Subatomic Physics

12.4. The Higgs Mechanism; Spontaneous Symmetry Breaking 399
The ground state is shifted from that of the globally gauge invariant example.
The lowest energy now occurs when
φ ∗ φ ≡ v′2
2
λ2 q2
= +
2η2 �2c2 1
2η 2 (A2 0 − A 2 ), (12.50)
and the expansion (12.44) can be used. Thus, for low excitations, where we keep
only linear terms in R and θ, the equation of motion for φ becomes
�
D 2 0 − D 2 +2λ 2 + 3q2
�
D 2 0 − D 2 + q2
�2c2 (A20 − A 2 �
) θ + q
�c
�2c2 (A20 − A 2 �
)
R =0,
� �
1 ∂A0
+ ∇·A =0.
c ∂t
(12.51)
The equation for R is somewhat more complicated than Eq. (12.45), but otherwise
there are no surprises. As anticipated, the R field has acquired a mass of √ 2λ�/c
and the θ field remains massless. Although there is an additional term that involves
the electromagnetic field, it can be eliminated by invoking the Lorentz condition,
Eq. (12.8). However, there is another change that has occurred, namely the electromagnetic
field quantum has acquired a mass. To see this fact explicitly, we return
to the charge and current, given by Eq. (12.47). With the substitution of ∂/∂t by
cD0 and ∇ by D, we find, for instance
ρ = iq[φ ∗ D0φ − φ(D0φ) ∗ ]. (12.52)
If we substitute Eq. (12.44) and keep only first-order terms in R and θ, weobtain
ρ = iq
c
�
∗ ∂φ
2iq
φ − φ∂φ∗ +
∂t ∂t � A0φ ∗ �
φ
(12.53)
′ q ∂θ 2q2
≈−v −
c ∂t �c A0v ′2 4q
− 2
�c A0v ′ R.
When this charge density is used in the equation of motion of A0, Eq. (12.7), we
find
1
c2 ∂2A0 ∂t2 −∇2A0 + 2q2
�c v′2A0 = iq
�
�
∗ ∂φ
φ − φ∂φ∗
c ∂t ∂t
≈ v′ q
c
∂θ 4q2
−
∂t �c A0v ′ R. (12.54)
By comparison with Eq. (12.10), we see that the new term (2q 2 /�c)v ′2 A0 corresponds
to a mass for the gauge “photon” of the electromagnetic field. The mass
is
�
m = 2 � q
c c v′ .