The hydrogen atom

A hydrogen atom consists of an electron, of charge $-e$ and mass $m_e$, and a proton, of charge $+e$ and mass $m_p$, moving in the Coulomb potential

$$V({\bf r}) = - \frac{e^2}{4\pi\,\epsilon_0\,\vert{\bf r}\vert},$$ (640)

where ${\bf r}$ is the position vector of the electron with respect to the proton. Now, according to the analysis in Sect. 6.4, this two-body problem can be converted into an equivalent one-body problem. In the latter problem, a particle of mass

$$\mu = \frac{m_e\,m_p}{m_e+m_p}$$ (641)

moves in the central potential

$$V(r) = - \frac{e^2}{4\pi\,\epsilon_0\,r}.$$ (642)

Note, however, that since $m_e/m_p\simeq 1/1836$ the difference between $m_e$ and $\mu$ is very small. Hence, in the following, we shall write neglect this difference entirely.

Writing the wave-function in the usual form,

$$\psi(r,\theta,\phi) = R_{n,l}(r)\,Y_{l,m}(\theta,\phi),$$ (643)

it follows from Sect. 9.2 that the radial function $R_{n,l}(r)$ satisfies

$$-\frac{\hbar^2}{2\,m_e}\left(\frac{d^2}{dr^2} + \frac{2}{r}\... ...} -\left(\frac{e^2}{4\pi\,\epsilon_0\,r}+E \right) R_{n,l}= 0.$$ (644)

Let $r = a\,z$, where

$$a = \sqrt{\frac{\hbar^2}{2\,m_e\,(-E)}}=\sqrt{\frac{E_0}{E}}\,a_0,$$ (645)

where $E_0$ and $a_0$ are defined in Eqs. (657) and (658), respectively. Here, it is assumed that $E<0$, since we are only interested in bound-states of the hydrogen atom. The above differential equation transforms to

$$\left(\frac{d^2}{dz^2} + \frac{2}{z}\frac{d}{dz}-\frac{l\,(l+1)}{z^2}+ \frac{\zeta}{z}-1\right) R_{n,l} = 0,$$ (646)

where

$$\zeta = \frac{2\,m_e\,a\,e^2}{4\pi\,\epsilon_0\,\hbar^2}=2\,\sqrt{\frac{E_0}{E}}.$$ (647)

Suppose that $R_{n,l}(r) = Z(r/a)\,\exp(-r/a)/(r/a)$. It follows that

$$\left(\frac{d^2}{dz^2} -2\,\frac{d}{dz} - \frac{l\,(l+1)}{z^2} + \frac{\zeta}{z}\right) Z = 0.$$ (648)

We now need to solve the above differential equation in the domain $z=0$ to $z= \infty$, subject to the constraint that $R_{n,l}(r)$ be square-integrable.

Let us look for a power-law solution of the form

$$Z(z) = \sum_k c_k\,z^k.$$ (649)

Substituting this solution into Eq. (648), we obtain

$$\sum_k c_k\left\{k\,(k-1)\,z^{k-2} - 2\,k\,z^{k-1} - l\,(l+1)\,z^{k-2} + \zeta\,z^{k-1}\right\} = 0.$$ (650)

Equating the coefficients of $z^{k-2}$ gives the recursion relation

$$c_k\,\left[k\,(k-1)-l\,(l+1)\right] = c_{k-1}\,\left[2\,(k-1) - \zeta\right].$$ (651)

Now, the power series (649) must terminate at small $k$, at some positive value of $k$, otherwise $Z(z)$ behaves unphysically as $z\rightarrow 0$ [i.e., it yields an $R_{n,l}(r)$ that is not square-integrable as $r\rightarrow 0$]. From the above recursion relation, this is only possible if $[k_{min}\,(k_{min}-1)-l\,(l+1)]=0$, where the first term in the series is $c_{k_{min}}\,z^{k_{min}}$. There are two possibilities: $k_{min}=-l$ or $k_{min}=l+1$. However, the former possibility predicts unphysical behaviour of $Z(z)$ at $z=0$. Thus, we conclude that $k_{min}=l+1$. Note that, since $R_{n,l}(r)\simeq Z(r/a)/(r/a)\simeq (r/a)^l$ at small $r$, there is a finite probability of finding the electron at the nucleus for an $l=0$ state, whereas there is zero probability of finding the electron at the nucleus for an $l>0$ state [i.e., $\vert\psi\vert^2=0$ at $r=0$, except when $l=0$].

For large values of $z$, the ratio of successive coefficients in the power series (649) is

$$\frac{c_k}{c_{k-1}} = \frac{2}{k},$$ (652)

according to Eq. (651). This is the same as the ratio of successive coefficients in the power series

$$\sum_k \frac{(2\,z)^k}{k!},$$ (653)

which converges to $\exp(2\,z)$. We conclude that $Z(z)\rightarrow \exp(2\,z)$ as $z\rightarrow\infty$. It thus follows that $R_{n,l}(r)\sim Z(r/a)\,\exp(-r/a)/(r/a)\rightarrow \exp(r/a)/(r/a)$ as $r\rightarrow\infty$. This does not correspond to physically acceptable behaviour of the wave-function, since $\int\vert\psi\vert^2\,dV$ must be finite. The only way in which we can avoid this unphysical behaviour is if the power series (649) terminates at some maximum value of $k$. According to the recursion relation (651), this is only possible if

$$\frac{\zeta}{2} = n,$$ (654)

where $n$ is an integer, and the last term in the series is $c_n\,z^n$. Since the first term in the series is $c_{l+1}\,z^{l+1}$, it follows that $n$ must be greater than $l$, otherwise there are no terms in the series at all. Finally, it is clear from Eqs. (645), (647), and (654) that

$$E = \frac{E_0}{n^2}$$ (655)

and

$$a = n\,a_0,$$ (656)

where

$$E_0 = -\frac{m_e\,e^4}{2\,(4\pi\,\epsilon_0)^2\,\hbar^2} = - \frac{e^2}{8\pi\,\epsilon_0\,a_0} = -13.6\,{\rm eV},$$ (657)

and

$$a_0 = \frac{4\pi\,\epsilon_0\,\hbar^2}{m_e\,e^2} = 5.3\times 10^{-11}\,{\rm m}.$$ (658)

Here, $E_0$ is the energy of so-called ground-state (or lowest energy state) of the hydrogen atom, and the length $a_0$ is known as the Bohr radius. Note that $\vert E_0\vert\sim \alpha^2\,m_e\,c^2$, where $\alpha = e^2/ (4\pi\,\epsilon_0\,\hbar\,c)\simeq 1/137$ is the dimensionless fine-structure constant. The fact that $\vert E_0\vert\ll m_e\,c^2$ is the ultimate justification for our non-relativistic treatment of the hydrogen atom.

We conclude that the wave-function of a hydrogen atom takes the form

$$\psi_{n,l,m}(r,\theta,\phi) = R_{n,l}(r)\,Y_{l,m}(\theta,\phi).$$ (659)

Here, the $Y_{l,m}(\theta,\phi)$ are the spherical harmonics (see Sect 8.7), and $R_{n,l}(z=r/a)$ is the solution of

$$\left(\frac{1}{z^2}\,\frac{d}{dz}\,z^2\,\frac{d}{dz}- \frac{l\,(l+1)}{z^2} + \frac{2\,n}{z}-1\right) R_{n,l} = 0$$ (660)

which varies as $z^l$ at small $z$. Furthermore, the quantum numbers $n$, $l$, and $m$ can only take values which satisfy the inequality

$$\vert m\vert \leq l < n,$$ (661)

where $n$ is a positive integer, $l$ a non-negative integer, and $m$ an integer.

Now, we expect the stationary states of the hydrogen atom to be orthonormal: i.e.,

$$\int \psi^\ast_{n',l',m'}\,\psi_{n,l,m}\,dV = \delta_{nn'}\,\delta_{ll'}\,\delta_{mm'},$$ (662)

where $dV$ is a volume element, and the integral is over all space. Of course, $dV = r^2\,dr\,d{\mit\Omega}$, where $d{\mit\Omega}$ is an element of solid angle. Moreover, we already know that the spherical harmonics are orthonormal [see Eq. (594)]: i.e.,

$$\oint Y_{l',m'}^{\,\ast}\,Y_{l,m}\,d\Omega = \delta_{ll'}\,\delta_{mm'}.$$ (663)

It, thus, follows that the radial wave-function satisfies the orthonormality constraint

$$\int_0^{\infty} R_{n',l}^\ast\,R_{n,l}\,r^2\,dr = \delta_{nn'}.$$ (664)

The first few radial wave-functions for the hydrogen atom are listed below:

$$R_{1,0}(r) = \frac{2}{a_0^{\,3/2}}\,\exp\left(-\frac{r}{a_0}\right),$$ (665)

$$R_{2,0}(r) = \frac{2}{(2\,a_0)^{3/2}}\left(1-\frac{r}{2\,a_0}\right) \exp\left(-\frac{r}{2\,a_0}\right),$$ (666)

$$R_{2,1}(r) = \frac{1}{\sqrt{3}\,(2\,a_0)^{3/2}}\,\frac{r}{a_0}\, \exp\left(-\frac{r}{2\,a_0}\right),$$ (667)

$$R_{3,0}(r) = \frac{2}{(3\,a_0)^{3/2}}\left(1- \frac{2\,r}{3\,a_0} + \frac{2\,r^2}{27\,a_0^{\,2}}\right)\exp\left(-\frac{r}{3\,a_0}\right),$$ (668)

$$R_{3,1}(r) = \frac{4\,\sqrt{2}}{9\,(3\,a_0)^{3/2}}\,\frac{r}{a_0} \left(1-\frac{r}{6\,a_0}\right)\,\exp\left(-\frac{r}{3\,a_0}\right),$$ (669)

$$R_{3,2}(r) = \frac{2\,\sqrt{2}}{27\,\sqrt{5}\,(3\,a_0)^{3/2}} \left(\frac{r}{a_0}\right)^2 \exp\left(-\frac{r}{3\,a_0}\right).$$ (670)

These functions are illustrated in Figs. 18 and 19.

Figure 18: The $a_0\,r^2\,\vert R_{n,l}(r)\vert^{\,2}$ plotted as a functions of $r/a_0$. The solid, short-dashed, and long-dashed curves correspond to $n,l=1,0$, and $2,0$, and $2,1$, respectively.

Figure 19: The $a_0\,r^2\,\vert R_{n,l}(r)\vert^{\,2}$ plotted as a functions of $r/a_0$. The solid, short-dashed, and long-dashed curves correspond to $n,l=3,0$, and $3,1$, and $3,2$, respectively.

Given the (properly normalized) hydrogen wave-function (659), plus our interpretation of $\vert\psi\vert^2$ as a probability density, we can calculate

$$\langle r^k\rangle = \int_0^\infty r^{2+k}\,\vert R_{n,l}(r)\vert^{\,2}\,dr,$$ (671)

where the angle-brackets denote an expectation value. For instance, it can be demonstrated (after much tedious algebra) that

$$\langle r^2\rangle = \frac{a_0^{\,2}\,n^2}{2}\,[5\,n^2+1-3\,l\,(l+1)],$$ (672)

$$\langle r\rangle = \frac{a_0}{2}\,[3\,n^2-l\,(l+1)],$$ (673)

$$\left\langle \frac{1}{r}\right\rangle = \frac{1}{n^2\,a_0},$$ (674)

$$\left\langle\frac{1}{r^2}\right\rangle = \frac{1}{(l+1/2)\,n^3\,a_0^{\,2}},$$ (675)

$$\left\langle\frac{1}{r^3}\right\rangle = \frac{1}{l\,(l+1/2)\,(l+1)\,n^3\,a_0^{\,3}}.$$ (676)

According to Eq. (655), the energy levels of the bound-states of a hydrogen atom only depend on the radial quantum number $n$. It turns out that this is a special property of a $1/r$ potential. For a general central potential, $V(r)$, the quantized energy levels of a bound-state depend on both $n$ and $l$ (see Sect. 9.3).

The fact that the energy levels of a hydrogen atom only depend on $n$, and not on $l$ and $m$, implies that the energy spectrum of a hydrogen atom is highly degenerate: i.e., there are many different states which possess the same energy. According to the inequality (661) (and the fact that $n$, $l$, and $m$ are integers), for a given value of $l$, there are $2\,l+1$ different allowed values of $m$ (i.e., $-l,-l+1, \cdots, l-1, l$). Likewise, for a given value of $n$, there are $n$ different allowed values of $l$ (i.e., $0,1,\cdots, n-1$). Now, all states possessing the same value of $n$ have the same energy (i.e., they are degenerate). Hence, the total number of degenerate states corresponding to a given value of $n$ is

$$1 + 3 + 5 + \cdots +2\,(n-1)+1 = n^2.$$ (677)

Thus, the ground-state ($n=1$) is not degenerate, the first excited state ($n=2$) is four-fold degenerate, the second excited state ($n=3$) is nine-fold degenerate, etc. [Actually, when we take into account the two spin states of an electron (see Sect. 10), the degeneracy of the $n$th energy level becomes $2\,n^2$.]