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Outgoing boundary condition
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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
/
2018
/
S18
/
1
/
b
/
Solution
2026-10-25
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Away from
x
=
x
′
, the
free
equation
is
G
′′
+
k
2
G
=
0
. The
outgoing boundary condition
selects
G
=
A
e
ik
∣
x
−
x
′
∣
, continuous at its source. Integrating the
differential equation
across the source gives the
derivative
jump
−
2
m
ℏ
2
[
G
′
(
x
′
+
,
x
′
)
−
G
′
(
x
′
−
,
x
′
)
]
=
1.
(1)
The
jump
for the proposed form is
2
ik
A
, so
A
=
im
/
(
k
ℏ
2
)
. Thus
G
(
x
,
x
′
)
=
k
ℏ
2
im
e
ik
∣
x
−
x
′
∣
.
(2)
For
x
far to either side, the
x
′
-dependent
phase
is absorbed into the outgoing
amplitude
. The outgoing prescription is equivalently the
boundary value
of the
resolvent
with
E
→
E
+
i
0
; changing that prescription would give
a
different
Green function
.
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