Question 3.14: The Magnetic Moment of the Proton Evaluate the right-hand si......
The Magnetic Moment of the Proton
Evaluate the right-hand side of (3.158) with the help of the wave functions (3.153) and (3.128).
{\mu}_{\mathrm{p}}=\int \mathrm{d}^3 r_1 \int \mathrm{d}^3 r_2 \int \mathrm{d}^3 r_3 \Psi_{\mathrm{p}}^{\dagger} \sum\limits_i\left(\frac{\hat{Q}_i}{2} \hat{{r}}_i \times \hat{{\alpha}}_i\right) \Psi_{\mathrm{p}} . (3.158)
\begin{gathered} \Psi_{\mathrm{p}}\left(m_s=\frac{1}{2}\right)=\frac{1}{3 \sqrt{2}}\left(2 u^{\uparrow}(1) u^{\uparrow}(2) d^{\downarrow}(3)-u^{\uparrow}(1) u^{\downarrow}(2) d^{\uparrow}(3)\right. \\ -u^{\downarrow}(1) u^{\uparrow}(2) d^{\uparrow}(3)-u^{\uparrow}(1) d^{\uparrow}(2) u^{\downarrow}(3)+u^{\uparrow}(1) 2 d^{\downarrow}(2) u^{\uparrow}(3) \\ -u^{\downarrow}(1) d^{\uparrow}(2) u^{\uparrow}(3)-d^{\uparrow}(1) u^{\uparrow}(2) d^{\downarrow}(3) \\ \left.-d^{\uparrow}(1) u^{\downarrow}(2) d^{\uparrow}(3)+2 d^{\downarrow}(1) u^{\uparrow}(2) d^{\uparrow}(3)\right) & (3.153) . \end{gathered}
\Psi=N\left(\begin{array}{c} j_{\ell}(E r) \chi_\kappa^\mu(\theta, \phi) \\ \mathrm{i} \operatorname{sgn}(\kappa) j_{\bar{\ell}}(E r) \chi_{-\kappa}^\mu(\theta, \phi) \end{array}\right) \mathrm{e}^{-\mathrm{i} E t} . (3.128)
The static magnetic moment is defined as
{m}=\frac{1}{2} \int {r} \times {j}(r) \mathrm{d}^{3} r. (1)
Using the definition of the electromagnetic current of quark i,
j_{i}^{\mu}=Q_{i} \bar{q}_{i} \gamma^{\mu} q_{i}, (2)
we get for the spatial component
\frac{1}{2} {r} \times {j}_{i}=\frac{Q_{i}}{2} \bar{q}_{i}({r} \times {\gamma}) q_{i}=\frac{Q_{i}}{2} q_{i}^{\dagger}({r} \times {\alpha}) q_{i} (3)
where \gamma^{0} {\gamma}={\alpha}, and therefore the operator of the magnetic moment, is
\frac{Q_{i}}{2}({r} \times {\alpha}) . (4)
Since the SU(6) wave function of the proton is symmetric under permutations of the indices 1,2 , and 3 , we have
\begin{aligned} \int \Psi_{\mathrm{p}}^{\dagger} & \left(\frac{\hat{Q}_{1}}{2} \hat{{r}}_{1} \times \hat{{\alpha}}_{1}\right) \Psi_{\mathrm{p}} \mathrm{d}^{3} r_{1} \mathrm{~d}^{3} r_{2} \mathrm{~d}^{3} r_{3} \\ & =\int \Psi_{\mathrm{p}}^{\dagger}\left(\frac{\hat{Q}_{2}}{2} \hat{{r}}_{2} \times \hat{{\alpha}}_{2}\right) \Psi_{\mathrm{p}} \mathrm{d}^{3} r_{1} \mathrm{~d}^{3} r_{2} \mathrm{~d}^{3} r_{3} \\ & =\int \Psi_{\mathrm{p}}^{\dagger}\left(\frac{\hat{Q}_{3}}{2} \hat{{r}}_{3} \times \hat{{\alpha}}_{3}\right) \Psi_{\mathrm{p}} \mathrm{d}^{3} r_{1} \mathrm{~d}^{3} r_{2} \mathrm{~d}^{3} r_{3} . & (5) \end{aligned}
Therefore (3.158) can in a first step be simplified to
\begin{aligned} {\mu}_{\mathrm{p}} & =3 \int \mathrm{d}^{3} r_{1} \mathrm{~d}^{3} r_{2} \mathrm{~d}^{3} r_{3} \Psi_{\mathrm{p}}^{\dagger}\left(\frac{\hat{Q}_{1}}{2} \hat{{r}}_{1} \times \hat{{\alpha}}_{1}\right) \Psi_{\mathrm{p}} \\ = & \frac{1}{6} \int \mathrm{d}^{3} r_{1}\left[10 u^{\uparrow}(1)^{\dagger}\left(\frac{\hat{Q}_{1}}{2} \hat{{r}}_{1} \times \hat{{\alpha}}_{1}\right) u^{\uparrow}(1)+2 u^{\downarrow}(1)^{\dagger}\left(\frac{\hat{Q}_{1}}{2} \hat{{r}}_{1} \times \hat{{\alpha}}_{1}\right) u^{\downarrow}(1)\right. \\ & \left.+4 d^{\downarrow}(1)^{\dagger}\left(\frac{\hat{Q}_{1}}{2} \hat{{r}}_{1} \times \hat{{\alpha}}_{1}\right) d^{\downarrow}(1)+2 d^{\uparrow}(1)^{\dagger}\left(\frac{\hat{Q}_{1}}{2} \hat{{r}}_{1} \times \hat{{\alpha}}_{1}\right) d^{\uparrow}(1)\right] . & (6) \end{aligned}
Here we have drawn on the orthogonality of the quark wave function
\int \mathrm{d}^{3} r q^{s \dagger} q^{\prime s^{\prime}}=\delta_{\mathrm{qq}^{\prime}} \delta_{\mathrm{ss}^{\prime}}, \quad q=u, d, \quad s=\uparrow, \downarrow . (7)
Next we insert the quark charges and make use of the fact that up and down quarks are described by the same spatial wave function. In addition we omit the index 1:
\begin{aligned} {\mu}_{\mathrm{p}}= & \frac{e}{6} \int \mathrm{d}^{3} r\left[\frac{10}{3} u^{\uparrow \dagger}(\hat{{r}} \times \hat{{\alpha}}) u^{\uparrow}+\frac{2}{3} u^{\downarrow \dagger}(\hat{{r}} \times \hat{{\alpha}}) u^{\downarrow}\right. \\ & \left.-\frac{2}{3} d^{\downarrow \dagger}(\hat{{r}} \times \hat{{\alpha}}) d^{\downarrow}-\frac{1}{3} d^{\uparrow \uparrow}(\hat{{r}} \times \hat{{\alpha}}) d^{\uparrow}\right] \\ = & \frac{e}{2} \int \mathrm{d}^{3} r u^{\uparrow \dagger}(\hat{{r}} \times \hat{{\alpha}}) u^{\uparrow} . & (8) \end{aligned}
In order to evaluate this expression we insert into it (3.128) in the form
u^{\uparrow}=N\left(\begin{array}{c} j_{0}(E r) \chi_{-1}^{\frac{1}{2}} \\ \mathrm{i} j_{1}(E r) \hat{\sigma}_{r} \chi_{-1}^{\frac{1}{2}} \end{array}\right) (9)
and obtain
\begin{aligned} \mu_{\mathrm{p}}= & \frac{e}{2} N^{2} \int_{0}^{R} \mathrm{~d} r r^{2} \int \mathrm{d} \Omega\left(j_{0}(E r) \chi_{-1}^{\frac{1}{2} \dagger},-\mathrm{i} j_{1}(E r) \chi_{-1}^{\frac{1}{2} \dagger} \hat{\sigma}_{r}\right) \\ & \times\left(\begin{array}{cc} 0 & \hat{{r}} \times \hat{\sigma} \\ \hat{{r}} \times \hat{\sigma} & 0 \end{array}\right)\left(\begin{array}{c} j_{0}(E r) \chi_{-1}^{\frac{1}{2}} \\ \mathrm{i} j_{1}(E r) \hat{\sigma}_{r} \chi_{-1}^{\frac{1}{2}} \end{array}\right) \\ = & \frac{e}{2} N^{2} \int_{0}^{R} \mathrm{~d} r r^{2} \int \mathrm{d} \Omega \mathrm{i} j_{0}(E r) j_{1}(E r) \chi_{-1}^{\frac{1}{2} \dagger}\left[(\hat{{r}} \times \hat{\sigma}) \hat{\sigma}_{r}-\hat{\sigma}_{r}(\hat{{r}} \times \hat{\sigma})\right] \chi_{-1}^{\frac{1}{2}} . & (10) \end{aligned}
By means of the commutation relations of the \sigma matrices, the term inside the brackets simplifies to
\begin{aligned} {\left[\varepsilon_{i j k} r_{j} \hat{\sigma}_{k}, \hat{\sigma}_{\ell} r_{\ell}\right] } & =\varepsilon_{i j k}\left[\hat{\sigma}_{k}, \hat{\sigma}_{\ell}\right] r_{j} r_{\ell} \\ & =\varepsilon_{i j k} 2 \mathrm{i} \varepsilon_{k \ell m} \hat{\sigma}_{m} r_{j} r_{\ell} \\ & =2 \mathrm{i}\left(\delta_{\ell i} \delta_{m j}-\delta_{\ell j} \delta_{i m}\right) r_{j} r_{\ell} \hat{\sigma}_{m} \\ & =2 \mathrm{i} r_{i} \hat{\sigma} \cdot {r}-2 \mathrm{i} r^{2} \hat{\sigma}_{i}. & (11) \end{aligned}
Hence
\begin{aligned} & \mu_{\mathrm{p}}=\frac{e}{2} N^{2} \int_{0}^{R} \mathrm{~d} r r^{2} \int \mathrm{d} \Omega \mathrm{i} j_{0}(E r) j_{1}(E r) \frac{1}{4 \pi}\left(\begin{array}{ll} 1 & 0 \end{array}\right)\left[2 \mathrm{i} {r} \sigma_{r}-2 \mathrm{i} r \sigma\right]\left(\begin{array}{l} 1 \\ 0 \end{array}\right) \\ & =\frac{e N^{2}}{4 \pi} \int_{0}^{R} \mathrm{~d} r r^{3} j_{0}(E r) j_{1}(E r) \int \mathrm{d} \Omega\left[{e}_{3}-\cos \theta\left(\begin{array}{c} \sin \theta \cos \phi \\ \sin \theta \sin \phi \\ \cos \theta \end{array}\right)\right], & (12) \end{aligned}
where we have used
\left(\sigma_{j}\right)_{11}=\delta_{j 3} (13)
and
\left(\sigma \cdot \frac{{r}}{r}\right)_{11}=\frac{r_{3}}{r}=\cos \theta . (14)
Now we can easily perform the \phi, \theta, and r integrations:
\begin{aligned} {\mu}_{\mathrm{p}} & =\frac{e N^{2}}{2} {e}_{3} \int\limits_{0}^{R} \mathrm{~d} r r^{3} j_{0}(E r) j_{1}(E r) \int\limits_{-1}^{1} \mathrm{~d} \cos \theta\left(1-\cos ^{2} \theta\right) \\ & ={e}_{3} \frac{2}{3} N^{2} e \int\limits_{0}^{R} \mathrm{~d} r r^{3} \frac{\sin (E r)}{E r}\left(\frac{\sin (E r)}{E^{2} r^{2}}-\frac{\cos (E r)}{E r}\right) \\ & ={e}_{3} \frac{2}{3} \frac{N^{2}}{E^{3}} e\left(-\frac{r}{2} \sin ^{2}(E r)-\frac{3}{4 E} \sin (E r) \cos (E r)+\frac{3}{4} r\right)_{0}^{R} \\ & ={e}_{3} \frac{2}{3} \frac{N^{2}}{E^{4}} e\left(-\frac{1}{2} \omega \sin ^{2} \omega-\frac{3}{4} \sin \omega \cos \omega+\frac{3}{4} \omega\right). & (15) \end{aligned}
Here \omega=E R. With the boundary condition
\omega \cos \omega=(1-\omega) \sin \omega (16)
the expression inside the brackets can further be simplified:
\begin{aligned} – & \frac{3}{4}\left(\omega \sin ^{2} \omega+\sin \omega \cos \omega-\omega\right)+\frac{1}{4} \omega \sin ^{2} \omega \\ & =-\frac{3}{4}\left(\sin ^{2} \omega-\omega \cos \omega \sin \omega+\sin \omega \cos \omega-\omega\right)+\frac{1}{4} \omega \sin ^{2} \omega \\ & =-\frac{3}{4}\left(\sin ^{2} \omega+\omega \cos ^{2} \omega-\omega\right)+\frac{1}{4} \omega \sin ^{2} \omega \\ & =-\frac{3}{4}(1-\omega) \sin ^{2} \omega+\frac{1}{4} \omega \sin ^{2} \omega \\ & =\frac{1}{4}(4 \omega-3) \sin ^{2} \omega . & (17) \end{aligned}
Finally we insert N^{2} from (3.142)
N_{\kappa=-1}^{2}={\frac{E R}{2R^{3}(E R-1)\;j_{0}^{2}(E R)}}\ , (3.142)
\begin{aligned} {\mu}_{\mathrm{p}} & ={e}_{3} \frac{\omega e}{3 R^{3}(\omega-1)} \frac{\omega^{2}}{\sin ^{2} \omega} \frac{1}{E^{4}} \frac{1}{4}(4 \omega-3) \sin ^{2} \omega \\ & ={e}_{3} \frac{e R}{\omega(\omega-1)} \frac{4 \omega-3}{12}=0.203 e {R}_{3} & (18) \end{aligned}
with \omega=2.04.
Clearly \mu_{\mathrm{p}} has to be parallel to the z direction, because in the spherical bag model the quantization axis of the angular momentum is the only direction of special importance.
Since \mu_{\mathrm{p}} defines a direction, this special direction can be chosen to be identical with the quantization axis mentioned. In units of the nuclear magneton \mu_{0}=e / 2 m_{\mathrm{p}} the absolute value of the magnetic moment of the proton is (with R=1 \mathrm{fm})
\begin{aligned} \frac{\left|\mu_{\mathrm{p}}\right|}{\mu_{0}} & =0.203 \times 1 \mathrm{fm} \times 2 \times 938.28 \mathrm{MeV}=1.9 \\ \mu_{0} & =\frac{e}{2 m_{\mathrm{p}}}. & (19) \end{aligned}