The Dirac Equation
Nonrelativistic quantum mechanics, which was developed in the previous chapters, is very successful when applied to problems like the hydrogen atom, where the typical velocity (speaking semiclassically) is small compared to c. (Recall v/c=β=α≅.1/ 137 in t
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20.1. The Free-Particle Dirac Equation Let us consider the simplest case, of a free particle. We start by stating the relation between classical mechanics and the free-particle Schrodinger equation in a way that facilitates generalization. If we start with the nonrelativistic relation (20.1.1)
and make the substitution p--+P
a .Yt -+illar
(20.1.2)
563
564
and let both sides act on a state vector I If/), we get Schrodinger's equation
CHAPTER 20
(20.1.3) A natural starting point for the relativistic equation is the corresponding relation due to Einstein (20.1.4) If we make the substitution mentioned above, we get (20.1.5) This equation is undesirable because it treats space and time asymmetrically. To see this, we first go to the momentum basis, where P is just p and the square root may be expanded in a series: (20.1.6) If we now transform to the coordinate basis, each p 2 becomes ( -1i 2 V2) and the asymmetry between space and time is manifest. What we want is an equation that is of the same order in both space and time. There are two ways out. One is to replace Eq. (20.1.4) by (20.1.7) and obtain, upon making the operator substitution, (20.1.8a) In the coordinate basis this becomes 1
[
a-
')2]
mc,--v +c ot rz 7
2
2
(
lfi=O
(20.1.8b)
This is called the Klein-Gordon equation and has the desired symmetry between space and time. But we move along, since 1f1 here is a scalar and cannot describe the electron. It is, however, a good candidate for pions, kaons, etc., which are spinless. The second alternative, due to Dirac, is the following. Let us suppose that the quantity in the square root in Eq. (20.1.5) can be written as a perfect square of a quantity that is linear in P. We can then take the square root (which will give us
our Hamiltonian) and obtain an equation that is of the first order in time and space. So let us write
c2 P 2 +m2 c4 = (caxPx+cayPy+cazPz+ f3mc 2 f =
(20.1.9)
(ca· P+ f3mc 2 ) 2
where a and f3 are to be determined by matching both sides of
c2 (P?, + P} + P;) + m 2 c4 =
[c 2 (a;P?c+ a;P}+ a;P;) + f3 2m2 c4 ]
+ [c2 PxPy(axay+ ayax) + and cyclic permutations] + [mc3Px(axf3 + f3ax) + x--+y + x--+z]
(20.l.IO)
(We have assumed that a and f3 are space independent, which is a reasonable assumption for a free particle.) These equations tell us that
(i=x, y, z) (i #j)
(20.1.11)
It is evident that a and f3 are not c numbers. They are matrices and furthermore
Hermitian (so that the Hamiltonian H=ca·P+f3mc 2 is Hermitian), traceless, and have eigenvalues ±1. (Recall the results of Exercise 1.8.8). They must also be even dimensional if the last two properties are to be compatible. They cannot be 2 x 2 matrices, since, as we saw in Exercise 14.3.8, the set of three Pauli matrices with these properties cannot be enlarged to include a fourth. So they must be 4 x 4 matrices. They are not unique (since a --+St aS, f3 --+St f3S preserves the desired properties if Sis unitary.) The following four are frequently used and will be used by us:
a=[~ ~l In the above,
CJ
(20.1.12)
and I ar
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