27.17 Lattice dispersion relation g=0.1

given the parametrization

$$\frac{k}{m} \cot{ \delta}=\frac{1}{am}+\frac{r_0 k^2}{2m}\,,$$ with $P[0]=am$ , $P[1]=r_0 m$ , $P[2]=P_2$. We solve the quantization condition finding the value of $k$ such that $$\cot{ \delta}=\frac{Z_{00}(1,q^2)}{\pi^{3/2}\gamma q}.$$

From k we compute the two particle energy $E_2$

$$k^2=\frac{E_{CM}^2}{4}-m^2 = \frac{E^2-\vec{P}^2}{4}-m^2.$$

and finally the energy shift $$\Delta E_2^{predicted} = E_2 - \sqrt{4m^2 + p_1^2}-\sqrt{4m^2 + p_2^2}$$

On the other hand we measure the two particle energy and we compute the energy shift with the lattice dispersion relation $$\Delta E_2^{latt}=E_2^{measured}-E_2^{free-latt}$$ $$\begin{gather} E_2^{free-latt} = \cosh^{-1}{\left( \cosh(m) +\frac{1}{2}\left( \sum_{i=1}^{2}4 \sin\left(\frac{ p_{1i}}{2}\right)^2\right)\right)} \\ + \cosh^{-1}{\left( \cosh(m) +\frac{1}{2}\left( \sum_{i=1}^{2}4 \sin\left(\frac{ p_{2i}}{2}\right)^2\right)\right)} \,. \tag{27.1} \end{gather}$$

Finally we minimise the $\chi^2$ $$\chi^2= \sum_i \frac{( \Delta E_2^{predicted} -\Delta E_2^{latt})^2}{\sigma^2}$$

$$\begin{gather} \chi^2/d.o.f.=0.912112 \\ P[0]=-0.120481\pm (0.0015) \\ P[1]=-4.24382\pm (0.13) \\ \end{gather}$$ { $$\begin{gather} C=\begin{pmatrix} 2.19e-06& 0.497\\ 0.497& 0.0159\\ \end{pmatrix} \end{gather}$$ }

png 2

the resulting $\delta$ determined is