A NEW VARIANT OF AN OLD PHYSICS PROBLEM

B. N. Rodimov

Derivation of the \(v\)-equation. The relativistic relation between the phase and group velocities of a particle, \(uv=c^2\), makes it possible to suggest that, since the phase velocity \(u\) determines the wave properties of particles described by the ordinary Schrödinger equation (we shall also call it the \(u\)-equation), the group velocity \(v\) must likewise determine these wave properties and must likewise yield its own wave equation (we shall call it the \(v\)-equation) [1].

The ordinary Schrödinger equation can be derived from the equation for a “modulated” wave propagating with phase velocity \(u\):

\[ \Delta \psi=\frac{1}{u^2}\frac{\partial^2\psi}{\partial t^2}. \tag{1} \]

From the function \(\psi\) one can extract the factor \(e^{i\frac{2\pi E}{h}t}\), where \(\omega=\frac{2\pi E}{h}\), which is obtained from the relations for a free particle

\[ u=\nu\lambda=\nu\frac{h}{mv}; \qquad \nu=\frac{muv}{h}=\frac{mc^2}{h}=\frac{E}{h} \]

and retains its form for bound systems as well. Here \(E\) is the total relativistic energy of the particle.

Differentiating

\[ \psi(r,t)=\psi(r)e^{i\frac{2\pi E}{h}t}, \]

we obtain

\[ \Delta \psi(r)+\frac{4\pi^2p^2}{h^2}\,\psi(r)=0, \tag{2} \]

where \(p=\frac{E}{u}\). Substituting for the momentum \(p\) its nonrelativistic value \(p^2=2m_0(E_{\text{kin}}-V(r))\), we obtain the ordinary Schrödinger equation

\[ \Delta \psi(r)+\frac{8\pi^2m_0}{h^2}\bigl(E_{\text{kin}}-V(r)\bigr)\psi(r)=0. \tag{3} \]

In the expression \(e^{i\frac{2\pi E}{h}t}\), since \(E \simeq m_0 c^2 + E_{\text{kin}}\), the factor \(e^{i\frac{2\pi}{h}m_0c^2t}\), since it appears in all terms of the equation, is discarded, and the factor \(e^{i\frac{2\pi}{h}E_{\text{kin}}t}\) remains.

The \(v\)-equation is obtained from the equation for the “modulated” wave, propagating with group velocity \(v\),

\[ \Delta \psi=\frac{1}{v^2}\frac{\partial^2\psi}{\partial t^2}. \tag{4} \]

Here too one may single out a time factor, in particular in the form \(e^{i\frac{4\pi E}{h}t}\). Here \(\Omega=\dfrac{4\pi E}{h}\) is the frequency of the modulating wave, and \(E\) is the nonrelativistic energy. In contrast to the preceding case, for a free particle we relate the de Broglie wavelength to the group velocity \(v\):

\[ \lambda=vT;\quad T=\frac{h}{mv^2};\quad \Omega=\frac{2\pi}{T}=\frac{2\pi mv^2}{h}=\frac{4\pi E}{h}. \]

For bound particles such an expression for \(\Omega\) will hold only in special cases. For \(v^2\) we take the usual nonrelativistic relation

\[ v^2=\frac{2}{m}(E-V(x)). \]

Thus, in the one-dimensional case and for the time factor in the form \(e^{i\frac{4\pi E}{h}t}\), we obtain

\[ \Delta\psi(x)+\frac{8\pi^2mE^2}{h^2(E-V(x))}\psi(x)=0. \tag{5} \]

We shall call this equation the \(v\)-equation.

In contrast to the \(u\)-equation, in the \(v\)-equation at the classical boundary (where \(E=V(x)\)) the coefficient of \(\psi\) becomes infinite. At this boundary the \(\psi\)-function of the \(v\)-equation becomes zero. For a free particle, for which \(V(x)\equiv0\), the two equations formally coincide, although one of them proceeds from the “modulated” wave and the other from the “modulating” one.

We shall next consider the hydrogen-atom problem from the standpoint of the \(v\)-equation.

\(s\)-states of the hydrogen atom. The \(v\)-equation for the \(s\)-states of the hydrogen atom is written in the form

\[ \frac{d^2R}{dr^2}+\frac{2}{r}\frac{dR}{dr}+ \frac{8\pi^2mE^2}{h^2\left(E+\dfrac{e^2}{r}\right)}R=0. \tag{6} \]

Introduce the notations

\[ A=-\frac{8\pi^2mE}{h^2}>0;\quad B=-\frac{e^2}{E}>0, \]

then

\[ R''+\frac{2}{r}R' - \frac{A}{1-\dfrac{B}{r}}\,R=0 \tag{7} \]

or

\[ R''+\frac{2}{x}R' + \frac{AB^2x}{1-x}\,R=0, \tag{8} \]

if \(x=\dfrac{r}{B}\). By the substitution \(R=\dfrac{y}{x}\) we bring equation (8) to the form

\[ (1-x)y''+\lambda xy=0, \tag{9} \]

where \(\lambda=AB^2\). This equation is solved by the substitution \(y=e^{\sqrt{\lambda}x}\cdot\eta(\xi)\), where \(1-x=\xi\),

\[ \xi\eta''-2\sqrt{\lambda}\,\xi\eta' + \lambda\eta=0. \tag{10} \]

The solution of this equation is sought in the form of a series \(\eta=\sum_m a_m\xi^m\). The coefficients of the series are related by the recurrence formula

\[ a_{n+1}=\frac{2\sqrt{\lambda}\,n-\lambda}{n(n+1)}\,a_n. \tag{11} \]

The series terminates when

\[ 2\sqrt{\lambda}\,n-\lambda=0. \]

Hence

\[ \lambda=4n^2. \tag{12} \]

Since \(\lambda=AB^2=-\dfrac{8\pi^2me^4}{h^2E}\), the usual values for the energy levels of the hydrogen atom are obtained,

\[ E=-\frac{2\pi^2me^4}{h^2n^2}. \tag{13} \]

The eigenfunctions for \(s\)-states will have the form

\[ R_n=\frac{e^2}{E_n r}\,e^{-\frac{2nE_n}{e^2}r}\cdot \sum_{m=1}^{n} a_m\left(1+\frac{rE_n}{e^2}\right)^m . \tag{14} \]

The first three functions are shown in Fig. 1 \((l=0)\).

Interpretation of the eigenfunctions of \(s\)-states. From Fig. 1 it is seen that the usual probabilistic interpretation of eigenfunctions is not suitable here, since the functions become infinite at the point \(r=0\) and terminate at the classical boundary for the given energy state. By their character the functions \(R_n\) may be interpreted as peculiar quantum potential functions. The complete \(\psi\)-function must also include a time factor. In contrast to ordinary quantum mechanics, this factor must be a real quantity.

Thus one may assume that in the \(s\)-state the electron moves along the radius from the proton to the classical boundary and excites a special spherical (electromagnetic?) wave, which, proceeding from the proton, reaches the classical boundary and is reflected back. Since the electron continuously interacts with this wave, its motion along the radius will be a complex oscillatory motion. The presence of this wave also prevents the “fall” of the electron onto the proton in the \(s\)-state¹.

The spherical symmetry of the wave is due to the spherical symmetry of the Coulomb field of the proton. Since the velocity of the wave is \(c\), i.e. the wave is “slow,” the hydrogen atom in the \(s\)-state resembles a high-frequency spherical resonator with a “load.”

Fig. 1

Although the wave velocity changes from point to point along the radius, its frequency must obviously be one and the same constant. This frequency must be determined by the mean values of the de Broglie wavelength \(\bar{\lambda}\) and the velocity \(\bar{\dot r}\) (averaging over the coordinate \(r\)),

\[ T_{\mathrm{kv}}=\frac{\bar{\lambda}}{\bar{\dot r}}. \]

Averaging \(\lambda\) and \(r\), we obtain for the \(s\)-states of the hydrogen atom

\[ \bar{\lambda}=\frac{h}{\sqrt{-2mE}}\cdot\frac{\pi}{2};\qquad \bar{r}=\sqrt{-\frac{2E}{m}}\cdot\frac{\pi}{2};\qquad T_{KB}=\frac{h}{2E}; \]

\[ \omega_{KB}=\frac{4\pi E}{h}. \tag{16} \]

In this case the same expression is obtained for the frequency as for a free particle.

Generalization to the case of rotational states.
The case of rotational \(l\)-states of the hydrogen atom must be described by two equations. The first equation will describe a spherical wave, only now with two spherical boundary surfaces corresponding to the two extreme values of the radius \(r_{\max}\) and \(r_{\min}\) in the classical Kepler problem for an electron in the field of a proton. The second equation must describe an azimuthal wave excited by the rotational motion of the electron in the direction of the coordinate \(\varphi\) in the spherical layer between the spheres of radii \(r_{\max}\) and \(r_{\min}\). We shall call the first wave the \(r_l\)-wave, and the second wave the \(\varphi\)-wave.

Equation for the \(r_l\)-wave. The frequency of the \(r_l\)-wave is found from the relation

\[ T_{KB,r}=\frac{\bar{\lambda}_r}{\bar{r}}. \]

In the present case

\[ \bar{\lambda}_r=-\frac{h}{m\left(r_{\max}-r_{\min}\right)} \int_{r_{\max}}^{r_{\min}}\frac{dr}{r} =\frac{h\pi}{2\varepsilon\sqrt{-2mE}}; \]

\[ \bar{r}=-\frac{1}{r_{\max}-r_{\min}} \int_{r_{\max}}^{r_{\min}} r\,dr = \sqrt{-\frac{2E}{m}}\cdot\pi\cdot \frac{1-\sqrt{1-\varepsilon^2}}{2\varepsilon}. \]

Hence

\[ T_{KB,r}=\frac{h}{2E\left(1-\sqrt{1-\varepsilon^2}\right)}; \qquad \omega_{KB,r}= \frac{4\pi E\left(1-\sqrt{1-\varepsilon^2}\right)}{h}. \tag{17} \]

Here

\[ \varepsilon=\sqrt{1+\frac{2E p_\varphi^{\,2}}{me^4}}; \qquad r_{\max}=-\frac{e^2}{2E}(1+\varepsilon); \]

\[ r_{\min}=-\frac{e^2}{2E}(1-\varepsilon). \tag{18} \]

The equation for the \(r_l\)-wave is obtained from the expression

\[ \Delta_r \psi_r+\frac{\omega_r^2}{\bar{v}_r^{\,2}}\psi_r=0; \tag{19} \]

\(\Delta_r\) is the part of the Laplace operator independent of the angles. After substituting

\[ \omega_r=\frac{4\pi E e^2}{h} \quad\text{and}\quad v_r^2=\dot r^{\,2}=\frac{2}{m}\left(E+\frac{e^2}{r}-\frac{p_\varphi^2}{2mr^2}\right), \]

we shall have

\[ \frac{d^2R}{dr^2}+\frac{2}{r}\frac{dR}{dr} +\frac{8\pi^2 m E^2\, [[unclear: factor]]}{h^2\left(E+\frac{e^2}{r}-\frac{p_\varphi^2}{2mr^2}\right)}R=0. \tag{20} \]

Equation for the \(\varphi\)-wave. From considerations of symmetry one may assume that the \(\varphi\)-wave at a given radius must travel with a constant velocity independent of \(\varphi\). Therefore

\[ \lambda_\varphi=\lambda_\zeta=\frac{h}{mr\dot\varphi}; \quad \overline{r\dot\varphi}=r\dot\varphi; \quad \omega_{\mathrm{kin},\varphi}=\frac{4\pi E e}{h} =\frac{2\pi}{h}\frac{p_\varphi^2}{mr^2}. \tag{21} \]

We obtain the equation for the \(\varphi\)-wave from the expression

\[ \Delta_\varphi \psi_\varphi+\frac{\omega_\varphi^2}{v_\varphi^2}\psi_\varphi=0. \tag{22} \]

Thus

\[ \frac{1}{r^2\sin^2\Theta}\frac{\partial^2\psi_\varphi}{\partial\zeta^2} +\frac{1}{r^2}\frac{\cos\Theta}{\sin\Theta}\frac{\partial\psi_\varphi}{\partial\Theta} +\frac{1}{r^2}\frac{\partial^2\psi_\varphi}{\partial\Theta^2} + \frac{8\pi^2 m}{h^2}E_\varphi\psi_\varphi=0. \tag{23} \]

As usual, we put \(\psi_\varphi=\Theta(\Theta)\cdot\Phi(\zeta)\) and obtain two equations

\[ \frac{d^2\Phi}{d\zeta^2}=-m^2\Phi, \]

\[ (1-x^2)\frac{d^2\Theta}{dx^2}-2x\frac{d\Theta}{dx} +\left(\alpha-\frac{m^2}{1-x^2}\right)\Theta=0, \tag{24} \]

where \(x=\cos\Theta;\ \alpha=\dfrac{4\pi^2 p_\varphi^2}{h^2}=l(l+1),\ |m|\le l\). The solutions of the equation for the \(\varphi\)-wave will be the associated Legendre functions

\[ P_l^m(x)=(1-x^2)^{\frac{|m|}{2}}\cdot \frac{d^{|m|}}{dx^{|m|}}P_l(x). \tag{25} \]

Solution of the equation for the \(r\)-wave. If one considers the equation for the \(r\)-wave by itself, it will be an eigenvalue equation with two parameters. Its solution will be complicated. In our case the problem is simplified by the fact that the values of the parameter \(P_\varphi\) are found independently from the equation for the \(\varphi\)-wave.

Let us write equation (20) in the form

\[ -\frac{d^2 R}{dr^2}+\frac{2}{r}\frac{dR}{dr} -\frac{A}{1-\dfrac{B}{r}+\dfrac{C}{r^2}}\,R=0, \tag{26} \]

where

\[ A=-\frac{8\pi^2 mE\left(1-\sqrt{1-\varepsilon^2}\right)}{h^2}; \qquad B=-\frac{e^2}{E}>0; \qquad C=-\frac{p^2}{2mE}>0. \]

Introduce the variable \(x=\dfrac{r}{B}\) and make the substitution \(R=\dfrac{y}{x}\); then

\[ y''-\frac{\lambda x^2}{(x-x_1)(x-x_2)}\,y=0. \tag{27} \]

Here \(\lambda=AB^2\), \(x_1=\dfrac{1}{2}(1-\varepsilon)\), \(x_2=\dfrac{1}{2}(1+\varepsilon)\). Obviously, the solution of the equation will break off at the points \(x=x_2\) and \(x=x_1\). The substitution \(y=e^{\sqrt{\lambda}x}\eta\) brings the equation to the form

\[ (x-x_1)(x-x_2)\eta_x'' +2\sqrt{\lambda}(x-x_1)(x-x_2)\eta_x' +\lambda[x_1x_2-x]\eta=0. \tag{28} \]

In passing to \(x_1\to0\), \(x_2\to1\), we obviously again obtain equation (10). Denote: \(x_2-x=z\), \(x-x_1=a-z\), \(a=x_2-x_1\); then

\[ z(a-z)\eta_z''-2\sqrt{\lambda}\,z(a-z)\eta_z' +\lambda(x_2^2-z)\eta=0. \tag{29} \]

We separate out the factor \((a-z)\), in order to ensure that the function \(\eta\) vanishes for \(x=x_1\): \(\eta=(a-z)W(z)\); then

\[ z(a-z)W''-2z\left[1+\sqrt{\lambda}(a-z)\right]W' +\left[(2\sqrt{\lambda}-\lambda)z+\lambda x_2^2\right]W=0. \tag{30} \]

We take the function \(W\) in the form of a series \(W=\sum_m b_m z^m\). The series begins with \(m=1\), in order to ensure that the solution breaks off at the second boundary \(x=x_2\). For the coefficient of the series we obtain the three-term recurrence formula

\[ b_{m+1}(m+1)ma +b_m\left[-m(m-1)-2(1+\sqrt{\lambda}a)m+\lambda x_2^2\right] \tag{31} \]

\[ {}+b_{m-1}\left[2\sqrt{\lambda}\,m-\lambda\right]=0. \]

If we denote the ratio \(\gamma_m=\dfrac{b_{m+1}}{b_m}\) and \(\gamma_{m-1}=\dfrac{b_m}{b_{m-1}}\), the formula can be obtained in the following form:

\[ \gamma_{m-1}= -\frac{2\sqrt{\lambda}\,m-\lambda} {\left[\lambda x_2^2-m(m-1)-2(1+\sqrt{\lambda}a)m\right] \left\{1+\dfrac{am(m+1)} {\lambda x_2^2-m(m-1)-2(1+\sqrt{\lambda}a)m}\,\gamma_m\right\}}. \tag{32} \]

or

\[ \gamma_{m-1}= \frac{q_m}{1+a_m\dfrac{q_{m+1}}{1+a_{m+1}\dfrac{q_{m+2}}{1+\ldots}}}, \tag{33} \]

where

\[ q_m=-\frac{2\sqrt{\lambda}\,m-\lambda} {\lambda x_2^{\,2}-m(m-1)-2(1+\sqrt{\lambda}\,a)m}; \]

\[ z_m=\frac{am(m+1)} {\lambda x_2^{\,2}-m(m-1)-2(1+\sqrt{\lambda}\,a)m}. \tag{34} \]

For the first two coefficients we obtain the relation

\[ b_2=\frac{2(1+\sqrt{\lambda})-\lambda x_2^{\,2}}{2a}\,b_1 . \tag{35} \]

To find \(\lambda\) (and hence also \(E\)), a transcendental equation is obtained.

To check our reasoning one may compute the functions \(R_{n,l}\), using the recurrence formula (31) and the known expressions for \(E\) in terms of \(n\) and for \(p_e\) in terms of \(l\). The results of the calculations are shown in Fig. 1 (the functions for which \(l=0\)).

Conclusions

The \(z\)-equation makes it possible to construct a new physical picture of the hydrogen atom. Although, in the sense of mathematical flexibility, the method of the \(z\)-equation can scarcely compete with the Schrödinger equation, nevertheless the new approach to problems of quantum mechanics may also yield new results.

LITERATURE

1. B. N. Rodimov. Atoms and molecules as self-oscillating systems. Proceedings of the 2nd Interuniversity Conference on Electron Accelerators, Tomsk, 1961.

Research Institute at Tomsk Polytechnic Institute.

1. A solution \(r_1=\sum a_m \xi^m\), if the recurrence formula (11) is satisfied, may contain either an infinite or a finite number of terms. But a solution with an infinite number of terms, and consequently with an infinite number of nodes, has no physical meaning. Solutions with a finite number of terms can be interpreted only in the manner just described, i.e. the \(\psi\)-function must represent a system of standing electromagnetic waves in a high-frequency resonator. The function \(R_n\) must have a finite number of nodes, and this requirement is expressed in the requirement that the series \(\eta=\sum a_m \xi^m\) be terminated. ↩