Question 3.13: The Mean Charge Radius of the Proton Problem. Evaluate the r......

Since r² does not change the quantum number m_s, all spin orientations yield the same value. Therefore we can restrict our calculation to one specific case,

\Psi_{\mathrm{p}}\left(m_{s}=\frac{1}{2}\right)\to\frac{1}{\sqrt{6}}\left(2u^{\uparrow}(1)u^{\uparrow}(2)d^{\downarrow }(3) -u^{\uparrow}(1)d^{\uparrow}(2)u^{\downarrow}(3)-d^{\uparrow}(1)u^{\uparrow}(2)u^{\downarrow}(3)\right)~~.    (2)

The charge operator {\hat{Q}}r^{2} acts on one quark at a time. For the proton it can be replaced by

\hat{Q}r^{2}\to Q_{1}r_{1}^{2}+Q_{2}r_{2}^{2}+Q_{3}r_{3}^{2}~.    (3)

Equation (1) then becomes

-u^{\uparrow}(1)d^{\uparrow}(2)u^{\downarrow}(3)-d^{\uparrow}(1)u^{\uparrow}(2)u^{\downarrow}(3)\Big)\;\;.     (4)

All quarks are assumed to be massless and in the same state (namely the 1s state), i.e., we do not distinguish between up and down quarks. Therefore r_{2}^{2} and r_{3}^{2} can be replaced in the integrand by r_{1}^{2}:

Q_{1}r_{1}^{2}+Q_{2}r_{2}^{2}+Q_{3}r_{3}^{2}\to(Q_{1}+Q_{2}+Q_{3})r_{1}^{2}=e r_{1}^{2}~.     (5)

Now the integrals over r_{2} and r_{3} can easily be evaluated by using the orthogonality relations

\int\mathrm{d}^{3}r_{2}\:q^{s\dagger}(2)\:q^{\prime s^{\prime}}(2)=\delta_{\mathrm{qq}^{\prime}}\,\delta_{s s^{\prime}}\ ,\quad\left\{\begin{array}{l l}{{q,\:\:q^{\prime}=u,\:d}}\\ {{s,\:s^{\prime}=\uparrow,\:\downarrow}}\end{array}\right.\ .   (6)

\langle r^{2}\rangle_{\mathrm{ch}}=\int\mathrm{d}^{3}r_{1}~r_{1}^{2}\frac{1}{6}\left(4u^{\uparrow \dagger}(1)u^{\uparrow}(1)+u^{\uparrow \dagger}(1)u^{\uparrow}(1)+d^{\uparrow \dagger}(1)d^{\uparrow}(1)\right)\ .   (7)

Owing to the assumption made above, the wave functions of u and d yield the same contributions, and (7) simplifies to

\langle r^{2}\rangle_{\mathrm{ch}}=\int d\Omega_{1}\int d r_{1}\,r_{1}^{4}\,u^{\uparrow \dagger}(1)u^{\uparrow}(1)\,\ .   (8)

Now we insert the explicit form of the wave function (3.128) and skip the index 1 in the remaining calculation:

\begin{array}{c}{{\displaystyle\mathcal{\Psi}=N\left(\begin{array}{c}{{j\ell\left(E r\right)\,\chi_{\kappa}^{\mu}\left(\theta,\phi\right)}}\\ {{\mathrm{i}\,\mathrm{sgn}(\kappa)\,j_{\bar{\ell}}\left(E r\right)\,\chi_{-\kappa}^{\mu}(\theta,\phi)}}\end{array}\right)\mathrm{e}^{-\mathrm{i}E t}~.}}\end{array}     (3.128)

\langle r^{2}\rangle_{\mathrm{ch}}=N^{2}\!\int\!\mathrm{d}r\!\int\!\mathrm{d}\Omega r^{4}\left[j_{0}^{2}(E r)\,\chi_{1}^{\frac{1}{2}\dagger}(\Omega)\,\chi_{1}^{\frac{1}{2}}(\Omega)+j_{1}^{2}(E r)\,\chi_{-1}^{\frac{1}{2}\dagger}(\Omega)\,\chi_{-1}^{\frac{1}{2}}(\Omega)\right]\ .   (9)

Because of the orthogonality of the spherical spinors and because of (3.142), (9) assumes the form

\langle r^{2}\rangle_{\mathrm{ch}}=\frac{E R}{2R^{3}(E R-1)j_{0}^{2}(E R)}\int_{0}^{R}d r\,r^{4}\left(j_{0}^{2}(E r)+j_{1}^{2}(E r)\right)\ .    (10)

The remaining integral can again be evaluated by means of the recursion relations or simply by inserting the explicit expressions

j_{1}(z)={\frac{\sin z}{z^{2}}}-{\frac{\cos z}{z}}~.    (11)

Hence

\times\Biggl[-r\sin^{2}(E r)+r-{\frac{1}{E}}\sin(E r)\cos(E r)+{\frac{1}{3}}r^{3}E^{2}\Biggr]_0^{R}   (12)

The boundary condition

⇒ ER cos(ER) = (1− ER) sin(ER)    (13)

simplifies (12) to give

=0.53\ R^{2}\ .   (14)

In the last step we have inserted the numerical value for ER, which, according to (3.131) is equal to 2.0428.

E R|_{\kappa=-1}\approx2.043,~~5.396,~~8.578,~\ldots~.    (3.131)