Implementation of \(Z_{00}(1,q^2)\)

In collab. with Max Hansen.

The following derivation in largely inspired by Flavor physics and lattice quantum chromodynamics, L.Lellouch

We want to evaluate the so-called Luescher Zeta function

\[\sqrt{4 \pi} Z_{00}(s,q^2) = \sum_{\vec n} \frac{1}{(|\vec n|^2 - q^2)^s} = \sum_k \nu_{k,D} \frac{1}{(k- q^2)^s}\]

We note that the proper regularized version is defined as follows

\[\sum_{\vec n} \frac{1}{(|\vec n|^2 - q^2)^s} - PV \int d^3 x \frac{1}{(|\vec x|^2 - q^2)^s}\]

but here we consider the analytic continuation of the former equation. We are interested in the derivatives (for error propagation)

\[\frac{\partial Z_{00}(s,q^2)}{\partial q^2} = \frac{1}{\sqrt{4\pi}} \sum_{\vec n} \frac{s}{(|\vec n|^2 - q^2)^{s+1}} = s Z_{00}(s+1, q^2)\]
\[\frac{\partial^2 Z_{00}(s,q^2)}{\partial (q^2)^2} = \frac{1}{\sqrt{4\pi}} \sum_{\vec n} \frac{s(s+1)}{(|\vec n|^2 - q^2)^{s+2}} = s(s+1) Z_{00}(s+2, q^2)\]

in the limit \(s \to 1\), and possibly for \(D\) dimensions.

Using the integral representation of the special Gamma funtion

\[\Gamma(s) a^{-s} = \int_0^\infty dt \, e^{-at} \, t^{s-1}\]

we obtain

\[ \sqrt{4 \pi} Z_{00}(s,q^2) \Gamma(s) = \sum_{\vec n} \int dt e^{-(|\vec n|^2 - q^2) t } t^{s-1} = \sum_{\vec n} \Big[ \int_0^\mu + \int_\mu^\infty \Big] dt e^{-(|\vec n|^2 - q^2) t } t^{s-1} = I_0(s,q^2) + I_1(s,q^2)\]

and we split the integral in two regions. Using this integral representation the higher derivatives become

\[\frac{\partial Z_{00}(s,q^2)}{\partial q^2} = \frac{s}{\sqrt{4 \pi} \Gamma(s+1)} [I_0(s+1, q^2) + I_1(s+1, q^2)]\]
\[\frac{\partial^2 Z_{00}(s,q^2)}{\partial (q^2)^2} = \frac{s(s+1)}{\sqrt{4 \pi} \Gamma(s+2)} [I_0(s+2, q^2) + I_1(s+2, q^2)]\]

The second piece

\[I_1(s,q^2) = \sum_{\vec n} \int_\mu^\infty dt e^{-(|\vec n|^2 - q^2) t } t^{s-1} = \sum_k \nu_{k,D} \int_\mu^\infty dt e^{-(k - q^2) t } t^{s-1}\]

is easily calculable for the limits \(s=1,2,3\) and for any dimension \(D\) (we drop the subscript in \(\nu_{k,D}\))

\[ \lim_{s \to 1} I_1(s,q^2) = \sum_{k=0}^\infty \nu_k e^{-\mu(k - q^2)}\frac{1}{k - q^2}\]
\[\lim_{s \to 2} I_1(s,q^2) = \sum_k \nu_k e^{-\mu (k^2 - q^2)} \frac{1 + (k-q^2) \mu}{(k - q^2)^2}\]
\[\lim_{s \to 3} I_1(s,q^2) = \sum_k \nu_k e^{-\mu (k^2 - q^2)} \frac{2 + 2 (k-q^2) \mu + (k-q^2)^2 \mu^2}{(k - q^2)^3}\]

where we have used $ k =0 ,1, 2, \dots `$. In our code, depending on the number of dimensions :math:`D we loop only over the values of \(k\) whose \(\nu_{k,D}\) is non-zero.

To evaluate the first integral \(I_0\) we use Poisson’s summation formula

\[\sum_{\vec n} f(\vec n) = \sum_{\vec n} \int d^D x f(\vec x) e^{2 \pi i \vec n \cdot \vec x}\]

to get

\[ I_0(s,q^2) = \sum_{\vec n} \int_0^\mu dt e^{q^2 t} e^{-|\vec n|^2 t} t^{s-1} = \int_0^\mu dt e^{q^2 t} t^{s-1} \sum_{\vec n} \int d^D x e^{-|\vec x|^2 t} e^{2 \pi i \vec n \cdot \vec x}\]

Using the following relation

\[\int d x e^{-x^2 a} e^{2 \pi i b x} = \sqrt{\pi/a} \, e^{-\frac{\pi^2 b^2}{a}}\]

we can solve the integral in \(D\) dimensions

\[\sum_{\vec n} \int d^D x e^{-|\vec x|^2 t} e^{2 \pi i \vec n \cdot \vec x} = \sum_k \nu_{k,D} \frac{\pi^{D/2}}{t^{D/2}} e^{- \pi^2 k/t}\]

leading to the following expression

\[I_0(s,q^2) = \sum_k \nu_{k,D} \pi^{D/2} \int_0^\mu dt e^{q^2 t} e^{- \pi^2 k/t} t^{s-1-\frac{D}{2}}\]

We note that for \(k=0\) and \(s-\frac{D}{2} \leq 0\) the integral is divergent because

\[\int_0^\mu dt e^{q^2 t} t^{s-1-\frac{D}{2}} = \int_0^\mu dt \Big[\frac1t + q^2 + O(t) \Big] t^{s-\frac{D}{2}}\]

Isolating the \(k=0\) case we regulate the integral

\[\int_0^\mu dt (e^{q^2 t} -1) t^{s-1-\frac{D}{2}} + \int_0^\mu dt t^{s-1-\frac{D}{2}}\]

and simply analytically continue the second part (which is finite for \(s-\frac{D}{2}>0\))

\[\int_0^\mu dt \ t^{s-1-\frac{D}{2}} = 2\frac{\sqrt{\mu}^{2s-D}}{2s-D}\]

The final result for the first integral is

\[\frac{1}{\sqrt{4 \pi}} I_0(s,q^2) = \frac{\pi^{(D-1)/2}}{2} \int_0^\mu dt (e^{q^2 t} -1) t^{s-1-\frac{D}{2}} + \pi \frac{\sqrt{\mu}^{2s-D}}{2s-D} + \frac{\pi}{2} \sum_{k \neq 0} \nu_{k,D} \int_0^\mu dt e^{q^2 t - \pi^2 k/t} t^{s-1-\frac{D}{2}}\]

Putting all results together we obtain the following representation of the L”uescher zeta function

\[Z_{00}(1,q^2) = \frac{1}{\sqrt{4 \pi}} [I_0(1,q^2) + I_1(1,q^2)]\]
\[\frac{\partial Z_{00}(s,q^2)}{\partial q^2} = \frac{1}{\sqrt{4 \pi}} [I_0(2, q^2) + I_1(2, q^2)]\]
\[\frac{\partial^2 Z_{00}(s,q^2)}{\partial (q^2)^2} = \frac{1}{\sqrt{4 \pi}} [I_0(3, q^2) + I_1(3, q^2)]\]

We remark that the freedom given by the parameter \(\mu\) is very important to keep the evaluation of \(Z_{00}\) for high values of \(q^2\) under control from the numerical point of view (rounding errors and convergence). We find beneficial to set \(\mu = (q^2)^{1/2}\).