198
7. Quantum Field Theory III
provided ˜
χ satisfies the massless KG equation
❗χ ˜ = 0.
(7.75)
This condition on ˜
χ ensures that, even after the further shift, the resulting
potential still satisfies ∂ μ A
μ = 0. For our plane-wave solutions, this residual
gauge freedom corresponds to changing ∈
μ by a multiple of k
μ :
∈
μ
→ ∈
μ + βk
μ
≡ ∈
′μ
(7.76)
which still satisfies ∈
′μ
· k = 0 since k
2 = 0 for these free-field solutions. The
condition k
2 = 0 is, of course, the statement that a free photon is massless.
This freedom has important consequences. Consider a solution with
k
μ = (k
0 , k)
(k
0 )
2 = k
2
(7.77)
and polarization vector
∈
μ = (∈
0 , ∈)
(7.78)
satisfying the Lorentz condition
k · ∈ = 0.
(7.79)
Gauge invariance now implies that we can add multiples of k
μ to ∈
μ and still
have a satisfactory polarization vector.
It is therefore clear that we can arrange for the time component of ∈
μ to
vanish so that the Lorentz condition reduces to the 3-vector condition
k · ∈ = 0.
(7.80)
This means that there are only two independent polarization vectors, both
transverse to k, i.e. to the propagation direction. For a wave travelling in the
z-direction (k
μ = (k
0 , 0, 0, k
0 )) these may be chosen to be
∈ (1) = (1, 0, 0)
(7.81)
∈ (2) = (0, 1, 0).
(7.82)
Such a choice corresponds to linear polarization of the associated E and B
fields – which can be easily calculated from (2.10) and (2.11), given
μ
A = N (0, ∈ (i) )e
−ik·x
i = 1, 2.
(7.83)
(i)
A commonly used alternative choice is
1
∈(λ = +1) = − √ (1, i, 0)
(7.84)
2
1
∈(λ = −1) = √ (1, −i, 0)
(7.85)
2
7. Quantum Field Theory III
provided ˜
χ satisfies the massless KG equation
❗χ ˜ = 0.
(7.75)
This condition on ˜
χ ensures that, even after the further shift, the resulting
potential still satisfies ∂ μ A
μ = 0. For our plane-wave solutions, this residual
gauge freedom corresponds to changing ∈
μ by a multiple of k
μ :
∈
μ
→ ∈
μ + βk
μ
≡ ∈
′μ
(7.76)
which still satisfies ∈
′μ
· k = 0 since k
2 = 0 for these free-field solutions. The
condition k
2 = 0 is, of course, the statement that a free photon is massless.
This freedom has important consequences. Consider a solution with
k
μ = (k
0 , k)
(k
0 )
2 = k
2
(7.77)
and polarization vector
∈
μ = (∈
0 , ∈)
(7.78)
satisfying the Lorentz condition
k · ∈ = 0.
(7.79)
Gauge invariance now implies that we can add multiples of k
μ to ∈
μ and still
have a satisfactory polarization vector.
It is therefore clear that we can arrange for the time component of ∈
μ to
vanish so that the Lorentz condition reduces to the 3-vector condition
k · ∈ = 0.
(7.80)
This means that there are only two independent polarization vectors, both
transverse to k, i.e. to the propagation direction. For a wave travelling in the
z-direction (k
μ = (k
0 , 0, 0, k
0 )) these may be chosen to be
∈ (1) = (1, 0, 0)
(7.81)
∈ (2) = (0, 1, 0).
(7.82)
Such a choice corresponds to linear polarization of the associated E and B
fields – which can be easily calculated from (2.10) and (2.11), given
μ
A = N (0, ∈ (i) )e
−ik·x
i = 1, 2.
(7.83)
(i)
A commonly used alternative choice is
1
∈(λ = +1) = − √ (1, i, 0)
(7.84)
2
1
∈(λ = −1) = √ (1, −i, 0)
(7.85)
2