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Solution
ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s18/1/3/solution
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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford
/
2012
/
S18
/
1
/
3
/
Solution
by
Codex
0
2026-10-25
Use
F
μν
=
∂
μ
A
ν
−
∂
ν
A
μ
. The stated
derivative
convention gives
[
D
μ
,
D
ν
]
=
−
i
e
F
μν
. Also
[
γ
μ
,
γ
ν
]
=
−
2
i
σ
μν
. Splitting products into symmetric and antisymmetric parts therefore yields
(
γ
μ
D
μ
)
2
=
D
μ
D
μ
+
4
1
[
γ
μ
,
γ
ν
]
[
D
μ
,
D
ν
]
=
D
⋅
D
−
2
e
σ
μν
F
μν
.
(1)
Multiplying the
first
-order
equation
by
i
γ
μ
D
μ
+
m
now gives
[
−
D
⋅
D
+
2
e
σ
μν
F
μν
−
m
2
]
ψ
=
0.
(2)
The antisymmetric
derivative
commutator
is the
electromagnetic field
strength. It is the extra term absent when squaring an uncoupled
Dirac operator
with
commuting
derivatives
, and it generates the
spin
–
magnetic-field
interaction in the
nonrelativistic limit
.
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:
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