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# Hydrogen Atom

A hydrogen atom consists of an electron, of charge and mass , and a proton, of charge and mass , moving in the Coulomb potential
 (661)

where 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
 (662)

moves in the central potential
 (663)

Note, however, that since the difference between and is very small. Hence, in the following, we shall write neglect this difference entirely.

Writing the wavefunction in the usual form,

 (664)

it follows from Sect. 9.2 that the radial function satisfies
 (665)

Let , with
 (666)

where and are defined in Eqs. (678) and (679), respectively. Here, it is assumed that , since we are only interested in bound-states of the hydrogen atom. The above differential equation transforms to
 (667)

where
 (668)

Suppose that . It follows that
 (669)

We now need to solve the above differential equation in the domain to , subject to the constraint that be square-integrable.

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

 (670)

Substituting this solution into Eq. (669), we obtain
 (671)

Equating the coefficients of gives the recursion relation
 (672)

Now, the power series (670) must terminate at small , at some positive value of , otherwise behaves unphysically as [i.e., it yields an that is not square-integrable as ]. From the above recursion relation, this is only possible if , where the first term in the series is . There are two possibilities: or . However, the former possibility predicts unphysical behaviour of at . Thus, we conclude that . Note that, since at small , there is a finite probability of finding the electron at the nucleus for an state, whereas there is zero probability of finding the electron at the nucleus for an state [i.e., at , except when ].

For large values of , the ratio of successive coefficients in the power series (670) is

 (673)

according to Eq. (672). This is the same as the ratio of successive coefficients in the power series
 (674)

which converges to . We conclude that as . It thus follows that as . This does not correspond to physically acceptable behaviour of the wavefunction, since must be finite. The only way in which we can avoid this unphysical behaviour is if the power series (670) terminates at some maximum value of . According to the recursion relation (672), this is only possible if
 (675)

where is an integer, and the last term in the series is . Since the first term in the series is , it follows that must be greater than , otherwise there are no terms in the series at all. Finally, it is clear from Eqs. (666), (668), and (675) that
 (676)

and
 (677)

where
 (678)

and
 (679)

Here, is the energy of so-called ground-state (or lowest energy state) of the hydrogen atom, and the length is known as the Bohr radius. Note that , where is the dimensionless fine-structure constant. The fact that is the ultimate justification for our non-relativistic treatment of the hydrogen atom.

We conclude that the wavefunction of a hydrogen atom takes the form

 (680)

Here, the are the spherical harmonics (see Sect 8.7), and is the solution of
 (681)

which varies as at small . Furthermore, the quantum numbers , , and can only take values which satisfy the inequality
 (682)

where is a positive integer, a non-negative integer, and an integer.

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

 (683)

where is a volume element, and the integral is over all space. Of course, , where is an element of solid angle. Moreover, we already know that the spherical harmonics are orthonormal [see Eq. (615)]: i.e.,
 (684)

It, thus, follows that the radial wavefunction satisfies the orthonormality constraint
 (685)

The first few radial wavefunctions for the hydrogen atom are listed below:
 (686) (687) (688) (689) (690) (691)

These functions are illustrated in Figs. 21 and 22.

Given the (properly normalized) hydrogen wavefunction (680), plus our interpretation of as a probability density, we can calculate

 (692)

where the angle-brackets denote an expectation value. For instance, it can be demonstrated (after much tedious algebra) that
 (693) (694) (695) (696) (697)

According to Eq. (676), the energy levels of the bound-states of a hydrogen atom only depend on the radial quantum number . It turns out that this is a special property of a potential. For a general central potential, , the quantized energy levels of a bound-state depend on both and (see Sect. 9.3).

The fact that the energy levels of a hydrogen atom only depend on , and not on and , 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 (682) (and the fact that , , and are integers), for a given value of , there are different allowed values of (i.e., ). Likewise, for a given value of , there are different allowed values of (i.e., ). Now, all states possessing the same value of have the same energy (i.e., they are degenerate). Hence, the total number of degenerate states corresponding to a given value of is

 (698)

Thus, the ground-state () is not degenerate, the first excited state () is four-fold degenerate, the second excited state () 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 th energy level becomes .]

Next: Rydberg Formula Up: Central Potentials Previous: Infinite Spherical Potential Well
Richard Fitzpatrick 2010-07-20