joccurs in both tensors (and is summed over) we can simplify things. Take the case thati= 1
andk= 2. We only have a nonzero term ifj= 3 so the other 2 terms in the sum are zero. But if
j= 3, we must have eitherm= 1 andn= 2 or vice versa. We also must not havei=ksince all
the indices have to be different on each epsilon. So we can write.
ǫmnjǫijk=ǫmnjǫkij= (δkmδin−δknδim)Applying this identity to the Maxwell equation above, we get.
∂
∂xi∂
∂xkAi−∂
∂xi∂
∂xiAk+1
c∂
∂xk∂φ
∂t+
1
c^2∂^2 Ak
∂t^2=
4 π
cJk∂
∂xk∇·~ A~−∇^2 Ak+^1
c∂
∂xk∂φ
∂t+
1
c^2∂^2 Ak
∂t^2=
4 π
cJk−∇^2 Ak+1
c^2∂^2 Ak
∂t^2+
∂
∂xk(
∇·~ A~+^1
c∂φ
∂t)
=
4 π
cJk−∇^2 A~+
1
c^2∂^2 A~
∂t^2+~∇
(
∇·~ A~+^1
c∂φ
∂t)
=
4 π
cJ~
The last two equations derived are wave equations with source terms obeyed by the potentials. As
discussed in the opening section of this chapter, they can be simplified with a choice of gauge.
20.5.2 The Lorentz Force from the Classical Hamiltonian
In this section, we wish to verify that the Hamiltonian
H=
1
2 m(
~p+
e
cA~
) 2
−eφgives the correct Lorentz Force law in classical physics. We will thenproceed to use this Hamiltonian
in Quantum Mechanics.
Hamilton’s equationsare
q ̇ =∂H
∂pp ̇ = −∂H
∂qwhere~q≡~rand the conjugate momentum is already identified correctly~p≡~p. Remember that
these are applied assuming q and p are independent variables.
Beginning with ̇q=∂H/∂p, we have
d~r
dt=
1
m(
~p+
e
cA~
)
m~v = ~p+e
cA~
~p = m~v−e
c