396
G. Green Functions
where s(r) is a known ‘source function’ (e.g. the charge density in (G.3)).
The solution of (G.18) may be written as
∫
′
ψ(r) = u(r) + G(r − r
′ )s(r
′ ) d
3
r
(G.19)
where u(r) is a solution of Ω r u(r) = 0. Thus once we know G, we have the
solution via (G.19).
Equation (G.19) has a simple physical interpretation. We know that G(r)
is the solution of (G.18) with s(r) replaced by δ(r). But by writing
∫
′
s(r) = δ(r − r
′ )s(r
′ ) d
3
r
(G.20)
we can formally regard s(r) as being made up of a superposition of point
′
sources, distributed at points r with a weighting function s(r
′ ). Then, since
the operator Ω r is (by assumption) linear, the solution for such a superposition of point sources must be just the same superposition of the point source
solutions, namely the integral on the right-hand side of (G.19). This integral
term is, in fact, the ‘particular integral’ of the differential equation (G.18),
while the u(r) is the ‘complementary function’.
Equation (G.19) can also be checked analytically. First note that it is
generally the case that the operator Ω r is translationally invariant, so that
Ω r = Ω r−r ' ;
(G.21)
′
the right-hand side of (G.21) amounts to shifting the origin to the point r .
Applying Ω r to both sides of (G.19), we find
∫
′
Ω r ψ(r) = Ω r u(r) + Ω r G(r − r
′ )s(r
′ ) d
3
r
∫
∫
′
′
= 0 + Ω r−r ' G(r − r
′ )s(r
′ ) d
3
r = δ(r − r
′ )s(r
′ ) d
3
r
= s(r)
as required in (G.18).
Finally, consider the Fourier transform of equation (G.10), defined as
∫
∫
−iq·r (∇
2
e
+ k
2 )G k (r) d
3
r = e
−iq·r δ(r) d
3
r.
The right-hand side is unity, by equation (G.4). On the left-hand side we may
use the result
∫
∫
u(r)∇
2 v(r) d
3
r = (∇
2 u(r))v(r) d
3
r
(proved by integrating by parts, assuming u and v go to zero sufficiently fast