The One-Electron Atom
125
similar analysis for charge conservation shows that the final state consists
of a negatively charged electron (with energy E n ) and a positively charged
particle (corresponding to the hole in the negative-charge sea). The hole
therefore has properties exactly opposite to those of the vacant negative
energy electron state, i.e. it has positive energy and positive charge (also
opposite momentum and spin). This hole state is called the positron, which
is an example of what are known as antiparticles. The overall process is
equivalent to two photons annihilating each other to produce an electron
and a positron. The process is known as pair creation. Similarly, pair
annihilation occurs when a positive energy electron drops into a vacancy
in the negative energy sea, giving out radiation [Fig. 4.3 (b)]. Pair creation
and annihilation are important processes in particle physics, though the
associated particles may not always be two photons or electrons.
4. The presence of an external charge polarizes the sea of negative charges
thus reducing the effective charge of the external particle. This is called
vacuum polarization. As a result, an s-wave electron in the hydrogen
atom, which is ‘nearer’ to the proton than a p-wave electron, sees a greater
charge for the proton. Hence, the vacuum polarization lowers the s-wave
levels compared to the p-wave levels. This contributes to the removal of
j-degeneracy, in particular, the degeneracy between 2p 1/2 and 2s 1/2 states.
However, there are additional contributions to the separation of these energy
levels called the Lamb shift, for other effects such as self interaction, field
fluctuations, etc. which can be treated within the framework of quantum
electrodynamics. The predictions of the theory for the Lamb shift are in
excellent agreement with the experimental observations.
4.9 EXAMPLES
A few examples to illustrate the properties of the one-electron atoms are discussed
here.
Example 1
Though the solutions to the radial equation, Eq. (4.13), are in general complicated,
the solutions for l = n – 1, are fairly simple.
Consider a solution of the form
R (r) = r
n–1
exp [– (–2m r E/
2
)
1/2
r]
(4.106)
substitution of which in Eq. (4.13) gives the relation
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