The series substitution of Section 9.5 and the Wronskian double integral of Section 9.6 provide the most general solution of the homogeneous, linear, second-order ODE. The specific solution,yp, linearly dependent on the source term (F (x)of Eq. (9.82)) may be cranked out by the variation of parameters method, Exercise 9.6.25. In this section we turn to a different method of solution — Green’s function.
For a brief introduction to Green’s function method, as applied to the solution of a non- homogeneous PDE, it is helpful to use the electrostatic analog. In the presence of charges
the electrostatic potentialψsatisfies Poisson’s nonhomogeneous equation (compare Sec- tion 1.14),
∇2ψ= −ρ
ε0 (mks units), (9.143)
and Laplace’s homogeneous equation,
∇2ψ=0, (9.144)
in the absence of electric charge(ρ=0). If the charges are point chargesqi, we know that the solution is
ψ= 1 4π ε0
i
qi
ri, (9.145)
a superposition of single-point charge solutions obtained from Coulomb’s law for the force between two point chargesq1andq2,
F= q1q2rˆ
4π ε0r2. (9.146)
By replacement of the discrete point charges with a smeared-out distributed charge, charge densityρ, Eq. (9.145) becomes
ψ (r=0)= 1 4π ε0
ρ(r)
r dτ (9.147)
or, for the potential at r=r1away from the origin and the charge at r=r2, ψ (r1)= 1
4π ε0
ρ(r2)
|r1−r2|dτ2. (9.148) We use ψ as the potential corresponding to the given distribution of charge and there- fore satisfying Poisson’s equation (9.143), whereas a functionG, which we label Green’s function, is required to satisfy Poisson’s equation with a point source at the point defined by r2:
∇2G= −δ(r1−r2). (9.149)
Physically, then,Gis the potential at r1corresponding to a unit source at r2. By Green’s theorem (Section 1.11, Eq. (11.104))
ψ∇2G−G∇2ψ dτ2=
(ψ∇G−G∇ψ )ãdσ. (9.150) Assuming that the integrand falls off faster than r−2 we may simplify our problem by taking the volume so large that the surface integral vanishes, leaving
ψ∇2G dτ2=
G∇2ψ dτ2, (9.151)
or, by substituting in Eqs. (9.143) and (9.149), we have
−
ψ (r2)δ(r1−r2) dτ2= −
G(r1,r2)ρ(r2) ε0
dτ2. (9.152)
Integration by employing the defining property of the Dirac delta function (Eq. (1.171b)) produces
ψ (r1)= 1 ε0
G(r1,r2)ρ(r2) dτ2. (9.153) Note that we have used Eq. (9.149) to eliminate∇2Gbut that the functionGitself is still unknown. In Section 1.14, Gauss law, we found that
∇2 1
r
dτ= 0,
−4π, (9.154)
0 if the volume did not include the origin and−4πif the origin were included. This result from Section 1.14 may be rewritten as in Eq. (1.170), or
∇2 1
4π r
= −δ(r), or ∇2 1
4π r12
= −δ(r1−r2), (9.155) corresponding to a shift of the electrostatic charge from the origin to the position r=r2. Here r12= |r1−r2|, and the Dirac delta function δ(r1−r2) vanishes unless r1=r2. Therefore in a comparison of Eqs. (9.149) and (9.155) the functionG(Green’s function) is given by
G(r1,r2)= 1
4π|r1−r2|. (9.156)
The solution of our differential equation (Poisson’s equation) is ψ (r1)= 1
4π ε0
ρ(r2)
|r1−r2|dτ2, (9.157) in complete agreement with Eq. (9.148). Actually ψ (r1), Eq. (9.157), is the particular solution of Poisson’s equation. We may add solutions of Laplace’s equation (compare Eq. (9.83)). Such solutions could describe an external field.
These results will be generalized to the second-order, linear, but nonhomogeneous dif- ferential equation
Ly(r1)= −f (r1), (9.158)
whereLis a linear differential operator. The Green’s function is taken to be a solution of LG(r1,r2)= −δ(r1−r2), (9.159) analogous to Eq. (9.149). The Green’s function depends on boundary conditions that may no longer be those of electrostatics in a region of infinite extent. Then the particular solution y(r1)becomes
y(r1)=
G(r1,r2)f (r2) dτ2. (9.160)
(There may also be an integral over a bounding surface, depending on the conditions spec- ified.)
In summary, Green’s function, often writtenG(r1,r2)as a reminder of the name, is a solution of Eq. (9.149)or Eq. (9.159)more generally. It enters in an integral solution of our differential equation, as in Eqs.(9.148)and (9.153). For the simple, but impor- tant, electrostatic case we obtain Green’s function,G(r1,r2), by Gauss’ law, comparing Eqs.(9.149)and(9.155). Finally, from the final solution(Eq.(9.157))it is possible to develop a physical interpretation of Green’s function. It occurs as a weighting function or propagator function that enhances or reduces the effect of the charge elementρ(r2) dτ2 according to its distance from the field point r1. Green’s function,G(r1,r2), gives the ef- fect of a unit point source at r2in producing a potential at r1. This is how it was introduced in Eq.(9.149);this is how it appears in Eq.(9.157).
Symmetry of Green’s Function
An important property of Green’s function is the symmetry of its two variables; that is, G(r1,r2)=G(r2,r1). (9.161) Although this is obvious in the electrostatic case just considered, it can be proved under more general conditions. In place of Eq. (9.149), let us require thatG(r,r1)satisfy21
∇ã
p(r)∇G(r,r1) +λq(r)G(r,r1)= −δ(r−r1), (9.162) corresponding to a mathematical point source at r=r1. Here the functionsp(r)andq(r) are well-behaved but otherwise arbitrary functions of r. The Green’s function,G(r,r2), satisfies the same equation, but the subscript 1 is replaced by subscript 2. The Green’s functions, G(r,r1)and G(r,r2), have the same values over a given surface S of some volume of finite or infinite extent, and their normal derivatives have the same values over the surfaceS, or these Green’s functions vanish onS(Dirichlet boundary conditions, Sec- tion 9.1).22ThenG(r,r2)is a sort of potential at r, created by a unit point source at r2.
We multiply the equation for G(r,r1) byG(r,r2)and the equation for G(r,r2) by G(r,r1)and then subtract the two:
G(r,r2)∇ã
p(r)∇G(r,r1) −G(r,r1)∇ã
p(r)∇G(r,r2)
= −G(r,r2)δ(r−r1)+G(r,r1)δ(r−r2). (9.163) The first term in Eq. (9.163),
G(r,r2)∇ã
p(r)∇G(r,r1) , may be replaced by
∇ã
G(r,r2)p(r)∇G(r,r1) −∇G(r,r2)ãp(r)∇G(r,r1).
21Equation (9.162) is a three-dimensional, inhomogeneous version of the self-adjoint eigenvalue equation, Eq. (10.8).
22Any attempt to demand that the normal derivatives vanish at the surface (Neumann’s conditions, Section 9.1) leads to trouble with Gauss’ law. It is like demanding that
Eãdσ=0 when you know perfectly well that there is some electric charge inside the surface.
A similar transformation is carried out on the second term. Then integrating over the vol- ume whose surface isSand using Green’s theorem, we obtain a surface integral:
S
G(r,r2)p(r)∇G(r,r1)−G(r,r1)p(r)∇G(r,r2) ãdσ
= −G(r1,r2)+G(r2,r1). (9.164)
The terms on the right-hand side appear when we use the Dirac delta functions in Eq. (9.163) and carry out the volume integration. With the boundary conditions earlier imposed on the Green’s function, the surface integral vanishes and
G(r1,r2)=G(r2,r1), (9.165) which shows that Green’s function is symmetric. If the eigenfunctions are complex, bound- ary conditions corresponding to Eqs. (10.19) to (10.20) are appropriate. Equation (9.165) becomes
G(r1,r2)=G∗(r2,r1). (9.166) Note that this symmetry property holds for Green’s functions in every equation in the form of Eq. (9.162). In Chapter 10 we shall call equations in this form self-adjoint. The symmetry is the basis of various reciprocity theorems; the effect of a charge at r2on the potential at r1is the same as the effect of a charge at r1on the potential at r2.
This use of Green’s functions is a powerful technique for solving many of the more difficult problems of mathematical physics.
Form of Green’s Functions
Let us assume thatLis a self-adjoint differential operator of the general form23 L1=∇1ã
p(r1)∇1 +q(r1). (9.167)
Here the subscript 1 onLemphasizes thatLoperates on r1. Then, as a simple generaliza- tion of Green’s theorem, Eq. (1.104), we have
(vL2uưuL2v) dτ2=
p(v∇2uưu∇2v)ãdσ2, (9.168) in which all quantities have r2as their argument. (To verify Eq. (9.168), take the divergence of the integrand of the surface integral.) We letu(r2)=y(r2)so that Eq. (9.158) applies and v(r2)=G(r1,r2)so that Eq. (9.159) applies. (Remember, G(r1,r2)=G(r2,r1).) Substituting into Green’s theorem we get
'−G(r1,r2)f (r2)+y(r2)δ(r1−r2)( dτ2
=
p(r2)'
G(r1,r2)∇2y(r2)−y(r2)∇2G(r1,r2)(
ãdσ2. (9.169)
23L1may be in 1, 2, or 3 dimensions (with appropriate interpretation of∇1).
When we integrate over the Dirac delta function y(r1)=
G(r1,r2)f (r2) dτ2
+
p(r2)'
G(r1,r2)∇2y(r2)−y(r2)∇2G(r1,r2)(
ãdσ2, (9.170) our solution to Eq. (9.158) appears as a volume integral plus a surface integral. Ify and Gboth satisfy Dirichlet boundary conditions or if both satisfy Neumann boundary con- ditions, the surface integral vanishes and we regain Eq. (9.160). The volume integral is a weighted integral over the source termf (r2)with our Green’s functionG(r1,r2)as the weighting function.
For the special case ofp(r1)=1 andq(r1)=0,Lis∇2, the Laplacian. Let us integrate
∇21G(r1,r2)= −δ(r1−r2) (9.171) over a small volume including the point source. Then
∇1ã∇1G(r1,r2) dτ1= −
δ(r1−r2) dτ1= −1. (9.172) The volume integral on the left may be transformed by Gauss’ theorem, as in the develop- ment of Gauss’ law — Section 1.14. We find that
∇1G(r1,r2)ãdσ1= −1. (9.173) This shows, incidentally, that it may not be possible to impose a Neumann boundary con- dition, that the normal derivative of the Green’s function,∂G/∂n, vanishes over the entire surface.
If we are in three-dimensional space, Eq. (9.173) is satisfied by taking
∂
∂r12G(r1,r2)= − 1
4π ã 1
|r1−r2|2, r12= |r1−r2|. (9.174) The integration is over the surface of a sphere centered at r2. The integral of Eq. (9.174) is
G(r1,r2)= 1 4π ã 1
|r1−r2|, (9.175)
in agreement with Section 1.14.
If we are in two-dimensional space, Eq. (9.173) is satisfied by taking
∂
∂ρ12
G(ρ1,ρ2)= 1
2π ã 1
|ρ1−ρ2|, (9.176)
with r being replaced byρ, ρ=(x2+y2)1/2, and the integration being over the circumfer- ence of a circle centered onρ2. Hereρ12= |ρ1−ρ2|. Integrating Eq. (9.176), we obtain
G(ρ1,ρ2)= − 1
2πln|ρ1−ρ2|. (9.177) ToG(ρ1,ρ2)(and toG(r1,r2)) we may add any multiple of the regular solution of the homogeneous (Laplace’s) equation as needed to satisfy boundary conditions.
Table 9.5 Green’s Functionsa
Laplace Helmholtz Modified Helmholtz
∇2 ∇2+k2 ∇2−k2
One-dimensional space No solution i
2kexp(ik|x1−x2|) 1
2kexp(−k|x1−x2|) for(−∞,∞)
Two-dimensional space − 1
2πln|ρ1−ρ2| i
4H0(1)(k|ρ1−ρ2|) 1
2πK0(k|ρ1−ρ2|) Three-dimensional space 1
4π ã 1
|r1−r2|
exp(ik|r1−r2|) 4π|r1−r2|
exp(−k|r1−r2|) 4π|r1−r2|
aThese are the Green’s functions satisfying the boundary conditionG(r1,r2)=0as r1→ ∞for the Laplace and modified Helmholtz operators. For the Helmholtz operator,G(r1,r2)corresponds to an outgoing wave.H0(1)is the Hankel function of Section 11.4.K0is he modified Bessel function of Section 11.5.
The behavior of the Laplace operator Green’s function in the vicinity of the source point r1=r2 shown by Eqs. (9.175) and (9.177) facilitates the identification of the Green’s functions for the other cases, such as the Helmholtz and modified Helmholtz equations.
1. For r1=r2, G(r1,r2)must satisfy the homogeneous differential equation
L1G(r1,r2)=0, r1=r2. (9.178) 2. As r1→r2(orρ1→ρ2),
G(ρ1,ρ2)≈ − 1
2πln|ρ1−ρ2|, two-dimensional space, (9.179) G(r1,r2)≈ 1
4π ã 1
|r1−r2|, three-dimensional space. (9.180) The term±k2 in the operator does not affect the behavior of Gnear the singular point r1=r2. For convenience, the Green’s functions for the Laplace, Helmholtz, and modified Helmholtz operators are listed in Table 9.5.
Spherical Polar Coordinate Expansion24
As an alternate determination of the Green’s function of the Laplace operator, let us assume a spherical harmonic expansion of the form
G(r1,r2)= ∞
l=0
l m=−l
gl(r1, r2)Ylm(θ1, ϕ1)Ylm∗(θ2, ϕ2), (9.181) where the summation indexl is the same for the spherical harmonics, as a consequence of the symmetry of the Green’s function. We will now determine the radial functions
24This section is optional here and may be postponed to Chapter 12.
gl(r1, r2). From Exercises 1.15.11 and 12.6.6, δ(r1−r2)= 1
r12δ(r1−r2)δ(cosθ1−cosθ2)δ(ϕ1−ϕ2)
= 1
r12δ(r1−r2) ∞
l=0
l m=−l
Ylm(θ1, ϕ1)Ylm∗(θ2, ϕ2). (9.182) Substituting Eqs. (9.181) and (9.182) into the Green’s function differential equation, Eq. (9.171), and making use of the orthogonality of the spherical harmonics, we obtain a radial equation:
r1 d2 dr12
r1gl(r1, r2) −l(l+1)gl(r1, r2)= −δ(r1−r2). (9.183)
This is now a one-dimensional problem. The solutions25 of the corresponding homoge- neous equation arer1l andr1−l−1. If we demand thatglremain finite asr1→0 and vanish asr1→ ∞, the technique of Section 10.5 leads to
gl(r1, r2)= 1 2l+1
r1l
r2l+1, r1< r2, r2l
r1l+1, r1> r2,
(9.184)
or
gl(r1, r2)= 1 2l+1 ã r<l
r>l+1. (9.185)
Hence our Green’s function is G(r1,r2)=
∞
l=0
l m=−l
1 2l+1
r<l
r>l+1Ylm(θ1, ϕ1)Ylm∗(θ2, ϕ2). (9.186) Since we already haveG(r1, r2)in closed form, Eq. (9.175), we may write
1
4π ã 1
|r1−r2| = ∞
l=0
l m=−l
1 2l+1
r<l
r>l+1Ylm(θ1, ϕ1)Ylm∗(θ2, ϕ2). (9.187) One immediate use for this spherical harmonic expansion of the Green’s function is in the development of an electrostatic multipole expansion. The potential for an arbitrary charge distribution is
ψ (r1)= 1 4π ε0
ρ(r2)
|r1−r2|dτ2
25Compare Table 9.2.
(which is Eq. (9.148)). Substituting Eq. (9.187), we get ψ (r1)= 1
ε0 ∞
l=0
l m=−l
1 2l+1
Ylm(θ1, ϕ1) r1l+1
ã
ρ(r2)Ylm∗(θ2, ϕ2)r2ldϕ2sinθ2dθ2r22dr2
, forr1> r2.
This is the multipole expansion. The relative importance of the various terms in the double sum depends on the form of the source,ρ(r2).
Legendre Polynomial Addition Theorem26
From the generating expression for Legendre polynomials, Eq. (12.4a), 1
4π ã 1
|r1−r2| = 1 4π
∞
l=0
r<l r>l+1
Pl(cosγ ), (9.188)
whereγ is the angle included between vectors r1and r2, Fig. 9.4. Equating Eqs. (9.187) and (9.188), we have the Legendre polynomial addition theorem:
Pl(cosγ )= 4π 2l+1
l m=−l
Ylm(θ1, ϕ1)Ylm∗(θ2, ϕ2). (9.189)
FIGURE9.4 Spherical polar coordinates.
26This section is optional here and may be postponed to Chapter 12.
It is instructive to compare this derivation with the relatively cumbersome derivation of Section 12.8 leading to Eq. (12.177).
Circular Cylindrical Coordinate Expansion27
In analogy with the preceding spherical polar coordinate expansion, we write δ(r1−r2)= 1
ρ1δ(ρ1−ρ2)δ(ϕ1−ϕ2)δ(z1−z2)
= 1 ρ1
δ(ρ1−ρ2) 1 4π2
∞
m=−∞
eim(ϕ1−ϕ2) ∞
−∞
eik(z1−z2)dk, (9.190) using Exercise 12.6.5 and Eq. (1.193c) and the Cauchy principal value. But why a summa- tion for theϕ-dependence and an integration for thez-dependence? The requirement that the azimuthal dependence be single-valued quantizesm, hence the summation. No such restriction applies tok.
To avoid problems later with negative values ofk, we rewrite Eq. (9.190) as δ(r1−r2)= 1
ρ1δ(ρ1−ρ2) 1 2π
∞
m=−∞
eim(ϕ1−ϕ2)1 π
∞
0
cosk(z1−z2) dk. (9.191) We assume a similar expansion of the Green’s function,
G(r1, r2)= 1 2π2
∞
m=−∞
gm(ρ1, ρ2)eim(ϕ1−ϕ2) ∞
0
cosk(z1−z2) dk, (9.192) with the ρ-dependent coefficients gm(ρ1, ρ2) to be determined. Substituting into Eq. (9.171), now in circular cylindrical coordinates, we find that ifg(ρ1, ρ2)satisfies
d dρ1
ρ1dgm
dρ1
−
k2ρ1+m2 ρ1
gm= −δ(ρ1−ρ2), (9.193) then Eq. (9.171) is satisfied.
The operator in Eq. (9.193) is identified as the modified Bessel operator (in self- adjoint form). Hence the solutions of the corresponding homogeneous equation areu1= Im(kρ), u2=Km(kρ). As in the spherical polar coordinate case, we demand thatG be finite atρ1=0 and vanish asρ1→ ∞. Then the technique of Section 10.5 yields
gm(ρ1, ρ2)= −1
AIm(kρ<)Km(kρ>). (9.194) This corresponds to Eq. (9.155). The constant A comes from the Wronskian (see Eq. (9.120)):
Im(kρ)Km′ (kρ)−Im′(kρ)Km(kρ)= A
P (kρ). (9.195)
27This section is optional here and may be postponed to Chapter 11.
From Exercise 11.5.10,A= −1 and
gm(ρ1, ρ2)=Im(kρ<)Km(kρ>). (9.196) Therefore our circular cylindrical coordinate Green’s function is
G(r1,r2)= 1
4π ã 1
|r1−r2|
= 1 2π2
∞
m=−∞
∞
0
Im(kρ<)Km(kρ>)eim(ϕ1−ϕ2)cosk(z1−z2) dk.
(9.197) Exercise 9.7.14 is a special case of this result.
Example 9.7.1 QUANTUMMECHANICALSCATTERING— NEUMANNSERIESSOLUTION
The quantum theory of scattering provides a nice illustration of integral equation tech- niques and an application of a Green’s function. Our physical picture of scattering is as follows. A beam of particles moves along the negativez-axis toward the origin. A small fraction of the particles is scattered by the potential V (r) and goes off as an outgoing spherical wave. Our wave functionψ (r)must satisfy the time-independent Schrửdinger equation
− ¯h2
2m∇2ψ (r)+V (r)ψ (r)=Eψ (r), (9.198a) or
∇2ψ (r)+k2ψ (r)= −
−2m
¯
h2V (r)ψ (r)
, k2=2mE
¯
h2 . (9.198b)
From the physical picture just presented we look for a solution having an asymptotic form
ψ (r)∼eik0ãr+fk(θ, ϕ)eikr
r . (9.199)
Hereeik0ãris the incident plane wave28with k0the propagation vector carrying the sub- script 0 to indicate that it is in theθ=0 (z-axis) direction. The magnitudesk0andkare equal (ignoring recoil), and eikr/r is the outgoing spherical wave with an angular (and energy) dependent amplitude factorfk(θ, ϕ).29Vector k has the direction of the outgoing scattered wave. In quantum mechanics texts it is shown that the differential probability of scattering,dσ/d, the scattering cross section per unit solid angle, is given by|fk(θ, ϕ|2.
Identifying[−(2m/h¯2)V (r)ψ (r)]withf (r)of Eq. (9.158), we have ψ (r1)= −
2m
¯
h2V (r2)ψ (r2)G(r1,r2) d3r2 (9.200)
28For simplicity we assume a continuous incident beam. In a more sophisticated and more realistic treatment, Eq. (9.199) would be one component of a Fourier wave packet.
29IfV (r)represents a central force,fkwill be a function ofθonly, independent of azimuth.
by Eq. (9.170). This does not have the desired asymptotic form of Eq. (9.199), but we may add to Eq. (9.200)eik0ãr1, a solution of the homogeneous equation, and putψ (r)into the desired form:
ψ (r1)=eik0ãr1− 2m
¯
h2V (r2)ψ (r2)G(r1,r2) d3r2. (9.201) Our Green’s function is the Green’s function of the operatorL=∇2+k2(Eq. (9.198)), satisfying the boundary condition that it describe an outgoing wave. Then, from Table 9.5, G(r1,r2)=exp(ik|r1−r2|)/(4π|r1−r2|)and
ψ (r1)=eik0ãr1− 2m
h¯2V (r2)ψ (r2) eik|r1−r2|
4π|r1−r2|d3r2. (9.202) This integral equation analog of the original Schrửdinger wave equation is exact. Employ- ing the Neumann series technique of Section 16.3 (remember, the scattering probability is very small), we have
ψ0(r1)=eik0ãr1, (9.203a) which has the physical interpretation of no scattering.
Substitutingψ0(r2)=eik0ãr2 into the integral, we obtain the first correction term, ψ1(r1)=eik0ãr1−
2m
h¯2V (r2) eik|r1−r2|
4π|r1−r2|eik0ãr2d3r2. (9.203b) This is the famous Born approximation. It is expected to be most accurate for weak poten- tials and high incident energy. If a more accurate approximation is desired, the Neumann
series may be continued.30
Example 9.7.2 QUANTUMMECHANICALSCATTERING— GREEN’SFUNCTION
Again, we consider the Schrửdinger wave equation (Eq. (9.198b)) for the scattering prob- lem. This time we use Fourier transform techniques and derive the desired form of the Green’s function by contour integration. Substituting the desired asymptotic form of the solution (withkreplaced byk0),
ψ (r)∼eik0z+fk0(θ, ϕ)eik0r
r =eik0z+(r), (9.204) into the Schrửdinger wave equation, Eq. (9.198b), yields
∇2+k02
(r)=U (r)eik0z+U (r)(r). (9.205a) Here
¯ h2
2mU (r)=V (r),
30This assumes the Neumann series is convergent. In some physical situations it is not convergent and then other techniques are needed.
the scattering (perturbing) potential. Since the probability of scattering is much less than 1, the second term on the right-hand side of Eq. (9.205a) is expected to be negligible (relative to the first term on the right-hand side) and thus we drop it. Note that we are approximat- ing our differential equation with
∇2+k02
(r)=U (r)eik0z. (9.205b) We now proceed to solve Eq. (9.205b), a nonhomogeneous PDE. The differential oper- ator∇2generates a continuous set of eigenfunctions
∇2ψk(r)= −k2ψk(r), (9.206) where
ψk(r)=(2π )−3/2eikãr.
These plane-wave eigenfunctions form a continuous but orthonormal set, in the sense that
ψk∗
1(r)ψk2(r) d3r=δ(k1−k2)
(compare Eq. (15.21d)).31We use these eigenfunctions to derive a Green’s function.
We expand the unknown function(r1)in these eigenfunctions, (r1)=
Ak1ψk1(r1) d3k1, (9.207) a Fourier integral with Ak1, the unknown coefficients. Substituting Eq. (9.207) into Eq. (9.205b) and using Eq. (9.206), we obtain
Ak
k20−k2
ψk(r) d3k=U (r)eik0z. (9.208) Using the now-familiar technique of multiplying byψk∗
2(r)and integrating over the space coordinates, we have
Ak1
k20−k12 d3k1
ψk∗
2(r)ψk1(r) d3r=Ak2
k20−k22
=
ψk∗
2(r)U (r)eik0zd3r. (9.209) Solving forAk2 and substituting into Eq. (9.207) we have
(r2)= k02−k22−1 ψk∗
2(r1)U (r1)eik0z1d3r1
ψk2(r2) d3k2. (9.210) Hence
(r1)=
ψk1(r1)
k02−k12−1 d3k1
ψk∗
1(r2)U (r2)eik0z2d3r2, (9.211)
31d3r=dx dy dz, a (three-dimensional) volume element in r-space.
replacing k2by k1and r1by r2to agree with Eq. (9.207). Reversing the order of integra- tion, we have
(r1)= −
Gk0(r1,r2)U (r2)eik0z2d3r2, (9.212) whereGk0(r1,r2), our Green’s function, is given by
Gk0(r1,r2)= ψk∗
1(r2)ψk1(r1)
k12−k20 d3k1, (9.213) analogous to Eq. (10.90) of Section 10.5 for discrete eigenfunctions. Equation (9.212) should be compared with the Green’s function solution of Poisson’s equation (9.157).
It is perhaps worth evaluating this integral to emphasize once more the vital role played by the boundary conditions. Using the eigenfunctions from Eq. (9.206) and
d3k=k2dksinθ dθ dϕ, we obtain
Gk0(r1,r2)= 1 (2π )3
∞
0
π 0
2π 0
eikρcosθ
k2−k20 dϕsinθ dθ k2dk. (9.214) Herekρcosθ has replaced kã(r1−r2), withρ=r1−r2indicating the polar axis ink- space. Integrating over ϕ by inspection, we pick up a 2π. The θ-integration then leads to
Gk0(r1,r2)= 1 4π2ρi
∞
0
eikρ−e−ikρ
k2−k20 k dk, (9.215) and since the integrand is an even function ofk, we may set
Gk0(r1,r2)= 1 8π2ρi
∞
−∞
(eiκ−e−iκ)
κ2−σ2 κdκ. (9.216)
The latter step is taken in anticipation of the evaluation ofGk(r1,r2)as a contour integral.
The symbolsκ andσ (σ >0)representkρandk0ρ, respectively.
If the integral in Eq. (9.216) is interpreted as a Riemann integral, the integral does not exist. This implies thatL−1does not exist, and in a literal sense it does not.L=∇2+k2 is singular since there exist nontrivial solutionsψ for which the homogeneous equation Lψ=0. We avoid this problem by introducing a parameterγ, defining a different opera- torL−γ1, and taking the limit asγ→0.
Splitting the integral into two parts so that each part may be written as a suitable contour integral gives us
G(r1,r2)= 1 8π2ρi
C1
κeiκdκ κ2−σ2+ 1
8π2ρi
C2
κe−iκdκ
κ2−σ2. (9.217) Contour C1 is closed by a semicircle in the upper half-plane, C2 by a semicircle in the lower half-plane. These integrals were evaluated in Chapter 7 by using appropriately chosen infinitesimal semicircles to go around the singular pointsκ= ±σ. As an alternative procedure, let us first displace the singular points from the real axis by replacingσ by σ+iγ and then, after evaluation, taking the limit asγ→0 (Fig. 9.5).
FIGURE9.5 Possible Green’s function contours of integration.
Forγ positive, contourC1encloses the singular pointκ=σ+iγ and the first integral contributes
2π iã1
2ei(σ+iγ ). From the second integral we also obtain
2π iã1
2ei(σ+iγ ),
the enclosed singularity beingκ= −(σ+iγ ). Returning to Eq. (9.217) and lettingγ→0, we have
G(r1,r2)= 1
4πρeiσ = eik0|r1−r2|
4π|r1−r2|, (9.218) in full agreement with Exercise 9.7.16. This result depends on starting withγ positive. Had we chosenγ negative, our Green’s function would have includede−iσ, which corresponds to an incoming wave. The choice of positiveγ is dictated by the boundary conditions we wish to satisfy.
Equations (9.212) and (9.218) reproduce the scattered wave in Eq. (9.203b) and consti- tute an exact solution of the approximate Eq. (9.205b). Exercises 9.7.18 and 9.7.20 extend
these results.
Exercises
9.7.1 Verify Eq. (9.168),
(vL2uưuL2v) dτ2=
p(v∇2uưu∇2v)ãdσ2.
9.7.2 Show that the terms+k2in the Helmholtz operator and−k2in the modified Helmholtz operator do not affect the behavior ofG(r1,r2)in the immediate vicinity of the singular point r1=r2. Specifically, show that
|r1−limr2|→0
k2G(r1,r2) dτ2=1.
9.7.3 Show that
exp(ik|r1−r2|) 4π|r1−r2|
satisfies the two appropriate criteria and therefore is a Green’s function for the Helmholtz equation.
9.7.4 (a) Find the Green’s function for the three-dimensional Helmholtz equation, Exer- cise 9.7.3, when the wave is a standing wave.
(b) How is this Green’s function related to the spherical Bessel functions?
9.7.5 The homogeneous Helmholtz equation
∇2ϕ+λ2ϕ=0
has eigenvaluesλ2i and eigenfunctionsϕi. Show that the corresponding Green’s function that satisfies
∇2G(r1,r2)+λ2G(r1,r2)= −δ(r1−r2) may be written as
G(r1,r2)= ∞
i=1
ϕi(r1)ϕi(r2) λ2i −λ2 .
An expansion of this form is called a bilinear expansion. If the Green’s function is available in closed form, this provides a means of generating functions.
9.7.6 An electrostatic potential (mks units) is ϕ(r)= Z
4π ε0ãe−ar r .
Reconstruct the electrical charge distribution that will produce this potential. Note that ϕ(r)vanishes exponentially for larger, showing that the net charge is zero.
ANS.ρ(r)=Zδ(r)−Za2 4π
e−ar r .