Table of Integrals and Lattice Sums

We give here a series of formulas and derivations involving Gaussian integrals, Gaussian expectation values and lattice sums, in the context the model discussed in this paper, which are used for the calculations presented. All these can be derived from the basic result in momentum space

$$\left\langle \widetilde\varphi _{i}'(k_{\mu}) \widetilde\... ... = \frac{1}{N^{d}}\, \frac{1}{\rho^{2}(k_{\mu})+\alpha_{i}},$$ (B.1)

where $\alpha_{i}$ is either $\alpha_{0}$ or $\alpha_{\mathfrak{N}}$, depending on the field component involved, and where $\rho^{2}(k_{\mu})$ are the eigenvalues of the discrete Laplacian on the lattice, which are given by

$$\rho^{2}(k_{\mu}) = 4 \left[ \sin^{2}\!\left(\frac{\pi ... ...ldots + \sin^{2}\!\left(\frac{\pi k_{d}}{N}\right) \right].$$

Since in the measure of $S_{0}[\vec{\varphi}']$ the modes are decoupled in momentum space, the same expectation value with two different momenta $k_{\mu}$ and $k_{\mu}'$ is zero by simple parity arguments. We use the notation for the two-point functions in position space,

$$\begin{eqnarray*} g_{0}(n_{\mu}-n_{\mu}') & = & \left\langle \varphi_{i}'(n_... ...}'(n_{\mu})\varphi_{\mathfrak{N}}'(n_{\mu}') \right\rangle_{0}, \end{eqnarray*}$$

for $i=1,\ldots,\mathfrak{N}-1$. These are, of course, the inverse Fourier transforms of the corresponding two-point functions in momentum space,

$$\begin{eqnarray*} g_{0}(n_{\mu}-n_{\mu}') & = & \sum_{k_{\mu}}^{N^{d}} e^{-\... ...detilde\varphi _{\mathfrak{N}}'^{*}(k_{\mu}) \right\rangle_{0}. \end{eqnarray*}$$

In order to write this explicitly we may use the Fourier transforms of the fields, for example in the case of the $\varphi_{\mathfrak{N}}'(n_{\mu})$ field component,

$$\begin{eqnarray*} g_{\mathfrak{N}}(n_{\mu}-n_{\mu}') & = & \left\langle \var... ...\widetilde\varphi _{\mathfrak{N}}'(k'_{\mu}) \right\rangle_{0}. \end{eqnarray*}$$

The expectation value in momentum space in non-zero only if we have $k_{\mu}'=-k_{\mu}$, in which case we have the result, which can be obtained from Equation (B.1) above,

$$\begin{eqnarray*} \left\langle \widetilde\varphi _{\mathfrak{N}}'(k_{\mu}) \w... ...{1}{N^{d}}\, \frac{1}{\rho^{2}(k_{\mu})+\alpha_{\mathfrak{N}}}. \end{eqnarray*}$$

This eliminates one of the momentum-space sums, and thus we get

$$g_{\mathfrak{N}}(n_{\mu}-n_{\mu}') = \frac{1}{N^{d}} \su... ..._{\mu}-n_{\mu}')}} {\rho^{2}(k_{\mu})+\alpha_{\mathfrak{N}}},$$ (B.2)

which is just the statement that $g_{\mathfrak{N}}(n_{\mu}-n_{\mu}')$ is the inverse Fourier transform of the momentum-space propagator. Note that this is necessarily real, and that therefore the imaginary part of the right-hand side vanishes. In a completely similar way, we have the corresponding result for the other field $\mathfrak{N}-1$ components, with $i\neq\mathfrak{N}$,

$$g_{0}(n_{\mu}-n_{\mu}') = \frac{1}{N^{d}} \sum_{k_{\mu}}... ...d}k_{\mu}(n_{\mu}-n_{\mu}')}} {\rho^{2}(k_{\mu})+\alpha_{0}}.$$

The following sum involving $g_{0}(n_{\mu}-n_{\mu}')$ can also be easily calculated, using once more the Fourier transforms,

$$\sum_{n_{\mu}}^{N^{d}}g_{\mathfrak{N}}(n_{\mu}-n_{\mu}') =... ...\mu}'-n_{\mu}'')}} {\rho^{2}(k_{\mu})+\alpha_{\mathfrak{N}}}.$$

The orthogonality relation can be used to simplify this expression, and thus we get

$$\begin{eqnarray*} \sum_{n_{\mu}}^{N^{d}}g_{\mathfrak{N}}(n_{\mu}-n_{\mu}') & =... ...ha_{\mathfrak{N}}} \\ & = & \frac{1}{\alpha_{\mathfrak{N}}}. \end{eqnarray*}$$

This is simply the zero-mode of the propagator. The same can be done for the other components of the field, so we conclude that

$$\sum_{n_{\mu}}^{N^{d}} g_{0}(n_{\mu}-n_{\mu}') = \frac{1}{\alpha_{0}},$$

$$\sum_{n_{\mu}}^{N^{d}} g_{\mathfrak{N}}(n_{\mu}-n_{\mu}') = \frac{1}{\alpha_{\mathfrak{N}}}.$$ (B.3)

A similar sum with two chained factors of $g_{\mathfrak{N}}(n_{\mu}-n_{\mu}')$ can be calculated in a similar way. Using the Fourier expressions of $g_{0}(n_{\mu}-n_{\mu}')$ and $g_{0}(n_{\mu}-n_{\mu}'')$ we get

$$\begin{eqnarray*} \lefteqn { \sum_{n_{\mu}}^{N^{d}} g_{0}(n_{\mu}-n_{\mu}')\... ...mu}''n_{\mu}'')} } { [\rho^{2}(k_{\mu}'')+\alpha_{0}]^{2} }. \end{eqnarray*}$$

We see therefore that we get the sum expressed as a Fourier transform, with the general structure of a two-point function,

$$\sum_{n_{\mu}}^{N^{d}} g_{0}(n_{\mu}-n_{\mu}')\, g_{0}(n_... ...mu}'-n_{\mu}'')} } { [\rho^{2}(k_{\mu})+\alpha_{0}]^{2} }.$$ (B.4)

A similar result is true, of course, for the remaining field component

$$\sum_{n_{\mu}}^{N^{d}} g_{\mathfrak{N}}(n_{\mu}-n_{\mu}')\... ...}'')} } { [\rho^{2}(k_{\mu})+\alpha_{\mathfrak{N}}]^{2} }.$$ (B.5)

The squared dispersions, also referred to as widths or variances of the fields at a given site, are denoted as

$$\begin{eqnarray*} \sigma_{0}^{2} & = & \left\langle \varphi_{i}'^{2}(n_{\mu}... ...langle \varphi_{\mathfrak{N}}'^{2}(n_{\mu}) \right\rangle_{0}, \end{eqnarray*}$$

for $i=1,\ldots,\mathfrak{N}-1$. Using the expression of the two-point function in terms of Fourier components we may write these explicitly as

$$\sigma_{0}^{2} = \frac{1}{N^{d}} \sum_{k_{\mu}}^{N^{d}} \frac{1}{\rho^{2}(k_{\mu})+\alpha_{0}},$$ (B.6)

$$\sigma_{\mathfrak{N}}^{2} = \frac{1}{N^{d}} \sum_{k_{\mu}}^{N^{d}} \frac{1}{\rho^{2}(k_{\mu})+\alpha_{\mathfrak{N}}}.$$ (B.7)

In terms of these quantities the following decompositions of higher-point functions can be established, always for $i=1,\ldots,\mathfrak{N}-1$,

$$\left\langle \varphi_{i}'^{4}(n_{\mu}) \right\rangle_{0} = 3\sigma_{0}^{4},$$

$$\left\langle \varphi_{\mathfrak{N}}'^{4}(n_{\mu}) \right\rangle_{0} = 3\sigma_{\mathfrak{N}}^{4},$$ (B.8)

$$\left\langle \varphi_{i}'^{3}(n_{\mu}) \varphi_{i}'(n_{\mu}') \right\rangle_{0} = 3\sigma_{0}^{2}\, g_{0}(n_{\mu}-n_{\mu}'),$$

$$\left\langle \varphi_{\mathfrak{N}}'^{3}(n_{\mu}) \varphi_{\mathfrak{N}}'(n_{\mu}') \right\rangle_{0} = 3\sigma_{\mathfrak{N}}^{2}\, g_{\mathfrak{N}}(n_{\mu}-n_{\mu}'),$$ (B.9)

$$\left\langle \varphi_{i}'^{2}(n_{\mu}) \varphi_{i}'(n_{\mu}') \varphi_{i}'(n_{\mu}'') \right\rangle_{0} = \sigma_{0}^{2}\, g_{0}(n_{\mu}'-n_{\mu}'') +$$

$$+ 2\, g_{0}(n_{\mu}-n_{\mu}')\, g_{0}(n_{\mu}-n_{\mu}''),$$ (B.10)

$$\left\langle \varphi_{\mathfrak{N}}'^{2}(n_{\mu}) \varphi_{\mathfrak{N}}'(n_{\mu}') \varphi_{\mathfrak{N}}'(n_{\mu}'') \right\rangle_{0} = \sigma_{\mathfrak{N}}^{2}\, g_{\mathfrak{N}}(n_{\mu}'-n_{\mu}'') +$$

$$+ 2\, g_{\mathfrak{N}}(n_{\mu}-n_{\mu}')\, g_{\mathfrak{N}}(n_{\mu}-n_{\mu}''),$$ (B.11)

$$\left\langle \varphi_{i}'^{4}(n_{\mu}) \varphi_{i}'(n_{\mu}') \varphi_{i}'(n_{\mu}'') \right\rangle_{0} = 3\sigma_{0}^{4}\, g_{0}(n_{\mu}'-n_{\mu}'') +$$

$$+ 12\sigma_{0}^{2}\, g_{0}(n_{\mu}-n_{\mu}')\, g_{0}(n_{\mu}-n_{\mu}''),$$ (B.12)

$$\left\langle \varphi_{\mathfrak{N}}'^{4}(n_{\mu}) \varphi_{\mathfrak{N}}'(n_{\mu}') \varphi_{\mathfrak{N}}'(n_{\mu}'') \right\rangle_{0} = 3\sigma_{\mathfrak{N}}^{4}\, g_{\mathfrak{N}}(n_{\mu}'-n_{\mu}'') +$$

$$+ 12\sigma_{\mathfrak{N}}^{2}\, g_{\mathfrak{N}}(n_{\mu}-n_{\mu}')\, g_{\mathfrak{N}}(n_{\mu}-n_{\mu}'').$$ (B.13)

It is also not difficult to expand and calculate the following sums,

$$\begin{eqnarray*} \left\langle \left[ \sum_{i=2}^{\mathfrak{N}-1} \varphi_{i... ...ule{0em}{4ex} (\mathfrak{N}+1)(\mathfrak{N}-1) \sigma_{0}^{4}, \end{eqnarray*}$$

so that we get the results

$$\left\langle \left[ \sum_{i=2}^{\mathfrak{N}-1} \varphi_{i}'^{2}(n_{\mu}) \right]^{2} \right\rangle_{0} = \mathfrak{N}(\mathfrak{N}-2) \sigma_{0}^{4},$$ (B.14)

$$\left\langle \left[ \sum_{i=1}^{\mathfrak{N}-1} \varphi_{i}'^{2}(n_{\mu}) \right]^{2} \right\rangle_{0} = (\mathfrak{N}+1)(\mathfrak{N}-1) \sigma_{0}^{4}.$$ (B.15)