More on Bessel Functions Infinite Domain, Δ-Function Normalization Consider Bessel's Equation on the Domain 0 <Ρ< As R

More on Bessel functions

Inﬁnite domain, δ-function normalization ½ Consider Bessel’s equation on the domain 0 <ρ< as R ! . Bessel’s equation, (3.75) or

(3.93), says

2 1 d dJν(kρ) 2 ν ρ + k J (kρ)=0. ρ dρ dρ ρ2 ν ′

As in class, multiply this equation by ρJν(k ρ) and integrate from ρ =0 to R:

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R 2 ′ 1 d dJν(kρ) 2 ν ρdρJν(k ρ) ρ + k 2 Jν(kρ) =0. 0 ρ dρ dρ ρ

Integrate the ﬁrst term by parts,

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Z R ′ 2 R

dJν(k ρ) dJν(kρ) 2 ν ′ ′ dJν(kρ)

ρ dρ + k 2 Jν(k ρ)Jν(kρ) = ρJν(k ρ) . 0 dρ dρ ρ dρ 0

Exchanging the roles of k and k′, and subtracting leads to

Z R ′ R

2 ′2 ′ dJν(k ρ) ′ dJν(kρ)

(k k ) ρdρJν(k ρ)Jν(kρ)= ρJν(kρ) ρJν(k ρ) . 0 dρ dρ 0

′ 2 ν As ρ ! 0, both terms on the right-hand side have the same leading behavior,(kk ρ ) /Γ(ν), 2 ′2 2 ′ 2 ν

and so cancel for any value of ν. The next-leading terms go as (k k )ρ (kk ρ ) , and so

do not cancel but vanish as ρ ! 0 for ν > 1. The surface term on the right-hand side then contains only the contribution from ρ = R,

Z R ′ ′ ′ ′ ′ ′ k RJν(kR) Jν(k R) kRJν(k R) Jν(kR) ρdρJν(k ρ)Jν(kρ)= 2 ′2 . (1a) 0 k k Using recursion relations, this can also be written

Z R ′ ′ ′ ′ k RJν(kR) Jν−1(k R) kR Jν(k R) Jν−1(kR) ρdρJν (k ρ)Jν(kρ)= 2 ′2 . (1b) 0 k k

−1/2 Although Jν(kR) vanishes as R for large R, the presence of the factor of R means we must investigate in detail the behavior of the Bessel functions at large argument.

The Bessel functions for large argument are given in (3.91),

2 νπ π

J (kρ) ! cos kR . ν πkR 2 4 2 1 Making use of the trig identity cos α cos β = 2[cos(α β)+cos(α + β)], either (1a) or (1b) then leads to

Z R ′ ′

′ sin[(k k )R] cos[(k + k )R νπ]

Ô Ô

ρdρJ (k ρ)J (kρ) ! . (2) ν ν ′ ′ ′ ′ 0 π kk (k k ) π kk (k + k )

′ ½ The ﬁrst term “oscillates” as R ! and so “averages” to zero, unless k = k , when it becomes large, behavior we expect for a δ-function; the second term “averages” to zero for all k, k′. In more detail, the function

sin(x/ǫ) δ (x)= ǫ πx

has value δǫ =1/ǫπ at x = 0, oscillates rapidly outside the interval x = ¦ǫπ, and has integral

Z ∞ dx δǫ(x)=1. −∞

Thus, the limit ǫ ! 0 is a representation of the Dirac δ-function (cf. entry [34] in the Math- World page on δ-functions, http://mathworld.wolfram.com/DeltaFunction.html). This functional form also appears in time-dependent perturbation theory in quantum mechanics,

where it gives the “energy-conserving” δ-function. The limit in (2) then becomes Z ∞ h

′ 1 ′ Ô ρdρJν (k ρ)Jν(kρ)= lim δ1/R(k k )

0 kk′ R→∞ i

′ ′

′ 1 1 δ(k k ) δ(k k )

δ1/R k + k (ν )π = = . R 2 kk′ k

The second term does not contribute, because the argument never vanishes. I was assigned this problem when I was a graduate student, and I have written solutions to it at various times in the past, but this version is more correct than any of those earlier attempts. The

ﬁgure shows both the exact Jν result of (1a/b) (red) and the cosine approximation (2) for large argument (dotted, blue) plotted as a function of k for k′ = 1, R = 100, and ν = 1. 3

0.5 1 1.5 2 2.5 3

-10 4 Green’s function

Green’s function constructions always follow a similar pattern. The Green’s function is a

2 ′2 ′

Ö Ü Ü ! solution to Ö G = G = 4πδ( ), with the boundary condition that G 0 as

′

½ ! ½ Ü r ! or r . Since the δ-function vanishes almost everywhere, we can expand G( ) 2 in the modes which are solutions to Ö G = 0,

∞ Z X ∞ imφ ±kz G(Ü)= dk Am(k) Jm(kρ) e e , m=−∞ 0 ′ ′ ′ where the coeﬃcients Am(k) will depend on ρ , φ , z . We could guess more about them, but

the result will follow systematically from this starting point. The Bessel function satisﬁes the

! ½ boundary condition G ! 0 as ρ , but for the z-dependence we need diﬀerent expressions for the two regions z > z′ and z < z′:

∞ Z X ∞ < < imφ +kz ′ G (Ü)= dk Am(k) Jm(kρ) e e (z < z ), m=−∞ 0

∞ Z X ∞ > > imφ −kz ′ G (Ü)= dk Am(k) Jm(kρ) e e (z > z ). m=−∞ 0 Except at ρ = ρ′, φ = φ′, the Green’s function must be continuous at z = z′, and so we must < +kz′ > −kz′ have Am(k) e = Am(k) e = Cm(k). This leads to the single expression

∞ Z X ∞ imφ −k(z>−z<) G(Ü)= dkCm(k) Jm(kρ) e e , m=−∞ 0 ′ where z< and z> are the smaller and larger of z, z . The remainder of the construction

2 ′

Ü Ü uses the δ-function information, Ö G = 4πδ( ). Integrate this equation over z from

′ ′ ! z = z ǫ to z = z + ǫ for ǫ 0 to obtain

′

Z z′+ǫ z=z +ǫ ′

2 ∂G δ(ρ ρ ) ′

£ dz Ö G = = 4π δ(φ φ ) ( ) z′−ǫ ∂z z=z′−ǫ ρ (the ρ and φ derivatives produce factors of k and m that are ﬁnite term by term, and so

those contributions vanish as ǫ ! 0). Exchanging the roles of k and ρ in part (a), we have

′ ′ imφ

an expansion of δ(ρ ρ ) in Bessel functions, and the representation of δ(φ φ ) in e is

elementary. Writing both sides of (£) as series expansions gives

Z ′ i ∞ h X ∞ z=z +ǫ imφ ∂ −k(z>−z<) dkCm(k) Jm(kρ) e e ∂z z=z′−ǫ m=−∞ 0

Z ∞ ∞ X ′ 1 im(φ−φ′) = 4π kdkJ (kρ) J (kρ ) e . m m 2π 0 m=−∞ 5

imφ Writing the derivative explicitly and identifying coeﬃcients of Jm(kρ)e on left and right

hand sides we obtain

∂ −k(z−z′) ∂ −k(z′−z)

Cm(k) e e = 2kCm(k) ∂z ∂z z=z′

′ 1 −imφ′ ′ −imφ′

= 4πkJ (kρ ) e = 2kJ (kρ ) e . m 2π m ′ −imφ′ Thus, Cm(k) = Jm(kρ ) e , and we have completed the Bessel function representation of the Green’s function for free space,

∞ Z X ∞ ′ 1 ′ im(φ−φ ) −k(z>−z<)

′ = dkJm(kρ)Jm(kρ ) e e .

Ü Ü j j m=−∞ 0 In class we constructed Green’s functions for the square (see also JDJ Problem 2.15) and for the volume between two spherical surfaces (see also JDJ Section 3.9); and Jackson does a diﬀerent Bessel function construction in Section 3.11.

Some Bessel function relations

′ ! Take the Green’s function and evaluate for Ü 0. On the left-hand side

1 1 1

Ô Ô

′ = ! ;

Ü Ü j

j 2 ′2 ′ ′ ′ 2 2 2

ρ + ρ 2ρρ cos(φ φ )+(z z ) ρ + z m while, since Jm(x) (x/2) , only m = 0 contributes to the series on the right-hand side as ′ ρ ! 0. Thus, Z ∞ −k|z| 1 dkJ0(kρ) e = Ô . 0 ρ2 + z2

Ô 2 ′2 ′ ′

Now, evaluate this ﬁrst result at ρ = ρ + ρ 2ρρ cos(φ φ ), Z

∞ Ô

1 2 ′2 ′ ′ −k|z|

= dkJ [k ρ + ρ 2ρρ cos(φ φ )] e .

2 ′2 ′ ′ 2 0

ρ + ρ 2ρρ cos(φ φ )+ z 0 But, this is also just the Green’s function evaluated at z′ = 0, but arbitrary ρ′,

∞ Z X ∞ 1 ′ im(φ−φ′) −k|z| Ô = dkJ (kρ)J (kρ ) e e .

2 ′2 ′ ′ 2 m m

ρ + ρ 2ρρ cos(φ φ )+ z m=−∞ 0 The integral over k amounts to a Laplace transform, and the theory of Laplace transforms assures us that the transformation is unique and invertible, and so the integrands on the right-hand side must be equal:

∞ X Ô ′

2 ′2 ′ ′ ′ im(φ−φ )

J0[k ρ + ρ 2ρρ cos(φ φ )]= Jm(kρ)Jm(kρ ) e . m=−∞ 6 ′ ′ Evaluate the relation at the bottom of the previous page at kρ = kρ = x, φ φ = 0; recall m that J−m(x)=( 1) Jm(x), and

∞ ∞ X X 2 2 2 Jm(x) = [J0(x)] +2 [Jk(x)] = J0(0) = 1. m=−∞ k=1

Take the result and evaluate for φ′ = 0 in the limit ρ′ becomes large. In this limit the square root becomes Õ 2

2 ′2 ′ ′ ρ

Ç ρ + ρ 2ρρ cos φ = ρ ρ cos φ + ( ), ρ′

and the large argument limit of the Bessel functions then gives

Ö Ö

∞

2 ′ π X 2 ′ mπ π imφ

cos kρ kρ cos φ = J (kρ) cos kρ e , πkρ′ 4 m πkρ′ 2 4 m=−∞ ′ correct up to terms of order 1/ρ . Use that cos(α β) = cos α cos β + sin α sin β, and equate

′ π ′ π separately the coeﬃcients of cos(kρ 4 ) and sin(kρ 4 ) on each side (equality must hold for all values of ρ′), ∞ X mπ cos(kρ cos φ)= J (kρ) cos( ) eimφ, m 2 m=−∞ ∞ X mπ sin(kρ cos φ)= J (kρ) sin( ) eimφ. m 2 m=−∞ Finally, add i times the second to the ﬁrst, ∞ ikρ cos φ X im π imφ e = Jm(kρ) e 2 e . m=−∞

Take the real and imaginary parts of the complex exponential evaluated at kρ = x, φ = 0: i

h ∞ ∞ ∞

X X

X m k k

Re i Jm(x) = ( 1) J 2k (x)= J0(x)+2 ( 1) J 2k (x) = cos x,

m=−∞ k=−∞ k=1 i

h ∞ ∞ X X m k+1 Im i Jm(x) =2 ( 1) J2k+1(x) = sin x. m=−∞ k=0

Integral representation

Take the complex exponential, evaluate at kρ = x, multiply by e−imφ, and integrate over φ:

Z Z i h ∞ 2π X 2π dφ −imφ m′ im′φ m dφ ix cos φ−imφ e i J ′ (x) e = i J (x)= e . 2π m m 2π 0 m′=−∞ 0