Section 13.6: Angular Solutions of the Schrödinger Equation

m =    

Animation 1 Animation 2
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Most potential energy functions in three dimensions are not often rectangular in form. In fact, they are most often in spherical coordinates (due to a spherical symmetry) and occasionally in cylindrical coordinates due to a cylindrical symmetry. We begin by considering the generalization of the time-independent Schrödinger equation to three-dimensional spherical coordinates, which is1

−(ħ2/2μ)[(1/r2)∂/∂r(r2∂/∂r) + (1/r2sin(θ))(∂/∂θ)(sin(θ)∂/∂θ) + (1/r2sin2(θ))(∂2/∂φ2)]ψ(r) + V(r)ψ(r) = Eψ(r) . (13.20)

The probability per unit volume, the probability density, is ψ*(r)ψ(r) and therefore we require ∫ ψ*(r)ψ(r) d3r = 1 (where d3r = dV = r2sin(θ)drdθdφ) to maintain a probabilistic interpretation of the wave function in three dimensions.

As in the two-dimensional case, we use separation of variables variables, but now using ψ(r) = R(r) Y(θ,φ), i.e., separate the radial part from the angular part. This substitution yields

[(1/R(r))d/dr(r2dR(r)/dr) + (1/Ysin(θ))(∂/∂θ)(sin(θ)∂Y/∂θ) + (1/Ysin2(θ))(∂2Y/∂φ2)] − (2μr2/ħ2)[V(r) − E] = 0 , (13.21)

as long as V(r) = V(r) only. Note that each term involves either r or θ and φ.  We can separate these equations using the technique of separation of variables to give

(1/R(r)) d/dr (r2 dR(r)/dr) − (2μr2/ħ2)[V(r) − E] = l(l + 1) ,                    (13.22)

and

(1/Ysin(θ)) (∂/∂θ)(sin(θ) ∂Y/∂θ) + (1/Ysin2(θ)) (∂2Y/∂φ2) = −l(l + 1) ,                    (13.23)

for the radial and angular parts, respectively. The constant l(l + 1) is the separation constant that allows us to separate one differential equation into two. We can do so because the only way for preceding equation to be true for all r, θ, and φ is for the angular part and the radial part to each be equal to a constant, ± l(l + 1). Despite the seemingly odd form of the separation constant, it is completely general and can be made to equal any complex number. For the angular piece, we can again separate variables using the substitution Y(θ,φ) = Θ(θ)Φ(φ).  This gives:

sin(θ)/Θ d/dθ(sin(θ) dΘ/dθ) + l(l + 1)sin2(θ) = m2 ,                    (13.24)

and

1/Φ d2Φ/dφ2 = −m2 ,                    (13.25)

where we have written the separation constant as ± m2, again without any loss of generality.  The Φ(φ) part of the angular equation is a differential equation, d2Φ/dφ2 = −m2Φ, we have solved before.  We get as its unnormalized solution

Φm(φ) = exp(imφ) ,                     (13.26)

where m is the separation constant which can be both positive and negative.  Since the angle φ ε {0, 2π}, we have that Φm(φ) = Φm(φ + 2π).  Like the ring problem in Section 13.5, in order for Φm(φ) to be single valued means that m = 0, ±1, ±2, ±3,….  We show these solutions in Animation 1.  The Θ(θ) part of the angular equation is harder to solve.  It has the unnormalized solutions

Θlm(θ) = A Plm(cos(θ)) ,                     (13.27)

where the Plm are the associated Legendre polynomials, where

Plm(x) = (1 − x2)|m|/2 (d/dx)|m| Pl(x),                    (13.28)

are calculated from the Legendre polynomials

Pl(x) = (1/2ll!) (d/dx)l(x2 − 1)l  .      (Rodriques' formula)                     (13.29)

The first few Legendre polynomials are

P0(x) = 1 ,        P1(x) = x ,        and    P2(x) = (1/2) (3x2 1) ,                    (13.30)

or in terms of cos(θ)

P0 = 1,        P1 = cos(θ),        and    P2 = (1/2) (3cos2(θ)−1) .                    (13.31)

We can also write the Plm(x) using the above formulas as:

P00 = 1,        P11 = sin(θ),    P01 = cos(θ) ,                    (13.32)

P02 = (1/2)(3cos2(θ)-1),    P12 = 3sin(θ)cos(θ),    P22 = 3sin2(θ) .                     (13.33)

We notice that l > 0 for Rodrigues' formula to be valid. In addition, |m| ≤ l since Pl|m|>l = 0.  (For |m| > l, the power of the derivative is larger than the order of the polynomial and hence the result is zero.) We also note that there must be 2l + 1 values for m, given a particular value of l.  Polar plots (zx plane) of associated Legendre polynomials are shown in Animation 2.  A positive angle θ is defined to be the angle down from the z axis toward the positive x axis. The length of a vector from the origin to the wave function, Plm, is the magnitude of the wave function at that angle. You may vary l and m to see how Plm varies.  We normalize Θlm(θ)Φm(φ) by normalizing the angular part separately from the radial part (which we have yet to consider):

∫∫ Ylm*(θ,φ)Ylm(θ,φ) sin (θ) dθdφ = 1    [θ integration from 0 to π, φ integration from 0 to 2π]    (13.34)

where Ylm(θ,φ) = Θlm(θ)Φm(φ).  When the Ylm(θ,φ) are normalized, they are called the spherical harmonics.The first few are

Y00(θ,φ) = (1/4π)1/2 ,                     (13.37)

Y ±11(θ,φ) = −/+ (3/8π)1/2 sin(θ) exp(±iφ)        Y 01(θ ,φ) = (3/4π)1/2 cos(θ) ,         (13.38)

and in general for m > 0,

Ylm(θ,φ) = (−1)m [(2l + 1)(lm)!/(4π(l + m)!)]1/2 exp(imφ) Plm cos(θ) ,                    (13.39)

and Yl−m(θ,φ) = (−1)mYlm*(θ,φ) for m < 0. When we represent the spherical harmonics this way, they are automatically orthogonal:

 ∫ Ylm*(θ,φ)Yl'm'(θ,φ) sin(θ) dθdφ = δm m' δl l' .                    (13.40)

Optional Visualization. You may also view Animation 3 and Animation 4 in three dimensions if you have Java 3d installed on your computer.3  Animation 3 shows Ylm(θ,φ) in the standard, but now three-dimensional, representation: The length of a vector from the origin to the wave function, Ylm(θ,φ), is the magnitude of the wave function at that angle. Here, the phase of the Ylm(θ,φ) is represented as color. Animation 4 shows Ylm(θ,φ) projected on a sphere such that brightness of the wave function indicates magnitude of the wave function and color again represents the phase.

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1To avoid future confusion, we hereafter use μ for mass, and reserve m for the azimuthal (or magnetic) quantum number.
2Classically, angular momentum is L = r × p. We can write  L using quantum-mechanical operators in rectangular coordinates as Lx = ypz zpy, Ly = zpx xpz, and Lz = xpyypx. We find that if we write L2 and Lz in spherical coordinates,

L2 = −ħ2 [(1/sin(θ)) (∂/∂θ)(sin(θ) ∂/∂θ) + (1/sin2(θ)) (∂2/∂φ2) ,                     (13.35)

 and

Lz = − (∂/∂φ) .                     (13.36)

To which we note L2Ylm = l(l + 1)ħ2Ylm and LzYlm  = Ylm; the spherical harmonics, the Ylm , are eigenstates of L2 and Lz.

3If you do not have Java 3D installed, go to the Sun Java 3D Web site for download: http://java.sun.com/products/java-media/3D/download.html.

 

 

© 2006 by Prentice-Hall, Inc. A Pearson Company