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Boundary Value Problems in Spherical Coordinatesweb.mst.edu/~hale/courses/m402/M402_notes/M402-Chapter5/Fall.2014/... · Boundary Value Problems in Spherical Coordinates Spherical

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Page 1: Boundary Value Problems in Spherical Coordinatesweb.mst.edu/~hale/courses/m402/M402_notes/M402-Chapter5/Fall.2014/... · Boundary Value Problems in Spherical Coordinates Spherical
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Page 2: Boundary Value Problems in Spherical Coordinatesweb.mst.edu/~hale/courses/m402/M402_notes/M402-Chapter5/Fall.2014/... · Boundary Value Problems in Spherical Coordinates Spherical

Boundary Value Problems in Spherical Coordinates

Spherical harmonics

Laplace’s and Poisson’s equation in spherical coordinates are encountered in a wide rangeof problems (E&M, thermodynamics, mechanics of continuous media, and quantummechanics). As might be expected the solutions of these equations have been well studied. Thesolutions to the angular part of the equation are the spherical harmonics. These are theappropriately normalized products of the associated Legendre polynomials in cos and theexpim

Yℓm, −1m 2ℓ 1

4ℓ − m!ℓ m!

Pℓmcoseim

The Yℓm, satisfy

L L Ylm ℓℓ 1Yℓ

m

and

LzYℓm mYℓ

m; where Lz z L

The associate Legendre functions are as follows:

Pℓmx constant x 1

x − 1

m/2 1 − x2

mFm − ℓ, ℓ m 1,m 1; 1 − x

2

Fa,b,c; z Γ1 − cΓaΓb

nΓa nΓb n

n!Γc nzn

Γa a − 1!

Some useful properties are:

Yℓ−m, −1mYℓ

m,∗

Yℓm − , 2 − −1ℓ Yℓ

−m,The orthonormality condition:

0

2 0

Yℓ′

m′,∗ Yℓ

m, sindd ℓℓ′mm′

Yℓ′m′

,Yℓm ℓℓ′mm′

and the completeness condition:

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Page 3: Boundary Value Problems in Spherical Coordinatesweb.mst.edu/~hale/courses/m402/M402_notes/M402-Chapter5/Fall.2014/... · Boundary Value Problems in Spherical Coordinates Spherical

∑ℓ0

∑m−ℓ

Yℓm ′, ′∗ Yℓ

m, − ′cos − cos ′

Green’s function for Laplace’s equation:

The Green’s function will satisfy:

∇2Gr,r ′ r − r ′

1r′ 2 r − r′cos − cos ′ − ′

We look for a solution of the form

Gr,r ′ ∑ℓ0

∑m−ℓ

fℓmr; r′, ′, ′Yℓm,

Inserting this into the differential equation we obtain (using Gaussian units)

∇2∑ℓ0

∑m−ℓ

fℓmr; r′, ′, ′Yℓm, 1

r′ 2 r − r′cos − cos ′ − ′

1r2 ∑

ℓ0

∑m−ℓ

ℓddr

r2 ddr

fℓm − ℓℓ 1fℓm Yℓm, 1

r′ 2 r − r′cos − cos ′ − ′

Multiply both sides by Yℓ′m′∗, and integrate over dcosd Using the orthonormality

of the spherical harmonics one finds:

1r2 ∑

ℓ0

∑m−ℓ

ℓddr

r2 ddr

fℓm − ℓℓ 1fℓm Yℓ′m′∗,Yℓ

m,dcosd

1r2 ∑

ℓ0

∑m−ℓ

ℓddr

r2 ddr

fℓ′m′ − ℓℓ 1fℓ′m′ ℓ′ℓm′m

1r′ 2 r − r′ − ′cos − cos ′Yℓ′

m′∗,dcosd

1r2

ddr

r2 ddr

fℓ′m′ − ℓℓ 1fℓ′m′ 1r′ 2 r − r′Yℓ′

m′ ′, ′∗

Thus the angular dependence of fℓmr; r′, ′, ′ is given by Yℓm ′, ′ and we define

fℓmr; r′, ′, ′ gℓr; r′Yℓm ′, ′∗.

The functions gℓr; r′ satisfy1r2

ddr

r2 ddr

gℓ − ℓℓ 1gℓ 1r′ 2 r − r′

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This will be the Green's function for all of space. That is, G = 0 at r = infinity.
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Page 4: Boundary Value Problems in Spherical Coordinatesweb.mst.edu/~hale/courses/m402/M402_notes/M402-Chapter5/Fall.2014/... · Boundary Value Problems in Spherical Coordinates Spherical

These equations are solved by :(1) obtaining the solutions to the homogeneous equations for r r′ and r r′,(2) making gℓr; r′ continuous at r r′, and(3) satisfying the condition placed on gℓr; r′ obtained by integrating

from r r′ − to r r′ .The solution to the homogeneous equation in r has the general form

gℓr; r′ ℓr′rℓ ℓr′r−ℓ−1.

For r r′ gℓr; r′ ℓr′rℓ and for r r′ gℓr; r′ ℓr′r−ℓ−1. Continuity at r r′

requires that

gℓr; r′ a rℓ

r′ ℓ1 , r r′

gℓr; r′ a r′ ℓ

rℓ1 , r r′

Integrating Eq. 4.29 from r r′ − to r r′ we obtain

r′−

r′ 1r2

ddr

r2 ddr

gℓ − ℓℓ 1gℓ dr 1r′ 2 r′−

r′r − r′dr

r′−

r′ 1r2

ddr

r2 ddr

gℓ − ℓℓ 1gℓ dr 1r′ 2

The second term on the left is continuous at r r′ and drops out of the equation.

r′−

r′ 1r2

ddr

r2 ddr

gℓ dr 1r′ 2

Integrating by parts twice,

1r2 r2 d

drgℓ |rr′−

rr′ − r′−

r′ −2r

ddr

gℓdr 1r′ 2

1r2 r2 d

drgℓ |rr′−

rr′ 2r gℓ|rr′−

rr′ r′−

r′ 2r2 gℓdr 1

r′ 2

Since gℓr, r′ is continuous at r r′, only the first term on the left is non-zero. Thus

r′ 2 dgℓr; r′dr

rr′

−dgℓr; r′

drrr′−

1

Now use the specific forms for gℓr; r′:

r′ 2a ddr

r′ ℓ

rℓ1rr′

− ddr

rℓ

r′ ℓ1rr′−

1

r′ 2a−ℓ 1 r′ ℓ

r′ℓ2 − ℓr′ℓ−1

r′ ℓ1 a−ℓ 1 − ℓ −a2ℓ 1

Thus

a −12ℓ 1

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Any term which is continuous vanishes in the limit as r --> r'.
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Page 5: Boundary Value Problems in Spherical Coordinatesweb.mst.edu/~hale/courses/m402/M402_notes/M402-Chapter5/Fall.2014/... · Boundary Value Problems in Spherical Coordinates Spherical

Using the above equations we have the expansion of the Green’s function as follows:

Gr,r ′ ∑ℓ0

∑m−ℓ

ℓ−1

2ℓ 1rℓ

rℓ1 Yℓm,Yℓ

m ′, ′∗.

∇2Gr,r ′ r − r ′ Gaussian units

Green’s Function and Expansion of 1|r − r ′|

Note that the Green’s function for Laplace’s equation (for V all space) satisfies thefollowing:

∇2 −14|r − r ′|

r − r ′

Thus .

Gr,r ′ ∑ℓ0

∑m−ℓ

ℓ−1

2ℓ 1rℓ

rℓ1 Yℓm,Yℓ

m ′, ′∗ −14|r − r ′|

∇2Gr,r ′ r − r ′The above equations yield a convenient expansion for 1

|r − r ′|in spherical harmonics

1|r − r ′|

∑ℓ0

∑m−ℓ

ℓ4

2ℓ 1rℓ

rℓ1 Yℓm,Yℓ

m ′, ′∗

and provides a convenient way to use the Helmholtz theorem to solve:

∇2r −r

r 14

r ′|r − r ′|

dV ′

14 r ′∑

ℓ0

∑m−ℓ

ℓ4

2ℓ 1rℓ

rℓ1 Yℓm,Yℓ

m ′, ′∗dV ′

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V-11

TABLE OF Y

m s

Y00(θ,φ) = 1/(4π)

Y10(θ,φ) = [3/(4π)]· cosθ

Y11(θ,φ) = - [3/(8π)]·sinθ eiφ

Y20(θ,φ) = [5/(4π)]·[(3/2)cos²θ - ½]

Y21(θ,φ) = - [15/(8π)]·sinθ·cosθ eiφ

Y22(θ,φ) = (1/4)[15/(2π)]·sin²θ ei2φ

Y30(θ,φ) = [7/(4π)]·[(5/2)cos3θ - (3/2)cosθ]

Y31(θ,φ) = - (1/4)[21/(4π)]·sinθ·[5cos²θ - 1] eiφ

Y32(θ,φ) = (1/4)[105/(2π)]·sin²θ·cosθ ei2φ

Y33(θ,φ) = - (1/4)[35/(4π)]·sin3θ ei3φ

Y40(θ,φ) = (3/4)[1/(16π)]·[35cos4θ - 30cos²θ + 3)]

Y41(θ,φ) = - (3/4)[1/(4π)]·sinθ·[7cos3θ - 3cosθ] eiφ

Y42(θ,φ) = (3/4)[5/(8π)]· [7cos4θ -8cos²θ +1] ei2φ

Y43(θ,φ) = - (3/4)[35/(4π)]·sinθ·[-cos3θ + cosθ] ei3φ

Y44(θ,φ) = (3/4)[35/(32π)]·sin4θ ei4φ

Note: Y

-m = (-1)mY

m*

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Green’s Function Solution to Poisson’s Equation∇2r r

∇2Gr,r ′ r − r ′

Consider the following integral:

all space

∇′ r ′∇Gr,r ′ − Gr,r ′∇′r ′d3r′

r′→

r ′∇′Gr,r ′ − Gr,r ′∇′r ′ dS′

Simply the integrand on the left side:

all space

∇′ r ′∇Gr,r ′ − Gr,r ′∇′r ′d3r′

all space

r ′∇2Gr,r ′ − Gr,r ′∇′2r ′d3r′

all space

r ′r − r ′ − Gr,r ′r ′d3r′

r − all space

Gr,r ′r ′d3r′

r − all space

Gr,r ′r ′d3r′

sphere with r′→

r ′∇Gr,r ′ − Gr,r ′∇′r ′ dS′

Thus the solution to Poisson’s equation is"

r all space

Gr,r ′r ′d3r′

sphere with r′→

r ′∇Gr,r ′ − Gr,r ′∇′r ′ dS′

If both r and Gr,r ′ 0 as r′ ,

r all space

Gr,r ′r ′d3r′

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Page V-45
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divergence theorem
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Use the Helmholtz theorem or the Green's function solution.
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Look this up in the table of spherical harmonics.
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Only the r' integration is left to be done.
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Reduces sum over l,m to 1,1 and 1,-1
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= 1/2 when r=4a
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y/r = [8pi/3]1/2 (1/2i)[-Y 1 1- Y 1 -1]
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vector!
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If a = -1, -2, -3, ... the series terminates for k = k1 = 0
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add 1-c to lambda
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(a-1+n)!/(a-1)!
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c is replaced by b here.
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confluent hypergeometric function
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x = z2 dx = 2zdz d/dz = 2x1/2 d/dx d2/dz2 = 2x1/2 d/dx [ 2x1/2 d/dx] = 4x d2 /dx2 +2d/dx
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series terminates
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series terminates
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This material will not be on the final exam. You can omit pages 56 - 58.
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The function R is redefined and a new, simpler differential equation results: the Laguerre equation.
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Solution to Laplace’s Equation in Cylindrical Coordinates

∇2,, z 1∂∂

∂∂

12

∂2

∂2 ∂2

∂z2 ,, z 0

Use separation of variables:

,, z RFHz ≠ 0

1R

1∂∂

∂∂

R 1F

12

∂2

∂2 F 1H∂2

∂z2 H 0

1R

1∂∂

∂∂

R 12

F ′′F

− H′′

H k2

The Hz thus satisfies:

H′′z −k2Hz

Hz Ckeikz Dke−ikz

leaving the equation in the form:Rdd

R ′ − 2k2 − F′′

F m2

The F thus satisfies:

F ′′ −m2F

F Ameim Bme−im

Finally, the equation for R

dd

R ′ − m2R − 2k2R 0

2R ′′ R ′ − m2R − 2k2R 0

Let x ik, and R Jx and finally one obtains Bessel’s differential equation:

x2J ′′ xJ ′ x2 − m2J 0

which has two linearly indepent solutions, Jx bJmx cNmx

The general solution for ,, z is:

∇2,, z 0

,, z ∑k,m

eimeikzckmJmik dkmNmik

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k can be real giving sin(kz) and cos(kz) dependence, or complex giving exp(kz) and exp(-kz) dependence. m is usually real, but depending on boundary conditions could be complex (in particle physics when angular momentum is analytically continued to complex values).
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note: x = aw x F ' (x) = w G'(w)
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m is usually an integer -- but it does not have to be. Also, x can be complex.
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Note the recursion relation relates: ao to a2, a2 to a4, ... a1 to a3, a3 to a5, ...
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If m = + 1/2 or - 1/2 a1 can be non-zero. But if m is not = 1/2 or - 1/2 then a1=0 and there are no odd powers of x.
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This Gamma function diverges for j < m. So series starts at j = m
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Series starts at j>m or j = s+m, with s = 0,1,2,3...
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k = 1/2
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k = -1/2
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Expand each exponential in a series. Let i = summation index in exp(-x/2t) and j the index for exp(xt/2). Then divide the product into two terms: one in which the power of t , n = (j-i), is negative (J<i) and one in which the power of t , n = (j-i), is positive (j>i). In the first term set i = j-n and i+j =2j-n. Rename j to L and sum over positive L and negative n. In the second term set j = i+n and i+j = 2i+n. Rename i to k and sum over positive k and positive n.
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Convert to form in example above.
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zeros of Jm
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diverges at x = 0
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The time independent wave equation (in spherical coordinates) gives rise to this differential equation in r. Note: n(n+1) would be l(l+1) where l = angular momentum.
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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 200.5

0.4

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0

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.5−

jo x( )

j1 x( )

j2 x( )

200 x

Plot of spherical Bessel Functions

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V-67Math 402 Table of Differential Equations

Equation (z can be complex) General Solution, w(z) Solution Form Comments/Misc

z(1-z)w’’ +[c-(a+b+1)z]w’- abw=0Hypergeometric

A F(a,b,c:z)+ B Φ(a,b,c;z)c …integer

F(a,b,c;z) =[Γ(c)/Γ(b)Γ(a)] Σ0,4

['(b+n)Γ(a+n) /n!Γ(c+n)]zn

Φ(a,b,c;z) = z1-c F(a+1-c,b+1-c,2-c;z)Hypergeometric functions of first and second kind

converges for |z|<1; c … 0,-1,-2,-3 ... see Abramowitz, p. 556F and Φ are linearly independent if c …integerNOTE: infinite series terminates if a or b are negative integers; this gives convergence of F

(1-z2)w’’-2zw’+[ R(R+1)-m2/(1-z2)]w=0 Associated Legendre transform: w(z) = [(z+1)/(z-1)]m/2 F( [1-z]/2 )

A PRm(z)+B Q Rm(z)

often z = cosθ

PRm(z)=(1/2m)[( R-m)!/(( R+m)!m!](1-z2)(m/2)C

F(m- R,m+ R-1,m+1; (1-z)/2)Q Rm(z)=[(z+1)/(z-1)Γ(1-m)]F(- R, R+1,1-m;(1-z)/2) Associated Legendre functions of 1st and 2nd kind

PRm(z) converges for |z| # 1 if R =0,1,2,3,...and |m| # R

Q Rm(z) diverges at z=-1for R =0,1,2,3,...PR

m(z)=(1-z2) m/2 dm /dzmP R (z) ; PR-m(z)=[R -m)!/[R+-m)!PR

m(z)PR

m(±1) = δm0

(1-z2)w’’-2zw’+[ R(R+1) ]w=0 Legendre

A PR(z)+B Φ R(z) z) PR(z)=PR0(z) (see above, m=0)

QR(z)= QR0(z) (see above, m=0) converges for z>1

Legendre polynomials of first and second kind

PR(z)=(2-R/ R!) dR/dzR[z2-1]R ; R=0,1,2,3.. ; P0=1; P1=z; P2= [3z2-1]/2 QR(z) diverges at z=1; I-1

+1 PR(z)PR’ (z)dz=δRR’[2/(2R+1]

zw’’ +[c-z]w’- aw=0 Confluent Hypergeometric

A F(a,c:z)+BΦ(a,c;z)c …integer

F(a,c;z)= Σ0,4 [Γ( c)/ Γ( a) ]@C[Γ( a+n) /n!Γ(c+n)]zn

Φ(a,b;z) = z1-c F(a+1-c,2-c;z)

erf(x)=[2x/π]F(1/2,3/2;-x2)=(2/π1/2)I0xexp(-t2)dt

e-x= F(a,a;z);

zw’’ +[k+1-z]w’+(n-k)w=0 Associated Laguerre

ALnk(z)+BFn

k(z) Lnk(z) =[n!n!/(n-k)!k!]F(k-n,k+1;z); terminates for integer k,n

Fnk(z) =z1-c F(a+1-c,2-c;z)

converges if |Z| <1; k…-1,-2,-3 ...; in general k and n don’t have to be integerLn

0(z)= Laguerre polynomials; L0=1, L1=(1-x)

z2 w’’ -2zw’ +2nw=0 Hermite

A Hn(z)+BΦn (z) H2m (z) = const. F(-m,1/2;z2) see Abramowitz p780Φn (z)= const Φ(-n/1,1/2;z2)

Hn (x) =(-1)nexp(x2)dn/dxn [exp(-x2)]; H0 =1; H1=2x; H2=4x2-2;I-4

4 exp(-x2)Hn (x)Hn’ (x)dx= π1/2 2nn!δnn’; see Korn&Korn p 853

z2 w’’ +zw’ +(z2 - n2 ) w=0Bessel

A Jn(z)+B Nn (z)n can be complex

Jn(z)=3j=0 to 4 [(-1)j/j!Γ( n+j+1)](z/2)2j+n; j…1/2Nn(z)= [Jn(z)cos(nπ) - J-n(z)]/sin(nπ) if z …0; diverges at z=0Jn is Bessel function; Nn is Neumann function

J-n(z)=(-1)n Jn(z) if n is an integer Abramowitz page 853Hankel functions: H(1)

n(z) = Jn(z) + i Nn(z); H(2)n(z) = Jn(z) - i Nn(z)

z2 w’’ +2zw’ +(z2 - n(n+1) ) w=0Spherical Bessel[L2+k2]Ψ=0; Ψ = 3aRm YRm(θ,φ) jR (kr)

A jn(z)+Bnn (z)

spherical Bessel functions

jn (z) = [ π/2z]1/2 Jn+1/2(z) nn (z)=[ π/2z]1/2 Nn+1/2(z) diverges at z =0

n = 0,±1,±2,±3... n can be complex ( see monographs Korn & Korn)j0(x)=sin(x)/x; n0(x)=cos(x)/x

L 2Ψ=[(1/r2)[M/Mr(r2 M/Mr)]-LCL/(r2 hG2)]Ψ=0Laplace ‹ 2 = -LCL/ hG 2=(1/sinθ)[M/Mθ(sinθ M/Mθ)] + (sin-2θ)[M2/Mφ2 ]

L2 1/|r-r’| = -4πδ(r-r’)

Ψ(r,θ,φ) =

3R,m cm RR(r)YRm(θ,φ)

RR (r)=ARrR+BRr-R-1; R=0,1,2,3.... |m|#RYR

m(θ,φ) =(-1)R[(2R+1)(R-m)!/4π(R+m)!]1/2 PRm(z)eimφ

LCL YRm(θ,φ)=hG 2R(R+1)YR

m(θ,φ)L=r x [hG/i]L = angular momentum operator

1/|r-r’| =3R,m[4π/(2R+1)][r<

R/r>R+1]YR

m(θ’,φ’)*YRm(θ,φ)

YR-m(θ,φ)=(-1)m YR

m(θ,φ)* ; Q Rm(z) diverges at cosθ=-1IIYR’

m’(θ,φ)*YRm(θ,φ)dΩ = δmm’δRR’

Y00 =[1/4π]1/2; Y1

0 =[3/4π]1/2cosθ; Y11 =[3/8π]1/2sinθeiφ

Y20 =[5/4π]1/2[(3/2)cos2θ-1/2]

I-1+1 PR’

m(z)PRm(z)dz = δRR’[2(R+m)!/(2R+1)(R-m)!]

Gamma Function Γ(z)=I04 tz-1 e-t dt

Re(z) >0; z…0,-1,-2,-3...Γ(z)=(z-1)!; z can real or complexdiverges at z = 0,-1,-2,-3,...

Γ(1)=1; Γ(1/2)=%π; Γ(z+1)=zΓ(z)

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