introduction to spherical harmonicslototsky/math681/lecture9.pdfintroduction to spherical harmonics...
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Introduction to Spherical Harmonics
Lawrence Liu
23 June 2014
1 Possibly useful information
1.1 Legendre polynomials
1. Rodrigues formula:
Pn(x) =1
2nn!
dn
dxn(x2 − 1
)n.
2. Generating function:
w(x, t) =(1− 2xt+ t2
)− 12 =
∞∑n=0
Pn(x)tn, (usually) |t| < 1.
3. Recurrence relations:(n+ 1)Pn+1(x)− (2n+ 1)xPn(x) + nPn−1(x) = 0.
P ′n+1(x)− xP ′n(x) = (n+ 1)Pn(x).
xP ′n(x)− P ′n−1(x) = nPn(x).
P ′n+1(x)− P ′n−1(x) = (2n+ 1)Pn(x).(1− x2
)P ′n(x) = nPn−1(x)− nxPn(x).
4. Differential equation: [(1− x2
)u′]′
+ n(n+ 1)u = 0.
5. Normalization:
φn(x) =
√n+
1
2Pn(x).
6. Orthonormality: ∫ 1
−1φm(x)φn(x) dx = δmn.
1.2 (Associated) Laguerre polynomials
1. Rodrigues formula:
Lαn(x) = exx−α
n!
dn
dxn(e−xxn+α
).
(When α = 0, these are the Laguerre polynomials; otherwise, they are called associated or generalizedLaguerre polynomials.)
1
2. Generating function:
w(x, t) = (1− t)−α−1e−xt1−t =
∞∑n=0
Lαn(x)tn, |t| < 1.
3. Recurrence relations:
(n+ 1)Lαn+1(x) + (x− α− 2n− 1)Lαn(x) + (n+ α)Lαn−1(x) = 0.
xd
dxLαn(x) = nLαn(x)− (n+ α)Lαn−1(x).
Lα+1n (x)− Lαn(x) = Lα−1n−1(x).
d
dxLαn(x) = −Lα+1
n−1(x).
4. Differential equation:xu′′ + (α+ 1− x)u′ + nu = 0.
5. Orthogonality: ∫ ∞0
e−xxα︸ ︷︷ ︸weight
Lαm(x)Lαn(x) dx =Γ(n+ α+ 1)
n!δmn.
6. Normalization:
φαn(x) = e−x/2xα/2
√n!
Γ(n+ α+ 1)Lαn(x).
7. Orthonormality: ∫ ∞0
φαm(x)φαn(x) dx = δmn.
1.3 Hypergeometric equation/function
1. Hypergeometric equation:
z(1− z)u′′ + [γ − (α+ β + 1)z]u′ − αβu = 0.
2. Solution:u = c1u1 + c2u2;
u1 = F (α, β; γ; z) =
∞∑k=0
(α)k(β)kk!(γ)k
zk, |z| < 1;
u2 = z1−γF (1− γ + α, 1− γ + β; 2− γ; z), |z| < 1, |arg z| < π.
Lemma 1. If u(z) is a solution to the hypergeometric differential equation with parameters α, β, and γ,then v(z) = u′(z) is a solution to the hypergeometric equation with parameters α+ 1, β + 1, and γ + 1.
Proof. Differentiate the hypergeometric equation:
z(1− z)u′′′ − zu′′ + (1− z)u′′ + [γ − (α+ β + 1)z]u′′ − (α+ β + 1)u′ − αβu′ = 0;
then write v = u′:
z(1− z)v′′ + [γ + 1− (α+ β + 3)z]v′ − (αβ + α+ β + 1)︸ ︷︷ ︸(α+1)(β+1)
v = 0;
and we concludev = F (α+ 1, β + 1; γ + 1; z).
2
1.4 Gamma and beta functions
1.Γ(z + 1) = zΓ(z).
2.Γ(z)Γ(1− z) =
π
sinπz.
3.
22z−1Γ(z)Γ
(z +
1
2
)=√π Γ(2z).
4.
B(x, y) =
∫ 1
0
tx−1(1− t)y−1 dt.
5.
B(x, y) =Γ(x)Γ(y)
Γ(x+ y).
1.5 Fundamental constants
1. Planck’s constant:~ = 1.05457× 10−34 J · s.
2. Electron mass:µe = 9.10938× 10−31 kg.
3. Fundamental charge:e = 1.60218× 10−19 C.
4. Coulomb constant:
k =1
4πε0= 8.98755× 109
N ·m2
C2.
5. Bohr radius:
a =4πε0~2
µee2= 5.29177× 10−11 m.
2 Legendre functions and associated Legendre functions
2.1 Legendre functions
By taking (1.1.4) to be our definition for the Legendre polynomials, we can extend the definition and letn→ ν ∈ C, x→ z ∈ C. The equation can then be written[(
1− z2)u′]′
+ ν(ν + 1)u = 0,(1− z2
)u′′ − 2zu′ + ν(ν + 1)u = 0,
which is called Legendre’s equation. Make the substitution t = 12 (1− z). Then
dt = −1
2dz =⇒ −1
2
du
dt=
du
dz;
−[1− (1− 2t)
2]· 1
4
d2u
dt2+ 2(1− 2t) · 1
2
du
dt+ ν(ν + 1)u = 0,
3
t(1− t)d2u
dt2+ (1− 2t)
du
dt+ ν(ν + 1)u = 0.
This has solution
F
(−ν, ν + 1; 1;
1− z2
)≡ Pν(z); (|z − 1| < 2) .
Notice that when ν = 0, 1, 2, . . ., the function reduces to a polynomial; these turn out to be precisely theLegendre polynomials and will be shown later.
We can also substitute t = z−2, u = z−1−νv. Then
du
dz= z−1−ν
dv
dz+ (−1− ν)z−2−νv,
d2u
dz2= z−1−ν
d2v
dz2+ 2(−1− ν)z−2−ν
dv
dz+ (−1− ν)(−2− ν)z−3−νv,
dt = −2z−3 dz,
dv
dz= −2t
32
dv
dt,
d2v
dz2= −2t
32
d
dt
(−2t
32
dv
dt
)= 6t2
dv
dt+ 4t3
d2v
dt2.
The equation becomes (1− z2
)u′′ − 2zu′ + ν(ν + 1)u = 0,
(1− z2
)[z−1−ν d2v
dz2+ 2(−1− ν)z−2−ν dv
dz+ (−1− ν)(−2− ν)z−3−ν
v
]− 2z
[z−1−ν dv
dz+ (−1− ν)z−2−ν
v
]+ ν(ν + 1)z
−1−νv = 0,
(1 −
1
t
) 2t 5+ν22t
d2v
dt2+ 3
dv
dt
− 4t5+ν2 (−1 − ν)
dv
dt+ t
3+γ2 (−2 − ν)(−1 − ν)v
− 2√t
[−2t
4+ν2
dv
dt+ t
2+ν2 (−1 − ν)v
]+ t
1+ν2 ν(1 + ν)v = 0,
t3+ν2
{4t(t− 1)
d2v
dt2+ 2[−3− 2ν + (5 + 2ν)t]
dv
dt+ (ν + 1)(ν + 2)v
}= 0,
t(1− t)d2v
dt2+
[(ν +
3
2
)−(ν +
5
2
)t
]dv
dt−(ν
2+ 1)(ν
2+
1
2
)v = 0.
The solution is
v(t) = F
(ν
2+ 1,
ν
2+
1
2; ν +
3
2; t
),
so that
u(z) =1
zν+1F
(ν
2+ 1,
ν
2+
1
2; ν +
3
2;
1
z2
).
We normalize this second solution:
Qν(z) ≡√π Γ(ν + 1)
Γ(ν + 3
2
)(2z)ν+1
F
(ν
2+ 1,
ν
2+
1
2; ν +
3
2;
1
z2
), |z| > 1, |arg z| < π, ν 6= −1,−2, . . .
so that the analytic continuation of Qν has
Qν(0) = −π
32 tan
(νπ2
)νΓ(12 −
ν2
)Γ(ν2
) .The Legendre functions satisfy the same recurrences as the Legendre polynomials; this is proven later.
4
2.2 Associated Legendre functions
Consider the equation
(1− z2
)u′′ − 2zu′ +
[ν(ν + 1)− m2
1− z2
]u = 0, m ∈ N.
Substitute u =(z2 − 1
)m2 v. Then
u′ =(z2 − 1
)m2 v′ +mz
(z2 − 1
)m2 −1 v,
u′′ =(z2 − 1
)m2 v′′ + 2mz
(z2 − 1
)m2 −1v′ + 2mz2
(m2− 1)(z2 − 1
)m2 −2v +m
(z2 − 1
)m2 −1v.
The differential equation becomes
−(z2 − 1
)m2
+1v′′ − 2mz
(z2 − 1
)m2 v′ −m2
z2(z2 − 1
)m2−1
v −m(z2 − 1
)m2 v − 2z
(z2 − 1
)m2 v′+
ν(ν + 1) −m2
1 − z2
(z2 − 1)m
2 v = 0,
(z2 − 1
)v′′ + 2mzv′ + 2zv′ +m2z2
(z2 − 1
)−1v +mv −
[ν(ν + 1) +m2
(z2 − 1
)−1]v = 0,(
z2 − 1)v′′ + 2(m+ 1)zv′ + [m(m+ 1)− ν(ν + 1)]v = 0,(
z2 − 1)v′′ + 2(m+ 1)zv′ + (m− ν)(m+ ν + 1)v = 0.
Substitute z = 2t− 1, dvdz = 1
2dvdt .
t(t− 1)d2v
dt2+ (2t− 1)(m+ 1)
dv
dt+ (m− ν)(m+ ν + 1)v = 0.
This is the hypergeometric differential equation with parameters α = m− ν, β = m+ ν + 1, and γ = m+ 1(expand the second term and reverse the sign of each term). Applying Lemma 1 m times, we can write thesolution v(t) as w(m)(t), where
t(t− 1)d2w
dt2+ (2t− 1)
dw
dt− ν(ν + 1)w = 0.
Changing back into z,
z + 1
2
(z + 1
2− 1
)4w′′ + 2
(2 · z + 1
2− 1
)w′ − ν(ν + 1)w = 0,
(z2 − 1
)w′′ + 2zw′ − ν(ν + 1)w = 0.
This is Legendre’s equation, so the solutions to the original equation, by going back through the substitutions,are
u1 =(z2 − 1
)m2
dm
dzmPν(z) ≡ Pmν (z),
u2 =(z2 − 1
)m2
dm
dzmQν(z) ≡ Qmν (z),
the associated Legendre functions. Occasionally, these are called associated Legendre polynomials, but ingeneral, they are not, since m odd gives the square root factor. Usually we are interested in real −1 < x < 1,and these can be written:
Pmν (x) = Pmν (x± i0) = e±mπi2
(1− x2
)m2
dm
dxmPν(x) = (−1)m
(1− x2
)m2
dm
dxmPν(x);
Qmν (x) = (−1)m(1− x2
)m2
dm
dxmQν(x), ν 6= −1,−2, . . .
5
Notice also that for ν ∈ N, m > ν, Pmν (z) ≡ 0. The differential equation is invariant under a change of signin m, so we might as well define, for m ∈ Z,
Pmν ≡(z2 − 1
) |m|2
d|m|
dz|m|Pν(z),
Qmν ≡(z2 − 1
) |m|2
d|m|
dz|m|Qν(z).
Sometimes, a different convention is used for negative m, which should be kept in mind when encounteringthis expression. Furthermore, Pmν may be generalized to Pµν , where µ ∈ C, but we will not be needing this,and integer m is the most commonly encountered situation. Also, notice that
Pmν (z) = Pm−1−ν(z).
This is easily proven since the hypergeometric function does not depend on the order of the first twoarguments.
3 Problem: Laplace’s equation
We want to find those functions which are harmonic in spherical coordinates. We begin with Laplace’sequation, written in spherical coordinates:
∇2u =1
r2∂
∂r
(r2∂u
∂r
)+
1
r2 sin θ
∂
∂θ
(sin θ
∂u
∂θ
)+
1
r2 sin2 θ
∂2u
∂φ2= 0.
Separate the radial and angular components of the equation, by guessing that the solution looks likeu(r, θ, φ) = R(r)Y (θ, φ). [N.B.: here we use the “physics convention” to spherical coordinates; that is,0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π. This is fairly standard for the spherical harmonics.]
∇2u =1
r2∂
∂r
(r2∂(RY )
∂r
)+
1
r2 sin θ
∂
∂θ
(sin θ
∂(RY )
∂θ
)+
1
r2 sin2 θ
∂2(RY )
∂φ2
=Y
r2∂
∂r
(r2∂R
∂r
)+
R
r2 sin θ
∂
∂θ
(sin θ
∂Y
∂θ
)+
R
r2 sin2 θ
∂2Y
∂φ2
=1
R
∂
∂r
(r2∂R
∂r
)+
1
Y sin θ
∂
∂θ
(sin θ
∂Y
∂θ
)+
1
Y sin2 θ
∂2Y
∂φ2= 0.
Now that we have separated the radial and angular components, each must be equal to a constant, which,for convenience, we write as l(l + 1):
1
R
d
dr
(r2
dR
dr
)= l(l + 1),
1
Y sin θ
∂
∂θ
(sin θ
∂Y
∂θ
)+
1
Y sin2 θ
∂2Y
∂φ2= −l(l + 1).
So far there is no loss of generality, since l could be any complex number at this point. Take the angularequation, and multiply by Y sin2 θ:
sin θ∂
∂θ
(sin θ
∂Y
∂θ
)+∂2Y
∂φ2= −l(l + 1)Y sin2 θ.
[You may have seen this equation if you’ve studied physics; it shows up in electrodynamics.] Again, separatevariables: Y (θ, φ) = Θ(θ)Φ(φ).
Φ sin θd
dθ
(sin θ
dΘ
dθ
)+ ΘΦl(l + 1) sin2 θ + Θ
d2Φ
dφ2= 0,
6
1
Θsin θ
d
dθ
(sin θ
dΘ
dθ
)+ l(l + 1) sin2 θ +
1
Φ
d2Φ
dφ2= 0.
The components are separated, so each must be a constant, which, without loss of generality, we write asm2.
1
Θsin θ
d
dθ
(sin θ
dΘ
dθ
)+ l(l + 1) sin2 θ = m2,
1
Φ
d2Φ
dφ2= −m2.
The Φ equation is easy:Φ = eimφ.
[There are actually two solutions corresponding to ±m, but this is fine if we allow m to be negative orpositive. There is also a possible constant in front, but we’ll absorb that into Θ.] Since Φ(φ) = Φ(φ + 2π),we must have that m is an integer. Now look at the Θ equation:
sin θd
dθ
(sin θ
dΘ
dθ
)+[l(l + 1) sin2 θ −m2
]Θ = 0.
Make a change of variables:
x = cos θ,
dx = − sin θ dθ,
df
dx=
1
sin θ
df
dθ,√
1− x2 df
dx=
df
dθ,
=⇒ d
dθ≡√
1− x2 d
dx.
Then: √1− x2
[√1− x2 d
dx
(√1− x2 ·
√1− x2 dΘ
dx
)]+[l(l + 1)
(1− x2
)−m2
]Θ = 0,
(1− x2
) d
dx
[(1− x2
) dΘ
dx
]+[l(l + 1)
(1− x2
)−m2
]Θ = 0,
d
dx
[(1− x2
) dΘ
dx
]+
[l(l + 1)− m2
(1− x2)
]Θ = 0.
This is an equation we’ve seen; the solution is
Θ = APml (x) = APml (cos θ).
The second solution, not of interest here because it is unbounded, is
Θ = BQml (cos θ).
The spherical harmonics are then
Y ml (θ, φ) = APml (cos θ)eimφ.
We can also normalize this so that∫ 2π
0
∫ π
0
Y ml (θ, φ)Y m′
l′ (θ, φ) sin θ dθ dφ = δll′δmm′ .
7
This determines A, which gives the normalized spherical harmonics:
Y ml (θ, φ) =
√2l + 1
4π
(l − |m|)!(l + |m|)!
Pml (cos θ)eimφ.
Occasionally the normalized functions are defined to be the spherical harmonics; this usually happens inphysics texts. Also, in other disciplines, several different normalization factors may appear.
Now look at the radial equation:d
dr
(r2
dR
dr
)= l(l + 1)R,
r2d2R
dr2+ 2r
dR
dr− l(l + 1)R = 0.
This is a Cauchy-Euler equation, and is easy to solve:
R = Arl +Br−l−1.
If we want the solution to be bounded at the origin, then we pick B = 0. The set of solutions to Laplace’sequation is then
ul,m =(Alr
l +Blr−l−1)Y ml (θ, φ).
The general solution is
u(r, θ, φ) =
∞∑l=0
l∑m=−l
(Alr
l +Blr−l−1)Y ml (θ, φ).
4 Application: hydrogen atoms
4.1 Brief introduction to quantum mechanics
The central idea of quantum mechanics is wave-particle duality. The wave function of a system, Ψ(x, t),contains all the information about the system, and is in general complex. Born’s interpretation of the wavefunction is that |Ψ|2 is a probability density; therefore, the wave function must be normalized:∫ ∞
−∞|Ψ|2 dx = 1.
(This is for a one-dimensional wave function.) Schrodinger’s equation describes the evolution of the wavefunction:
i~∂Ψ
∂t= − ~2
2µ
∂2Ψ
∂x2+ VΨ,
where V , usually depending only on x, is a potential to be specified, µ is a mass, and ~ is Planck’s constant.By separation of variables, we obtain Ψ = ψ(x)φ(t), where
φ(t) = e−iEt~
and ψ satisfies the time-independent Schrodinger equation:
− ~2
2µ
∂2ψ
∂x2+ V ψ = Eψ.
8
4.2 Hydrogen wave functions
It is easy to generalize the Schrodinger equation to R3:
i~∂Ψ
∂t= − ~2
2µ∇2Ψ + VΨ.
The probability of finding a particle with wave function Ψ(r, t) in the volume d3r = dxdy dz is |Ψ(r, t)|2 d3r,so the normalization condition is ∫
|Ψ|2 d3r = 1.
Again, by separation of variables, we can obtain a complete set of stationary states:
Ψn(r, t) = ψn(r)e−iEnt
~ ,
and the ψn satisfy
− ~2
2µ∇2ψ + V ψ = Eψ.
The general solution is just a linear combination of these states. Now, if we rewrite the time-independentSchrodinger equation in spherical coordinates,
− ~2
2µ
[1
r2∂
∂r
(r2∂ψ
∂r
)+
1
r2 sin θ
∂
∂θ
(sin θ
∂ψ
∂θ
)+
1
r2 sin2 θ
(∂2ψ
∂φ2
)]+ V ψ = Eψ.
As usual, we begin by separating variables. Here we will split the solutions into a product of radial andangular components:
ψ(r, θ, φ) = R(r)Y (θ, φ).
Substituting into the Schrodinger equation,
− ~2
2µ
[Y
r2d
dr
(r2
dR
dr
)+
R
r2 sin θ
∂
∂θ
(sin θ
∂Y
∂θ
)+
R
r2 sin2 θ
∂2Y
∂φ2
]+ V RY = ERY.
Divide by RY , multiply by −2µr2/~2:
1
R
d
dr
(r2
dR
dr
)− 2µr2
~2[V (r)− E] +
1
Y
[1
sin θ
∂
∂θ
(sin θ
∂Y
∂θ
)+
1
sin2 θ
∂2Y
∂φ2
]= 0.
Since we have separated the angular and radial components, each is a constant, which we write l(l + 1):
1
R
d
dr
(r2
dR
dr
)− 2µr2
~2[V (r)− E] = l(l + 1),
1
Y
[1
sin θ
∂
∂θ
(sin θ
∂Y
∂θ
)+
1
sin2 θ
∂2Y
∂φ2
]= −l(l + 1).
The angular equation once again gives spherical harmonics: Y = Y ml (θ, φ), l ∈ Z, m ∈ Z.[N. B.: In the following, e is used for both the elementary charge and the logarithmic base. The usage is
fairly clear from context, and there should be no confusion as long as you are aware of it.] For a hydrogenatom, the potential V is the Coulomb potential:
V (r) = − e2
4πε0r,
9
where e is the elementary charge, k = 14πε0
is the Coulomb constant. After changing variables, u(r) ≡ rR(r),the differential equation becomes
− ~2
2µ
d2u
dr2+
[V +
~2
2µ
l(l + 1)
r2
]u = Eu,
− ~2
2µ
d2u
dr2+
[− e2
4πε0r+
~2
2µ
l(l + 1)
r2
]u = Eu.
Let κ ≡√−2µE~ (which is real because E is negative for states of interest), and divide by E:
1
κ2d2u
dr2=
[1− µe2
2πε0~2κ2r+l(l + 1)
κ2r2
]u.
Then let ρ ≡ κr, ρ0 ≡ µe2
2πε0~2κ :
d2u
dρ2=
[1− ρ0
ρ+l(l + 1)
ρ2
]u.
Now consider the asymptotics. For large ρ the equation is approximately
d2u
dρ2= u.
The general solution is u = Ae−ρ + Beρ, but eρ is not normalizable, so B = 0. For small ρ, the equation isapproximately
d2u
dρ2=l(l + 1)
ρ2u,
which has solution Cρl+1 + Dρ−l, but ρ−l is not normalizable, so D = 0. Factoring out the asymptoticbehavior, we have, without loss of generality,
u(ρ) = ρl+1e−ρv(ρ).
Putting this into the differential equation,
ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v = 0.
If we attempt a series solution for v, the asymptotic behavior is v ≈ c0e2ρ, which makes u not normalizable.Therefore, the series must terminate; that is, v must be a polynomial. Letting s = 2ρ,
sd2v
ds2+ (2l + 2− s)dv
ds+(ρ0
2− l − 1
)v = 0.
Writing ρ0 = 2n, n ∈ N (it is easily shown that this is the case), we recognize the solution to be associatedLaguerre polynomials [see (1.2.4)]:
v(ρ) = L2l+1n−l−1(2ρ).
ρ0 determines E:
En = − µe4
32π2ε20~2n2.
Going back through all the substitutions, we have the solution for ψ = ψnlm:
ψnlm = C0e− rna
(2r
na
)l[L2l+1n−l−1
(2r
na
)]Y ml (θ, φ).
10
Once C0 is determined to normalize the probability, we finally have the normalized hydrogen wave functions:
ψnlm =
√(2
na
)3(n− l − 1)!
2n[(n+ l)!]3e−
rna
(2r
na
)l[L2l+1n−l−1
(2r
na
)]Y ml (θ, φ).
They are all orthonormal: ∫ ∫ ∫ψnlmψn′l′m′r
2 sin θ dr dθ dφ = δnn′δll′δmm′ .
Adding on the time dependence gives a complete set of stationary states; the general solution is a linearcombination of these:
Ψnlm(r, θ, φ, t) = ψnlm(r, θ, φ)e−iEnt
~ .
5 Application: Helmholtz’s equation
Consider Helmholtz’s equation,∇2u+ k2u = 0.
Once again separate variables: u = R(r)Θ(θ)Φ(φ). We obtain:
d2Φ
dφ2= −m2Φ,
sin θd
dθ
(sin θ
dΘ
dθ
)+[l(l + 1) sin2 θ −m2
]Θ = 0,
d
dr
(r2
dR
dr
)+[k2r2 − ν(ν + 1)
]R = 0.
In the radial equation, substitute R = r−12 v. The result is
v′′ +1
rv′ +
[k2 −
(ν + 1
2
)2r2
]v = 0,
which has solutionv = AJν+ 1
2(kr) +BH
(2)
ν+ 12
(kr),
where Jν is Bessel’s function of the first kind and H(2)ν is the second Hankel function, H
(2)ν = Jν − iYν , Yν
being Bessel’s function of the second kind. The particular solutions to Helmholtz’s equation are then
u = r−12
[AJν+ 1
2(kr) +BH
(2)
ν+ 12
(kr)]
[CPmν (cos θ) +DQmν (cos θ)] eimφ.
Usually, the situation of interest is rotationally symmetric and requires a bounded solution; this forces D = 0,m = 0:
u = r−12
[AJν+ 1
2(kr) +BH
(2)
ν+ 12
(kr)]
[CPν(cos θ)] .
A solution can then be constructed from a superposition of these.
6 Properties of spherical harmonics
(Really, properties of Legendre/associated Legendre functions, but since the spherical harmonics are definedthis way, the properties transfer somewhat.)
11
6.1 Analytic continuation of Pν(z)
Using ∫ π2
0
sinν θ dθ =
√π
2
Γ(ν+12
)Γ(ν2 + 1
) , <ν > −1,
and properties of the gamma function, we arrive at
2
π
∫ π2
0
sin2k φdφ =
(12
)k
k!, k = 0, 1, 2, . . . ,
and write
Pν(z) =
∞∑k=0
(−ν)k(ν + 1)k
(k!)2
(1− z
2
)k=
2
π
∞∑k=0
(−ν)k(ν + 1)k(12
)kk!
(1− z
2
)k ∫ π2
0
sin2k φ dφ
=2
π
∫ π2
0
dφ
∞∑k=0
(−ν)k(ν + 1)k(12
)kk!
(1− z
2sin2 φ
)k=
2
π
∫ π2
0
F
(−ν, ν + 1;
1
2;
1− z2
sin2 φ
)dφ,
where uniform convergence justifies reversing the order of integration and summation. Using the identity
F
(−ν, ν + 1,
1
2,−w
)=
(√1 + w +
√w)2ν+1
+(√
1 + w +√w)−2ν−1
2√
1 + w
≡ fν(w), |w| < 1,
and some arguments from complex analysis, the analytic continuation of Pν(z) is
Pν(z) =2
π
∫ π2
0
fν
(z − 1
2sin2 φ
)dφ, |arg(z + 1)| < π,
fν defined as above. Usually in applications we look for bounded solutions, so that Qν is not as frequentlyused as Pν . Since the analytic continuation of Qν is much more tedious, and we won’t be using it here, theanalytic continuation, without proof, is
Qν(z) =
∫∞
0
(√1 + z−1
2 cosh2 ψ +√
z−12 cosh2 ψ
)−2ν−1√
1 + z−12 cosh2 ψ
dψ, <ν > −1, |arg(z − 1)| < π.
6.2 Some integral representations
To derive these integral representations, the general technique is to assume z > 1 through a substitution,then use analytic continuation to show that the formula is valid in an extended region. To illustrate this,begin with
Pν(z) =2
π
∫ π2
0
fν
(z − 1
2sin2 φ
)dφ, |arg(z + 1)| < π,
fν defined as before. Let z = coshα, α > 0, and in the integral, substitute
sinhθ
2= sinh
α
2sinφ.
12
The result is
Pν(coshα) =2
π
∫ α
0
cosh(ν + 1
2
)θ
√2 coshα− 2 cosh θ
dθ.
Similarly, by substituting z = cos θ, sin ψ2 = sin θ
2 sinφ,
Pν(cos θ) =2
π
∫ θ
0
cos(ν + 1
2
)ψ
√2 cosψ − 2 cos θ
dψ,
which gives an integral representation for −1 < z < 1.By an analogous procedure, we can also derive an integral representation for Qν(z):
Qν(coshα) =
∫ ∞α
e−(ν+ 12)θ
√2 cosh θ − 2 coshα
dθ.
Finally we state without proof the following for associated Legendre functions of the first kind, whichmay be useful in numerical computation of the spherical harmonics:
1.
Pmν (z) =Γ(ν +m+ 1)
(z2 − 1
)m2
2m√π Γ(m+ 1
2
)Γ(ν −m+ 1)
∫ π
0
(z +
√z2 − 1 cosψ
)ν−msin2mψ dψ, Re z > 0, m = 0, 1, 2, . . .
2.
Pmν (z) =Γ(ν +m+ 1)
πΓ(ν + 1)
∫ π
0
(z +
√z2 − 1 cosψ
)νcosmψ dψ, Re z > 0, m = 0, 1, 2, . . .
3.
Pmν (cos θ) =(−1)m2Γ(ν +m+ 1)√π Γ(m+ 1
2
)Γ(ν −m+ 1)
1
(2 sin θ)m
∫ θ
0
cos(ν + 1
2
)φ
(2 cosφ− 2 cos θ)12−m
dφ, 0 < θ < π, m = 0, 1, 2, . . .
6.3 Recurrence relations
Using from above,
Pν(coshα) =2
π
∫ α
0
cosh(ν + 1
2
)θ
√2 coshα− 2 cosh θ
dθ,
and
Qν(coshα) =
∫ ∞α
e−(ν+ 12)θ
√2 cosh θ − 2 coshα
dθ,
we can derive recurrence relations that relate the functions with varying (integer) degrees. For example:
Pν+1(coshα) + Pν−1(coshα) =4
π
∫ α
0
cosh(ν + 1
2
)θ cosh θ
√2 coshα− 2 cosh θ
dθ
=4
π
∫ α
0
coshα cosh(ν + 1
2
)θ
√2 coshα− 2 cosh θ
dθ − 2
π
∫ α
0
√2 coshα− 2 cosh θ cosh
(ν +
1
2
)dθ
= 2 coshαPν(coshα)− 4
(2ν + 1)π
∫ α
0
√2 coshα− 2 cosh θ d sinh
(ν +
1
2
)θ
= 2 coshαPν(coshα)− 4
(2ν + 1)π
∫ α
0
sinh(ν + 1
2
)θ sinh θ
√2 coshα− 2 cosh θ
dθ
= 2 coshαPν(coshα)− 2
(2ν + 1)π
∫ α
0
cosh(ν + 3
2
)θ − cosh
(ν − 1
2
)θ
√2 coshα− 2 cosh θ
dθ
= 2 coshαPν(coshα)− 1
2ν + 1[Pν+1(coshα)− Pν−1(coshα)] ,
13
and going back to coshα = z, we have
(ν + 1)Pν+1(z)− (2ν + 1)zPν(z) + νPν−1(z) = 0,
analogous to (1.1.3). The other recurrence relations can be derived similarly. If we start from Qν , it turnsout that the exact same recurrence relations are obtained. Finally, we can prove that Pν reduces to theLegendre polynomials defined by the Rodrigues formula (1.1.1) for ν ∈ N. Use ν = 0 and ν = 1 in thehypergeometric series definition of Pν to find that P0(z) = 1 and P1(z) = z. Then since the same recurrencerelations hold, the functions constructed by putting these into the recurrence must generate the exact samepolynomials.
The Legendre functions of the second kind are not polynomials, but using
Q0(z) =1
2log
z + 1
z − 1,
Q1(z) =z
2log
z + 1
z − 1− 1,
which can be calculated from previously given formulas, the recurrence relations, and induction, it can beshown that
Qn(z) =1
2Pn(z) log
z + 1
z − 1− fn−1(z), n = 0, 1, 2, . . .
where fn−1 is a polynomial of degree n− 1, and f−1(z) ≡ 0.Without proof, we also give the following:
1.(ν −m+ 1)Pmν+1 − (2ν + 1)zPmν + (ν +m)Pmν−1 = 0, m = 0, 1, 2, . . .
2. (z2 − 1
) dPmνdz
= νzPmν − (ν +m)Pmν−1, m = 0, 1, 2, . . .
These also hold exactly in the same way for Qmν .
6.4 Relations to other special functions
A value of interest is Pν(0). We begin computing this by using the hypergeometric series:
Pν(0) = F
(−ν, ν + 1; 1;
1
2
)=
∞∑k=0
(−ν)k(ν + 1)k(1)kk!
1
2k
=1
Γ(−ν)Γ(ν + 1)
∞∑k=0
Γ(k − ν)Γ(k + ν + 1)
2k(k!)2
= − sin νπ
π
∞∑k=0
Γ(k − ν)Γ(k + ν + 1)
2k(k!)2 ,
14
where the last step is the reflection formula. Replace one of the k! by Γ(k + 1), and we have
Pν(0) = − sin νπ
π
∞∑k=0
Γ(k + ν + 1)
2kk!Γ(ν + 1)
Γ(k − ν)Γ(ν + 1)
Γ(k + 1)︸ ︷︷ ︸B(k−ν, ν+1)
= − sin νπ
π
∞∑k=0
Γ(k + ν + 1)
2kk!Γ(ν + 1)
∫ 1
0
tk−ν−1(1− t)ν dt
= − sin νπ
π
∫1
0
t−ν−1(1− t)ν∞∑k=0
Γ(k + ν + 1)
k!Γ(ν + 1)
(t
2
)kdt
= − sin νπ
π
∫1
0
t−ν−1(1− t)ν(
1− t
2
)−ν−1dt,
where absolute convergence justifies switching the order of integration and summation. Another change ofvariables, t = 1−
√s, gives
Pν(0) = −2ν sin νπ
π
∫ 1
0
s12 (ν−1)(1− s)−ν−1 ds︸ ︷︷ ︸
B( ν2+12 ,−ν)
= −2ν sin νπ
π
Γ(−ν)Γ(ν2 + 1
2
)Γ(12 −
ν2
) ,
and finally, we have
Pν(0) =
√π
Γ(12 −
ν2
)Γ(ν2 + 1
) .It also turns out that Legendre functions of degree ν = n − 1
2 , n ∈ N, can be written as elliptic integrals.First, for ν = − 1
2 , reduce, using a formula given previously,
P− 12(coshα) =
2
π cosh α2
K(
tanhα
2
),
where K is the elliptic integral of the first kind:
K(k) =
∫ π2
0
dφ√1− k2 sin2 φ
.
Then, the recurrence relations can be used to obtain the other functions. Similarly, using
Q− 12(coshα) = 2e−
α2 K(e−α
),
and the recurrence relations, we have the Legendre functions of the second kind, for half-integer degrees.
6.5 Other facts
1. Y ml (θ, φ) are eigenfunctions of the differential operator
(r×∇)2
=1
sin θ
∂
∂θ
(sin θ
∂
∂θ
)+
1
sin2 θ
∂2
∂φ2;
(r×∇)2Y ml = l(l + 1)Y ml .
2. ∫ 1
−1x2mP2n(x) dx = 22n+1 Γ(2m+ 1)Γ(m+ n+ 1)
Γ(m− n+ 1)Γ(2m+ 2n+ 2).
15
7 Pictures
The pictures following the bibliography are from [2].
References
[1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions. NIST, Maryland, 1964.
[2] S. Brandt, H. D. Dahmen, The Picture Book of Quantum Mechanics. Springer, New York, 4th edition,2012.
[3] D. J. Griffiths, An Introduction to Quantum Mechanics. Prentice Hall, New Jersey, 2nd edition, 2005.
[4] G. Kristensson, Second Order Differential Equations. Springer, New York, 2010.
[5] N. N. Lebedev, Special Functions & Their Applications. Dover Publications, USA, 1972.
16
10.3 Angular Momentum, Spherical Harmonics 193
Fig.10.3. The first ten Legendre polynomials P�(u) = 12��!
d�
du�[(u2 −1)�
].
194 10. Wave Packet in Three Dimensions
Fig.10.4. Graphs of the associated Legendre functions Pm� (u), top, and of the absolute
squares of the spherical harmonics Y�m(ϑ ,φ), bottom. Except for a normalization fac-tor, the absolute squares of the spherical harmonics are the squares of the associatedLegendre functions.
196 10. Wave Packet in Three Dimensions
Fig.10.5. The spherical harmonics Y�m are complex functions of the polar angle ϑ , with0 ≤ ϑ ≤ π , and the azimuth φ, with 0 ≤ φ < 2π . They can be visualized by showing theirreal and imaginary parts and their absolute square over the ϑ ,φ plane. Such graphs areshown here for �= 3 and m = 0, 1, 2, 3.
sphere. For all possible values � and m the functions |Y�m|2 are rotationallysymmetric around the z axis. They can vanish for certain values of ϑ . Theseare called ϑ nodes if they occur for values of ϑ other than zero or π . It shouldbe noted that |Y��|2 does not have nodes, whereas |Y�m|2 possesses �− |m|nodes.
10.3 Angular Momentum, Spherical Harmonics 197
Fig.10.5. (continued)
The Legendre polynomials possess the following orthonormality proper-ties: ∫ 1
−1P�(u)P�′(u)du = 2
2�+1δ��′ .
Here δ��′ is the Kronecker symbol
δ��′ ={
1 , �= �′0 , � = �′ .
198 10. Wave Packet in Three Dimensions
Fig.10.6. Polar diagrams of the absolute squares of the spherical harmonics. The distancefrom the origin of the coordinate system to a point on the surface seen under the anglesϑ and φ is equal to |Y�m(ϑ ,φ)|2. Different scales are used for the individual parts of thefigure.