The hydrogen atom consists of a proton and an electron. If $e$ symbolizes the charge on the proton $\left(e=+1.6 \times 10^{-19} \mathrm{C}\right)$, then the electron's charge is $-e$.
A few scientists have speculated that the proton and electron charges might not be exactly equal in magnitude. Experiments show that the magnitudes of the electron and proton charges are equal to within one part in $10^{21}$. See G. Bressi et al., Phys. Rev. A, 83, 052101 (2011) (available online at arxiv.org/abs/1102.2766).
We shall assume the electron and proton to be point masses whose interaction is given by Coulomb's law. In discussing atoms and molecules, we shall usually be considering isolated systems, ignoring interatomic and intermolecular interactions.
Instead of treating just the hydrogen atom, we consider a slightly more general problem: the hydrogenlike atom, which consists of one electron and a nucleus of charge $Z e$. For $Z=1$, we have the hydrogen atom; for $Z=2$, the $\mathrm{He}^{+}$ion; for $Z=3$, the $\mathrm{Li}^{2+}$ ion; and so on. The hydrogenlike atom is the most important system in quantum chemistry. An exact solution of the Schrödinger equation for atoms with more than one electron cannot be obtained because of the interelectronic repulsions. If, as a crude first approximation, we ignore these repulsions, then the electrons can be treated independently. (See Section 6.2.) The atomic wave function will be approximated by a product of one-electron functions, which will be hydrogenlike wave functions. A one-electron wave function (whether or not it is hydrogenlike) is called an orbital. (More precisely, an orbital is a one-electron spatial wave function, where the word spatial means that the wave function depends on the electron's three spatial coordinates $x, y$, and $z$ or $r, \theta$, and $\phi$. We shall see in Chapter 10 that the existence of electron spin adds a fourth coordinate to a one-electron wave function, giving what is called a spin-orbital.) An orbital for an electron in an atom is called an atomic orbital. We shall use atomic orbitals to construct approximate wave functions for atoms with many electrons (Chapter 11). Orbitals are also used to construct approximate wave functions for molecules.
For the hydrogenlike atom, let $(x, y, z)$ be the coordinates of the electron relative to the nucleus, and let $\mathbf{r}=\mathbf{i} x+\mathbf{j} y+\mathbf{k} z$. The Coulomb's law force on the electron in the hydrogenlike atom is [see Eq. (1.37)]
\(
\begin{equation}
\mathbf{F}=-\frac{Z e^{2}}{4 \pi \varepsilon_{0} r^{2}} \frac{\mathbf{r}}{r} \tag{6.56}
\end{equation}
\)
where $\mathbf{r} / r$ is a unit vector in the $\mathbf{r}$ direction. The minus sign indicates an attractive force.
The possibility of small deviations from Coulomb's law has been considered. Experiments have shown that if the Coulomb's-law force is written as being proportional to $r^{-2+s}$, then $|s|<10^{-16}$. A deviation from Coulomb's law can be shown to imply a nonzero photon rest mass. No evidence exists for a nonzero photon rest mass, and data indicate that any such mass must be less than $10^{-51} \mathrm{~g}$; A. S. Goldhaber and M. M. Nieto, Rev. Mod. Phys., 82, 939 (2010) (arxiv.org/abs/0809.1003); G. Spavieri et al., Eur. Phys. J. D, 61, 531 (2011) (link.springer.com/content/pdf/10.1140/epjd/ e2011-10508-7).
The force in (6.56) is central, and comparison with Eq. (6.4) gives $d V(r) / d r=$ $Z e^{2} / 4 \pi \varepsilon_{0} r^{2}$. Integration gives
\(
\begin{equation}
V=\frac{Z e^{2}}{4 \pi \varepsilon{0}} \int \frac{1}{r^{2}} d r=-\frac{Z e^{2}}{4 \pi \varepsilon{0} r} \tag{6.57}
\end{equation}
\)
where the integration constant has been taken as 0 to make $V=0$ at infinite separation between the charges. For any two charges $Q{1}$ and $Q{2}$ separated by distance $r_{12}$, Eq. (6.57) becomes
\(
\begin{equation}
V=\frac{Q{1} Q{2}}{4 \pi \varepsilon{0} r{12}} \tag{6.58}
\end{equation}
\)
Since the potential energy of this two-particle system depends only on the relative coordinates of the particles, we can apply the results of Section 6.3 to reduce the problem to two one-particle problems. The translational motion of the atom as a whole simply adds some constant to the total energy, and we shall not concern ourselves
FIGURE 6.5 Relative spherical coordinates.
with it. To deal with the internal motion of the system, we introduce a fictitious particle of mass
\(
\begin{equation}
\mu=\frac{m{e} m{N}}{m{e}+m{N}} \tag{6.59}
\end{equation}
\)
where $m{e}$ and $m{N}$ are the electronic and nuclear masses. The particle of reduced mass $\mu$ moves subject to the potential-energy function (6.57), and its coordinates $(r, \theta, \phi)$ are the spherical coordinates of one particle relative to the other (Fig. 6.5).
The Hamiltonian for the internal motion is [Eq. (6.43)]
\(
\begin{equation}
\hat{H}=-\frac{\hbar^{2}}{2 \mu} \nabla^{2}-\frac{Z e^{2}}{4 \pi \varepsilon_{0} r} \tag{6.60}
\end{equation}
\)
Since $V$ is a function of the $r$ coordinate only, we have a one-particle central-force problem, and we may apply the results of Section 6.1. Using Eqs. (6.16) and (6.17), we have for the wave function
\(
\begin{equation}
\psi(r, \theta, \phi)=R(r) Y_{l}^{m}(\theta, \phi), \quad l=0,1,2, \ldots, \quad|m| \leq l \tag{6.61}
\end{equation}
\)
where $Y_{l}^{m}$ is a spherical harmonic, and the radial function $R(r)$ satisfies
\(
\begin{equation}
-\frac{\hbar^{2}}{2 \mu}\left(R^{\prime \prime}+\frac{2}{r} R^{\prime}\right)+\frac{l(l+1) \hbar^{2}}{2 \mu r^{2}} R-\frac{Z e^{2}}{4 \pi \varepsilon_{0} r} R=E R(r) \tag{6.62}
\end{equation}
\)
To save time in writing, we define the constant $a$ as
\(
\begin{equation}
a \equiv \frac{4 \pi \varepsilon_{0} \hbar^{2}}{\mu e^{2}} \tag{6.63}
\end{equation}
\)
and (6.62) becomes
\(
\begin{equation}
R^{\prime \prime}+\frac{2}{r} R^{\prime}+\left[\frac{8 \pi \varepsilon_{0} E}{a e^{2}}+\frac{2 Z}{a r}-\frac{l(l+1)}{r^{2}}\right] R=0 \tag{6.64}
\end{equation}
\)
Solution of the Radial Equation
We could now try a power-series solution of (6.64), but we would get a three-term rather than a two-term recursion relation. We therefore seek a substitution that will lead to a twoterm recursion relation. It turns out that the proper substitution can be found by examining the behavior of the solution for large values of $r$. For large $r$, (6.64) becomes
\(
\begin{equation}
R^{\prime \prime}+\frac{8 \pi \varepsilon_{0} E}{a e^{2}} R=0, \quad r \text { large } \tag{6.65}
\end{equation}
\)
which may be solved using the auxiliary equation (2.7). The solutions are
\(
\begin{equation}
\exp \left[ \pm\left(-8 \pi \varepsilon_{0} E / a e^{2}\right)^{1 / 2} r\right] \tag{6.66}
\end{equation}
\)
Suppose that $E$ is positive. The quantity under the square-root sign in (6.66) is negative, and the factor multiplying $r$ is imaginary:
\(
\begin{equation}
R(r) \sim e^{ \pm i \sqrt{2 \mu E} r / \hbar}, \quad E \geq 0 \tag{6.67}
\end{equation}
\)
where (6.63) was used. The symbol $\sim$ in (6.67) indicates that we are giving the behavior of $R(r)$ for large values of $r$; this is called the asymptotic behavior of the function. Note the resemblance of (6.67) to Eq. (2.30), the free-particle wave function. Equation (6.67) does not give the complete radial factor in the wave function for positive energies. Further study (Bethe and Salpeter, pages 21-24) shows that the radial function for $E \geq 0$ remains finite for all values of $r$, no matter what the value of $E$. Thus, just as for the free particle, all nonnegative energies of the hydrogen atom are allowed. Physically, these eigenfunctions correspond to states in which the electron is not bound to the nucleus; that is, the atom is ionized. (A classical-mechanical analogy is a comet moving in a hyperbolic orbit about the sun. The comet is not bound and makes but one visit to the solar system.) Since we get continuous rather than discrete allowed values for $E \geq 0$, the positive-energy eigenfunctions are called continuum eigenfunctions. The angular part of a continuum wave function is a spherical harmonic. Like the free-particle wave functions, the continuum eigenfunctions are not normalizable in the usual sense.
We now consider the bound states of the hydrogen atom, with $E<0$. (For a bound state, $\psi \rightarrow 0$ as $x \rightarrow \pm \infty$.) In this case, the quantity in parentheses in (6.66) is positive. Since we want the wave functions to remain finite as $r$ goes to infinity, we prefer the minus sign in (6.66), and in order to get a two-term recursion relation, we make the substitution
\(
\begin{gather}
R(r)=e^{-C r} K(r) \tag{6.68}\
C \equiv\left(-\frac{8 \pi \varepsilon_{0} E}{a e^{2}}\right)^{1 / 2} \tag{6.69}
\end{gather}
\)
where $e$ in (6.68) stands for the base of natural logarithms, and not the proton charge. Use of the substitution (6.68) will guarantee nothing about the behavior of the wave function for large $r$. The differential equation we obtain from this substitution will still have two linearly independent solutions. We can make any substitution we please in a differential equation; in fact, we could make the substitution $R(r)=e^{+C r} J(r)$ and still wind up with the correct eigenfunctions and eigenvalues. The relation between $J$ and $K$ would naturally be $J(r)=e^{-2 C r} K(r)$.
Proceeding with (6.68), we evaluate $R^{\prime}$ and $R^{\prime \prime}$, substitute into (6.64), multiply by $r^{2} e^{C r}$, and use (6.69) to get the following differential equation for $K(r)$ :
\(
\begin{equation}
r^{2} K^{\prime \prime}+\left(2 r-2 C r^{2}\right) K^{\prime}+\left[\left(2 Z a^{-1}-2 C\right) r-l(l+1)\right] K=0 \tag{6.70}
\end{equation}
\)
We could now substitute a power series of the form
\(
\begin{equation}
K=\sum{k=0}^{\infty} c{k} r^{k} \tag{6.71}
\end{equation}
\)
for $K$. If we did we would find that, in general, the first few coefficients in (6.71) are zero. If $c_{s}$ is the first nonzero coefficient, (6.71) can be written as
\(
\begin{equation}
K=\sum{k=s}^{\infty} c{k} r^{k}, \quad c_{s} \neq 0 \tag{6.72}
\end{equation}
\)
Letting $j \equiv k-s$, and then defining $b{j}$ as $b{j} \equiv c_{j+s}$, we have
\(
\begin{equation}
K=\sum{j=0}^{\infty} c{j+s} r^{j+s}=r^{s} \sum{j=0}^{\infty} b{j} r^{j}, \quad b_{0} \neq 0 \tag{6.73}
\end{equation}
\)
(Although the various substitutions we are making might seem arbitrary, they are standard procedure in solving differential equations by power series.) The integer $s$ is evaluated by substitution into the differential equation. Equation (6.73) is
\(
\begin{gather}
K(r)=r^{s} M(r) \tag{6.74}\
M(r)=\sum{j=0}^{\infty} b{j} r^{j}, \quad b_{0} \neq 0 \tag{6.75}
\end{gather}
\)
Evaluating $K^{\prime}$ and $K^{\prime \prime}$ from (6.74) and substituting into (6.70), we get
\(
\begin{equation}
r^{2} M^{\prime \prime}+\left[(2 s+2) r-2 C r^{2}\right] M^{\prime}+\left[s^{2}+s+\left(2 Z a^{-1}-2 C-2 C s\right) r-l(l+1)\right] M=0 \tag{6.76}
\end{equation}
\)
To find $s$, we look at (6.76) for $r=0$. From (6.75), we have
\(
\begin{equation}
M(0)=b{0}, \quad M^{\prime}(0)=b{1}, \quad M^{\prime \prime}(0)=2 b_{2} \tag{6.77}
\end{equation}
\)
Using (6.77) in (6.76), we find for $r=0$
\(
\begin{equation}
b_{0}\left(s^{2}+s-l^{2}-l\right)=0 \tag{6.78}
\end{equation}
\)
Since $b_{0}$ is not zero, the terms in parentheses must vanish: $s^{2}+s-l^{2}-l=0$. This is a quadratic equation in the unknown $s$, with the roots
\(
\begin{equation}
s=l, \quad s=-l-1 \tag{6.79}
\end{equation}
\)
These roots correspond to the two linearly independent solutions of the differential equation. Let us examine them from the standpoint of proper behavior of the wave function. From Eqs. (6.68), (6.74), and (6.75), we have
\(
\begin{equation}
R(r)=e^{-C r} r^{s} \sum{j=0}^{\infty} b{j} r^{j} \tag{6.80}
\end{equation}
\)
Since $e^{-C r}=1-C r+\ldots$, the function $R(r)$ behaves for small $r$ as $b_{0} r^{s}$. For the root $s=l, R(r)$ behaves properly at the origin. However, for $s=-l-1, R(r)$ is proportional to
\(
\begin{equation}
\frac{1}{r^{l+1}} \tag{6.81}
\end{equation}
\)
for small $r$. Since $l=0,1,2, \ldots$, the root $s=-l-1$ makes the radial factor in the wave function infinite at the origin. Many texts take this as sufficient reason for rejecting this root. However, this is not a good argument, since for the relativistic hydrogen atom, the $l=0$ eigenfunctions are infinite at $r=0$. Let us therefore look at (6.81) from the standpoint of quadratic integrability, since we certainly require the bound-state eigenfunctions to be normalizable.
The normalization integral [Eq. (5.80)] for the radial functions that behave like (6.81) looks like
\(
\begin{equation}
\int{0}|R|^{2} r^{2} d r \approx \int{0} \frac{1}{r^{2 l}} d r \tag{6.82}
\end{equation}
\)
for small $r$. The behavior of the integral at the lower limit of integration is
\(
\begin{equation}
\left.\frac{1}{r^{2 l-1}}\right|_{r=0} \tag{6.83}
\end{equation}
\)
For $l=1,2,3, \ldots,(6.83)$ is infinite, and the normalization integral is infinite. Hence we must reject the root $s=-l-1$ for $l \geq 1$. However, for $l=0$, (6.83) is finite, and there is no trouble with quadratic integrability. Thus there is a quadratically integrable solution to the radial equation that behaves as $r^{-1}$ for small $r$.
Further study of this solution shows that it corresponds to an energy value that the experimental hydrogen-atom spectrum shows does not exist. Thus the $r^{-1}$ solution must be rejected, but there is some dispute over the reason for doing so. One view is that the $1 / r$ solution satisfies the Schrödinger equation everywhere in space except at the origin and hence must be rejected [Dirac, page 156; B. H. Armstrong and E. A. Power, Am. J. Phys., 31, 262 (1963)]. A second view is that the $1 / r$ solution must be rejected because the Hamiltonian operator is not Hermitian with respect to it (Merzbacher, Section 10.5). (In Chapter 7 we shall define Hermitian operators and show that quantum-mechanical operators are required to be Hermitian.) Further discussion is given in A. A. Khelashvili and T. P. Nadareishvili, Am. J.Phys., 79, 668 (2011) (see arxiv.org/abs/1102.1185) and in Y. C. Cantelaube, arxiv.org/abs/1203.0551.
Taking the first root in (6.79), we have for the radial factor (6.80)
\(
\begin{equation}
R(r)=e^{-C r_{r}^{l} M(r)} \tag{6.84}
\end{equation}
\)
With $s=l$, Eq. (6.76) becomes
\(
\begin{equation}
r M^{\prime \prime}+(2 l+2-2 C r) M^{\prime}+\left(2 Z a^{-1}-2 C-2 C l\right) M=0 \tag{6.85}
\end{equation}
\)
From (6.75), we have
\(
\begin{gather}
M(r)=\sum{j=0}^{\infty} b{j} r^{j} \tag{6.86}\
M^{\prime}=\sum{j=0}^{\infty} j b{j} r^{j-1}=\sum{j=1}^{\infty} j b{j} r^{j-1}=\sum{k=0}^{\infty}(k+1) b{k+1} r^{k}=\sum{j=0}^{\infty}(j+1) b{j+1} r^{j} \
M^{\prime \prime}=\sum{j=0}^{\infty} j(j-1) b{j} r^{j-2}=\sum{j=1}^{\infty} j(j-1) b{j} r^{j-2}=\sum{k=0}^{\infty}(k+1) k b{k+1} r^{k-1} \
M^{\prime \prime}=\sum{j=0}^{\infty}(j+1) j b{j+1} r^{j-1} \tag{6.87}
\end{gather}
\)
Substituting these expressions in (6.85) and combining sums, we get
\(
\sum{j=0}^{\infty}\left[j(j+1) b{j+1}+2(l+1)(j+1) b{j+1}+\left(\frac{2 Z}{a}-2 C-2 C l-2 C j\right) b{j}\right] r^{j}=0
\)
Setting the coefficient of $r^{j}$ equal to zero, we get the recursion relation
\(
\begin{equation}
b{j+1}=\frac{2 C+2 C l+2 C j-2 Z a^{-1}}{j(j+1)+2(l+1)(j+1)} b{j} \tag{6.88}
\end{equation}
\)
We now must examine the behavior of the infinite series (6.86) for large $r$. The result of the same procedure used to examine the harmonic-oscillator power series in (4.42) suggests that for large $r$ the infinite series (6.86) behaves like $e^{2 C r}$. (See Prob. 6.20.) For large $r$, the radial function (6.84) behaves like
\(
\begin{equation}
R(r) \sim e^{-C r} r^{l} e^{2 C r}=r^{l} e^{C r} \tag{6.89}
\end{equation}
\)
Therefore, $R(r)$ will become infinite as $r$ goes to infinity and will not be quadratically integrable. The only way to avoid this "infinity catastrophe" (as in the harmonicoscillator case) is to have the series terminate after a finite number of terms, in which case the $e^{-C r}$ factor will ensure that the wave function goes to zero as $r$ goes to infinity. Let the
last term in the series be $b{k} r^{k}$. Then, to have $b{k+1}, b{k+2}, \ldots$ all vanish, the fraction multiplying $b{j}$ in the recursion relation (6.88) must vanish when $j=k$. We have
\(
\begin{equation}
2 C(k+l+1)=2 Z a^{-1}, \quad k=0,1,2, \ldots \tag{6.90}
\end{equation}
\)
$k$ and $l$ are integers, and we now define a new integer $n$ by
\(
\begin{equation}
n \equiv k+l+1, \quad n=1,2,3, \ldots \tag{6.91}
\end{equation}
\)
From (6.91) the quantum number $l$ must satisfy
\(
\begin{equation}
l \leq n-1 \tag{6.92}
\end{equation}
\)
Hence $l$ ranges from 0 to $n-1$.
Energy Levels
Use of (6.91) in (6.90) gives
\(
\begin{equation}
C n=Z a^{-1} \tag{6.93}
\end{equation}
\)
Substituting $C \equiv\left(-8 \pi \varepsilon_{0} E / a e^{2}\right)^{1 / 2}$ [Eq. (6.69)] into (6.93) and solving for $E$, we get
\(
\begin{equation}
E=-\frac{Z^{2}}{n^{2}} \frac{e^{2}}{8 \pi \varepsilon{0} a}=-\frac{Z^{2} \mu e^{4}}{8 \varepsilon{0}^{2} n^{2} h^{2}} \tag{6.94}
\end{equation}
\)
where $a \equiv 4 \pi \varepsilon_{0} \hbar^{2} / \mu e^{2}$ [Eq. (6.63)]. These are the bound-state energy levels of the hydrogenlike atom, and they are discrete. Figure 6.6 shows the potential-energy curve [Eq. (6.57)] and some of the allowed energy levels for the hydrogen atom $(Z=1)$. The crosshatching indicates that all positive energies are allowed.
It turns out that all changes in $n$ are allowed in light absorption and emission. The wavenumbers [Eq. (4.64)] of H -atom spectral lines are then
\(
\begin{equation}
\widetilde{\nu} \equiv \frac{1}{\lambda}=\frac{v}{c}=\frac{E{2}-E{1}}{h c}=\frac{e^{2}}{8 \pi \varepsilon{0} a h c}\left(\frac{1}{n{1}^{2}}-\frac{1}{n{2}^{2}}\right) \equiv R{\mathrm{H}}\left(\frac{1}{n{1}^{2}}-\frac{1}{n{2}^{2}}\right) \tag{6.95}
\end{equation}
\)
where $R_{\mathrm{H}}=109677.6 \mathrm{~cm}^{-1}$ is the Rydberg constant for hydrogen.
Degeneracy
Are the hydrogen-atom energy levels degenerate? For the bound states, the energy (6.94) depends only on $n$. However, the wave function (6.61) depends on all three quantum numbers $n, l$, and $m$, whose allowed values are [Eqs. (6.91), (6.92), (5.104), and (5.105)]
FIGURE 6.6 Energy levels of the hydrogen atom.
\(
\begin{gather}
n=1,2,3, \ldots \tag{6.96}\
l=0,1,2, \ldots, n-1 \tag{6.97}\
m=-l,-l+1, \ldots, 0, \ldots, l-1, l \tag{6.98}
\end{gather}
\)
Hydrogen-atom states with different values of $l$ or $m$, but the same value of $n$, have the same energy. The energy levels are degenerate, except for $n=1$, where $l$ and $m$ must both be 0 . For a given value of $n$, we can have $n$ different values of $l$. For each of these values of $l$, we can have $2 l+1$ values of $m$. The degree of degeneracy of an H -atom bound-state level is found to equal $n^{2}$ (spin considerations being omitted); see Prob. 6.16. For the continuum levels, it turns out that for a given energy there is no restriction on the maximum value of $l$; hence these levels are infinity-fold degenerate.
The radial equation for the hydrogen atom can also be solved by the use of ladder operators (also known as factorization); see Z. W. Salsburg, Am. J. Phys., 33, 36 (1965).