Store the bond list separately from the basis. An open chain contains $(0,1),\ldots,(L-2,L-1)$. A periodic chain adds $(L-1,0)$, but for very small $L$ the intended graph must be specified: blindly wrapping $L=2$ can count the only pair twice.

Check basis uniqueness, sector dimension, Hermiticity, and invariance of the chosen sector. A transverse field $\sum_j h_xS_j^x$ changes magnetization, so a fixed-$N_\uparrow$ basis is not closed. Projecting it into that basis silently discards the field rather than approximating its dynamics.

For higher spin, use digits $m_j=-S,\ldots,S$ and ladder coefficients $\sqrt{S(S+1)-m_j(m_j\pm1)}$. Binary flip rules do not generalize unchanged.

Exercise

Construct the one-up-spin sector of a periodic exchange chain and determine the hopping amplitude of the up spin.

Solution

The basis has $L$ states labeled by the up-spin position. Each antiparallel bond moves the up spin to a neighbor with amplitude $J/2$. For $L\geq3$ with each undirected bond counted once, two bonds are antiparallel, giving diagonal energy $J(L-4)/4$. Plane waves then have energy $J(L-4)/4+J\cos k$. This checks the bond list and momentum convention.

References

• Weinberg, P., and Bukov, M., “QuSpin: a Python package for dynamics and exact diagonalisation of quantum many body systems. Part I: spin chains,” SciPost Physics 2, 003 (2017), arXiv:1610.03042 — basis construction and symmetry-resolved examples.