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Химия

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

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302

Appendix E

1 M ( 1)á(2l

2á)!ìlÿ2á

g(ì, s)

0 2lá!(l ÿ á)!(l ÿ

2á)!

in the expansion (E.1) of

Since the Legendre polynomials are the coef®cients of s

g(ì, s), we have

(ÿ1)á(2l ÿ 2á)!

ìlÿ2á

(ì)

(E:2)

2lá!(l ÿ á)!(l ÿ 2á)!

We see from equation (E.2) that Pl(ì) for even l is a polynomial with only even powers of ì, while for odd l only odd powers of ì are present.

The ®rst few Legendre polynomials may be readily obtained from equation (E.2)

and are
P0(ì) 1	P3(ì) 21(5ì3 ÿ 3ì)
P1(ì) ì	P4(ì) 81(35ì4	ÿ 30ì2	3)
P2(ì) 21(3ì2 ÿ 1)	P5(ì) 81(63ì5	ÿ 70ì3	15ì)

We observe that Pl(1) 1, which can be shown rigorously by setting ì 1 in

equation (E.1) and noting that	X	X	l
	X	X	l
g(1, s) (1 ÿ s)ÿ1	1	1
g(1, s) (1 ÿ s)ÿ1	sl	Pl(1)sl
	l 0	l 0
Since Pl(ì) is either even or odd in ì, it follows that Pl(ÿ1) (ÿ1)			and that
Pl(0) 0 for l odd.

Recurrence relations

We next derive some recurrence relations for the Legendre polynomials. Differentiation of the generating function g(ì, s) with respect to s gives

	@ g		ì ÿ s		(ì ÿ s) g	1 lP (ì)s lÿ1				(E:3)
	@s	(1 ÿ 2ì s2)3=2				1 lP (ì)s lÿ1				(E:3)
	@s	(1 ÿ 2ì s2)3=2		1 ÿ 2ì s2 l				1	l
						X
The term with l 0 in the summation vanishes, so that the summation now begins
with the l 1 term. We may write equation (E.3) as						1
	(ì ÿ s)		1			1	lPl(ì)s lÿ1
	(ì ÿ s)		Pl(ì)sl (1 ÿ 2ìs s2)				lPl(ì)s lÿ1
			l 0			l 1
			X			X
If we equate coef®cients of s lÿ1 on each side of the equation, we obtain
ìPlÿ1(ì) ÿ Plÿ2(ì) lPl(ì) ÿ 2(l ÿ 1)ìPlÿ1(ì) (l ÿ 2)Plÿ2(ì)
or		lPl(ì) ÿ (2l ÿ 1)ìPlÿ1(ì) (l ÿ 1)Plÿ2(ì) 0
		lPl(ì) ÿ (2l ÿ 1)ìPlÿ1(ì) (l ÿ 1)Plÿ2(ì) 0								(E:4)

The recurrence relation (E.4) is useful for evaluating Pl(ì) when the two preceding polynomials are known.

Differentiation of the generating function g(ì, s) in equation (E.1) with respect to ì

yields
@ g			sg

@ì			1 ÿ 2ìs s2

which may be combined with equation (E.3) to give

Legendre and associated Legendre polynomials

303

@ g

(ì ÿ s)

@ g

@ì

so that

1 dPl

lPl(ì)sl (ì ÿ s)

l 0

dì

l 1

Equating coef®cients of s

on each side of this equation yields a second recurrence

relation

dPl

dPlÿ1

ÿ lPl

(ì) 0

(E:5)

dì

A third recurrence relation may be obtained by differentiating equation (E.4) to give

l	dPl	ÿ (2l ÿ 1)ì	dPlÿ1	ÿ (2l ÿ 1)Plÿ1(ì) (l ÿ 1)	dPlÿ2	0
	dì		dì		dì

and then eliminating dPlÿ2=dì by the substitution of equation (E.5) with l replaced by l ÿ 1. The result is

dPl	ÿ ì	dPlÿ1	ÿ lPlÿ1(ì) 0	(E:6)
dì		dì

Differential equation

To ®nd the differential equation satis®ed by the polynomials Pl(ì), we ®rst multiply equation (E.5) by ÿì and add the result to equation (E.6) to give

(1 ÿ ì2) ddPìl lìPl(ì) ÿ lPlÿ1(ì) 0

We then differentiate to obtain
2		d2 Pl			dPl		dPl		dPlÿ1
(1 ÿ ì	)	dì2	ÿ 2	ì	dì	lì	dì	lPl(ì) ÿ l	dì	0

The third and last terms on the left-hand side may be eliminated by means of equation (E.5) to give Legendre's differential equation

(1 ÿ ì2)	d2 Pl	ÿ 2ì	dPl	l(l 1)Pl(ì) 0	(E:7)
	dì2		dì

Rodrigues' formula

Rodrigues' formula for the Legendre polynomials may be derived as follows. Consider the expression

v (ì2 ÿ 1)l

The derivative of v is

ddvì 2lì(ì2 ÿ 1)lÿ1 2lìv(ì2 ÿ 1)ÿ1 which is just the differential equation

(1 ÿ ì2) ddvì 2lìv 0 If we differentiate this equation, we obtain

304	Appendix E

(1 ÿ ì2) d2v 2(l ÿ 1)ì dv 2lv 0 dì2 dì

We now differentiate r times more and obtain

(1 ÿ ì2)

dr 2v

ÿ r ÿ 1)ì

dr 1v

2(l

(r 1)(2l ÿ r)

dìr 2

dìr 1

If we let r l and de®ne w as

dlv

(ì2 ÿ 1)l

dìl

then equation (E.8) reduces to

(1 ÿ ì2)

d2 w

2ì

l(l 1)w 0

dì2

dì

drv	0	(E:8)
dìr

which is just Legendre's differential equation (E.7). Since the polynomials Pl(ì) represent all of the solutions of equation (E.7), these polynomials must be multiples of w, so that

Pl(ì) cl ddìl l (ì2 ÿ 1)l

The proportionality constants cl may be evaluated by setting the term in ìl, namely

cl ddìl l ì2l cl (2ll!)! ìl

equal to the term in ìl in equation (E.2), i.e.,

	(2l)!			ìl
	2l(l!)2
Thus, we have
	cl		1

			2l l!
and
	1		dl
Pl(ì)					(ì2 ÿ 1)l	(E:9)
Pl(ì)	2l l!	dìl			(ì2 ÿ 1)l	(E:9)

This expression (equation (E.9)) is Rodrigues' formula.

Associated Legendre polynomials

The associated Legendre polynomials Pml (ì) are de®ned in terms of the Legendre polynomials Pl(ì) by

Plm(ì) (1 ÿ ì2)m=2	dm Pl(ì)	(E:10)
	dìm

where m is a positive integer, m 0, 1, 2, . . . , l. If m 0, then the corresponding associated Legendre polynomial is just the Legendre polynomial of degree l. If m . l, then the corresponding associated Legendre polynomial vanishes.

The generating functions g(m)(ì, s) for the associated Legendre polynomials may be found from equation (E.1) by letting

Legendre and associated Legendre polynomials

305

g(m)(ì, s)

(1 ÿ ì2)m=2

dm g(ì, s)

dìm

Since

dm g(ì, s)

: 5 : :

(2m ÿ 1)sm(1 ÿ 2ìs s2)ÿ(m 2)

dìm

(2m)!

sm(1 ÿ 2ìs s2)ÿ(m 2)

2m m!

we have

g(m)(ì, s)

Pm(ì)sl

(2m)!(1 ÿ ì2)m=2 sm

(E:11)

l m

2m m!(1 2ìs s2)m 2

We can also write an explicit series for Pl

(ì) by differentiating equation (E.2) m

times

M9 ( 1)á(2l

2á)!ìlÿmÿ2á

Pm(ì)

ì2)m=2

(E:12)

á 0 2l

á!(l ÿ á)!(l ÿ m ÿ 2á)!

where M9 (l ÿ m)=2 or (l ÿ m ÿ 1)=2, whichever is an integer. Furthermore,

combining equation (E.10) with Rodrigues' formula (E.9), we see that

Plm(ì)

ÿ ì2)m=2

dl m

(ì2 ÿ 1)l

(E:13)

2l l!

dìl m

The ®rst few associated Legendre polynomials are

P00(ì) P0(ì) 1

P10(ì) ì

P11(ì) (1 ÿ ì2)1=2

P20(ì) P2(ì) 21(3ì2 ÿ 1)

P21(ì) 3ì(1 ÿ ì2)1=2

P22(ì) 3(1 ÿ ì2)

Differential equation
The differential equation satis®ed by the polynomials Pm(ì) may be obtained as
							l
follows. Let r l m in equation (E.8) and de®ne wm as
			dl m					dm Pl
		wm			(ì2 ÿ 1)l		2l l!		(E:14)
			dìl m					dìm
so that
		wm 2l l!(1 ÿ ì2)ÿm=2 Plm(ì)							(E:15)
Equation (E.8) then becomes
d2 wm				dwm
(1 ÿ ì2)		ÿ 2(m 1)ì				[l(l 1)	ÿ m(m 1)]wm 0		(E:16)
	dì2			dì

306 Appendix E

We then substitute equation (E.15) for wm and take the ®rst and second derivatives as indicated to obtain

(1 ÿ ì2)	d2 Pm		dPm	"l(l 1) ÿ	m2	#Plm(ì) 0
	l	ÿ 2ì	l				(E:17)
	dì2		dì		1 ÿ ì2
Equation (E.17) is the associated Legendre differential equation.

Orthogonality

Equation (E.17) as satis®ed by Pm(ì) and by Pm(ì) may be written as

ì2)

dPm

"l(l 1) ÿ

#Plm(ì) 0

(1 ÿ

dì

ì2

and

dPm

(1 ÿ ì2)

"l9(l9 1) ÿ

#Plm9 (ì) 0

dì

1 ÿ ì2

If we multiply the ®rst by Pm(ì) and the second by Pm

(ì) and then subtract, we have

dPm

Plm9

(1 ÿ

ì2)

ÿ Plm

(1 ÿ ì2)

[l9(l9 1) ÿ l(l 1)]Plm Plm9

dì

We then add to and subtract from the left-hand side the term

dPlm dPlm9

(1 ÿ ì

)

dì dì

so as to obtain

dPm

ÿ ì2) Plm9

ÿ Plm

[l9(l9 1) ÿ l(l 1)]Plm Plm9

dì

We next integrate with respect to ì from ÿ1 to 1 and note that

dPm

(1 ÿ ì2) Plm9

ÿ Plm

ÿ1 0

dì

giving

[l9(l9 1) ÿ l(l 1)]…ÿ1 Plm Plm9

dì 0

If l9 6l, then the integral must vanish

…ÿ1 Plm(ì)Plm9 (ì) dì 0

(E:18)

so that the associated Legendre polynomials Pml (ì) with ®xed m form an orthogonal set of functions. Since equation (E.18) is valid for m 0, the Legendre polynomials Pl(ì) are also an orthogonal set.

Normalization

We next wish to evaluate the integral Ilm

…1

Ilm [Pml (ì)]2 dì

ÿ1

As a ®rst step, we evaluate I l0

Legendre and associated Legendre polynomials

307

…1

I l0 [Pl(ì)]2 dì

ÿ1

We solve the recurrence relation (E.4) for Pl(ì), multiply both sides by Pl(ì), integrate with respect to ì from ÿ1 to 1, and note that one of the integrals vanishes according to the orthogonality relation (E.18), so that

1		1
…ÿ1[Pl(ì)]2 dì	2l ÿ	1	…ÿ1 ìPl(ì)Plÿ1(ì) dì
	l

Replacing l by l 1 in equation (E.4), we can substitute for ìPl(ì) on the right-hand side. Again applying equation (E.18), we ®nd that

[P (ì)]2 dì

2l ÿ 1

(ì)]2 dì

…ÿ1

2l 1

…ÿ1

lÿ1

This relationship can then be applied successively to obtain

[P (ì)]2 dì

(2l ÿ 1)(2l ÿ 3)

(ì)]2 dì

…ÿ1

lÿ2

(2l 1)(2l ÿ 1)

(2l ÿ

1)(2l ÿ 3) 1

[P (ì)]2 dì

(2l 1)(2l ÿ 1)(2l ÿ 3) 3

…ÿ1

…ÿ1[P0(ì)]2 dì

2l 1

Since P0(1) 1, the desired result is

…ÿ1

[Pl(ì)]2 dì

…ÿ1 dì

(E:19)

2l 1

We are now ready to evaluate Ilm. From equation (E.10) we have

Ilm …ÿ1(1 ÿ ì2)m dìm

dì

…ÿ1(1 ÿ ì2)m

dìm

dì

dìmÿ1 ! dì

dm Pl

d dmÿ1 Pl

Integration by parts gives

dm Pl

dmÿ1 Pl

1 dmÿ1 Pl d

dm Pl

Ilm (1 ÿ ì2)m dìm

dìmÿ1

…ÿ1 d

ìmÿ1 dì

(1 ÿ ì2)m dìm dì (E:20)

ì )

0 at ì

The integrated part vanishes because

To evaluate the integral on the right-hand side of equation (E.20), we replace m by

m ÿ 1 in (E.16) and multiply by (1 ÿ ì2)mÿ1 to obtain

ì2)m

d2 wmÿ1

2mì(1

ì2)mÿ1

dwmÿ1

dì2

dì

which can be rewritten as

[l(l 1) ÿ m(m ÿ 1)](1 ÿ ì2)mÿ1 wmÿ1 0

dwm 1

ÿ ì2)m

dìÿ

ÿ(l m)(l ÿ m 1)(1 ÿ ì2)mÿ1 wmÿ1 0

dì

From equation (E.14) we see that

308

Appendix E

wmÿ1 2

dmÿ1 Pl

dìmÿ1

and

mÿ1

dm P

wm 2

dì

dìm

so that

(1 ÿ ì2)m

dm P

ÿ(l

m)(l ÿ m 1)(1 ÿ ì2)mÿ1

dì

dìm

Thus, equation (E.20) takes the form

dmÿ1 Pl

Ilm (l m)(l ÿ m

1)…ÿ1(1 ÿ ì2)mÿ1"

dìmÿ1

Using equation (E.10) to introduce Pmÿ1(ì), we have

l …1

Ilm (l m)(l ÿ m 1) [Pml ÿ1(ì)]2 dì

ÿ1

dmÿ1 Pl

dìmÿ1

dì

(l m)(l ÿ m 1)I l,mÿ1

which relates Ilm to I l,mÿ1. This process can be repeated until I l0 is obtained Ilm [(l m)(l m ÿ 1)][(l ÿ m 1)(l ÿ m 2)]I l,mÿ2

...

[(l m)(l m ÿ 1) (l 1)][(l ÿ m 1)(l ÿ m 2) l]I l0

(l m)! l!

l! (l ÿ m)! I l0

so that

1		2(l m)!
[Pm(ì)]2 dì
…ÿ1 l	(2l 1)(l ÿ m)!

Completeness

The set of associated Legendre polynomials Pml (ì) with m ®xed and l m,

m 1, . . . , form a complete orthogonal set1 in the range ÿ1 < ì < 1. Thus, an arbitrary function f (ì) can be expanded in the series

f (ì) almPml (ì)

l m

with the expansion coef®cients given by

1The proof of completeness may be found in W. Kaplan (1991) Advanced Calculus, 4th edition (AddisonWesley, Reading, MA) p. 537 and in G. Birkhoff and G.-C. Rota (1989) Ordinary Differential Equations, 4th edition (John Wiley & Sons, New York), pp. 350±4.

Legendre and associated Legendre polynomials

309

2l 1

(l ÿ m)!

Pm(ì) f (ì) dì

(l m)!

…ÿ1m

The completeness relation for the polynomials Pl

(ì) is

2l 1

(l ÿ m)!

Pm(ì)Pm(ì9)

ä(ì

ì9)

m)!

Appendix F

Laguerre and associated Laguerre polynomials

Laguerre polynomials

The Laguerre polynomials Lk (r) are de®ned by means of the generating function g(r, s)

	eÿrs=(1ÿs)			X
g(r, s)	1	ÿ	s	k	0	Lk (r)	k!	(F:1)

where 0 < r < 1 and where jsj , 1 in order to ensure convergence of the in®nite series. Since the right-hand term is a Taylor series expansion of g(r, s), the Laguerre polynomials are given by

@k g( , s)

rs=(1ÿs)

s 0

Lk (r)

(F:2)

@sk

To evaluate Lk (r) from equation (F.2), we ®rst factor out e

the generating

function and expand the remaining exponential

function in a Taylor

series

1 ( 1)á

1 ( 1)árá

(1 ÿ s)ÿ(á 1)

g(r, s) 1 ÿ s eÿr=(1ÿs)

1 ÿ s á 0

ÿá!

1 ÿ s

er á 0

á!

We then take k successive derivatives of g(r, s) with respect to s

@ g(r, s)

(ÿ1)árá

(á

1)(1

s)ÿ(á 2)

á 0

@2 g(r, s)

(ÿ1)árá

(á

1)(á

2)(1

s)ÿ(á 3)

@s2

á 0

@k g(r, s)

(ÿ1)árá

(á k)!

s)ÿ(á k 1)

@sk

á 0

When the kth derivative is evaluated at s 0, we have

L (

)

(ÿ1)á(á k)!

(F:3)

á 0

(á!)2

310

Laguerre and associated Laguerre polynomials

Using equation (A.1) we note that

(

k eÿr)

(ÿ1)á

á k

á 0

(ÿ1)á(á k)!

drk

drk á 0 á!

(á!)2

Combining equations (F.3) and (F.4), we obtain the formula for the Laguerre polynomials

311

(F:4)

dk
Lk (r) er drk (rk eÿr)	(F:5)

Another relationship for the polynomials Lk (r) can be obtained by expanding the generating function g(r, s) in equation (F.1) using (A.1)

g( , s)	eÿrs=(1ÿs)		X			sá
g( , s)	eÿrs=(1ÿs)		1		(ÿ1)árá	sá
r		1 ÿ s	á	0	á!	(1 ÿ s)á 1

The factor (1 ÿ s)ÿ(á 1) may be expanded in an in®nite series using equation (A.3) to obtain

				X
	(1	ÿ	s)ÿ(á 1)	1	(á â)!	sâ
so that g(r, s) becomes	(1		s)ÿ(á 1)	â 0	á!â!	sâ
				â 0	á!â!
			XX
			XX
g(	, s)		1 1	(ÿ1)á(á â)!rá			sá â
g(	, s)			(ÿ1)á(á â)!rá			sá â
r			á 0 â 0	(á!)2â!
			á 0 â 0

We next collect all the coef®cients of sk for an arbitrary k, so that á â k, and replace the summation over á by a summation over k. When á k, the index â equals zero; when á k ÿ 1, the index â equals one; and so on until we have á 0 and

â k. Thus, the result of the summation is

g(r, s) k! X1 Xk (ÿ1)kÿârkÿâ sk k 0 â 0 [(k ÿ â)!]2â!

Since the Laguerre polynomial Lk (r) divided by k! is the coef®cient of sk in the expansion (F.1) of the generating function, we have

		X
		k		(ÿ1)kÿâ		kÿâ
L ( )	(k!)2	â	0 [(k ÿ â)!]2		â! r	kÿâ
k r		â	0 [(k ÿ â)!]2		â! r

If we let k ÿ â ã and replace the summation over â by a summation over ã, we obtain the desired result

(ÿ1)ã

(

)

(k!)2

0 (ã!)2(k ÿ ã)! r

(F:6)

A third relationship for the polynomials Lk (r) can be obtained by expanding the

derivative in equation (F.5), using (A.4), to give

dá

dkÿáeÿr

( 1)kÿá k! dá

(

)

0 á!(k ÿ á)! drá

drkÿá

0 á!(k ÿ á)! drá

We now observe that the operator [(d=dr) ÿ 1]k may be expanded according to the binomial theorem (A.2) as

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