Fitts D.D. - Principles of Quantum Mechanics[c] As Applied to Chemistry and Chemical Physics (1999)(en)

142	Angular momentum

…2ð…ð

Y lm(è, j)Ylm(è, ö) sin è dè dj

00

	…0		…0
	ð		2ð
		Èlm(è)Èlm(è) sin è dè		Öm(j)Öm(j) dj		1

where the è- and j-dependent parts of the volume element dô are included in the integration. For convenience, we require that each of the two factors Èlm(è) and Öm(j) be normalized. Writing Öm(j) as

Öm(j) Aeimj

we ®nd that	(Aeimj) (Aeimj) dj jAj2				…0	ð dj 1
…0ð
2				= ••••••	2
giving
	A		eiá	p2ð
	Öm(j) p1 eimj					(5:40)
where we have arbitrarily set á equal to				••••••		associated

2ð

zero in the phase factor eiá

with the normalization constant.

The function Èl,ÿl(è) is given by equation (5.39) and the value of the constant of integration Al is determined by the normalization condition

ð

ð

…0

[Èl,ÿl(è)] Èl,ÿl(è) sin è dè jAlj2

…0

sin2l 1è dè 1

(5:41)

We need to evaluate the integral Il

ÿ1

ð

1

Il

…0

sin2l 1è dè ÿ…1

(1 ÿ ì2)l dì …ÿ1

(1 ÿ ì2)l dì

where we have de®ned the variable ì by the relation

so that

ì cos è

(5:42)

1 ÿ ì2 sin2è,

dì ÿsin è dè

The integral Il may be transformed as follows

1

1

1 ì

Il …ÿ1

(1 ÿ ì2)lÿ1 dì ÿ

…ÿ1

(1 ÿ ì2)lÿ1 ì2 dì I lÿ1 …ÿ1

d(1

ÿ ì2)l

2l

I

1

(1 ÿ ì2)l

dì

I

1

I

lÿ1

ÿ …ÿ1 2l

lÿ1 ÿ 2l

l

where we have integrated by parts and noted that the integrated term vanishes. Solving for Il, we obtain a recurrence relation for the integral

5.3 Application to orbital angular momentum

143

Il

2l

I lÿ1

(5:43)

Since I0 is given by

2l 1

I0

…ÿ1 dì 2

1

we can obtain Il by repeated application of equation (5.43) starting with I0

I

l

(2l)(2l ÿ 2)(2l ÿ 4) . . . 2

I

22l 1(l!)2

3

0

(2l

1)!

(2l

1)(2l

ÿ

1)(2l

ÿ

3) . . .

where we have noted that

(2l)(2l ÿ 2) . . . 2 2ll! (2l 1)(2l ÿ 1)(2l ÿ 3) . . . 3

(2l 1)(2l)(2l ÿ 1)(2l ÿ 2)(2l ÿ 3) . . . 3 3 2 3 1 (2l 1)! (2l)(2l ÿ 2) . . . 2 2l l!

Substituting this result into equation (5.41), we ®nd that r•••••••••••••••••

jAlj

1

(2l 1)!

2l l!

2

It is customary to let á equal zero in the phase factor eiá for È

l,ÿl

(è), so that

2

! r•••••••••••••••••

È

(è)

1

(2l 1)!

sinl è

(5:44)

l,ÿl

l l

2

Combining equations (5.35), (5.40) and (5.44), we obtain the normalized

eigenfunction					! r•••••••••••••••••
Yl,		l(è, j)		1			(2l 1)!	sinl è eÿilj	(5:45)
	ÿ			2l l			4ð
Spherical harmonics
The functions Ylm(è, j) are			known				as spherical harmonics and		may	be
obtained from Yl,ÿl(è, j) by repeated application of the raising operator										^
										L

according to (5.38a). By this procedure, the spherical harmonics Yl,ÿl 1(è, j), Yl,ÿl 2(è, j), . . . , Yl,ÿ1(è, j), Yl0(è, j), Yl1(è, j), . . . , Yll(è, j) may be determined. Since the starting function Yl,ÿl(è, j) is normalized, each of the spherical harmonics generated from equation (5.38a) will also be normalized.

We may readily derive a general expression for the spherical			harmonic
		^	(è, j). We
Ylm(è, j) which results from the repeated application of L to Yl,ÿl
begin with equation (5.38a) with m set equal to ÿl
1		^
	•••••
Yl,ÿl 1 p2l " L Yl,ÿl			(5:46)

144

Angular momentum

For m equal to ÿl 1, equation (5.38a) gives

Yl,ÿl 2

1

^

1

^

2

2(2l ÿ 1) " L Yl,ÿl 1

2(2l)(2l ÿ 1) "2 L Yl,ÿl

(5.46) has been introduced in the last term. If we continue in

where equation

p•••••••••••••••••••

p••••••••••••••••••••••••••

the same pattern, we ®nd

Yl,ÿl 3

1

^

1

^

3

3(2l ÿ 2) "

L Yl,ÿl 2

2 : 3(2l)(2l ÿ 1)(2l ÿ 2) "3 L Yl,ÿl

.

p•••••••••••••••••••

p•••••••••••••••••••••••••••••••••••••••••••••••

.

.

Y

(2l ÿ k)!

1

^Lk

Y

l,ÿl k

l,ÿl

k (2l) "k

s••••••••••••••••••! !

where k is the number of steps in this sequence. We now set k l m in the last expression to obtain

Y		(l ÿ m)!			1			^Ll m Y		(5:47)
	lm		m) (2l)		"l		m		l,ÿl
		s•••••••••••••••••••••••••(l ! !

If the number of steps k is less than the value of l, then the integer m is negative; if k equals l, then m is zero; if k is greater than l, then m is positive;

and ®nally if k equals 2l, then m equals its largest value of l.
The next step in this derivation is the evaluation of ^Ll m Y		l,ÿl	using equation
^

(5.37a). If the operator L in (5.37a) acts on Yl,ÿl(è, j) as given in (5.45), we have

^ "eij @ i cot è @ sinlè eÿilj

L Yl,ÿl cl è ö

@ @

cl"eÿi( lÿ1)j		d	l cot è sinlè
	dè
		d	sin 2lè
cl"eÿi( lÿ1)j			l cot è
		dè		sinlè

1

d

cl"eÿi( lÿ1)j

sin2lè ÿ l sinlÿ1è cos è l sinlÿ1 è cos è

sinlè

dè

1

d

ÿcl"eÿi(lÿ1)j

sin2lè

sinlÿ1è

d(cos è)

where for brevity we have de®ned cl as

! r•••••••••••••••••

cl

1

(2l 1)!

(5:48)

2l l

4ð

5.3 Application to orbital angular momentum

145

^

We then operate on this result with L to obtain

^L2 Yl,ÿl ÿcl"2eij

@

i cot è

@

eÿi( lÿ1)j

d

@è

@j

sinlÿ1è d(cos è) sin2lè!

ÿcl"2eÿi(lÿ2)j) ddè (l ÿ 1) cot è

1

d

sin 2lè

sinlÿ1

è d(cos è)

cl"2eÿi(lÿ2)j

1

d2

sin 2lè

sinl

ÿ

2è d(cos è)2

^

After k such applications of L to the function Yl,ÿl(è, j), we have

^Lk

Y

l,ÿl

(

ÿ

")k c

eÿi(lÿk)j

1

dk

sin2lè

sinlÿk è d(cos è)k

l

If we set k l m in this expression, we obtain the desired result

^Ll m Y

( ")l m c

eimj sinmè

dl m

sin2lè

(5:49)

d(cos è)l m

l,ÿl ÿ

l

The general expression for Ylm(è, j) is obtained by substituting equation

(5.49) into (5.47) with cl

given by equation (5.48)

!

s•••••••••••••••••••••••••••••••••!

è

Ylm(è, j)

(ÿ1)l m

(2l 1)

(l ÿ m)!

eimj sinmè

dl m

sin2lè

l

m

l

4ð (l

m)

2 l

d(cos )

(5:50) When Ylm(è, j) is decomposed into its two normalized factors according to

equations (5.35) and (5.40), we have

2 l!

s•••••••••••••••••••••••••••••••••!

d(cos è )

Èlm(è)

(ÿ1)l m

(2l 1)

(l ÿ m)!

sinmè

dl m

sin2lè

(5:51)

l

m

l

2

(l

m)

Öm(j) p1

eimj

(5:52)

The spherical harmonics for l 0, 1, ••••••

2ð

2, 3 are listed in Table 5.1. We note

that the function Èl,ÿm(è) is related to Èlm(è) by

Èl,ÿm(è) (ÿ1)mÈlm(è)

(5:53)

and that the complex conjugate

Y lm(è, j) is related to Ylm(è, j) by

(5:54)

Y

(è, j)

( 1)m Yl,

ÿ

m

ÿ

lm

^2

^

are hermitian, the spherical harmonics Ylm(è, j)

Because both L

and

Lz

form an orthogonal set, so that

…2ð…ðY l9m9(è, j)Ylm(è, j) sin è dè dj äll9ämm9

(5:55)

00

146 Angular momentum

Table 5.1. Spherical harmonics Ylm(è, j) for l 0, 1, 2, 3

1

1=2

7

1=2

Y00

Y30

(5 cos3è ÿ 3 cos è)

4ð

16ð

3

1=2

21

1=2

Y10

cos è

Y3, 1

sin è(5 cos2 è ÿ 1)e ij

4ð

64ð

3

1=2

105

1=2

Y1, 1

sin è e ij

Y3, 2

sin2è cos è e 2ij

8ð

32ð

5

1=2

Y3, 3

35

1=2

Y20

(3 cos2 è ÿ 1)

sin3è e 3ij

16ð

64ð

15

1=2

Y2, 1

sin è cos è e ij

8ð

Y2, 2

15

1=2

sin2 è e 2ij

32ð

If we introduce equation (5.35) into (5.55), we have
	…0	l9m9			…0	9
	ð				2ð
	È (è)Èlm(è) sin è dè					Öm (j)Öm(j) dj					äll9ämm9
The integral over the angle j is				…0				2ð …0
…0	9		2ð
2ð			1	2ð				1		2ð
	Öm (j)Öm(j) dj				eÿim9jeimj dj						ei(mÿm9)j dj

where equation (5.52) has been introduced. Since m and m9 are integers, this

integral vanishes unless m m9, so that
…0
	2ð
		Öm9(j)Öm(j) dj ämm9		(5:56)
from which it follows that
ð
…0	Èl9m(è)Èlm(è) sin è dè äll9			(5:57)

Note that in equation (5.57) the same value for m appears in both Èl9m(è) and Èlm(è). Thus, the functions Èlm(è) and Èl9m(è) for l 6l9 are orthogonal, but the functions Èlm(è) and Èl9m9(è) are not orthogonal. However, for m 6m9, the spherical harmonics Ylm(è, j) and Yl9m9(è, j) are orthogonal because of equation (5.56).

5.3 Application to orbital angular momentum

147

Relationship of spherical harmonics to associated Legendre polynomials

The functions Èlm(è) and consequently the spherical harmonics Ylm(è, j) are related to the associated Legendre polynomials, whose de®nition and properties are presented in Appendix E. To show this relationship, we make the substitu-

tion of equation (5.42) for cos è in equation (5.51) and obtain

Èlm

(ÿ1)m

(2l 1)

(l ÿ m)!

(1

ÿ

ì2)m=2

dl m

(ì2

ÿ

1)l

(5:58)

2

l

l

2

(l

m)

d

l m

!

s•••••••••••••••••••••••••••••••••!

ì

polynomial Pm(ì)

Equation (E.13)

relates

the

associated

Legendre

to the

(l m)th-order derivative in equation (5.58)

l

m

1

2

m=2 dl m

2

l

Pl

(ì)

(1

ÿ ì )

(ì

ÿ

1)

2l l!

dìl m

where l and

m are positive integers (l, m > 0) such that

m < l. Thus, for

positive m we have the relation

Èlm(è)

(

1)m

(2l 1)

(l ÿ m)!

Plm(cos è),

m > 0

ÿ

2

(l

m)

s••••••••••••••••••••••••••••••••••!

For negative m, we may write m ÿjmj and note that equation (5.53) states Èl,ÿjmj(è) (ÿ1)mÈl,jmj(è)

so that we have

È

l,ÿjmj

(è)

(2l 1)

(l ÿ jmj)!

Pjmj(cos è)

2 (l

m )

l

s•••••••••••••••••••••••••••••••••••!

j

j

These two results may be combined as

Èlm(è)

å

(2l 1)

(l ÿ jmj)!

Pjlmj

(cos è)

2 (l m )

s•••••••••••••••••••••••••••••••••••!

j

j

where å (ÿ1)m for m . 0 and å 1 for m < 0. Accordingly, the spherical harmonics Ylm(è, j) are related to the associated Legendre polynomials by

Ylm(è, j)

å

(2l 1)

(l ÿ jmj)!

Pjlmj(cos è)eimj

4ð

(l

m )

s•••••••••••••••••••••••••••••••••••!

j

j

å (ÿ1)m,

m . 0

(5:59)

1,

m < 0

The eigenvalues

and

eigenfunctions

of

the orbital

angular momentum

operator

^2

may

also

be

obtained by

solving the

differential equation

L

^2

2

L

ø ë" ø using the Frobenius or series solution method. The application of

this method is presented in Appendix G and, of course, gives the same results

148 Angular momentum

as the procedure using ladder operators. However, the Frobenius method may not be used to obtain the eigenvalues and eigenfunctions of the generalized

^2	and	^	because their eigenfunctions do not
angular momentum operators J		J z
have a spatial representation.
5.4	The rigid rotor

The motion of a rigid diatomic molecule serves as an application of the quantum-mechanical treatment of angular momentum to a chemical system. A rigid diatomic molecule consists of two particles of masses m1 and m2 which rotate about their center of mass while keeping the distance between them ®xed at a value R. Although a diatomic molecule also undergoes vibrational motion in which the interparticle distance oscillates about some equilibrium value, that type of motion is neglected in the model being considered here; the interparticle distance is frozen at its equilibrium value R. Such a rotating system is called a rigid rotor.

We begin with a consideration of a classical particle i with mass mi rotating in a plane at a constant distance ri from a ®xed center as shown in Figure 5.2. The time ô for the particle to make a complete revolution on its circular path is

equal to the distance traveled divided by its linear velocity vi
ô	2ðri			(5:60)
		vi

The reciprocal of ô gives the number of cycles per unit time, which is the

frequency í of the rotation. The velocity vi may then be expressed as
vi	2ðri			2ðíri ùri	(5:61)
		ô
where ù 2ðí is the angular		velocity. According to equation			(5.1), the
angular momentum Li of particle i is
Li ri		3 pi mi(ri 3 vi)			(5:62)

Since the linear velocity vector vi is perpendicular to the radius vector ri, the magnitude Li of the angular momentum is

vi

mi

ri

Figure 5.2 Motion of a rotating particle.

5.4 The rigid rotor	149
Li mirivi sin (ð=2) mirivi ùmir2i	(5:63)

where equation (5.61) has been introduced.

We next apply these classical relationships to the rigid diatomic molecule. Since the molecule is rotating freely about its center of mass, the potential energy is zero and the classical-mechanical Hamiltonian function H is just the

kinetic energy of the two particles,
	p2		p2		1	2		1	2
H	1		2
					2m1v1			2m2v2		(5:64)
	2m1		2m2

If we substitute equation (5.61) for each particle into (5.64) while noting that the angular velocity ù must be the same for both particles, we obtain

H 21ù2(m1 r12 m2 r22) 21Iù2	(5:65)
where we have de®ned the moment of inertia I by
I m1 r12 m2 r22	(5:66)

In general, moments of inertia are determined relative to an axis of rotation. In this case the axis is perpendicular to the interparticle distance R and passes

through the center of mass. Thus, we have
r1 r2 R
and
m1 r1 m2 r2
or, upon inversion
r1		m2		R
	m1 m2				(5:67)
r2		m1		R
	m1 m2
Substitution of equations (5.67) into (5.66) gives
I ìR2					(5:68)
where the reduced mass ì is de®ned by
ì		m1 m2			(5:69)
		m1 m2
The total angular momentum L for the two-particle system is given by
L L1 L2 ù(m1 r12 m2 r22) Iù					(5:70)

where equations (5.63) and (5.66) are used. A comparison of equations (5.65) and (5.70) shows that

H	L2	(5:71)
	2I

150			Angular momentum
Accordingly, the quantum-mechanical Hamiltonian operator							^
							H for this system
							^2
is proportional to the square of the angular momentum operator L
				^	1	^2
				H	2I L			(5:72)
^	and	^2		have	the same eigenfunctions,			namely, the
Thus, the operators H		L

spherical harmonics YJm(è, j) as given in equation (5.50). It is customary in discussions of the rigid rotor to replace the quantum number l by the index J in the eigenfunctions and eigenvalues.

^

The eigenvalues of H are obtained by noting that

H^ Y

(è,

)

1

^L2 Y

(è,

j

)

J(J 1)"2

Y

(è,

)

(5:73)

2I

Jm

j

2I

Jm

Jm

j

where l is replaced by J in equation (5.28a). Thus, the energy levels EJ for the rigid rotor are given by

EJ J(J 1)	"2	J 0, 1, 2, . . .
	2I J(J 1)B,		(5:74)

where B "2=2I is the rotational constant for the diatomic molecule. The energy levels EJ are shown in Figure 5.3. We observe that as J increases, the difference between successive levels also increases.

Energy

J 5 4, g4 5 9

20B

J 5 3, g3 5 7

12B

J 5 2, g2 5 5

6B

J 5 1, g1 5 3

2B

J 5 0, g0 5 1

0

Figure 5.3 Energy levels of a rigid rotor.

5.5 Magnetic moment

151

To ®nd the degeneracy of the eigenvalue EJ , we note that for a given value of J, the quantum number m has values m 0, 1, 2, . . . , J. Accordingly, there are (2J 1) spherical harmonics for each value of J and the energy level EJ is (2J 1)-fold degenerate. The ground-state energy level E0 is nondegenerate.

5.5 Magnetic moment

Atoms are observed to have magnetic moments. To understand how an electron circulating about a nuclear core can give rise to a magnetic moment, we may apply classical theory. We consider an electron of mass me and charge ÿe bound to a ®xed nucleus of charge Ze by a central coulombic force F(r) with

potential V (r)
F(r)		dV (r)				ÿZe2	(5:75)
	ÿ		dr			4ðå0 r2
	V (r)			ÿZe2			(5:76)
				4ðå0 r

Equation (5.75) is Coulomb's law for the force between two charged particles separated by a distance r. In SI units, the charge e is expressed in coulombs (C), while å0 is the permittivity of free space with the value

å0 8:854 19 3 10ÿ12 Jÿ1 C2 mÿ1

According to classical mechanics, a stable circular orbit of radius r and angular

velocity ù is established for the electron if the centrifugal force me rù2
balances the attractive coulombic force
me rù2	Ze2
	4ðå0 r2

This assumption is the basis of the Bohr model for the hydrogen-like atom. When solved for ù, this balancing equation is

Ze2		1=2
ù 4ðå0 me r3		(5:77)

An electron in a circular orbit with an angular velocity ù passes each point in the orbit ù=2ð times per second. This electronic motion constitutes an electric current I, de®ned as the amount of charge passing a given point per

second, so that
I	eù	(5:78)
	2ð

From the de®nition of the magnetic moment in electrodynamics, a circulat-