Separation of Variables The Angular Equation

4.1.1 Separation of Variables

1 Spherical coordinates Cartesian coordinates: z y x , , φ θ , , r Spherical coordinates: , sin 2 φ θ θ τ d drd r d = P x y z o , , z y x M θ r • • ϕ ϕ ϕ ϕ z y x A = = , cos sin φ θ φ θ r x In spherical coordinates the Laplacian takes the form ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ = ∇ 2 2 2 2 2 2 2 2 sin 1 sin sin 1 1 φ θ θ θ θ θ r r r r r r , sin 2 φ θ θ τ d drd r d = = = . cos , sin sin θ φ θ r z r y The time-independent Schrodinger equation in Cartesian coordinates In spherical coordinates ψ ψ ψ ψ ψ E V z y x m = + ∂ ∂ + ∂ ∂ + ∂ ∂ − 2 2 2 2 2 2 2 2 ψ ψ φ ψ θ θ ψ θ θ θ ψ E V r r r r r r m = + ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ − 2 2 2 2 2 2 2 2 sin 1 sin sin 1 1 2 We begin by looking for solutions that are separable into products: , , , φ θ φ θ ψ Y r R r = Putting this into above equation, we have ERY VRY Y r R Y r R dr dR r dr d r Y m = + ∂ ∂ + ∂ ∂ ∂ ∂ + − 2 2 2 2 2 2 2 sin sin sin 2 φ θ θ θ θ θ 2 2 2mr − Dividing by RY and multiplying by 2 sin 1 sin sin 1 1 2 2 2 2 2 2 = − − ∂ ∂ + ∂ ∂ ∂ ∂ + E r V mr Y Y Y Y dr dR r dr d R φ θ θ θ θ θ ∂ ∂ + ∂ ∂ ∂ ∂ − = − − 2 2 2 2 2 2 sin 1 sin sin 1 2 1 φ θ θ θ θ θ Y Y Y Y E r V mr dr dR r dr d R The term on the left hand depends only on r , whereas the right depends only on ; accordingly, each must be a constant, which is in the form ll+1 : ; 1 2 1 2 2 2 + = − − l l E r V mr dr dR r dr d R . 1 sin 1 sin sin 1 2 2 2 + − = ∂ ∂ + ∂ ∂ ∂ ∂ l l Y Y Y Y φ θ θ θ θ θ r V V = Spherically symmetric potential

4.1.2 The Angular Equation

Solution of Y : Equation 4.17 determines Y function as Multiplying above equation by Ysin 2 , it becomes: . 1 sin 1 sin sin 1 2 2 2 + − = ∂ ∂ + ∂ ∂ ∂ ∂ l l Y Y Y Y φ θ θ θ θ θ . sin 1 sin sin 2 2 2 Y l l Y Y θ φ θ θ θ θ + − = ∂ ∂ + ∂ ∂ ∂ ∂ As always, we try separation of variables: , φ θ φ θ Φ ⋅ Θ = Y Plugging this in, and dividing by Y, we find . 1 sin 1 sin sin 1 2 2 2 = Φ Φ + + + Θ Θ φ θ θ θ θ θ d d l l d d d d The first term is a function only of , and the second is a function only of , so each must be a constant. This time I will call the separation constant m 2 : φ ; sin 1 sin sin 1 2 2 m l l d d d d = + + Θ Θ θ θ θ θ θ d d Θ θ θ . 1 2 2 2 m d d − = Φ Φ φ 1 The equation is easy to solve: . 2 2 2 φ φ φ im e m d d = Φ Φ − = Φ φ Now, when advances by 2 , we return to the same point in space, so it is natural to require that φ . 2 φ φ π = + In other words, 1 2 2 = = + m i im im e e e π φ π φ From this it follows that m must be an integer : , 3 , 2 , 1 , ± ± ± = m 2 The equation is not simple. θ [ ] . sin 1 sin sin 2 2 = Θ − + + Θ m l l d d d d θ θ θ θ θ Turn into x by: θ θ cos = x . 1 1 2 1 2 2 2 2 2 = Θ − − + + Θ − Θ − x m l l d d x d d x θ θ Above equation is l th-order associated Legendre equation . Therefore, the solution of is , cos θ θ m l m l AP x AP = = Θ where P m l is the associated Legendre function, defined by , 1 | | 2 | | 2 x P dx d x x P l m m m l − ≡ and P l x is the l th Legendre polynomial , defined by the Rodrigues formula : , 1 2 1 2 l l l l x dx d l x P − ≡ , 1 3 1 , , 1 2 1 − = = = x x P x x P x P 3 30 35 8 1 , 3 5 2 1 , 1 3 2 2 4 4 3 3 2 2 + − = − = − = x x x P x x x P x x P P l x is a polynomial of degree l in x , and is even or odd according to the parity of l . is not, in general, a polynomial——if m is odd it carries a factor of , 1 | | 2 | | 2 x P dx d x x P l m m m l − ≡ But for associated Legendre function P m l : . 1 2 x − l m ≤ , 1 = = x P x P , 1 1 x x P x P = = , 1 1 2 1 2 1 2 1 1 x x P dx d x x P − = − = , 1 3 2 1 2 2 2 − = = x x P x P , 1 3 1 2 2 2 1 2 1 2 x x x P dx d x x P − = − = , 1 3 1 2 2 2 2 2 2 2 x x P dx d x x P − = − = not a polynomial As , cos θ = x , 1 = r , cos θ = r P x y z o , , z y x M θ r • • ϕ ϕ ϕ ϕ z y x A , sin θ = r cos θ m l P r = Plot: , 1 = x P , cos 1 θ = x P , sin 1 1 θ = x P , 1 cos 3 2 1 2 2 − = θ x P , cos sin 3 1 2 θ θ = x P , sin 3 2 2 2 θ = x P cos θ m l P r = Plot: P x y z o , , z y x M θ r • • ϕ ϕ ϕ ϕ z y x A , 1 cos 3 1 2 − = θ r , cos sin 3 θ θ = r , sin 3 2 θ = r , 1 cos 3 2 − = θ r , cos sin 3 θ θ = r , sin 3 θ = r , cos 3 cos 5 2 1 3 3 θ θ − = x P , 1 cos 5 sin 2 3 2 1 3 − = θ θ x P , cos sin 15 2 2 3 θ θ = x P , cos 1 sin 15 2 3 3 θ θ − = x P cos θ m l P r = Plot: , cos 3 cos 5 2 1 3 θ θ θ − = r 1 , 1 − = = π r r 1 Notice that l must be a nonnegative integer , for the Rodrigues formula to make any sense. Some Notes: , 1 | | 2 | | 2 x P dx d x x P l m m m l − ≡ l m ≤ If |m|l , then P l m =0. Therefore, for any given l , there are 2l+1 possible values of m : values of m : , 2 , 1 , = l l l l l m , 1 , 1 , , 1 , , 1 , − − + − − = , , , , 1 1 x P x P x P x P l l l l l l ± − ± ± 2 Now, the volume element in spherical coordinates is so the normalization condition becomes 1 sin sin 2 2 2 2 2 = ⋅ = φ θ θ φ θ θ ψ d d Y dr r R d drd r , sin 2 3 φ θ θ d drd r r d = It is convenient to normalize R and Y separately: It is convenient to normalize R and Y separately: , 1 2 2 = ∞ dr r R , 1 sin 2 2 = π π φ θ θ d d Y where R determined by Vr and Y can be obtained. ; 1 2 1 2 2 2 + = − − l l E r V mr dr dR r dr d R The normalized angular wave functions are called spherical harmonics : where , cos | | | | 4 1 2 , θ π ε φ θ φ m l im m l P e m l m l l Y + − + = ≤ ≥ − = . for , 1 , for , 1 m m m ε Now we list here some few spherical harmonics: See book for more φ θ θ π φ θ i e Y ± ± − = 1 cos 5 sin 64 21 , 2 1 3 Actually, the Y s are automatically orthogonal, so [ ] , sin , , 2 m m l l m l m l d d Y Y ′ ′ ′ ′ = δ δ φ θ θ φ θ φ θ π π Notice that , 1 , φ θ φ θ m l m m l Y Y − = − For historical reasons, l is called the azimuthal quantum number, and m the magnetic quantum number. Notice that the angular part of the wave function, Y , , is the same for all spherically symmetric potentials; the actual shape of the potential, Vr , affects only the radial part of the wave function, Rr , which is determined by Equation 4.16:

4.1.3 The Radial Equation