In order to calculate the spin angular momentum and magnetic moment, we first make some assumptions for the electron structure. Enlightenment can be achieved from the mass distribution of quarks. Particle physics tells us that the mass of up/down current quark (bare quark) is only several Mev, while the constituent mass of the quark is more than 300 Mev [13], so most of the quark mass comes from the effective masses of the sea quarks (virtual quark-antiquark pairs) and the virtual gluons. Similarly, we think that the electron is composed of point electric charge at the center, virtual electron-positron pairs inside the electron and virtual photons both inside and outside the electron, as shown in Fig. 2.

Figure 2 Illustration of the electron structure. -e is the point electric charge

We assume that the rest energy of the electron originates from the kinetic energies of the virtual electron-positron pairs and virtual photons. The virtual electron-positron pairs are induced by the point electric charge at the center of the electron; while the virtual photons are induced by the motion of the virtual electron-positron pairs. The virtual photons constitute the electromagnetic fields of the electron. There also exist virtual electron-positron pairs outside the electron due to vacuum polarization of the electromagnetic fields. But there is difference between the two instances. Inside the electron, there is a steady distribution of virtual electron-positron pairs due to the induction of the point electric charge. While outside the electron there is no inducing source, the presence of the virtual electron-positron pairs relies on the vacuum polarization, whose probability may be very small (as calculated in the later section). In the case of *R* = *ƛ*_{c}, the energy of the electromagnetic fields is the order of *α* (fine structure constant) of the electron rest energy, so the rest energy of electron mainly comes from the energy of the virtual electron-positron pairs.

Let the effective mass of the virtual electron-positron pairs be *m*_{p} and their speeds be *c*. For the motion of the virtual electron-positron pairs, it results that
$$\begin{array}{}{p}^{2}{c}^{2}={m}_{p}^{2}{c}^{4}.\end{array}$$(3.1)

The above equation seems plausible at first glance. However, it is not the case. The electromagnetic fields are generated by the motion of the virtual electron-positron pairs, so the masses of the electromagnetic fields and the virtual electron-positron pairs are tied together and cannot be separated from each other. To change the motion state of the virtual electron-positron pairs implies to change the distribution of electromagnetic fields, so the effective mass of the electromagnetic fields should also be included in the inertial mass of the virtual electron-positron pairs. This is just like the instance where the earth moves with its aerosphere whose mass should also be included in the inertial mass of the earth. Thus the mass of the virtual electron-positron pairs in Eq. (3.1) should be replaced by the observable mass of the electron *m*_{0}.

In fact, if we think that the overall rest energy of the electron originates from the motion of the virtual particles, it directly leads to
$$\begin{array}{}{p}^{2}{c}^{2}={m}_{0}^{2}{c}^{4},\end{array}$$(3.2)

without considering the internal structure of the electron. Quantizing the equation, i.e., substituting *p*^{2} with operator −ℏ^{2}*Δ*, we find
$$\begin{array}{}-{\hslash}^{2}\mathit{\Delta}\psi ={m}_{0}^{2}{c}^{2}\psi ,\end{array}$$(3.3)

that is,
$$\begin{array}{}{\displaystyle \frac{{\mathrm{\partial}}^{2}\psi}{\mathrm{\partial}{x}^{2}}+\frac{{\mathrm{\partial}}^{2}\psi}{\mathrm{\partial}{y}^{2}}+\frac{{\mathrm{\partial}}^{2}\psi}{\mathrm{\partial}{z}^{2}}+\frac{{m}_{0}^{2}{c}^{2}}{{\hslash}^{2}}\psi =0.}\end{array}$$(3.4)

This is the wave equation for the electron spin. To solve the above equation, we refer to the wave equation of the orbital motion of the electron in hydrogen atom, which is
$$\begin{array}{}{\displaystyle \mathit{\Delta}\psi +\frac{2{m}_{0}}{{\hslash}^{2}}(E+\frac{{e}^{2}}{4\pi {\epsilon}_{0}}\frac{1}{r})\psi =0.}\end{array}$$(3.5)

It can be seen that the wave equation of the spin motion is much simpler that that of the orbital motion. Following the solution to the wave equation of the orbital motion, we adopt separation of variables method. The spin motion is decomposed into angular motions and radial motion. The solutions to the spin angular wave equations are the same as those of orbital angular wave equations. The radial wave equation for the orbital motion is [1]
$$\begin{array}{}{\displaystyle \frac{{d}^{2}R(r)}{d{r}^{2}}+\frac{2}{r}\frac{dR(r)}{dr}+[\frac{2{m}_{0}}{{\hslash}^{2}}E+\frac{2{m}_{0}}{{\hslash}^{2}}\frac{{e}^{2}}{4\pi {\epsilon}_{0}}\frac{1}{r}-\frac{l(l+1)}{{r}^{2}}]R(r)=0.}\end{array}$$(3.6)

Similarly, the wave equation for the spin motion can be written as
$$\begin{array}{}{\displaystyle \frac{{d}^{2}R(r)}{d{r}^{2}}+\frac{2}{r}\frac{dR(r)}{dr}+\frac{{m}_{0}^{2}{c}^{2}}{{\hslash}^{2}}R(r)-\frac{l(l+1)}{{r}^{2}}R(r)=0.}\end{array}$$(3.7)

Eq. (3.7) is spherical Bessel equation. Before starting solving the equation, we first see how to derive the values of angular momentum and magnetic moment with wave function method. In the spherical coordinate system, the gradient operator is
$$\begin{array}{}{\displaystyle \mathrm{\nabla}={\mathbf{e}}_{\mathbf{r}}\frac{\mathrm{\partial}}{\mathrm{\partial}r}+{\mathbf{e}}_{\theta}\frac{1}{r}\frac{\mathrm{\partial}}{\mathrm{\partial}\theta}+{\mathbf{e}}_{\varphi}\frac{1}{r\mathrm{sin}\theta}\frac{\mathrm{\partial}}{\mathrm{\partial}\varphi}.}\end{array}$$(3.8)

The probability current density in quantum mechanics is
$$\begin{array}{}{\displaystyle j=\frac{\hslash}{2{m}_{0}i}[{\psi}^{\ast}\mathrm{\nabla}\psi -\psi \mathrm{\nabla}{\psi}^{\ast}].}\end{array}$$(3.9)

The above expression applies to non-relativistic particles, but it can be verified that it also holds for the particles with the speed of light. In this case, *m*_{0} denotes the effective mass. As a particle rotates around the specific z axis, as shown in Fig. 3, we obtain
$$\begin{array}{}{\displaystyle {j}_{\varphi}=\frac{\hslash}{2{m}_{0}i}[R(r){Y}_{lm}^{\ast}(\theta ,\varphi )\frac{1}{r\mathrm{sin}\theta}\frac{\mathrm{\partial}}{\mathrm{\partial}\varphi}(R(r){Y}_{lm}(\theta ,\varphi ))}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-R(r){Y}_{lm}(\theta ,\varphi )\frac{1}{r\mathrm{sin}\theta}\frac{\mathrm{\partial}}{\mathrm{\partial}\varphi}(R(r){Y}_{lm}^{\ast}(\theta ,\varphi ))]}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\frac{m\hslash}{{m}_{0}r\mathrm{sin}\theta}{R}^{2}(r)|{Y}_{lm}(\theta ,\varphi ){|}^{2}.}\end{array}$$(3.10)

Figure 3 The rotation of a particle around z axis

It can be seen from Fig. 3 that the differential angular momentum is *dJ*_{z} = *m*_{0} *j*_{ϕ} r sin *θ dV*, where *dV* is the differential volume of the thin circular ring with the cross-sectional area of *dσ* = *rdθ dr* and the perimeter of 2*π r* sin *θ*. The overall angular momentum is
$$\begin{array}{}{\displaystyle {J}_{z}=\underset{V}{\int}d{J}_{z}=m\hslash \underset{V}{\int}|\psi {|}^{2}dV=m\hslash}\end{array}$$(3.11)

We then turn to the calculation of magnetic moment. When *ej*_{ϕ} is multiplied by the differential area *dσ*, where *e* is the charge of the electron, we get the differential current *dI* = *erj*_{ϕ} drdθ, which generates the magnetic moment of *dμ*_{z} = *π r*^{2}sin^{2}*θ dI*. The overall magnetic moment for the spinning electron is
$$\begin{array}{}{\displaystyle {\mu}_{z}=\int e\pi {r}^{2}{\mathrm{sin}}^{2}\theta \frac{m\hslash}{{m}_{0}r\mathrm{sin}\theta}|\psi {|}^{2}d\sigma}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle =\frac{em\hslash}{2{m}_{0}}\int 2\pi r\mathrm{sin}\theta |\psi {|}^{2}d\sigma}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle =\frac{em\hslash}{2{m}_{0}}\underset{V}{\int}|\psi {|}^{2}dV=\frac{em\hslash}{2{m}_{0}}=m{\mu}_{B}.}\end{array}$$(3.12)

The above arguments may refer to Ref. [1]. For our electron spin model, if we assume a quantum number *l* = 1/2 for angular momentum while *l* = 1 for magnetic moment, we get the correct values. The reason that we use two different quantum numbers for the electron spin is that the mass distribution and the charge distribution inside the electron are different based on the model of Fig. 2. In contrast, the quantum numbers of the angular momentum and the magnetic moment can be regarded as equal for the orbital motion of the electron. This is because the electron radius is much smaller compared to the orbital radius; the orbital motion around the nucleus may be regarded as the motion of a point particle; while the spin magnetic moment is due to the rotation of electric dipole, which is the deep reason for a Landé *g*-factor of 2. For the electron structure model of Fig. 2, our goal is to solve for the distributions of mass and charge inside the electron. As we are more interested in the charge distribution, we solve the spherical Bessel equation for *l* = 1. In this case, the two specific solutions are
$$\begin{array}{}{\displaystyle {J}_{1}(r)=\frac{\mathrm{sin}r}{{r}^{2}}-\frac{\mathrm{cos}r}{r},\phantom{\rule{1em}{0ex}}{n}_{1}(r)=-\frac{\mathrm{cos}r}{{r}^{2}}-\frac{\mathrm{sin}r}{r},}\end{array}$$(3.13)

respectively. Now consider the boundary condition at *r* = 0, where we should have *rψ* (*r*) → 0. It can be seen that *rJ*_{1} (*r*) → 0 and *rn*_{1} (*r*) → ∞, so we may assume *R*(*r*) = *k*_{1} *J*_{1} (*r*), where *k*_{1} is a constant and can be derived by the normalization condition of the wave function. As
$$\begin{array}{}{\displaystyle \int {r}^{2}|R(r){|}^{2}dr}\\ ={k}_{1}[r(3+{r}^{2})/6+3r\mathrm{cos}2r/4+(2{r}^{2}-5)\mathrm{sin}2r/8+C],\end{array}$$(3.14)

if the upper limit of the integral is infinity, the above expression will diverge, so the electron must have a definite radius, which is obviously different from the orbital motion that may extend to infinity. We assume a radius of *ƛ*_{c} for the electron. As *r* is very small, we expand *R*(*r*) into power series. As
$$\begin{array}{}{\displaystyle \mathrm{sin}r\approx r-\frac{{r}^{3}}{6},\phantom{\rule{1em}{0ex}}\mathrm{cos}r\approx 1-\frac{{r}^{2}}{2},}\end{array}$$(3.15)

we obtain
$$\begin{array}{}{\displaystyle \underset{0}{\overset{{\lambda}_{c}^{\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}-}}{\int}}{k}_{1}^{2}\frac{{r}^{4}}{9}dr=1.}\end{array}$$(3.16)

It follows that
$\begin{array}{}{k}_{1}^{2}=45/{{\lambda}^{\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}-}}_{c}^{5},\end{array}$ and
$$\begin{array}{}{\displaystyle R(r)=\sqrt{\frac{45}{{{\lambda}^{\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}-}}^{5}}}(\frac{\mathrm{sin}r}{{r}^{2}}-\frac{\mathrm{cos}r}{r})\approx \sqrt{\frac{5}{{{\lambda}^{\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}-}}^{5}}}r.}\end{array}$$(3.17)

The wave function for the charge distribution inside the electron is
$$\begin{array}{}{\displaystyle \psi (r,\theta ,\varphi )=\frac{\sqrt{3}}{2\sqrt{2\pi}}R(r)\mathrm{sin}\theta {e}^{i\varphi}\approx \frac{1}{2}\sqrt{\frac{15}{2\pi {{\lambda}^{\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}-}}^{5}}}r\mathrm{sin}\theta {e}^{i\varphi}.}\end{array}$$(3.18)

It can be seen that inside the electron the charge density increases with *r* and reaches a maximum value at the surface. We know that the classical motion of a particle corresponds to the motion state with a maximum probability in quantum mechanics, so the thin circular plate model can obtain the correct magnetic moment. In the meanwhile, it is interesting to see that the charge density becomes zero at the center, which shows that the electric charge of the point charge is totally shielded by the virtual positrons.

Now consider the instance of *l* = 0, which represents the non-polarized state of the electron. In this case, Eq. (3.7) can be written as
$$\begin{array}{}{\displaystyle \frac{{d}^{2}R(r)}{d{r}^{2}}+\frac{2}{r}\frac{dR(r)}{dr}+\frac{{m}_{0}^{2}{c}^{2}}{{\hslash}^{2}}R(r)=0,}\end{array}$$(3.19)

whose two specific solutions are
$\begin{array}{}\frac{1}{r}\mathrm{cos}\frac{r}{{{\lambda}^{\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}-}}_{c}}\text{\hspace{0.17em}and\hspace{0.17em}}\frac{1}{r}\mathrm{sin}\frac{r}{{{\lambda}^{\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}-}}_{c}}\end{array}$ respectively. Due to the boundary condition of *rψ* (*r*) → 0 at *r* = 0, we can only take the latter. It results that
$$\begin{array}{}{\displaystyle R(r)=\frac{{C}_{1}}{r}\mathrm{sin}\frac{r}{{{\lambda}^{\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}-}}_{c}}.}\end{array}$$(3.20)

From the normalization condition of the wave function we find
$\begin{array}{}{C}_{1}=1/\sqrt{{{\lambda}^{\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}-}}_{c}(\frac{1}{2}-\frac{\mathrm{sin}2}{4})}.\end{array}$ In the case of *l* = 0, *Y*(*θ*,*ϕ* ) =
$\begin{array}{}{Y}_{00}(\theta ,\varphi )=\frac{1}{2\sqrt{\pi}},\end{array}$ thus the spin wave function is
$$\begin{array}{}{\displaystyle \psi (r,\theta ,\varphi )=\frac{{C}_{1}}{2\sqrt{\pi}}\frac{1}{r}\mathrm{sin}\frac{r}{{{\lambda}^{\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}-}}_{c}}.}\end{array}$$(3.21)

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