The Chemistry Maths Book, Second Edition

(Grace) #1

14.6 The hydrogen atom 403


To separate the angular variables we now write


Y(θ, φ) 1 = 1 Θ(θ) 1 × 1 Φ(φ) (14.57)


Substitution of this product and its derivatives in the angular equation, division by


Y 1 = 1 ΘΦ, and multiplication by sin


2

1 θgives


(14.58)


so that, with separation constant−m


2

(14.59)


(14.60)


The separation of the variables is complete. It is now only necessary to solve the three


boundary value problems represented by equation (14.55), (14.59), and (14.60), with


appropriate boundary conditions.


The Φequation


(14.59)


The function Φis defined in the interval 01 ≤ 1 φ 1 ≤ 12 πand must satisfy the condition


Φ(2π) 1 = 1 Φ(0) (14.61)


for continuity round the circle. We therefore have the same boundary value problem


as that discussed in Section 12.7 for the particle in a ring and in 14.5 for angular


motion of the particle in a circular box. The normalized solutions are


m 1 = 1 0, ±1, ±2, = (14.62)


or, in real form (equations (12.71)),


(14.63)


form 1 = 1 0, 1, 2, =. The functions form an orthonormal set, with property


(14.64)


Z


0

2 π

ΦΦ


mm mm

*( ) ( )φφφδd


′ , ′

=


1


2


1


2


1


2


1


2


()cos ()sinΦΦ ΦΦ


mm mm

m


i


+= , −= m


−−

ππ


φφ


Φ


m

im

()φ e


φ

=,


1


2 π


d


d


m


2

2

2

Φ


Φ


()


()


φ


φ


=− φ


1


1


2

2

sin


sin ( )


sin


θθ


θ


θ
θ

d


d


d


d


ll


Θ m








++−












Θ= 0


d


d


m


2

2

2

Φ


Φ


()


()


φ


φ


=− φ


sin


sin ( ) sin


θ


θ


θ


θ


θ


Θ


d Θ


d


d


d


ll








++










1 +


2

11


0


2

2

Φ












=

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