Quantization of the Maxwell Field: Photons
We will now start to quantize the Maxwell field Aμ(x) = {− Φ(x)∕c, A(x)} similar to the quantization of the Schrödinger field. The fact that electromagnetism has a gauge invariance implies that there are more components than actual dynamical degrees of f
- PDF / 703,484 Bytes
- 60 Pages / 439.36 x 666.15 pts Page_size
- 21 Downloads / 395 Views
Quantization of the Maxwell Field: Photons
We will now start to quantize the Maxwell field Aμ (x) = {− (x)/c, A(x)} similar to the quantization of the Schrödinger field. The fact that electromagnetism has a gauge invariance implies that there are more components than actual dynamical degrees of freedom in the Maxwell field. This will make quantization a little more challenging than for the Schrödinger field, but we will overcome those difficulties. Electromagnetic field theory is implicitly relativistic, and quantized Maxwell theory therefore also provides us with a first example of a relativistic quantum field theory. Appendix B provides an introduction to 4-vector and tensor notation in electromagnetic theory.
18.1 Lagrange Density and Mode Expansion for the Maxwell Field The equations of motion for the Maxwell field are the inhomogeneous Maxwell equations,1 ∂μ F μν = ∂μ ∂ μ Aν − ∂ ν Aμ = − μ0 j ν .
(18.1)
These equations can be written as jν +
∂L ∂L 1 ∂μ ∂ μ Aν − ∂ ν Aμ = − ∂μ =0 μ0 ∂Aν ∂(∂μ Aν )
(18.2)
1 Recall
from electrodynamics that the homogeneous Maxwell equations, viz. Gauss’ law of absence of magnetic monopoles and Faraday’s law of induction, were solved through the introduction of the potentials Aμ .
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 R. Dick, Advanced Quantum Mechanics, Graduate Texts in Physics, https://doi.org/10.1007/978-3-030-57870-1_18
431
432
18 Quantization of the Maxwell Field: Photons
if we use the Lagrange density L = j ν Aν −
1 0 1 2 Fμν F μν = E 2 − B + j · A − . 4μ0 2 2μ0
(18.3)
This Lagrangian provides us with the canonically conjugate momentum for the vector potential A: ˙ = 0 (A ˙ + ∇) = − 0 E, A = ∂L/∂ A
(18.4)
˙ =0 = ∂L/∂
(18.5)
but
vanishes identically! Therefore we cannot simply impose canonical commutation relations between the four components Aμ of the 4-vector potential and four conjugate momenta ν . To circumvent this problem we revisit the pertinent Maxwell equations (18.1), i.e. Coulomb’s law, ˙ =− + ∇ · A
1 , 0
(18.6)
and Ampère’s law, ∇(∇ · A) − A +
1 ∂ 1 ∂2 A + 2 ∇ = μ0 j . 2 2 c ∂t c ∂t
(18.7)
One way to solve the problem with = 0 is to eliminate ∇ · A from the equations of motion through the gauge freedom (x, t) → f (x, t) = (x, t) − f˙(x, t),
(18.8)
A(x, t) → Af (x, t) = A(x, t) + ∇f (x, t),
(18.9)
i.e. we impose the gauge condition ∇ · Af = 0. The equation f (x, t) = − ∇ · A(x, t)
(18.10)
can be solved with the Green’s function G(r) = (4π r)−1 for the Laplace operator,
1 = − δ(x − x ), 4π |x − x |
see Eqs. (11.14) and (11.24) for E = 0. This yields
(18.11)
18.1 Lagrange Density and Mode Expansion for the Maxwell Field
f (x, t) =
1 4π
d 3x
1 ∇ · A(x , t). |x − x |
433
(18.12)
This gauge is denoted as Coulomb gauge. We denote the gauge transformed fields again with and A, i.e. we have ∇ · A(x, t) = 0.
(18.13)
and = −
1 , 0
(18.14)
1 ∂2 1 ∂ A − A + 2 ∇ = μ0 j . c2 ∂t 2 c ∂t
(18.15)
W
Data Loading...
