Past exam of the theoretical physics major option course of the University of Oxford / 2016 / 1

Let $\beta =(k_{B}T{)}^{-1}$ and sum over all $2^{2L}$ configurations of the Ising spins. The canonical partition function and Helmholtz free energy per site areBoth spins in each cell count as sites. In particular the denominator is $2L$, rather than $L$.

The mean spin magnetization per site is the Boltzmann distribution averageEquivalently introduce a uniform magnetic field through $E_h=E-h{\sum }_j({\sigma }_j+{\tau }_j)$ and differentiate:This convention absorbs the microscopic magnetic moment into $h$. Simultaneous reversal of every spin leaves $E$ invariant and changes the sign of the numerator, so $⟨m⟩=0$ at zero field for finite $L$.

For each pair of adjacent spins, define the transfer matrix for a classical spin chain by $T_{s{s}^{\prime }}=e^{-\beta E_0(s,s^{\prime })}$. Multiplying its entries gives the Boltzmann factor of a whole configuration. The periodic boundary condition identifies the last index with the first, soThe assumed interchange symmetry of $E_0$ makes $T$ a real symmetric matrix, hence it has an orthonormal eigenbasis. At finite positive temperature, every matrix entry is strictly positive, so the Perron–Frobenius theorem supplies a unique positive largest eigenvalue ${\lambda }_{+}$, larger in modulus than every other one. Therefore $L^{-1}\log Z_L\to \log {\lambda }_{+}$.

Local observables become matrix insertions. For example, with $S_{s{s}^{\prime }}=s{\delta }_{s{s}^{\prime }}$,Thus partition functions and correlation functions reduce to a fixed-dimensional eigenvalue problem, rather than a sum over exponentially many microstates.

For the sawtooth Ising chain, each apex spin occurs in only one triangle. Put $K=\beta J$. Summing that spin first gives the effective transfer matrixIn the ordered basis $s=+1,-1$ this isIts symmetric and antisymmetric eigenvectors are $(1,1)/\sqrt{2}$ and $(1,-1)/\sqrt{2}$, with eigenvaluesConsequentlyAt $J=0$, the result is $Z_L=4^L$, correctly counting $2L$ independent Ising spins. As $T\to 0$, $f_{\infty }\to -3J/2$, the energy of three satisfied ferromagnetic bonds per two sites.

The printed summand lacks grouping brackets. The result above uses the translation-invariant common-$J$ interpretation. If base and apex bonds instead have independent couplings $J_b,J_a$, the same derivation gives ${\lambda }_{\pm }=2e^{\beta J_b}\cosh (2\beta J_a)\pm 2e^{-\beta J_b}$; specifying those couplings is essential before giving a different answer.

Let $S=\mathrm{diag}(1,-1)$ be the base-spin insertion matrix. In the eigenbasis from solution, its diagonal matrix elements vanish: $⟨+|S|+⟩=⟨-|S|-⟩=0$. Hencenot only at large $L$ but for every finite ring at zero field.

This vanishing is not merely an averaging artifact hiding a finite-temperature ordered phase. The two-point correlation function is $⟨{\sigma }_1{\sigma }_{1+r}⟩=({\lambda }_{-}^r{\lambda }_{+}^{L-r}+{\lambda }_{+}^r{\lambda }_{-}^{L-r})/Z_L$. In the thermodynamic limit it becomes $({\lambda }_{-}/{\lambda }_{+}{)}^r$, with finite correlation length ${\xi }^{-1}=\log ({\lambda }_{+}/{\lambda }_{-})$ for every $T>0$ and $J>0$. Thermal domain walls destroy long-range alignment in this one-dimensional short-range system. At zero temperature, the two fully aligned ground states are degenerate; that limiting case must be distinguished from any fixed positive temperature.