# Discovery of the Relativistic Schrödinger equation

Kamal Barley  Applied Mathematics and Statistics, Stony Brook University, NY
11794-3600, U. S. A.  ,  José Vega-Guzmán  Lamar University, 200 Lucas
Building, P. O. Box 10047, Beaumont, TX 77710, U. S. A.  and  Sergei K. Suslov
Arizona State University, P. O. Box 871804, Tempe, AZ 85287-1804, U. S. A.

###### Abstract.

We discuss the discovery of the relativistic wave equation for a spin-zero
charged particle in the Coulomb field by Erwin Schrödinger (and elaborate on
why he didn’t publish it).

###### Key words and phrases:

Schrödinger equation, Klein–Gordon–Fock equation, Dirac equation, hydrogen
atom, WKB-approximation

###### 1991 Mathematics Subject Classification:

Primary 81Q05, 35C05; Secondary 01A05, 34E20

##  1\. Introduction

There are many great recollections and reviews of remarkable events of the
early days of quantum mechanics. Among those [2], [4], [12], [19], [24], [25],
[26], [27], [37], [38] (and references therein), as well as great biographies
of the ‘pioneers’ of the quantum revolution, namely, Werner Heisenberg [6],
[7], Erwin Schrödinger [18], [28], and Paul Dirac [13].

A traditional story of the discovery of a wave equation – “rumor has it”– goes
this way [2], [18], [28]: Early in November 1925, Peter Debye, at the
ETH,111Swiss Federal Institute of Technology in Zürich. asked Schrödinger to
prepare a talk for the Zürich physicists about de Broglie’s work just
published in Annales de Physique. The date on which this particular colloquium
took place has not been recorded, but it must have been in late November or
early December, before the end of the academic term [18], [26]. According to
Felix Bloch [2], at the end of Schrödinger’s talk, Debye casually remarked
that (de Broglie’s) way of talking was rather childish and someone had to
think about an equation for these (de Broglie’s) “matter waves”. As a student
of Sommerfeld, Debye had learned that to deal properly with waves, one had to
have a wave equation. Indeed, at his second talk, just a few weeks later,
Schrödinger mentioned: “My colleague Debye suggested that one should have a
wave equation; well, I have found one!”

For a sequence of those remarkable events, a direct quote is quite appropriate
[18]: “Schrödinger’s first thought was to attempt to find a wave equation,
moving on from de Broglie’s work, that would describe the behavior of an
electron in the simplest atom, hydrogen. He naturally included in his
calculations allowance for the effects described by the special theory of
relativity, deriving, probably early in December 1925, what became known as
the relativistic Schrödinger (not hydrogen—our correction!) equation.
Unfortunately, it didn’t work. The predictions of the relativistic equation
did not match up with observations of real atoms. We now know that this is
because Schrödinger did not allow for the quantum spin of the electron, which
is hardly surprising, since the idea of spin had not been introduced into
quantum mechanics at that time. But it is particularly worth taking note of
this false start, since it highlights the deep and muddy waters in which
quantum physicists go swimming—for you need to take account of spin, a
property usually associated with particles, in order to derive a wave equation
for the electron!222The concept of electron spin was introduced by G. E.
Uhlenbeck and S Goudsmit in a letter published in Die Naturwissenschaften; the
issue of 20 November 1925; see [23] for more details.

With the Christmas break coming up, he had an opportunity to get away from
Zürich and think things over in the clean air of Arosa … [18] .”(See also [26]
and [28] for more details333Arosa, an Alpine _Kurort_ at about 1800 m
altitude, not far from ski-resort Davos, and overlooked by the great peak of
the Weisshorn. For a modern video, see:
https://www.news.uzh.ch/en/articles/2017/Schroedinger.html . Here, among other
things, a female physicist meets the grandson of Dr. Herwig and he shows the
entry of the payment in a guest book, done by Schrödinger. (The authors are
thankful to Priv.-Doz. Dr. Andreas Ruffing who found this video.).)

Whereas the honorable historians of science are arguing about the sequence of
some events [19], [26], [38], our remarks are dedicated to a somewhat missing,
mathematical, part of this story, namely, the discovery of a relativistic
(spin-zero) wave equation by Erwin Schrödinger (and why he did report and
publish only the centennial nonrelativistic version for the hydrogen atom).

Figure 1. The Villa Herwig–Frisia (right), Arosa, where it is believed wave
mechanics was discovered during the Christmas holidays 1925–26. Figure 2. The
Villa Herwig–Frisia postcard.

##  2\. Introducing the relativistic Schrödinger equation

Let us start our discussion by following original Schrödinger’s idea — “rumor
has it”! For a classical relativistic particle in the electromagnetic field
one has to replace

$E^{2}=c^{2}\mathbf{p}^{2}+m^{2}c^{4}$ (2.1)

by

$\left(E-e\varphi\right)^{2}=\left(c\mathbf{p}-e\mathbf{A}\right)^{2}+m^{2}c^{4}.$
(2.2)

In relativistic quantum mechanics, we replace the four-vector of energy $E$
and linear momentum $\mathbf{p}$ with the following operators

$E\rightarrow\widehat{E}=i\hbar\frac{\partial}{\partial
t},\qquad\mathbf{p}\rightarrow\widehat{\mathbf{p}}=-i\hbar\nabla.$ (2.3)

As a result, we arrive at the relativistic Schrödinger equation [30]:

$\displaystyle\left(-\hbar^{2}\frac{\partial^{2}}{\partial
t^{2}}-2ie\hbar\varphi\frac{\partial}{\partial
t}-ie\hbar\frac{\partial\varphi}{\partial t}+e^{2}\varphi^{2}\right)\psi$
(2.4)
$\displaystyle\qquad=\left(-\hbar^{2}c^{2}\Delta+2ic\hbar\mathbf{A}\cdot\nabla+2iec\hbar\,\nabla\cdot\mathbf{A}+e^{2}{\mathbf{A}}^{2}+m^{2}c^{4}\right)\psi\/.$

In the special case of Coulomb field,

$e\varphi=-\frac{Ze^{2}}{r},\qquad\mathbf{A}=\mathbf{0},$ (2.5)

one can separate the time and spatial variables

$\psi\left(\mathbf{r},t\right)=\chi(\mathbf{r})e^{-i(E\;t)/\hbar}.$ (2.6)

Finally, the stationary relativistic Schrödinger equation takes the form

$\left(E+\frac{Ze^{2}}{r}\right)^{2}\chi=\left(-\hbar^{2}c^{2}\Delta+m^{2}c^{4}\right)\chi.$
(2.7)

(As one can see, it is NOT a traditional eigenvalue problem!) The standard
rumor says that Schrödinger invented this equation sometime in late 1925,
found the corresponding relativistic energy levels, and derived the fine
structure formula, different from Sommerfeld’s one [11]. This is why he had to
abandon this equation and consider its nonrelativistic limit. Moreover, they
say that he had to withdraw the original relativistic article from a journal,
which had never been found, and start all over with his centennial articles on
the nonrelativistic Schrödinger equation [31]. Our goal here is to follow the
original relativistic idea. Our analysis, nonetheless, will suggest that it is
hardly possible to believe that a complete relativistic solution (for a
charged spin-zero particle) had been found at that time (see, nonetheless,
[26] for an opposite opinion and [14] for the first published solution of this
problem).

##  3\. Solving the relativistic Schrödinger equation

We follow [29] with somewhat different details (see also [9], [14], [30] for a
different approach) and, first of all, separate the variables in spherical
coordinates, $\chi\left(r,\theta,\varphi\right)=R(r)Y_{lm}(\theta,\varphi),$
where $Y_{lm}(\theta,\varphi)$ are the spherical harmonics with familiar
properties [36]. As a result,

$\frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{dR}{dr}\right)+\left[\frac{\left(E+Ze^{2}/r\right)^{2}-m^{2}c^{4}}{\hbar^{2}c^{2}}-\frac{l\left(l+1\right)}{r^{2}}\right]R=0\qquad(l=0,1,2,\;...)$
(3.1)

Figure 3. A page from Notebook $N1\/$, with the first record of the wave
equation [28].

(see Figure 3 for the original version). In the dimensionless quantities,

$\varepsilon=\frac{E}{mc^{2}},\qquad x=\beta
r=\frac{mc}{\hbar}r,\qquad\mu=\frac{Ze^{2}}{\hbar c},$ (3.2)

for the new radial function,

$R(r)=F(x)=\frac{u(x)}{x},$ (3.3)

one gets

$\frac{1}{x^{2}}\frac{d}{dx}\left(x^{2}\frac{dF}{dx}\right)+\left[\left(\varepsilon+\frac{\mu}{x}\right)^{2}-1-\frac{l\left(l+1\right)}{x^{2}}\right]F=0.$
(3.4)

Given the identity,

$\frac{1}{x^{2}}\frac{d}{dx}\left(x^{2}\frac{dF}{dx}\right)=\frac{1}{x}\frac{d^{2}}{dx^{2}}\left(xF\right),$
(3.5)

we finally obtain

$u^{\prime\prime}+\left[\left(\varepsilon+\frac{\mu}{x}\right)^{2}-1-\frac{l\left(l+1\right)}{x^{2}}\right]u=0.$
(3.6)

This is the generalized equation of hypergeometric form (A.1) below, when

$\sigma\left(x\right)=x,\qquad\widetilde{\tau}(x)\equiv
0,\qquad\widetilde{\sigma}(x)=\left(\varepsilon^{2}-1\right)x^{2}+2\mu\varepsilon
x+\mu^{2}-l(l+1)$ (3.7)

(see also Table 2 provided in Appendix A below).

In view of the normalization condition,

$\int_{0}^{\infty}R^{2}(r)r^{2}\
dr=1,\qquad\text{or\quad}\int_{0}^{\infty}u^{2}(x)\;dx=\beta^{3},\quad\beta=\frac{mc}{\hbar^{2}},$
(3.8)

it’s useful to start from asymptotic as $r\rightarrow 0$ $($or $x\rightarrow
0).$ In this limit, from (3.6), one gets Euler’s equation:

$x^{2}u^{\prime\prime}+\left[\mu^{2}-l\left(l+1\right)\right]u=0,$ (3.9)

which can be solved by the standard substitution $u=x^{\alpha},$ resulting in
the following quadratic equation:

$\alpha^{2}-\alpha+\left[\mu^{2}-l\left(l+1\right)\right]=0.$ (3.10)

Thus the general solution of Euler’s equation has the form

$u=C_{1}x^{\nu+1}+C_{2}x^{-\nu};\qquad\nu=-\frac{1}{2}+\sqrt{\left(l+\frac{1}{2}\right)^{2}-\mu^{2}},\quad\mu=\frac{Ze^{2}}{\hbar
c};$ (3.11)

which is bounded as $x\rightarrow 0$ with $C_{2}=0:$ $\ u\simeq
C_{1}x^{\nu+1}.$

We transform (A.1) into the simpler form (A.3) by using
$u\left(x\right)=\varphi\left(x\right)y\left(x\right),$ where

$\frac{\varphi^{\prime}}{\varphi}=\frac{\pi\left(x\right)}{\sigma\left(x\right)},\qquad\pi(x)=\frac{\sigma^{\prime}-\widetilde{\tau}}{2}\pm\sqrt{\left(\frac{\sigma^{\prime}-\widetilde{\tau}}{2}\right)^{2}-\widetilde{\sigma}+k\sigma},$
(3.12)

or, in this particular case,

$\displaystyle\pi(x)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\pm\sqrt{\left(l+\frac{1}{2}\right)^{2}-\mu^{2}-2\mu\varepsilon
x+\left(1-\varepsilon^{2}\right)x^{2}+kx}$ (3.13) $\displaystyle=$
$\displaystyle\frac{1}{2}\pm\left\\{\begin{array}[]{cc}\sqrt{1-\varepsilon^{2}}\;x+\nu+\frac{1}{2},&k=2\mu\varepsilon+(2\nu+1)\sqrt{1-\varepsilon^{2}};\\\
\sqrt{1-\varepsilon^{2}}\;x-\nu-\frac{1}{2},&k=2\mu\varepsilon-(2\nu+1)\sqrt{1-\varepsilon^{2}}.\end{array}\right.$
(3.16)

Here, one has to choose the second possibility, when
$\tau^{\prime}=2\pi^{\prime}<0,$ namely$:$

$\tau\left(x\right)=2\left(\nu+1-ax\right),\qquad a=\sqrt{1-\varepsilon^{2}}.$
(3.17)

Thus

$\displaystyle\pi\left(x\right)$ $\displaystyle=$
$\displaystyle\nu+1-ax,\qquad\qquad\quad\varphi\left(x\right)=x^{\nu+1}e^{-ax};$
(3.18) $\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
2\left(\mu\varepsilon-(\nu+1)a\right),\qquad\rho\left(x\right)=x^{2\nu+1}e^{-2ax}.$

For example,

$\displaystyle\left(\ln\varphi\right)^{\prime}$ $\displaystyle=$
$\displaystyle\frac{\nu+1}{x}-a,\qquad\ln\varphi=\left(\nu+1\right)\ln
x-ax,\qquad\text{etc.}$ $\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
k+\pi^{\prime}=2\left(\mu\varepsilon-(\nu+1)a\right).$

(All these data are collected in Table 2, Appendix A, for the reader’s
convenience.)

In Nikiforov–Uvarov’s approach, the energy levels can be obtained from the
quantization rule [29]:

$\lambda+n\tau^{\prime}+\frac{1}{2}n\left(n-1\right)\sigma^{\prime\prime}=0\qquad(n=0,1,2,\
...),$ (3.19)

which gives Schrödinger’s fine structure formula (for a charged spin-zero
particle in the Coulomb field) [14]:

$E=E_{n_{r}}=\frac{mc^{2}}{\sqrt{1+\left(\dfrac{\mu}{n_{r}+\nu+1}\right)^{2}}}\qquad\left(n=n_{r}=0,1,2,\
...\right).$ (3.20)

Here,

$\mu=\frac{Ze^{2}}{\hbar
c},\qquad\nu=-\frac{1}{2}+\sqrt{\left(l+\frac{1}{2}\right)^{2}-\mu^{2}}.$
(3.21)

The corresponding eigenfunctions are given by the Rodrigues-type formula

$y_{n}\left(x\right)=\frac{B_{n}}{x^{2\nu+1}e^{-2ax}}\frac{d^{n}}{dx^{n}}\left(x^{n+2\nu+1}e^{-2ax}\right)=C_{n}L_{n}^{2\nu+1}\left(2ax\right).$
(3.22)

Up to a constant, they are Laguerre polynomials [29]. By (3.8),

$\displaystyle\beta^{3}$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}u^{2}(x)\;dx=C_{n}^{2}\int_{0}^{\infty}\left[\varphi\left(x\right)L_{n}^{2\nu+1}\left(2ax\right)\right]^{2}\
dx$ (3.23) $\displaystyle=$
$\displaystyle\frac{C_{n}^{2}}{\left(2a\right)^{2\nu+3}}\int_{0}^{\infty}\xi^{2\nu+2}\left(L_{n}^{2\nu+1}\left(\xi\right)\right)^{2}\
d\xi,\quad\xi=2ax.$

The corresponding integral is given by [35]:

$I_{1}=\int_{0}^{\infty}e^{-x}x^{\alpha+1}\left(L_{n}^{\alpha}\left(x\right)\right)^{2}\
dx=\left(\alpha+2n+1\right)\frac{\Gamma\left(\alpha+n+1\right)}{n!}.$ (3.24)

As a result,

$C_{n}=2\left(a\beta\right)^{3/2}\left(2a\right)^{\nu}\sqrt{\frac{n!}{\Gamma\left(2\nu+n+2\right)}}.$
(3.25)

The normalized radial eigenfunctions, corresponding to the relativistic energy
levels (3.20), are explicitly given by

$R\left(r\right)=R_{n_{r}}\left(r\right)=2\left(a\beta\right)^{3/2}\sqrt{\frac{n_{r}!}{\Gamma\left(2\nu+n_{r}+2\right)}}\;\xi^{\nu}e^{-\xi/2}\;L_{n}^{2\nu+1}\left(\xi\right),$
(3.26)

where

$\xi=2ax=2a\beta r=\sqrt{1-\varepsilon^{2}}\frac{mc}{\hbar}r.$ (3.27)

Our consideration here, or somewhat similar to that in [9], [14], [30],
outlines the way which Schrödinger had to complete (during his Xmas-vacation
in 1925–26?) to discover the above solution of the relativistic spin-zero
charged particle in the central Coulomb field. One can see that it’s a bit
harder than the corresponding nonrelativistic case originally published in
Ref. [31] (see also [1], [21], [29], [35] for modern treatments).

##  4\. Nonrelativistic limit

According to Dirac [11] (see also [2], [18], and [26]), Schrödinger was able
to solve his relativistic wave equation (and submitted an article to a
journal, which original had never been found?). But later he had to abandon
the whole idea after discovering an inconsistency of his fine structure
formula with Sommerfeld’s one, which was in agreement with the available
experimental data [34]. During his Christmas 1925-vacation at Arosa,
Schrödinger introduced, eventually, his famous nonrelativistic stationary wave
equation and solved it for the hydrogen atom [31] (thus ignoring the fine
structure). The relativistic Schrödinger equation for a free spin-zero
particle was later rediscovered by Klein, Gordon, and Fock.444According to
[19], Louis de Broglie discovered the relativistic wave equation for a spin-
zero free particle on February 1925 [5], which did not attract any interest at
that time. The relativistic wave equation for a free electron, spin-$1/2$
particle, was discovered by Dirac only in 1928 [10].555For a modern definition
of the spin, in terms of representations of the Poincaré group, see [20]. Its
solution in the central Coulomb field, when Sommerfeld–Dirac’s fine structure
formula had been finally derived from relativistic quantum mechanics, was
later found by Darwin [8] and Gordon [16]. As a result, the relativistic
energy levels of an electron in the central Coulomb field are given by

$E=E_{n_{r},\,j}=\frac{mc^{2}}{\sqrt{1+\mu^{2}/\left(n_{r}+\nu\right)^{2}}},\quad\mu=\frac{Ze^{2}}{\hbar
c}\quad(n_{r}=0,1,2,\ ...).$ (4.1)

In Dirac’s theory,

$\nu=\sqrt{\left(j+1/2\right)^{2}-\mu^{2}},$ (4.2)

where $j=1/2,3/2,5/2,\ ...\ $ is the total angular momentum including the spin
of the relativistic electron. A detailed solution of this problem, including
the nonrelativistic limit, can be found in [29], [35] (in Nikiforov–Uvarov’s
paradigm), or in classical sources [1], [9], [15], [30].

Let us analyze a nonrelativistic limit of Schrödinger’s fine structure formula
(3.20)–(3.21) and compare with the corresponding results for
Sommerfeld–Dirac’s one (4.1)–(4.2). For the relativistic Schrödinger equation,
in the limit $c\rightarrow\infty$ (or $\mu\rightarrow 0),$ one gets [9], [30]:

$\displaystyle\frac{E_{n_{r},\,l}}{mc^{2}}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{1+\dfrac{\mu^{2}}{\left(n_{r}+\frac{1}{2}+\sqrt{\left(l+\frac{1}{2}\right)^{2}-\mu^{2}}\right)}}}$
(4.3) $\displaystyle=$ $\displaystyle
1-\frac{\mu^{2}}{2n^{2}}-\frac{\mu^{4}}{2n^{4}}\left(\frac{n}{l+1/2}-\frac{3}{4}\right)+\text{O}\left(\mu^{6}\right),\quad\mu\rightarrow
0,$

which can be derived by a direct Taylor’s expansion and/or verified by a
computer algebra system. Here, $n=n_{r}+l+1$ is the corresponding
nonrelativistic principal quantum number. The first term in this expansion is
simply the rest mass energy $E_{0}=mc^{2}$ of the charged spin-zero particle,
the second term coincides with the energy eigenvalue in the nonrelativistic
Schrödinger theory and the third term gives the so-called fine structure of
the energy levels.

In Dirac’s theory of the relativistic electron, the corresponding limit has
the form [1], [9], [30], [35]:

$\frac{E_{n_{r},\,j}}{mc^{2}}=1-\frac{\mu^{2}}{2n^{2}}-\frac{\mu^{4}}{2n^{4}}\left(\frac{n}{j+1/2}-\frac{3}{4}\right)+\text{O}\left(\mu^{6}\right),\quad\mu\rightarrow
0,$ (4.4)

where $n=n_{r}+j+1/2$ is the principal quantum number of the nonrelativistic
hydrogenlike atom. Once again, the first term in this expansion is the rest
mass energy of the relativistic electron, the second term coincides with the
energy eigenvalue in the nonrelativistic Schrödinger theory and the third term
gives the so-called fine structure of the energy levels — the correction
obtained for the energy in the Pauli approximation which includes the
interaction of the spin of the electron with its orbital angular momentum (see
[1], [9], [15], [30], [34] for further discussion of the hydrogenlike energy
levels including comparison with the experimental data). For an alternative
discussion of the discovery of Sommerfeld’s formula, see [17].

Let us also consider a nonrelativistic limit of the radial wave functions
(3.26)–(3.27) of the relativistic Schrödinger equation. From (3.21),
$\nu\rightarrow l$ as $c\rightarrow\infty.$ Moreover,

$1-\varepsilon^{2}=\left(1-\varepsilon\right)(1+\varepsilon)\rightarrow\frac{\mu^{2}}{2n^{2}}\cdot
2=\frac{\mu^{2}}{n^{2}},\quad\mu\rightarrow 0,$ (4.5)

and, therefore,

$\xi=\sqrt{1-\varepsilon^{2}}\frac{mc}{\hbar}r\rightarrow
2\frac{\mu}{n}\cdot\frac{mc}{\hbar}r=\frac{2Z}{n}\left(\frac{r}{a_{0}}\right)=\eta,\quad\mu\rightarrow
0,$ (4.6)

where $a_{0}=\hbar^{2}/me^{2}$ is Bohr’s radius. As a result, from
(3.26)–(3.27) one gets the nonrelativistic radial wavefunctions of the
hydrogenlike atom [1], [21], [31], [35]:

$R\left(r\right)\rightarrow
R_{nl}\left(r\right)=\frac{2}{n^{2}}\left(\frac{Z}{a_{0}}\right)^{3/2}\sqrt{\frac{\left(n-l-1\right)!}{\left(n+l\right)!}}\
\eta^{l}\ e^{-\eta/2}\ L_{n-l-1}^{2l+1}\left(\eta\right)$ (4.7)

with

$\eta=\frac{2Z}{n}\left(\frac{r}{a_{0}}\right),\qquad
a_{0}=\frac{\hbar^{2}}{me^{2}}\quad\text{(Bohr's radius)}$ (4.8)

and

$\int_{0}^{\infty}R_{nl}^{2}\left(r\right)\ dr=1.$

In Table 1, provided in Appendix A, we present the corresponding data for the
nonrelativistic hydrogenlike atom for the reader’s convenience. (More details
can be found in [29], [35].)

##  5\. Semiclassical approximation: WKB-method for Coulomb fields

Phenomenological quantization rules of ‘old’ quantum mechanics [34] are
derived in modern physics from the corresponding wave equations in the so-
called semiclassical approximation [3], [9], [21] (WKB-method). In this
approximation, for a particle in the central field, one can use a generic
radial equation of the form [22], [29]:

$u^{\prime\prime}(x)+q(x)\,u=0\/,$ (5.1)

where $x^{2}\,q(x)$ is continuous together with its first and second
derivatives for $0\leq x\leq b<\infty\/.$ As is well-known, the traditional
semiclassical approximation cannot be used in a neighborhood of $x=0\/.$
However, the change of variables $x=e^{z}\/$, $u=e^{z/2}\,v(z)$ transforms the
equation into a new form,

$v^{\prime\prime}(z)+q_{1}(z)\,v=0\/,$ (5.2)

where

$q_{1}(z)=-\dfrac{1}{4}+\left(x^{2}\,q(x)\right)_{x=e^{z}}\/$ (5.3)

(Langer’s transformation [22]). As $z\to-\infty$ (or $x\to 0$), the new
function $q_{1}(z)$ is changing slowly near the constant $-1/4+\lim_{x\to
0}x^{2}q(x)\/$ and $\lim_{z\to-\infty}q_{1}^{(k)}(z)=0\/$ ($k=1,2$). Hence the
function $q_{1}(z)$ and its derivatives are changing slowly for large negative
$z\/.$ The WKB-method can be applied to (5.2) and, as a result, in the
original equation (5.1) one should replace $q(x)$ with $q(x)-1/(4x^{2})\/$
(see [22], [29] for more details).

For example, in the well-known case of nonrelativistic Coulomb’s problem, one
gets

$u^{\prime\prime}+\left[2\left(\varepsilon_{0}+\frac{Z}{x}\right)-\frac{\left(l+1/2\right)^{2}}{x^{2}}\right]u=0$
(5.4)

(in dimensional units, see Table 1 or [35]). The Bohr–Sommerfeld quantization
rule takes the form

$\int_{r_{1}}^{r_{2}}p(r)\,dr=\pi\left(n+\dfrac{1}{2}\right)\qquad\qquad(n=0\/,1\/,2\/,\,\dots)$
(5.5)

with

$p(r)=\left[2\left(\varepsilon_{0}+\frac{Z}{r}\right)-\frac{\left(l+1/2\right)^{2}}{r^{2}}\right]^{1/2},\qquad
p(r_{1})=p(r_{2})=0\/$ (5.6)

(see, for instance, [29] for more details).

For all Coulomb problems under consideration, we utilize the following generic
integral, originally evaluated by Sommerfeld [34] performing complex
integration: If

$p(r)=\sqrt{-A+\dfrac{B}{r}-\dfrac{C}{r^{2}}}\;\qquad A\/,C>0\/.$ (5.7)

Then

$\int_{r_{1}}^{r_{2}}p(r)\,dr=\pi\left(\dfrac{B}{2\sqrt{A}}-\sqrt{C}\right)$
(5.8)

provided $p(r_{1})=p(r_{2})=0\/.$ (See Appendix B for an elementary evaluation
of this integral.)

In (5.4), $A=-2\varepsilon_{0}\/,$ $B=2Z\/,$ and $C=(l+1/2)^{2}.\/$ In view of
the quantization rule (5.5) ,

$\dfrac{Z}{\sqrt{-2\varepsilon_{0}}}-l-\dfrac{1}{2}=n_{r}+\dfrac{1}{2}\/.$
(5.9)

As a result, we obtain exact energy levels for the nonrelativistic
hydrogenlike problem:

$\varepsilon_{0}=\dfrac{E}{E_{0}}=-\dfrac{Z^{2}}{2(n_{r}+l+1)^{2}}\/\qquad\quad\left(E_{0}=\dfrac{e^{2}}{a_{0}},\quad
a_{0}=\frac{\hbar^{2}}{me^{2}}\right)\/.$ (5.10)

Here, $n=n_{r}+l+1$ is known as the principal quantum number.

Our main goal is to analyze the corresponding relativistic problems. In the
case of the relativistic Schrödinger equation (3.6)–(3.7), one should write

$u^{\prime\prime}+\left[\left(\varepsilon+\frac{\mu}{x}\right)^{2}-1-\frac{\left(l+1/2\right)^{2}}{x^{2}}\right]u=0$
(5.11)

and

$p(x)=\left[\left(\varepsilon+\frac{\mu}{x}\right)^{2}-1-\frac{\left(l+1/2\right)^{2}}{x^{2}}\right]^{1/2}\/.$
(5.12)

Here, $A=1-{\varepsilon}^{2}\/,$ $B=2\mu\varepsilon\/,$ and
$C=(l+1/2)^{2}-{\mu}^{2}\/.$ The quantization rule (5.5) implies

$\dfrac{\mu\varepsilon}{\sqrt{1-{\varepsilon}^{2}}}=n_{r}+\nu+1\/$ (5.13)

and the exact formula for the relativistic energy levels (3.20)–(3.21) follows
once again.666This is, possibly, how Schrödinger was able to derive his fine
structure formula so quickly. But a true reason why this algebraic equation,
obtained in semiclassical approximation, coincides with the exact one in
relativistic quantum mechanics still remains a mystery!

Sommerfeld’s fine structure formula (4.1)–(4.2), for the relativistic Coulomb
problem, can be thought of as the main achievement of the ‘old’ quantum
mechanics [17], [34]. Here, we will derive this result in a semiclassical
approximation for the radial Dirac equations (separation of variables in
spherical coordinates is discussed in detail in Refs. [29] and [35]). In the
dimensionless units, one of these second-order differential equations has the
form

$v^{\prime\prime}_{1}+\dfrac{({\varepsilon}^{2}-1)x^{2}+2\varepsilon\mu
x-\nu(\nu+1)}{x^{2}}\,v_{1}=0\/$ (5.14)

and the second equation can be obtained from the first one by replacing
$\nu\to-\nu\/$ (see Eqs. (3.81)–(3.82) in Ref. [35]). By Langer’s
transformation, we obtain

$p(x)=\left[\left(\varepsilon+\frac{\mu}{x}\right)^{2}-1-\frac{\left(\nu+1/2\right)^{2}+\mu^{2}}{x^{2}}\right]^{1/2}\/.$
(5.15)

Hence, for the Dirac equation, $A=1-{\varepsilon}^{2}\/,$
$B=2\mu\varepsilon\/,$ and $C=(\nu+1/2)^{2}.\/$ The quantization rule (5.5)
implies (5.13). (One has to replace $n_{r}\to n_{r}-1\/$, see [35] for more
details.) Thus the fine structure formula for the energy levels (4.1)–(4.2)
has been derived here in semiclassical approximation for the corresponding
radial Dirac equations. This explains the quantization rules of the ‘old’
quantum mechanics of Bohr and Sommerfeld [34] (see also [17]).

Conclusions. Schrödinger’s diaries for the years 1925–26 are not available.
This is why we may never know for sure all the details of his revolutionary
discovery. Nonetheless, our analysis of the relativistic Schrödinger equation
allows suggesting that he, probably, didn’t have a chance to solve this
equation for Coulomb potential (from the viewpoint of the current standards
when eigenvalues and the normalized eigenfunctions should be explicitly
presented? – it was certainly done by Fock just a few months later [14]).
Instead, he seems found the corresponding fine structure formula for a charged
spin-zero particle by using the Bohr–Sommerfeld quantization rule (correcting
a mistake in the original work [17]?) and observed discrepancies with the
experimental data [26]. Analyzing this “disaster”, he switched to the
nonrelativistic case and introduced the stationary Schrödinger equation, which
he was able to solve around December-January of 1925–26 (consulting with
Hermann Weyl?). His main legacy, namely, the time-dependent Schrödinger
equation (Figure 4), that was, de facto, required in an article dedicated to
the coherent states [32], was published only about six months later [33]. We
hope that our consideration will help the readers to draw their own
conclusions.

Figure 4. The Schrödinger equation — University of Vienna (left) and Alpbach
(right), respectively.

##  Appendix A Summary of the Nikiforov and Uvarov approach

Generalized equation of the hypergeometric form

$u^{\prime\prime}+\frac{\widetilde{\tau}(x)}{\sigma(x)}u^{\prime}+\frac{\widetilde{\sigma}(x)}{\sigma^{2}(x)}u=0$
(A.1)

($\sigma,$ $\widetilde{\sigma}$ are polynomials of degrees at most
$2,\widetilde{\tau}$ is a polynomial at most first degree) by the substitution

$u=\varphi(x)y(x)$ (A.2)

can be reduced to the form

$\sigma(x)y^{\prime\prime}+\tau(x)y^{\prime}+\lambda y=0$ (A.3)

if:

$\frac{\varphi^{\prime}}{\varphi}=\frac{\pi(x)}{\sigma(x)},\qquad\pi(x)=\frac{1}{2}\left(\tau(x)-\widetilde{\tau}(x)\right)$
(A.4)

(or, $\tau(x)=\widetilde{\tau}+2\pi,$ for later),

$k=\lambda-\pi^{\prime}(x)\qquad(\text{or, }\lambda=k+\pi^{\prime}),$ (A.5)

and

$\pi(x)=\frac{\sigma^{\prime}-\widetilde{\tau}}{2}\pm\sqrt{\left(\frac{\sigma^{\prime}-\widetilde{\tau}}{2}\right)^{2}-\widetilde{\sigma}+k\sigma}$
(A.6)

is a linear function. (Use the choice of constant $k$ to complete the square
under the radical sign; see [29] for more details.)

In Nikiforov–Uvarov’s approach, the energy levels can be obtained from the
quantization rule:

$\lambda+n\tau^{\prime}+\frac{1}{2}n\left(n-1\right)\sigma^{\prime\prime}=0\qquad(n=0,1,2,\
...)$ (A.7)

and the corresponding square-integrable solutions are classical orthogonal
polynomials, up to a factor. They can be found by the Rodrigues-type formula
[29]:

$y_{n}(x)=\dfrac{B_{n}}{\rho(x)}\left(\sigma^{n}(x)\rho(x)\right)^{(n)},\qquad(\sigma\rho)^{\prime}=\tau\rho,$
(A.8)

where $B_{n}$ is a constant (see also [35]).

$\sigma(x)$ | $x$
---|---
$\widetilde{\sigma}(x)$ | $2\varepsilon_{0}\;x^{2}+2Z\;x-l\left(l+1\right)\/,\quad{\varepsilon_{0}=E/E_{0}}$
$\widetilde{\tau}(x)$ | $0$
$k$ | $2Z-\left(2l+1\right)\sqrt{-2\varepsilon_{0}\/}$
$\pi\left(x\right)$ | $l+1-\sqrt{-2\varepsilon_{0}}\;x$
$\tau\left(x\right)=\widetilde{\tau}+2\pi\qquad$ | $2\left(l+1-\sqrt{-2\varepsilon_{0}}\;x\right)$
$\lambda=k+\pi^{\prime}$ | $2\left(Z-(l+1)\sqrt{-2\varepsilon_{0}}\right)$
$\varphi\left(x\right)$ | $x^{l+1}e^{-x\sqrt{-2\varepsilon_{0}}}$
$\rho(x)$ | $x^{2l+1}e^{-(2Z\ x)/n},\quad x=r/a_{0}$
$y_{n_{r}}(x)$ | $C_{n_{r}}L_{n_{r}}^{2l+1}\left(\eta\right),\quad\eta=\dfrac{2Z}{n}x=\dfrac{2Z}{n}\left(\dfrac{r}{a_{0}}\right)$
$C_{n_{r}}^{2}$ | $\dfrac{Z}{a_{0}n^{2}}\dfrac{\left(n-l-1\right)!}{\left(n+l\right)!},\quad n_{r}=n-l-1$
Table 1. Schrödinger equation for the Coulomb potential
$V(r)=-\dfrac{Ze^{2}}{r}.$
Dimensionless quantities:
$r=a_{0}x,\quad a_{0}=\frac{\hbar^{2}}{m_{e}e^{2}}\simeq 0.5\cdot
10^{-8}\;\text{cm},\quad E_{0}=\frac{e^{2}}{a_{0}};\quad
R\left(r\right)=F\left(x\right)=\frac{u\left(x\right)}{x}.$

$\sigma(x)$ | $x$
---|---
$\widetilde{\sigma}(x)$ | $\left(\varepsilon^{2}-1\right)\;x^{2}+2\mu\varepsilon\;x+\mu^{2}-l\left(l+1\right)$
$\widetilde{\tau}(x)$ | $0$
$k$ | $2\mu\varepsilon-\left(2\nu+1\right)\sqrt{1-\varepsilon^{2}},\quad\nu=-\dfrac{1}{2}+\sqrt{\left(l+\dfrac{1}{2}\right)^{2}-\mu^{2}}$
$\pi\left(x\right)$ | $\nu+1-a\;x,\qquad a=\sqrt{1-\varepsilon^{2}}$
$\tau\left(x\right)=\widetilde{\tau}+2\pi$ | $2\left(\nu+1-a\;x\right)$
$\lambda=k+\pi^{\prime}$ | $2\left(\mu\varepsilon-(\nu+1)a\right)$
$\varphi\left(x\right)$ | $x^{\nu+1}e^{-a\ x}$
$\rho(x)$ | $x^{2\nu+1}e^{-2a\ x}$
$y_{n_{r}}(x)$ | $C_{n_{r}}L_{n_{r}}^{2\nu+1}\left(\xi\right),\quad\xi=2ax=2a\beta r=2\sqrt{1-\varepsilon^{2}}\dfrac{mc}{\hbar}r$
$C_{n_{r}}$ | $2\left(a\beta\right)^{3/2}\left(2a\right)^{\nu}\sqrt{\dfrac{n_{r}!}{(\nu+n_{r}+1)\Gamma\left(2\nu+n_{r}+2\right)}}$
Table 2. Relativistic Schrödinger equation for the Coulomb potential
$V(r)=-\dfrac{Ze^{2}}{r}.$
Dimensionless quantities:
$\varepsilon=\frac{E}{mc^{2}},\quad x=\beta
r=\frac{mc}{\hbar}r,\quad\mu=\frac{Ze^{2}}{\hbar c};\qquad
R\left(r\right)=F\left(x\right)=\frac{u\left(x\right)}{x}.$

##  Appendix B Evaluation of the integral

Integrating by parts in the right-hand side of (5.8), one gets

$\displaystyle{\left.\int_{r_{1}}^{r_{2}}p(r)\,dr=rp(r)\right|}_{r_{1}}^{r_{2}}-\int_{r_{1}}^{r_{2}}\dfrac{r\left[-(B/r^{2})+2(C/r^{3})\right]}{2p(r)}\,dr$
(B.1)
$\displaystyle\quad=\dfrac{B}{2}\int_{r_{1}}^{r_{2}}\dfrac{dr}{\sqrt{-Ar^{2}+Br-C}}-\int_{r_{1}}^{r_{2}}\dfrac{(C/r^{2})\,dr}{\sqrt{-A+(B/r)-(C/r^{2})}}\/.$

For the last but one integral, we can write

$\int_{r_{1}}^{r_{2}}\dfrac{dr}{\sqrt{\left(\frac{B^{2}}{4{A}}-C\right)-\left(r\sqrt{A}-\frac{B}{2\sqrt{A}}\right)^{2}}}\/={\left.\dfrac{1}{\sqrt{A}}\arcsin\dfrac{r\sqrt{A}-\frac{B}{2\sqrt{A}}}{{\sqrt{\frac{B^{2}}{4A}-C}}}\right|}_{r_{1}}^{r_{2}}=\dfrac{\pi}{\sqrt{A}}\
.$ (B.2)

The following substitution $r=1/x\/,$ in the last integral of (B.1), results
in

$-\int_{x_{1}=1/{r_{1}}}^{x_{2}=1/{r_{2}}}\dfrac{C\,dx}{\sqrt{-A+Bx-
Cx^{2}}}=-\sqrt{C}\int_{x_{1}}^{x_{2}}\dfrac{\sqrt{C}\,dx}{\sqrt{\left(\frac{B^{2}}{4C}-A\right)-\left(x\sqrt{C}-\frac{B}{2\sqrt{C}}\right)^{2}}}=\pi\,\sqrt{C}\/,$
(B.3)

where we have completed the square and performed a familiar integral
evaluation. (One may also interchange $A$ and $C$ in the previous integral
evaluation.) Combining the last two integrals, one can complete the proof.

Acknowledgments. We are grateful to Prof. Dr. Ruben Abagyan, Prof. Dr. Yakov
I. Granovskiĭ, Dr. Sergey I. Kryuchkov, Prof. Dr. Peter Paule, Priv.-Doz. Dr.
Andreas Ruffing, Dr. Eugene Stepanov, and Prof. Dr. Alexei Zedanov for
support, help, and valuable comments.

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