Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 72 / 4

Write the finite-chain spin vector as $\sigma$ and the exchange matrix as $\mathsf{J}$. For $J>0$ and $\kappa >0$ this matrix is positive definite, as can be seen from the positive Fourier symbol in part (b), or from the explicit positive open-chain inverse. The linear-source multivariate Gaussian integral, obtained by completing the square, givesIndeed the exponent is $-(m-\mathsf{J}\sigma {)}^T{\mathsf{J}}^{-1}(m-\mathsf{J}\sigma )+{\sigma }^T\mathsf{J}\sigma$. Insert this Hubbard–Stratonovich transformation into the spin partition function. The field term remains $h{\sum }_i{\sigma }_i$, and each independent spin sum isConsequentlyThe interaction convention counts the full ordered pair sum, with no factor of one-half, so the auxiliary coupling is $2m_i{\sigma }_i$. Diagonal terms $J_{ii}{\sigma }_i^2$ only contribute a constant to the spin Hamiltonian; retaining them makes the given positive exchange matrix and normalization convenient. The auxiliary field is not itself the physical spin magnetization: the conditional spin expectation is $\tanh (2m_i+h)$, so $⟨{\sigma }_i⟩=⟨\tanh (2m_i+h)⟩$ in the auxiliary-field integral.

Use the infinite translation-invariant chain first. Set $\rho =e^{-\kappa }$ and Fourier-transform the exchange matrix:For positive $J,\kappa$ this is strictly positive. Its reciprocal is a linear polynomial in $\cos q$, proving the nearest-neighbour form of the exponentially coupled Ising inverse kernel:Expanding the squared difference givesSince $(\cosh \kappa -1)/\sinh \kappa =\tanh (\kappa /2)$, substitution into part (a) yieldsThis is the exact bulk identity. A finite periodic version uses the periodized coupling $J_{ij}=J{\sum }_{\ell \in \mathbb{Z}}{e}^{-\kappa |i-j+\ell N|}$; its discrete Fourier symbols are the same and the squared differences wrap around the ring. Simply truncating the exponential at open ends gives a different inverse at those ends.

For an open chain of $N\ge 2$ sites with the literal finite matrix $J{\rho }^{|i-j|}$, the open-chain endpoint corrections for an exponential Ising kernel are explicit:The inverse has endpoint diagonal entries $1/[J(1-{\rho }^2)]$, interior diagonal entries $(1+{\rho }^2)/[J(1-{\rho }^2)]$ and adjacent entries $-\rho /[J(1-{\rho }^2)]$. Multiplication by the original geometric matrix verifies every entry. These boundary terms affect boundary amplitudes, but not the bulk free-energy density in the thermodynamic limit at fixed $\kappa >0$.

For an auxiliary field varying slowly over lattice distances, measure $x$ in lattice-spacing units and replace the squared difference by $(dm/dx{)}^2$. The long-wavelength action becomesThis is the Euclidean path integral of a one-coordinate quantum particle, with the classical spatial coordinate serving as imaginary time and $\hslash =1$. To see the correspondence at the operator level, time-slice it by $ϵ$ and use the symmetric kernelActing on a smooth test function, Gaussian integration gives $K_{ϵ}\psi =\psi +ϵ{\psi }^{\prime \prime }/(2M_Q)-ϵU\psi +O({ϵ}^2)$. Thus the continuum auxiliary-field transfer operator for an Ising chain is $e^{-ϵ{\hat{H}}_Q}$, whereMultiplying kernels and integrating over intermediate fields gives $⟨m_f|e^{-L{\hat{H}}_Q}|m_i⟩$ for fixed endpoints, and $\mathrm{tr}{e}^{-L{\hat{H}}_Q}$ for periodic fields, up to the normalization already absorbed in $C$. This is the precise classical Ising chain as a quantum particle correspondence. It concerns an imaginary-time transition kernel; it is not the squared modulus of a real-time transition amplitude. The continuum approximation retains smooth long-distance modes, rather than asserting exact equality of the finite-spacing transfer operator with this differential operator.

At $h=0$, set $A=\tanh (\kappa /2)/J$. Expansion of the local potential givesThe potential is even and grows as $A{m}^2-2|m|$ at large $|m|$. It is a symmetric double well precisely whenThe nonzero stationary points satisfy $Am=\tanh (2m)$. For $m>0$, the ratio $\tanh (2m)/m$ decreases strictly from 2 to 0: writing $z=2m$, the sign follows from $\tanh z-z{\mathrm{sech}}^{2}z>0$, whose derivative is $2z{\mathrm{sech}}^{2}z\tanh z>0$. Hence there is one minimum on each side when $A<2$. At $A=2$ the potential instead has a single quartic minimum, and for $A>2$ a single quadratic minimum. Small $\kappa$ with fixed positive $J$ falls in the double-well regime, but the printed claim requires this coupling condition; $\kappa \ll 1$ alone does not imply it if $J$ scales to zero too quickly.

A weak magnetic field givesThe linear term is odd in $m$. For positive $h$, the positive well becomes deeper. If the zero-field minima are $\pm m_{\ast }$, their depth difference to first order is $U(m_{\ast },h)-U(-m_{\ast },h)=-2h\tanh (2m_{\ast })$. The magnetic field tilts the symmetric double well and favors the matching spin orientation. A large enough field can eliminate the metastable well.

Finally, two minima do not imply a true finite-temperature phase transition in this one-dimensional chain at fixed finite interaction range $1/\kappa$. A finite quantum barrier permits quantum tunnelling; the confining one-dimensional Schrödinger operator has a unique symmetric ground state at zero field and a nonzero first excitation gap. The quantum gap corresponds to a finite classical correlation length. A very small gap can mimic ordering on a finite sample, but genuine degeneracy requires a singular limit such as vanishing tunnelling. This distinction is central to interpreting the quantum mapping.