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THE DIRAC EQUATION
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We then iterate the process. A second transformation H"
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where (!)" = O(m-2). Finally, a third step with S" = -(if3(!)"/2m) eliminates this odd term, leaving the desired hamiltonian
H'" =
13 [m + (p 8m
eA - (p)4 2m 8m 3
eAo - -e-f3a B 2m 8m
e e . + ( - -ie2 curl E - --a E x p) - --dlV E -a 2 2 4m
(2-82)
The interpretation of the various terms deserves some comments. The term in the bracket is the expansion (to the required order) of [(p - eA)2 + m 2]1/2. The second term eAo is the electrostatic energy of a point-like charge, whereas the third one represents the energy of a magnetic dipole for 9 = 2. The term inside parentheses may be seen to correspond to a spin-orbit (s.o.) interaction. Indeed, for a static spherically symmetric potential, curl E = 0 and E = - VAa, Therefore
1 dAo 1 dAo a . (E x p) = - - a (r x p) = - - a L
r dr
r dr
and the term in parentheses is H
=----a L 2
e dAo 4m r dr
The magnetic field B' = - v x E acting on the particle is responsible for this additional magnetic energy. Its interaction with the magnetic moment p (2-72) would give e , e H 8.0. = - -2m a B = - 2m 2 a (E x p) but due to the Thomas precession, this result is reduced by a factor 2. Finally the last term in Eq. (2-82), referred to as the Darwin term -(eI8m 2) div E, may be traced to the zitterbewegung. The electron position fluctuates by an amount (5r such that <(5r2 ) ~ 11m 2 , and its effective electrostatic energy is the average
From spherical symmetry, this random fluctuation is (5ii (5ij <(5r i (5ri) = - <(5r2 ) ~ - 2
3 3m
QUANTUM FIELD THEORY
and the correction to eAO(r) is
in good agreement with the sign and magnitude of the Darwin term. The reader will have noticed that as the Foldy-Wouthuysen transformation is time dependent, the expectation values of H' in the state l/J' are, in general, different from those of H in the corresponding state l/J.
2-3 HYDROGEN-LIKE ATOMS
An important application of the Dirac equation is the discussion of the fine structure of atomic spectra. This is a field notorious for the successes of quantum mechanics and its relativistic generalizations, including radiative corrections. In view of the intricacy of the Foldy-Wouthuysen method, it is quite remarkable that an exact solution of the Dirac equation in a static Coulomb field exists and leads to an excellent agreement with the observed results on hydrogen-like atoms. The difficulties discussed in Sec. 2-2 are not expected to playa significant role. In atomic physics, the relevant length scale, the Bohr radius ao = hjmecex = 0.53 A, is 137 times larger than the electron wavelength. For heavy atoms these difficulties do arise and other methods have to be developed.
2-3-1 Nonrelativistic Versus Relativistic Spectrum
Recall the nonrelativistic result obtained from the Schrodinger equation, a triumph for early quantum mechanics. The wave equation is 2 - ~ - Ze _ en l]l/Jn l(r) = [ 2m 4nr . .
where
-Ll=
- or2- - - + r2 r or
L l/Jn.1 = 1(1
+ l)l/Jn.1
m; 1
1=0,1, ...
and m is the reduced mass of the electron nucleus system: m1 =
+ mN 1 ~ m; 1
The quantization condition requires that n' = n - (l + 1) must be a nonnegative integer (number of zeros of the wave function), and the energy levels are
en.l
m(Zex 2n 2
n = 1,2, ...
(2-83)
Numerically, the Rydberg constant mex 2 j2 is equal to 13.6 eV.
THE DIRAC EQUATION
The s-state (I = 0) wave function at the origin is
1jJ(0)
1[1/2
/ 1 (mZa)3 2
(2-84)
Since for a given n ;;::: 1, I can take integer values from 0 to n - 1, each level has a degeneracy equal to L~-l (21 + 1) = n2 This is related to a dynamical symmetry of the Coulomb problem according to a group 0(4) of rotations in four-dimensional space, which was used by Pauli and Fock in the early days of quantum mechanics to give an algebraic derivation of the Balmer spectrum (2-83). This degeneracy disappears in the relativistic treatment leading to the fine splitting of the spectrum. Ignoring for a while the effects of spin, let us see the predictions of the Klein-Gordon theory when the electromagnetic coupling is introduced in a minimal way. Let E denote the total energy equal to the rest energy mc 2 plus negative binding energy e :
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