80
3. Relativistic Quantum Mechanics
and ρ ≥ 0 always. We nevertheless want to set up a correspondence so that
positive-energy solutions describe electrons (taken to be the ‘particle’, by convention, in this case) and negative-energy solutions describe positrons, if we
reverse the sense of incoming and outgoing waves. For the KG case this
was straightforward, since the probability current was proportional to the
4-momentum:
μ
j
μ (KG) ∼ p .
(3.95)
We were therefore able to set up the correspondence for the electromagnetic
current of π
+ and π
− :
μ
π
+
j
μ
:
em
∼ ep
positive energy π
+
(3.96)
μ
π
− : j
μ
∼ (−e)p
positive energy π
−
(3.97)
em
≡ (+e)(−p
μ )
negative energy π
+ .
(3.98)
This simple connection does not hold for the Dirac case since ρ ≥ 0 for
both signs of the energy. It is still possible to set up the correspondence,
but now an extra minus sign must be inserted ‘by hand’ whenever we have a
negative-energy fermion in the final state. We shall make use of this rule in
section 8.2.4. We therefore state the Feynman hypothesis for fermions:
The invariant amplitude for the emission (absorption) of an antifermion
μ
of 4-momentum p and spin projection s z in the rest frame is equal to
the amplitude (minus the amplitude) for the absorption (emission) of a
μ
fermion of 4-momentum −p and spin projection −s z in the rest frame.
As we shall see in chapters 5–7, the Feynman interpretation of the negativeenergy solutions is naturally embodied in the field theory formalism.
3.5 Inclusion of electromagnetic interactions via the
gauge principle: the Dirac prediction of g = 2
for the electron
Having set up the relativistic spin-0 and spin1 free-particle wave equations,
2
we are now in a position to use the machinery developed in chapter 2, in
order to include electromagnetic interactions. All we have to do is make the
replacement
∂
μ
→ D
μ
≡ ∂
μ + iqA
μ
(3.99)
for a particle of charge q. For the spin-0 KG equation (3.10) we obtain, after
some rearrangement (problem 3.9),
(❗ + m
2 )φ = −iq(∂ μ A
μ + A
μ ∂ μ )φ + q
2 A
2 φ
(3.100)
= −V ˆ KG φ.
(3.101)
3. Relativistic Quantum Mechanics
and ρ ≥ 0 always. We nevertheless want to set up a correspondence so that
positive-energy solutions describe electrons (taken to be the ‘particle’, by convention, in this case) and negative-energy solutions describe positrons, if we
reverse the sense of incoming and outgoing waves. For the KG case this
was straightforward, since the probability current was proportional to the
4-momentum:
μ
j
μ (KG) ∼ p .
(3.95)
We were therefore able to set up the correspondence for the electromagnetic
current of π
+ and π
− :
μ
π
+
j
μ
:
em
∼ ep
positive energy π
+
(3.96)
μ
π
− : j
μ
∼ (−e)p
positive energy π
−
(3.97)
em
≡ (+e)(−p
μ )
negative energy π
+ .
(3.98)
This simple connection does not hold for the Dirac case since ρ ≥ 0 for
both signs of the energy. It is still possible to set up the correspondence,
but now an extra minus sign must be inserted ‘by hand’ whenever we have a
negative-energy fermion in the final state. We shall make use of this rule in
section 8.2.4. We therefore state the Feynman hypothesis for fermions:
The invariant amplitude for the emission (absorption) of an antifermion
μ
of 4-momentum p and spin projection s z in the rest frame is equal to
the amplitude (minus the amplitude) for the absorption (emission) of a
μ
fermion of 4-momentum −p and spin projection −s z in the rest frame.
As we shall see in chapters 5–7, the Feynman interpretation of the negativeenergy solutions is naturally embodied in the field theory formalism.
3.5 Inclusion of electromagnetic interactions via the
gauge principle: the Dirac prediction of g = 2
for the electron
Having set up the relativistic spin-0 and spin1 free-particle wave equations,
2
we are now in a position to use the machinery developed in chapter 2, in
order to include electromagnetic interactions. All we have to do is make the
replacement
∂
μ
→ D
μ
≡ ∂
μ + iqA
μ
(3.99)
for a particle of charge q. For the spin-0 KG equation (3.10) we obtain, after
some rearrangement (problem 3.9),
(❗ + m
2 )φ = −iq(∂ μ A
μ + A
μ ∂ μ )φ + q
2 A
2 φ
(3.100)
= −V ˆ KG φ.
(3.101)