170
Chapter 3. Wave optics
smaller for an unbound particle. This will be discussed in more detail later. In principle, we substitute the terms S 0 , S 1 , S 2 , . . . into
(3.160) to form the wave function ψ(x, t).
We now turn our attention to the important special case where
the potentials A and φ have no explicit time dependence. In this
case the potentials can be written as A(x) and φ(x), respectively.
The earlier analysis showed that the Hamiltonian has no explicit
time dependence in this case, from which it follows that the total
energy H is conserved. According to Hamilton–Jacobi theory, the
function S 0 can be expressed as
S 0 (x, t) = W 0 (x) − H t,
(3.174)
where W 0 is Hamilton’s characteristic function. Noting that
vW 0 = vS 0 , we obtain
(vW 0 − qA)
2 = 2m (H − qφ) .
(3.175)
We recognize the right side as the square of the kinetic momentum
[ p(x) ]
2 , where
[ p(x) ]
2 = 2m [ H − qφ(x) ].
(3.176)
This is satisfied by
vW 0 − q A = ± p(x),
(3.177)
Retaining only the positive (right-propagating) root, and ignoring
the negative (left-propagating) root, we obtain
vW 0 = P(x),
(3.178)
recalling that P = p + q A is the canonical momentum. Integrating, we obtain
x b
W 0 (x b , t b ) − W 0 (x a , t a ) =
P · dx,
(3.179)
xa
Chapter 3. Wave optics
smaller for an unbound particle. This will be discussed in more detail later. In principle, we substitute the terms S 0 , S 1 , S 2 , . . . into
(3.160) to form the wave function ψ(x, t).
We now turn our attention to the important special case where
the potentials A and φ have no explicit time dependence. In this
case the potentials can be written as A(x) and φ(x), respectively.
The earlier analysis showed that the Hamiltonian has no explicit
time dependence in this case, from which it follows that the total
energy H is conserved. According to Hamilton–Jacobi theory, the
function S 0 can be expressed as
S 0 (x, t) = W 0 (x) − H t,
(3.174)
where W 0 is Hamilton’s characteristic function. Noting that
vW 0 = vS 0 , we obtain
(vW 0 − qA)
2 = 2m (H − qφ) .
(3.175)
We recognize the right side as the square of the kinetic momentum
[ p(x) ]
2 , where
[ p(x) ]
2 = 2m [ H − qφ(x) ].
(3.176)
This is satisfied by
vW 0 − q A = ± p(x),
(3.177)
Retaining only the positive (right-propagating) root, and ignoring
the negative (left-propagating) root, we obtain
vW 0 = P(x),
(3.178)
recalling that P = p + q A is the canonical momentum. Integrating, we obtain
x b
W 0 (x b , t b ) − W 0 (x a , t a ) =
P · dx,
(3.179)
xa
