The Hydrogen Atom
We have here a two-body problem, of an electron of charge -e and mass m, and a proton of charge +e and mass M. By using CM and relative coordinates and working in the CM frame, we can reduce the problem to the dynamics of a single particle whose mass µ =
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f/>=e/r
(13.1.1)
due to the proton is V = -e2I r, the Schrodinger equation (13.1.2) determines the energy levels in the rest frame of the atom, as well as the wave functionst If/ Etm(r, (),
4>) =
REt(r) Y'!'( (),
4>) =
UEl(r) Y/( (),
r
4>)
(13.1.3)
It is clear upon inspection of Eq. ( 13.1.2) that a power series ansatz will lead to a three-term recursion relation. So we try to factor out the asymptotic behavior.
t It should be clear from the context whether angular momentum.
m
stands for the electron mass or the z component of
353
354 CHAPTER 13
We already know from Section 12.6 that up to (possibly fractional) powers of r [Eq. (12.6.19)],
Um r--+~ exp[-(2mW/fi 2 ) 1/ 2r]
(13.1.4)
~;r:_
where
W=-E is the binding energy (which is the energy it would take to liberate the electron) and that (13.1.5)
Equation (13.1.4) suggests the introduction of the dimensionless variable (13.1.6)
and the auxiliary function v m defined by (13.1.7)
The equation for v is then ( 13.1.8) where (13.1.9) and the subscripts on v are suppressed. You may verify that if we feed in a series into Eq. (13.1.8), a two-term recursion relation will obtain. Taking into account the behavior near p=O [Eq. (13.1.5)] we try (13.1.10) k~O
and obtain the following recursion relation between successive coefficients: ck+l
ck
-e 2JL+2(k+/+ 1) ---
--
(k+l+2)(k+l+l)-··l(l+l)
(13.1.11)
The Energy Levels
Since ck+l ------------ -----+ ·w
ck k
2 k ·~
(13.1.12)
is the behavior of the series pme2P, and would lead to u~e-pv~pme-pe2 P-pmep as p-HfJ, we demand that the series terminate at some k. This will happen if (13.1.13) or [from Eq. (13.1.9)]
k=O, 1, 2, ... : l=O, 1, 2,...
(13.1.14)
In terms -.:>f the principal quantum number
n=k+l+ 1
( 13.1.15)
the allowed energies are
n=1,2,3, ...
(13.1.16)
and at each n the allowed values of I are, according to Eq. (13.1.15), l = n - k - 1 = n - 1, n - 2, ... , 1, 0
( 13.1.17)
That states of different I should be degenerate indicates that H contains more symmetries besides rotational invariance. We discuss these later. For the present, let us note that the degeneracy at each n is n-1
L
(21+ l)=n 2
(13.1.18)
t~o
It is common to refer to the states with l = 0, 1, 2, 3, 4, ... as s, p, d,j, g, h, ... states. In this spectroscopic notation, Is denotes the state (n = l, l = 0); 2s and 2p the l = 0 and l= 1 states at n=2; 3s, 3p, and 3d the /=0, 1, and 2 states at n=3, and so on. No attempt is made to keep track of m. It is convenient to employ a natural unit of energy, called a Rydberg (Ry), for measuring the energy levels of hydrogen: 4
R _me y- 21i2
(13.1.19)
355 THE HYDROGEN ATOM
356 CHAPTER 13
E
t0
-1/16 -1/9
-1/4
-1
- -d
f
-n=4 -n=3
-n=2
Figure 13.1. The first few eigenstates of hydrogen. The energy is measured in Rydbergs and the states are labelled in the spectroscopic notation.
-n=l
in terms of which -Ry
E n = 2-
n
(13.1.20)
Figure 13.1 shows some of the lowest-energy states of hydrogen.
The Wave Functions Given the recursion relations, it is a strai
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