Klein Gordon and the Bohr model

We will derive an alternative way modelling Hydrogene like atoms or ions, at least for simple energy levels, using the classical relativistic quantum mechanics as building block and show that they produce Energy level estimates as well as indicating a connection to more classical approaches like the Bohr model and Randell Mills GUTCP. The paper uses the Klein Gordon equation nad divedes the space in a interior sphere $$r<r_0$$ where there is no charge and all charge at the sphere's surface. Then demanding that the solution to the equation sould have a node lead to the classical result that the kinetic energy is half the potential and a formula for the radius and finally very exact energy levels. We will compare with Meassurements, GUTCP, Dirac solution.

We will keep it simplistic (and avoid spinors) and use The Klein Gordon equation for hydrogen. Let

$$C = \frac{Ze^2}{4\pi\epsilon_0}.$$

Then the modified Klein Gordon equation for the electrical potential is

$$\Big( -(\hat h \frac{\partial}{\partial t} + i C/r)^2 + \hat h^2\frac{\partial^2}{\partial x_1^2} + \hat h^2c^2\frac{\partial^2 }{\partial x_2^2} + \hat h^2\frac{\partial^2}{\partial x_3^2} \Big) \varphi = m^2c^4\varphi,$$

with $$\hat h = \hbar/2/\pi$$

The spherical symmetric solutions (for constant $$V$$) can be expressed as the following solution,

$$\varphi = \exp(i t E/\hat h) \int_{|e|=1} \exp(i k (x \cdot e)) \, de$$

This solves the Klein Gordon equation if,

$$\hat h^2 c^2 k_2^2 = (E_2+C/r_0)^2-m^2c^4.$$

If the charge is zero below $$r$$ then in stead we have the equation

$$\hat h^2 c^2 k_1^2 = E_1^2-m^2c^4.$$

The solution is actually of the form

$$\varphi = Aj_0(kr)+ B\eta_0(kr),$$

but the innner part must be regular at 0 hence the solution is essentially

$$Aj_0(kr) = A\frac{\sin(kr)}{kr}$$

Now the frequencyes will be assumed different internally and externally, but we will wave that away by claiming that the jump in the derivative will yield the charge density (associating to the Maxwells equations) and that by selecting $$kr=2\pi n, \quad n=1,2,3,4,...$$ we will not radiate and be able to conseve energy.

Hence the condition

$$|k_1|r_0 = 2\pi n_1, \quad n_1 = 1, 2, 3, \ldots .$$

Similarly for the charge part we will have,

$$|k_2|r_0 = 2\pi n_2, \quad (n_2 = 1, 2, 3, \ldots .$$

For $$n_1=n_2$$ we get the condition that

$$k_1=k_2.$$

And hence,

$$E_1 = m_e c^2 - E',$$
$$E_2 = m_e c^2 + E',$$

and

$$E' = C/r_0/2.$$

So we conclude then that

$$k_2r_0 = \sqrt{(m_ec^2 + C/2/r_0)^2-(m_ec^2)^2}/(\hat h c) = 2 \pi n_2.$$

Or using $$\hat h = \hbar/2/\pi)$$, $$C'=C/2$$,

$$\sqrt{(m_ec^2 + C'/r_0)^2-(m_ec^2)^2}/(\hbar c) = n_2 = n.$$

Solving for $$r_0$$ now leads to the equation

$$r_0 = \frac{(n \hbar c)^2-C'^2}{2C'm_{e}c^2}$$

Now if $$m_n$$ is the mass of the nucleus, then we find that as usually done, reduce the mass, via

$$m_\mu = \frac{m_e}{1+\frac{m_e}{m_n}}$$

And now we can express the radius with the help of the reduced mass as

$$r_0 = \frac{(n \hbar c)^2-C'^2}{2C'm_{\mu}c^2}.$$

The energy then of the shell is,

$$E' = C'/r_0.$$

To compare we will use GUTCP calculated values, Meassured values and values clauclated from Dirac via

$$E_{dirac} = m_\mu c^2\Big (1 - \frac{1}{\sqrt{1 + \frac{Z^2 \alpha^2}{\Big (n - j + 1/2 + \sqrt{(j + 1/2)^2 - Z^2\alpha^2} \Big)^2}}}\Big )$$

Lets find values for some of the nature constants. Take $$m_e=9.1093837015\cdot 10^{-31}$$ and $$m_p=1.67262192369\cdot 10^{-27}$$. It is standard to use the reduced mass which then is,

$$m_\mu = \frac{m_e}{1+\frac{m_e}{m_p}} = 9.104425276523571\cdot 10^{31}.$$

Electron charge $$e=1.602176634\cdot 10^{-19}$$. Plank constant $$h=6.62607015\cdot 10^{-34}$$. And hence $$\hbar = h/(2\pi) = 1.0545718176\cdot 10^{-34}$$ Measured ground state, $$|V|/2 \approx 13.59844eV$$, $$c=299792458$$. The bohr radius, $$a_0=5.29177210903\cdot 10^{-11}$$. Take $$\epsilon_0=8.8541878128 \cdot 10^{-12}$$. Then we will find that the theoretical value for $$r_0$$, when $$n=1$$ and then calculate the energy via,

$$E'_{theoretical} = 13.59847eV,$$

with

$$E'_{measured} = 13.59844eV,$$

So this looks better than the usual calculation with Dirac that result in

$$E'_{old theoretical} = 13.5991eV.$$

Mills also calculate this using classical means and get,

$$E'_{GUTCP} = 13.59847eV.$$

For this calculation we had $$r_0$$,

$$r_0 = \frac{(n\hbar c)^2-C'^2}{2C'm_{mu}c^2} = 5.294583611\cdot 10^{-11} \approx a_0 = 5.29177210903\cdot 10^{-11}.$$

We can look at different excited states,

\begin{array}{l|l|l|l} \textbf{} & \textbf{Dirac [eV]} & \textbf{Theoretical [eV]} & \textbf{Meassured (NIST) [eV]}\\ \hline s1 & 13.5992 & 13.598468 & 13.598434 \\ s2 & 3.39972 & 3.39958 & 3.39962 \\ s3 & 1.510968 & 1.510923 & 1.510940 \\ s4 & 0.849913 & 0.849894 & 0.849902 \\ \end{array}

And also compare for various energy levels for ions,

\begin{array}{l|l|l|l|l} \textbf{} & \textbf{Dirac [eV]} & \textbf{This [eV]} & \textbf{GUTCP [eV]} & \textbf{Meassured [eV]}\\ \hline H & 13.598468 & 13.598468 & 13.59847& 13.598434 \\ He^+ & 54.418266 & 54.418266 & 54.41826 & 54.4178 \\ Li^{2+} & 122.45630 & 122.45630 & 122.45637 & 122.4543 \\ Be^{3+} & 217.7243 & 217.7243 & 217.72427 & 217.7186 \\ B^{4+} & 340.2385 & 340.2384 & 340.23871 & 340.2260 \\ C^{5+} & 490.0177 & 490.0176 & 489.01759 & 489.9931 \\ N^{6+} & 667.0885 & 667.0882 & 667.08834 & 667.0461 \\ O^{7+} & 871.4778 & 871.4772 & 871.47768 & 871.3410 \\ U^{91+} & 132.2k & 129.8k & 132.3k & 131.8k \\ \end{array}

Anyhow using the correspondance principle for the shell we are in a sense motivate why a classical approch can be quite successful and why Bohr succeded as well as why Randell Mills GUTCP work out. Note also how the Rydberg series is expained through this calculation as well.

To calculate the magnetic interaction from the nucleus we could start with assuming two concentric current loops a small one for the proton and a large one for the electron. Now we could consider the proton as a Magnet. And find the dipole moment as

$$p = \pi * r_p^2I_p$$

Now we also have

$$m_p r_p v_p = \hbar,$$

or

$$v_p = \hbar/r_p/m_p.$$

And hence

$$I_p = v_p e / (2 \pi r) = \hbar e /(r_p^2 m_p 2 \pi).$$

Finally

$$p = \hbar e /(m_p 2).$$

Now, the magnetic field are like

$$B_e = \mu_0I_e/2/r_e$$

And we get the energy as (assuming the inclination between the loops are $$\theta$$)

$$|W| = \frac{\mu_0I_p I_e}{2 r_e}\pi r_p = \frac{\mu_0 e^2\hbar^2}{m_e m_p 8 pi r_e^3}.$$

Actually if we assume $$E\to 2\pi E$$ as the reason for using $$n$$ in stead of $$2\pi n$$ we conclude that we need to correct the formula with some multiple factor of $$2\pi$$. The method of calculation is that we use the following correction to $$C$$

$$B = -\frac{p(Z)}{p(0)}\mu_n\frac{\mu_0 e \hbar}{m_e 16 \pi r^2}(2\pi)^3,$$

with,

$$\mu_n = \frac{e \hbar}{2 m_p}$$

and $$p(Z)$$ the measured magnetic moment for the most popular isotope.

Using this correction to the energy lead to the table,

\begin{array}{l|l|l|l} Z & \textbf{This [eV]} & \textbf{This + mag [eV]} & \textbf{Meassured [eV]}\\ \hline H & 13.598468 & 13.598419 & 13.598434 \\ He^+ & 54.418266 & 54.41797 & 54.41777 \\ Li^{2+} & 122.45630 & 122.45476 & 122.45436 \\ Be^{3+} & 217.7243 & 217.722 & 217.71859 \\ B^{4+} & 340.2384 & 340.232 & 340.22602 \\ C^{5+} & 490.0176 & 490.007 & 489.99320 \\ N^{6+} & 667.0882 & 667.0858 & 667.04612 \\ O^{7+} & 871.4772 & 871.4622 & 871.4099 \\ \end{array}