Past exam of the theoretical physics major option course of the University of Oxford / 2015 / 4

Let $\beta =(k_{B}T{)}^{-1}$, and sum over all $2^N$ configurations satisfying the periodic identification. The canonical partition function and finite-size free energy per site areThe plus sign multiplying $J$ in the energy means that $J>0$ favors opposite neighboring spins: this is an antiferromagnetic Ising model. Even $N$ permits an alternating configuration to close without frustration. The finite-size formula does not require taking a thermodynamic limit; that limit is taken only after evaluating $Z_N$.

Share each on-site field energy equally between its two neighboring bonds. In the ordered basis $\sigma =+1,-1$, the symmetric transfer matrix for a classical spin chain isThe product ${\prod }_j{T}_{{\sigma }_j{\sigma }_{j+1}}$ recovers each bond and each field term once. Summing the periodic indices closes a trace, giving $Z_N=\mathrm{Tr}{T}^N$. The two eigenvalues areThe square-root argument is positive. Moreover $\det T=e^{-2K}-e^{2K}<0$, so ${\lambda }_{-}<0$ even at nonzero field. Its sign must not be discarded merely because all matrix entries are positive. The Perron–Frobenius theorem ensures that the positive eigenvalue ${\lambda }_{+}$ has the larger modulus, and the infinite-chain free energy is $f_{\infty }=-{\beta }^{-1}\log {\lambda }_{+}$. For even $N$, both terms in the finite-size partition function are positive.

Insert the two observed spins into the configuration sum for the transfer matrix for a classical spin chain trace. Since the diagonal matrix ${\sigma }^z$ has entry $s$ on state $s$, multiplication by it at positions $j$ and $j+r$ produces the factor ${\sigma }_j{\sigma }_{j+r}$. The trace representation is consequentlyThere are exactly $N$ transfer factors: $(j-1)+r+(N+1-j-r)=N$. The distance between the two insertions is $r$ bonds, not $r-1$. Cyclicity of the trace combines the outside factors and gives the equivalent expressionwhich makes independence of the starting site manifest. No assumption that $T$ commutes with ${\sigma }^z$ is used; in general it does not. For $r=0$ the two insertions multiply to the identity and $g(0)=1$, a useful normalization check.

At zero field the normalized transfer matrix for a classical spin chain eigenvectors and eigenvalues areThe spin insertion exchanges the two eigenvectors: ${\sigma }^z|+⟩=|-⟩$ and conversely. Evaluating the trace in c therefore gives, with $x={\lambda }_{-}/{\lambda }_{+}=-\tanh K$,This is the finite-ring result. In the thermodynamic limit, or for $1\ll r\ll N$ away from the wrap-around image, it becomesHere $\xi$ is in lattice-site units. The alternating sign is antiferromagnetic order at short distances; the exponential envelope shows the absence of long-range order at every positive temperature in this one-dimensional finite-range model. As $T\to 0$, $\xi$ diverges, asymptotically $\xi \sim \frac{1}{2}{e}^{2\beta J}$. For the even ring the zero-temperature result is $g(r)=(-1{)}^r$, consistent with the two alternating ground configurations. At fixed finite $N$, “large distance” must still respect periodicity: $g_N(r)=g_N(N-r)$ rather than decaying indefinitely beyond the ring.

At $B=0$ a global reversal ${\sigma }_j\mapsto -{\sigma }_j$ leaves the energy and statistical weight unchanged but changes the sign of the one-spin observable. Pairing reversed configurations givesEquivalently, $⟨\pm |{\sigma }^z|\pm ⟩=0$ in the transfer eigenbasis, so $\mathrm{Tr}({\sigma }^z{T}^N)=0$. This holds for every finite even ring and all finite temperatures, and for the zero-field Gibbs limit. A symmetry-selected zero-temperature alternating state can have nonzero staggered magnetization; its uniform magnetization is still zero. The classical spin variable in this model has no separate quantum spin-operator dynamics despite the conventional matrix notation ${\sigma }^z$ used for the insertion.