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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.

With lattice spacing set to one and a smoothly varying auxiliary field, the discrete action becomesThis is precisely the Euclidean path integral of a quantum particle with $\hslash =1$, mass $M$ and potential $U$. With fixed endpoints it gives the transition amplitude $⟨m_f|e^{-\mathcal{L}{\hat{H}}_Q}|m_i⟩$, while periodic classical boundary conditions give its trace. The corresponding operator isThis is classical Ising chain as a quantum particle. It is the continuum approximation to the exact positive transfer kernel, so its numerical spectrum is accurate when the relevant auxiliary configurations vary slowly. The transfer-kernel reasoning and the domain-wall argument below also support the qualitative conclusion without treating a continuum approximation as an exact finite-lattice identity.

At zero field write $a=\tanh (\kappa /2)/J>0$. Expansion givesA double well exists when $a<2$; the stationary points satisfy $am=\tanh (2m)$. For $a\ge 2$ there is a single minimum at zero, with the quadratic curvature marginal at equality. Thus a double well is the low-effective-temperature regime of the model, not an unconditional property for every $J$. The exact potential remains confining because $U(m)=a{m}^2-2|m|+O(1)$ at large $|m|$.

Even when two wells exist, a quantum particle with finite mass and a finite barrier tunnels between them. Its one-dimensional confining Schrödinger operator has a unique, strictly positive ground-state wavefunction. At $h=0$ reflection symmetry makes that state even, so every odd observable has zero expectation. An antisymmetric first excited state is split from it by a positive gap. A spectral expansion of an odd-observable correlation has factors $e^{-x(E_n-E_0)}$; the smallest contributing gap therefore gives a finite classical correlation length. The original spin observable, after decoupling, is $\tanh (2m+h)$, which is odd at zero field. HenceA deep well can give an exponentially small quantum tunnelling gap and an extremely long correlation length, which can mimic order in a finite sample, but it does not produce true long-range order.

The classical counterpart is the finite cost of a domain wall. Because the pair sum is doubled, a sharp wall has dimensionless energyAt any finite temperature such walls have nonzero fugacity, divide a long chain into domains and destroy spontaneous magnetization. The range is large for $\kappa \ll 1$ but finite. The zero-temperature limit and an infinite-range limit are singular and must not be interchanged casually with the thermodynamic limit; an infinite-range interaction also requires an extensive normalization before comparing models.

A nonzero $h$ tilts the potential: to first order $U_h(m)=U_0(m)-h\tanh (2m)+O(h^2)$. A positive field favors the positive well and produces positive magnetization. For fixed finite parameters the ground state and classical free energy vary smoothly, so there is no zero-field discontinuity due merely to a double-well bare potential. A discrete broken-symmetry state appears only in an appropriate singular limit, not by ignoring quantum tunnelling.

More generally a local $d$-dimensional classical statistical model can be sliced along one coordinate and represented by a transfer operator. That coordinate becomes imaginary time; the other $d-1$ coordinates become the spatial coordinates of a quantum system. Infinite extent in the transfer direction corresponds to its ground-state, or zero-quantum-temperature, limit. The classical Ising model is thus related to a quantum Ising system in $d-1$ spatial dimensions, with transverse dynamics generated by the transfer operator. For $d=1$ this is quantum mechanics of finitely many local variables, where quantum tunnelling restores symmetry. For $d\ge 2$ the quantum system has infinitely many spatial degrees of freedom and can have distinct symmetry-broken ground states, allowing a classical finite-temperature ordering transition. Locality and anisotropy of the mapping determine the quantum couplings; the dimension correspondence alone does not fix them.