146
5. Quantum Field Theory I: The Free Scalar Field
and this can be expressed in terms of the a ˆ’s and the a ˆ
† ’s using the expansion
ˆ
for φ and π ˆ and the commutator
[ˆ a(k), a ˆ
† (k
′ )] = (2π)
3 δ
3 (k − k
′ )
(5.158)
with all others vanishing. The result is, as expected,
∫
1
d
3
k
ˆ
H KG =
[ˆ a
† (k)ˆ a(k) + a ˆ(k)ˆ a
† (k)]ω
(5.159)
2
(2π) 3
and, normally ordering as usual, we arrive at
∫ d
3
k
ˆ
H KG =
a ˆ
† (k)ˆ a(k)ω.
(5.160)
(2π) 3
This supports the physical interpretation of the mode operators a ˆ
† and a ˆ as
ˆ
creation and destruction operators for quanta of the field φ as before, except
that now the energy–momentum relation for these particles is the relativistic
one, for particles of mass m.
ˆ
ˆ
ˆ
Since φ is real (φ = φ
† ) and has no spin degrees of freedom, it is called
a real scalar field. Only field quanta of one type enter – those created by
a ˆ
†
ˆ
and destroyed by a ˆ. Thus φ would correspond physically to a case where
there was a unique particle state of a given mass m – for example the π
0 field.
Actually, of course, we would not want to describe the π
0 in any fundamental
sense in terms of such a field, since we know it is not a point-like object (‘φ’
is defined only at the single space–time point (x, t)). The question of whether
true ‘elementary’ scalar fields exist in nature is an interesting one: in the
Standard Model, as we shall eventually see in volume 2, the Higgs field is a
scalar field (though it contains several components with different charge). It
remains to be seen if this field – and the associated quantum, the Higgs boson
– is a scalar, and if so whether it is elementary or composite.
We have learned how to describe free relativistic spinless particles of finite
mass as the quanta of a relativistic quantum field. We now need to understand
interactions in quantum field theory.
Problems
5.1 Verify equation (5.36).
5.2 Consider one-dimensional motion under gravity so that V (x) = −mgx in
(5.39). Evaluate S of (5.38) for t 1 = 0, t 2 = t 0 , for three possible trajectories:
(a) x(t) = at,
(b) x(t) =
1 gt
2 (the Newtonian result) and
2
(c) x(t) = bt
3
5. Quantum Field Theory I: The Free Scalar Field
and this can be expressed in terms of the a ˆ’s and the a ˆ
† ’s using the expansion
ˆ
for φ and π ˆ and the commutator
[ˆ a(k), a ˆ
† (k
′ )] = (2π)
3 δ
3 (k − k
′ )
(5.158)
with all others vanishing. The result is, as expected,
∫
1
d
3
k
ˆ
H KG =
[ˆ a
† (k)ˆ a(k) + a ˆ(k)ˆ a
† (k)]ω
(5.159)
2
(2π) 3
and, normally ordering as usual, we arrive at
∫ d
3
k
ˆ
H KG =
a ˆ
† (k)ˆ a(k)ω.
(5.160)
(2π) 3
This supports the physical interpretation of the mode operators a ˆ
† and a ˆ as
ˆ
creation and destruction operators for quanta of the field φ as before, except
that now the energy–momentum relation for these particles is the relativistic
one, for particles of mass m.
ˆ
ˆ
ˆ
Since φ is real (φ = φ
† ) and has no spin degrees of freedom, it is called
a real scalar field. Only field quanta of one type enter – those created by
a ˆ
†
ˆ
and destroyed by a ˆ. Thus φ would correspond physically to a case where
there was a unique particle state of a given mass m – for example the π
0 field.
Actually, of course, we would not want to describe the π
0 in any fundamental
sense in terms of such a field, since we know it is not a point-like object (‘φ’
is defined only at the single space–time point (x, t)). The question of whether
true ‘elementary’ scalar fields exist in nature is an interesting one: in the
Standard Model, as we shall eventually see in volume 2, the Higgs field is a
scalar field (though it contains several components with different charge). It
remains to be seen if this field – and the associated quantum, the Higgs boson
– is a scalar, and if so whether it is elementary or composite.
We have learned how to describe free relativistic spinless particles of finite
mass as the quanta of a relativistic quantum field. We now need to understand
interactions in quantum field theory.
Problems
5.1 Verify equation (5.36).
5.2 Consider one-dimensional motion under gravity so that V (x) = −mgx in
(5.39). Evaluate S of (5.38) for t 1 = 0, t 2 = t 0 , for three possible trajectories:
(a) x(t) = at,
(b) x(t) =
1 gt
2 (the Newtonian result) and
2
(c) x(t) = bt
3