201
7.3. The Maxwell field A
μ (x)
μν
ones available are g and k
μ k
ν . So the general form of (M
−1 )
νμ must be
(M
−1 )
νμ
νμ
= A(k
2 )g + B(k
2 )k
ν k
μ .
(7.92)
Now the inverse is defined by
ν
(M
−1 )
νμ M μσ = g σ .
(7.93)
Putting (7.92) and (7.90) into (7.93) yields (problem 7.11(b))
ν
ν
−k
2 A(k
2 )g σ + A(k
2 )k
ν k σ = g
(7.94)
σ
which cannot be satisfied. So we are thwarted again.
Nothing daunted, the attentive reader may have an answer ready for the
propagator problem. Suppose that, instead of (7.68), we start from the much
simpler equation
❗A
ν = 0
(7.95)
which results from imposing the Lorentz condition (7.70). Then, in momentum–
space, (7.95) becomes
−k
2 A ˜ ν = 0.
(7.96)
The ‘−k
2 ’ on the left-hand side certainly has an inverse, implying that the
Feynman propagator for the photon is (proportional to) g μν /k
2 . This form
is indeed plausible, as it is very much what we would expect by taking the
massless limit of the spin-0 propagator and tacking on g μν to account for the
Lorentz indices in <0|T (A ˆ μ (x 1 )A ˆ ν (x 2 ))|0> (but then why no term in k μ k ν ? –
see the final two paragraphs of this section!).
Perhaps this approach helps with the ‘no canonical momentum π
0 ’ problem
too. Let us ask: What Lagrangian leads to the field equation (7.95)? The
answer is (problem 7.12)
1
L L = − F μν F
μν
−
1
2 (∂ μ A
μ )
2 .
(7.97)
4
This form does seem to offer better prospects for quantization, since at least
all our π
μ ’s are non-zero; in particular
π
0
∂L
−∂ μ A
μ
=
=
.
(7.98)
∂A ˙0
The other π’s are unchanged by the addition of the extra term in (7.97) and
are given by
π
i
−A ˙ i + ∂
i A
0
=
.
(7.99)
Interestingly, these are precisely the electric fields E
i (see (2.10)). Let us see,
then, if all our problems are solved with L L .
7.3. The Maxwell field A
μ (x)
μν
ones available are g and k
μ k
ν . So the general form of (M
−1 )
νμ must be
(M
−1 )
νμ
νμ
= A(k
2 )g + B(k
2 )k
ν k
μ .
(7.92)
Now the inverse is defined by
ν
(M
−1 )
νμ M μσ = g σ .
(7.93)
Putting (7.92) and (7.90) into (7.93) yields (problem 7.11(b))
ν
ν
−k
2 A(k
2 )g σ + A(k
2 )k
ν k σ = g
(7.94)
σ
which cannot be satisfied. So we are thwarted again.
Nothing daunted, the attentive reader may have an answer ready for the
propagator problem. Suppose that, instead of (7.68), we start from the much
simpler equation
❗A
ν = 0
(7.95)
which results from imposing the Lorentz condition (7.70). Then, in momentum–
space, (7.95) becomes
−k
2 A ˜ ν = 0.
(7.96)
The ‘−k
2 ’ on the left-hand side certainly has an inverse, implying that the
Feynman propagator for the photon is (proportional to) g μν /k
2 . This form
is indeed plausible, as it is very much what we would expect by taking the
massless limit of the spin-0 propagator and tacking on g μν to account for the
Lorentz indices in <0|T (A ˆ μ (x 1 )A ˆ ν (x 2 ))|0> (but then why no term in k μ k ν ? –
see the final two paragraphs of this section!).
Perhaps this approach helps with the ‘no canonical momentum π
0 ’ problem
too. Let us ask: What Lagrangian leads to the field equation (7.95)? The
answer is (problem 7.12)
1
L L = − F μν F
μν
−
1
2 (∂ μ A
μ )
2 .
(7.97)
4
This form does seem to offer better prospects for quantization, since at least
all our π
μ ’s are non-zero; in particular
π
0
∂L
−∂ μ A
μ
=
=
.
(7.98)
∂A ˙0
The other π’s are unchanged by the addition of the extra term in (7.97) and
are given by
π
i
−A ˙ i + ∂
i A
0
=
.
(7.99)
Interestingly, these are precisely the electric fields E
i (see (2.10)). Let us see,
then, if all our problems are solved with L L .