Construct a small spin-chain Hamiltonian

How does a local interaction become a quantum many-body matrix? You will construct the periodic spin-$1/2$ XXX Hamiltonian, reduce it to a fixed number of down spins, and check its action without diagonalizing a $2^N\times2^N$ matrix. The key is a two-site rule: an unequal neighboring pair contributes a diagonal term and a spin exchange.

Required background. Multiply small matrices and interpret a tensor-product basis; the linear algebra bridge develops both operations and explains invariant sectors. The entry repair supplies the spin-operator rules needed here.

Helpful background. The course introduction explains the destination of the calculation; the convention reference keeps spin, coupling, and translation conventions together.

The periodic XXX Hamiltonian

Take $N\geq3$ sites on a ring, with Hilbert space $(\mathbb C^2)^{\otimes N}$ and orthonormal basis

$$|\sigma_1,\ldots,\sigma_N\rangle,\qquad \sigma_n\in\{\uparrow,\downarrow\}.$$

Set $\hbar=1$ and the lattice spacing to one. The spin operators are $S_n^\alpha=\sigma_n^\alpha/2$, where $\sigma^\alpha$ are Pauli matrices acting at site $n$. In particular, $S^z|\uparrow\rangle=\tfrac12|\uparrow\rangle$. Our ferromagnetic coupling is $J\gt0$, with units of energy:

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

There are $N$ bonds, including $(N,1)$. The constant makes the all-up state $|F\rangle=|\uparrow\cdots\uparrow\rangle$ have energy zero. The model record fixes this regime. Karbach and Müller instead use $H_{\rm KM}=-J\sum_n\mathbf S_n\cdot\mathbf S_{n+1}$, so

$$H=H_{\rm KM}+\frac{JN}{4}\,\mathbf 1.$$

Their excitation energy $E-E_0$ is our energy $E$; the eigenvectors are unchanged. See Karbach and Müller 1997, pp. 1–2, equation (1) and vacuum energy, arXiv v1 PDF.

Entry check and repair

Apply $S^+\otimes S^-$ and $S^z\otimes S^z$ to $|\downarrow\uparrow\rangle$. Does either operation change the number of down spins?

Repair. With $S^\pm=S^x\pm iS^y$,

$$\begin{aligned} S^+|\downarrow\rangle&=|\uparrow\rangle,& S^-|\uparrow\rangle&=|\downarrow\rangle,\\ S^+|\uparrow\rangle&=0,& S^-|\downarrow\rangle&=0. \end{aligned}$$

Operators on different tensor factors act separately. Thus $(S^+\otimes S^-)|\downarrow\uparrow\rangle=|\uparrow\downarrow\rangle$ and $(S^z\otimes S^z)|\downarrow\uparrow\rangle=-\tfrac14|\downarrow\uparrow\rangle$. Both retain one down spin. The factor $1/4$ comes from multiplying two spin eigenvalues; using Pauli eigenvalues $\pm1$ would change the Hamiltonian.

Derive the two-site exchange rule

For one bond, use

$$\mathbf S_n\cdot\mathbf S_{n+1} =S_n^zS_{n+1}^z+ \frac12\left(S_n^+S_{n+1}^-+S_n^-S_{n+1}^+\right).$$

In the ordered basis $|\uparrow\uparrow\rangle,|\uparrow\downarrow\rangle, |\downarrow\uparrow\rangle,|\downarrow\downarrow\rangle$, the bond matrix is

$$h=\frac J2 \begin{pmatrix} 0&0&0&0\\ 0&1&-1&0\\ 0&-1&1&0\\ 0&0&0&0 \end{pmatrix}.$$

Equal spins contribute zero. Unequal spins obey

$$h|\uparrow\downarrow\rangle =\frac J2\left(|\uparrow\downarrow\rangle- |\downarrow\uparrow\rangle\right),$$

and the analogous rule with the arrows interchanged. If $P_{n,n+1}$ swaps the two spins, the same result is

$$h_{n,n+1}=\frac J2(1-P_{n,n+1}), \qquad H=\frac J2\sum_{n=1}^N(1-P_{n,n+1}).$$

This identity is special to spin $1/2$ in this form. It also supplies a check: $P$ is Hermitian and satisfies $P^2=1$, so each $h$ has eigenvalues $0$ and $J$. Consequently $H$ is positive semidefinite. The individual bond terms need not commute.

Build only the sector you need

Let $M$ be the number of down spins. The total magnetization is

$$S_{\rm tot}^z=\sum_nS_n^z=\frac N2-M.$$

Every exchange preserves $M$, hence $[H,S_{\rm tot}^z]=0$. The fixed-$M$ sector has dimension $\binom NM$, and its basis can be labeled by ordered down-spin positions:

$$|x_1,\ldots,x_M\rangle,\qquad 1\leq x_1\lt\cdots\lt x_M\leq N.$$

For any basis column, inspect every bond. Add $J/2$ to its diagonal entry for each unequal pair, and add $-J/2$ to the row obtained by exchanging that pair. Accumulate contributions if different bonds reach the same row. This rule avoids constructing unnecessary tensor-product matrices.

For example, on four sites the state $|1,2\rangle=|\downarrow\downarrow\uparrow\uparrow\rangle$ has unequal pairs only on bonds $(2,3)$ and $(4,1)$. Therefore

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

The diagonal coefficient counts domain walls, not down spins. Two adjacent down spins have two domain walls; two separated down spins can have four.

Work through a four-site matrix

In the one-down-spin basis $|1\rangle,|2\rangle,|3\rangle,|4\rangle$,

$$H_{M=1}=\frac J2 \begin{pmatrix} 2&-1&0&-1\\ -1&2&-1&0\\ 0&-1&2&-1\\ -1&0&-1&2 \end{pmatrix}.$$

For the first column, bonds $(1,2)$ and $(4,1)$ move the down spin to sites $2$ and $4$. This checks the corner entries. The matrix is real symmetric, preserves the sector, and annihilates $(1,1,1,1)^{\mathsf T}$. These are useful independent checks of an implementation.

Its eigenvalues are $0,J,J,2J$, but the construction did not require guessing them. In the next lesson, discrete plane waves will explain the spectrum for every $N$.

Exercises

Guided practice: the energy of a bond

Find the bond energies of

$$|t\rangle=\frac{|\uparrow\downarrow\rangle+ |\downarrow\uparrow\rangle}{\sqrt2}, \qquad |s\rangle=\frac{|\uparrow\downarrow\rangle- |\downarrow\uparrow\rangle}{\sqrt2}.$$

Explain why the ferromagnetic sign favors the symmetric combination.

Hint

Find the eigenvalue of the swap $P$ before applying $h=J(1-P)/2$.

Solution

The swap gives $P|t\rangle=|t\rangle$ and $P|s\rangle=-|s\rangle$. Hence $h|t\rangle=0$ and $h|s\rangle=J|s\rangle$. The parallel states also have zero bond energy. These three symmetric states form the triplet; the antisymmetric state is the singlet. Since $J\gt0$, the triplet costs less energy. This is a statement about one bond; overlapping bonds must still be assembled into the many-site operator.

Independent practice: the two-down-spin block

Construct the four-site $M=2$ matrix in the order $|1,2\rangle,|1,3\rangle,|1,4\rangle,|2,3\rangle,|2,4\rangle,|3,4\rangle$. Check its symmetry and its action on the vector with every component equal to one.

Hint

The alternating configuration $|1,3\rangle$ has four unequal bonds. The adjacent configuration $|1,4\rangle$ has two, even though the positions straddle the displayed ends of the chain.

Solution

The exchange rule gives

$$H_{M=2}=\frac J2 \begin{pmatrix} 2&-1&0&0&-1&0\\ -1&4&-1&-1&0&-1\\ 0&-1&2&0&-1&0\\ 0&-1&0&2&-1&0\\ -1&0&-1&-1&4&-1\\ 0&-1&0&0&-1&2 \end{pmatrix}.$$

For instance, $H|1,3\rangle=2J|1,3\rangle- \tfrac J2(|1,2\rangle+|1,4\rangle+|2,3\rangle+|3,4\rangle)$. The matrix equals its transpose. Every row sums to zero, so the uniform vector is a zero-energy eigenvector. There are six states, agreeing with $\binom42=6$.

Transfer: remove the closing bond

Replace the four-site ring by an open chain, retaining only bonds $(1,2)$, $(2,3)$, and $(3,4)$. Construct the $M=1$ block. Which previous checks survive, and why do periodic plane waves no longer follow from the boundary condition?

Hint

At site $1$, only one bond can exchange the down spin. Removing a bond changes diagonal entries as well as off-diagonal entries.

Solution

The matrix becomes

$$H_{\rm open}=\frac J2 \begin{pmatrix} 1&-1&0&0\\ -1&2&-1&0\\ 0&-1&2&-1\\ 0&0&-1&1 \end{pmatrix}.$$

The endpoint diagonals are $J/2$, not $J$. Hermiticity, fixed $M$, positivity, and the uniform zero-energy vector survive because their bond-by-bond arguments survive. Translation around the ring is no longer a symmetry: it would move an endpoint to an interior site. There is no condition $\psi(n+4)=\psi(n)$ to impose. The open-chain eigenvectors must satisfy the endpoint equations of this new matrix.

What the construction establishes

You can now build finite matrices with the correct spin normalization, closing bond, and magnetization sectors. These symmetry reductions apply well beyond integrable models; they do not by themselves establish quantum integrability. Continue to solve the one-magnon sector and then test interacting two-magnon states.

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.