Monte Carlo Evaluation of the Continuum Limit of the Two-Point Function of the Euclidean Free Real Scalar Field Subject to Affine Quantization

Abstract

We study canonical and affine versions of the quantized covariant Euclidean free real scalar field-theory on four dimensional lattices through the Monte Carlo method. We calculate the two-point function near the continuum limit at finite volume. Our investigation shows that affine quantization is able to give meaningful results for the two-point function for which is not available an exact analytic result and therefore numerical methods are necessary.

Appendix: Field Configurations in the Vicinity of the Two Degenerate Minima in the Affine Version

Classically, the affine version of the free Hamiltonian has two degenerate minima, $\phi = \pm \Phi$. If the path integral is dominated by those field configurations that are located in the vicinity of one of these everywhere on the entire lattice or in the vicinity of the other, then it consists of two equal pieces

$$\begin{aligned} Z= & {} \int \mathcal{D}\phi \,\exp (-S[\phi ]),\\ Z_+= & {} \int \mathcal{D}\phi \,\exp (-S[\phi ]),\;\;\text{ integral } \text{ only } \text{ over }\;\;\phi (x)\approx \Phi ,\\ Z_-= & {} \int \mathcal{D}\phi \,\exp (-S[\phi ]),\;\;\text{ integral } \text{ only } \text{ over }\;\;\phi (x)\approx -\Phi , \end{aligned}$$

and $Z_+ = Z_-$. Under a broken symmetry $\phi \rightarrow -\phi$ one would get either $Z \approx Z_+$ or $Z \approx Z_-$. This has to be expected in the present case of a real field since in order to move the field $\phi (x)$ at a single x from around $\Phi$ to around $-\Phi$ in the MC path integral one has to overcome a large kinetic cost. This is not true for a complex field where one can go “slowly” “around” the “mountain” at $\phi =0$.

The expectation value of the field

$$\begin{aligned} \langle \phi (x)\rangle= & {} \int \mathcal{D}\phi \,\phi (x)\exp (-S[\phi ])/Z,\\ \langle \phi (x)\rangle _+= & {} \int \mathcal{D}\phi \,\phi (x)\exp (-S[\phi ])/Z_+,\;\;\text{ over }\;\;\phi (x)\approx \Phi \\ \langle \phi (x)\rangle _-= & {} \int \mathcal{D}\phi \,\phi (x)\exp (-S[\phi ])/Z_-,\;\;\text{ over }\;\;\phi (x)\approx -\Phi , \end{aligned}$$

with $\langle \phi (x)\rangle _+ \approx \Phi , \langle \phi (x)\rangle _- \approx -\Phi$, and under the broken symmetry, $\langle \phi (x)\rangle \approx \langle \phi (x)\rangle _\pm \approx \pm \Phi$ where the simulation, starting from $\phi =0$, will choose among the two different cases just after the first equilibration steps.

For the two-point function

$$\begin{aligned} D_+(x-y)= & {} \int \mathcal{D}\phi \,\phi (x)\phi (y)\exp (-S[\phi ])/Z_+-\langle \phi (x)\rangle _+^2,\;\;\text{ over }\;\;\phi (x)\approx \Phi ,\\ D_-(x-y)= & {} \int \mathcal{D}\phi \,\phi (x)\phi (y)\exp (-S[\phi ])/Z_+-\langle \phi (x)\rangle _-^2,\;\;\text{ over }\;\;\phi (x)\approx -\Phi , \end{aligned}$$

so that $D_+(z)\approx 0, D_-(z)\approx 0$, and $D(z)\approx D_\pm (z)\approx 0$.

Moreover one can see how in the broken symmetry configuration in which $\phi ^2(x) \approx \Phi ^2\sim a^{-3}$, the “3/8” term in the Hamiltonian density is also of the same order in the continuum limit $a\rightarrow 0$. This will lead to a convergent two-point function for $\phi -\langle \phi \rangle$ in the continuum limit.