n') =  \int_{-\pi}^{\pi} \frac{\mathrm{d}^d q}{(2 \pi)^d} \,
\mathrm{e}^{\mathrm{i} q \cdot (n- n')} z_b (q),$$ with $q \cdot n=q_\mu n_\mu = q_1 n_1 + \cdots q_d n_d$ and $q_\mu$ the dimensionless momentum variable, yielding $$z_{b+1} (q) =  \sum_\mu \cos(q_\mu) z_b (q).$$ With $z_0 (q) = 1$ as initial condition, this gives for $z_b(n-n')$ $$\label{prob}
z_b (n-n') = \int_{-\pi}^{\pi} \frac{\mathrm{d}^d q}{(2
\pi)^d} \, \mathrm{e}^{\mathrm{i} q \cdot (n- n')} \left[
2\sum_\mu \cos (q_\mu) \right]^b.$$ In the limit where the lattice spacing $a$ is small but still finite, so that the particles still take discrete steps, we have $$\begin{aligned}
\left[2K\sum_\mu \cos (q_\mu) \right]^b &\rightarrow&
\left(\frac{1 - q^2/2d}{1+m^2 a^2/2d} \right)^b
\nonumber \\ &\rightarrow& \exp \left[- (q^2 + m^2 a^2) b /2 d \right].\end{aligned}$$ The lattice correlation function (\[Sigma\]) then becomes $$\begin{aligned}
\label{discrete}  
\!\!\!\!\!\!\!\! \left\langle \phi^\alpha_n \phi^\beta_{n'}
\right\rangle &\rightarrow& \delta_{\alpha, \beta} \frac{1}{2}
\sum_{b=0}^{\infty} \int_{-\pi}^{\pi} \frac{\mathrm{d}^d q}{(2 \pi)^d}
\, \mathrm{e}^{\mathrm{i} q \cdot (n-n')} \, \mathrm{e}^{- (q^2 + m^2
a^2) b /2d} \nonumber \\ &\rightarrow& \delta_{\alpha, \beta}
\frac{1}{2} \sum_{b=1}^{\infty}  \left( \frac{d}{2 \pi b}
\right)^{d/2} \exp \left[- \frac{d}{2} \frac{(n - n')^2}{ b}\right] \,
\mathrm{e}^{- m^2 a^2 b/2d},\end{aligned}$$ showing that the end-to-end vector $n-n'$ is Gaussian distributed with an extra Boltzmann weight $\exp(-m^2 a^2/2d)$ associated with each step taken. Because 