Classical Ising chain as a quantum particle (${\hat{H}}_Q=-\frac{J\sinh \kappa }{2}\frac{d^2}{d{m}^2}+U(m)$)

Hubbard-Stratonovich decoupling of an exponentially coupled Ising chain gives a scalar-field action with a nearest-neighbor gradient term and $U(m)=\tanh (\kappa /2)m^2/J-\log [2\cosh (2m+h)]$. Its continuum partition function is the Euclidean path integral of the displayed quantum particle. At zero field the potential can have two wells, but a finite barrier still permits quantum tunnelling and gives a unique symmetric ground state. The excitation gap produces a finite classical correlation length, so there is no finite-temperature spontaneous magnetization for fixed positive $\kappa$. One classical spatial coordinate becomes imaginary time; a $d$-dimensional classical model corresponds to a quantum model with $d-1$ spatial dimensions.

For a nearest-neighbor auxiliary-field representation of an Ising model, sharing each local potential between adjacent bonds gives the displayed symmetric integral operator. Its periodic partition function is the trace of its $N$th power, up to Gaussian normalization constants. In the slowly varying continuum limit, a slice of length $ϵ$ has kernel $[M/(2\pi ϵ){]}^{1/2}\exp [-M(m^{\prime }-m{)}^2/(2ϵ)-ϵ\{U(m^{\prime })+U(m)\}/2]$. Expanding its action on a smooth function gives $1-ϵ\hat{H}+O({ϵ}^2)$, with $\hat{H}=-(2M{)}^{-1}{\partial }_m^2+U$. Thus fixed endpoints give an imaginary-time transition kernel and periodic endpoints give a Euclidean path integral trace. The finite-spacing integral operator and the continuum differential operator need not have identical high-energy spectra.

The zero-field Hubbard–Stratonovich transformation of an exponentially coupled Ising model produces $U(m)=A{m}^2-\log [2\cosh (2m)]$, where $A=\tanh (\kappa /2)/J>0$. Since $U^{\prime \prime }(0)=2A-4$ and $U$ tends to positive infinity, a symmetric double well exists exactly when $A<2$. Its nonzero stationary points solve $Am=\tanh (2m)$; the ratio $\tanh (2m)/m$ decreases from 2 to 0 for $m>0$, proving that there is exactly one positive minimum. At $A=2$ the minimum is a single quartic minimum. A small conjugate field adds $-h\tanh (2m)+O(h^2)$ and tilts the wells, while a sufficiently large field can remove the metastable well. A finite barrier permits quantum tunnelling; two classical minima alone do not establish spontaneous symmetry breaking.