Gaussian field theory

A Gaussian field theory has a free energy quadratic in its field modes. If $H_0={\sum }_{\mathbf{q}}^{+}J(q)|{\varphi }_{\mathbf{q}}{|}^2$ for a real field, then $⟨|{\varphi }_{\mathbf{q}}{|}^2⟩=J(q{)}^{-1}$ and its partition function is a product of elementary Gaussian integrals.

The uniform mode of a pure Gaussian field theory with negative squared mass has an unbounded quadratic energy. Its partition function diverges even with a finite-volume ultraviolet regulator, so it supplies no stable ordered branch. At positive squared mass the zero-source average is zero. Thus the scaling dimension of the field can yield a formal order-parameter critical exponent without establishing a negative-reduced-temperature magnetization law. A stabilizing quartic interaction is relevant below four dimensions and ordinarily changes the critical theory to the Wilson-Fisher fixed point; above four its coefficient is a dangerously irrelevant coupling for the ordered phase.

For the positive quadratic kernel $\tilde{\Delta }(p)={\kappa }^{-1}{p}^2+m^2$, $\kappa ,m>0$, the connected correlation function is its inverse. Fourier inversion gives $G(r)=\kappa (\sqrt{\kappa }m/r{)}^{D/2-1}{K}_{D/2-1}(\sqrt{\kappa }mr)/(2\pi {)}^{D/2}$. The Modified Bessel function of the second kind has an exponentially decaying large-argument tail, so the correlation length is $1/(\sqrt{\kappa }m)$ at fixed positive gradient coefficient. In three dimensions this becomes $\kappa e^{-\sqrt{\kappa }mr}/(4\pi r)$. The statement concerns the continuum long-distance theory; regulator-dependent contact terms do not define the physical correlation length.

For a real field, ${\varphi }_{-\mathbf{q}}={\varphi }_{\mathbf{q}}^{\ast }$. A sum ${\sum }_{\mathbf{q}}^{+}$ takes one representative from each nonzero pair $\{\mathbf{q},-\mathbf{q}\}$ so that every independent complex Fourier amplitude is counted once.

A Gaussian variational approximation chooses a quadratic trial kernel and minimizes a variational upper bound on the interacting free energy. The optimized kernel often equals the bare kernel plus a self-consistent, wavevector-independent mass shift.

For $H=\int [a{\varphi }^2/2+\kappa |\nabla \varphi {|}^2/2+\gamma ({\nabla }^2\varphi {)}^2/2+B{\varphi }^2|\nabla \varphi {|}^2]$ and an even positive trial kernel $J(q)$, define $S_0={\sum }_q^{+}1/J(q)$ and $S_2={\sum }_q^{+}{q}^2/J(q)$ using a positive-wavevector sum for a real field. Isserlis theorem gives $⟨H_4{⟩}_0=4B{S}_0{S}_2/V$. The Feynman-Bogoliubov inequality therefore has stationary kernelsThus both the mass and gradient coefficient shift. In the thermodynamic limit, their self-consistency equations areA coarse-graining ultraviolet cutoff $\Lambda$ is necessary for the first integral in three dimensions: its large-$k$ radial integrand tends to a constant. Physical solutions require $J(q)>0$.

For an exact dimensionless Hamiltonian $H$ and trial Hamiltonian $H_0$, the free energies obeywhere $F_0=-\log Z_0$ and the expectation uses the trial Gibbs distribution $Z_0^{-1}{e}^{-H_0}$.

The identity $e^{-F}=e^{-F_0}⟨e^{-(H-H_0)}{⟩}_0$ and the Jensen inequality for the convex exponential function give $\log ⟨e^{-(H-H_0)}{⟩}_0\ge -⟨H-H_0{⟩}_0$. Taking minus the logarithm proves the bound.