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

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Казанский национальный исследовательский технологический университет

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Химия

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á j ( j í)!(á ÿ j)!

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312

Appendix F

(ÿ1)á k!

dá

[(d=d )

1]k

( 1)k [1

(d=d

)]k

( 1)k

á 0 á!(k ÿ á)! drá

so that

Lk (r) [(d=dr) ÿ 1]k rk where we have noted that (ÿ1)á (ÿ1)ÿá.

From equation (F.2), (F.5), (F.6), or (F.7), we observe that the polynomial Lk (r) degree k and we may readily obtain the ®rst few polynomials of the set

L0(r) 1

L1(r) 1 ÿ r

L2(r) 2 ÿ 4r r2

L3(r) 6 ÿ 18r 9r2 ÿ r3

We also note that Lk (0) k!.

Differential equation

(F:7)

is of

Equation (F.5) can be used to ®nd the differential equation satis®ed by the polynomials Lk (r). We note that the function f (r) de®ned as

satis®es the relation			f (r) rk eÿr
satis®es the relation
		d f
		r		(r ÿ k) f 0
		r	dr	(r ÿ k) f 0
If we differentiate this expression k 1 times, we obtain
	d2 f (k)			d f (k)
r		(1 r)				(k 1) f (k) 0
r	dr2	(1 r)			dr	(k 1) f (k) 0

where f (k) is the kth derivative of f (r). Since from equation (F.5) we have f (k) eÿr Lk (r)

the Laguerre polynomials Lk (r) satisfy the differential equation

r	d2 Lk	(1 ÿ r)	dLk	kLk (r) 0	(F:8)
	dr2		dr

Associated Laguerre polynomials

The associated Laguerre polynomials Lkj (r) are de®ned in terms of the Laguerre polynomials by

Lkj	d j
Lkj	(r) drj Lk (r)	(F:9)

Since Lk (r) is a polynomial of degree k, Lkk (r) is a constant and Lkj (r) 0 for j . k. The generating function g(r, s; j) for the associated Laguerre polynomials with ®xed j is readily obtained by differentiation of equation (F.1)

Laguerre and associated Laguerre polynomials

313

d j

eÿrs=(1ÿs)

( 1) j sj

g(r, s;

eÿrs=(1ÿs) k j

(r)

(F:10)

drj

(1 s) j 1

The summation in the right-hand term begins with k j, since j cannot exceed k. We can write an explicit series for Lkj (r) by substituting equation (F.6) into (F.9)

(ÿ1)ã

d j

(ÿ1)ã

ãÿ j

L j

(

)

(k!)2

0 (ã!)2(k ÿ ã)! drj r

(k!)2

j ã!(k ÿ ã)!(ã ÿ j)! r

The summation over ã now begins with the term ã j because the earlier terms vanish in the differentiation. If we let ã ÿ j á and replace the summation over ã by a summation over á, we have

L j ( )	(k!)2	kÿ j		(ÿ1) j árá	(F:11)
k r		X
k r		á	0 á!(k ÿ j ÿ á)!( j á)!

For the purpose of deriving some useful relationships involving the polynomials Lkj (r), we de®ne the polynomial Ëij(r) as

Ëi

(r)

Li j(r)

(F:12)

(i j)!

If we replace the dummy index of summation k in equation (F.10) by i, where

i k ÿ j, then (F.10) takes the form

Lij

(r)

Ëj(r)

si j

g( , s; j)

s j

0 (i j)!

i 0

Thus, Ëij(r) are just the coef®cients in a Taylor series expansion of the function

sÿ j g(r, s; j) and are, therefore, given by

Ëij

(r)

sÿ j g(r, s; j) s 0

@si

Substituting for g(r, s; j) using equation (F.10), we obtain

Ëij(r) (ÿ1) j @si

"(1ÿ

s)ÿj 1# s

rs=(1

ere

r=(1ÿs)

(ÿ1) j

# s 0

@si

s)j 1

(

(ÿ1) @si "er á

0 á!(1 ÿs)á j 1# s

(á j)!

s 0

á 0

ÿá!

( 1) jer

r) (á j i)! (1

s)ÿ(á j i 1)

1 (

(

1) jer

(á j

i)!

(

ÿr

)á

(F:13)

0 á!(á j)!

We next note that

314

Appendix F

(

i jeÿr)

á j i

á j

dri

0 á!

á 0

á!

(á j)!

(á j i)!

(

ÿr

)á

0 á!(á j)!

Comparison of equations (F.13) and (F.14) yields the result that

Ëj( )

(

1) j

ÿ jer

(

i jeÿr)

dri

From equation (F.12) we obtain

(

)

( 1) j

(i j)!

ÿ jer

(

i jeÿr)

i j

dri

Finally, replacing i by the original index k ( i j), we have

L j

( )

(

1) j

ÿ jer

dkÿ j

(

k eÿr)

(k ÿ j)! r

drkÿ j

(F:14)

(F:15)

Equation (F.15) for the associated Laguerre polynomials is the analog of (F.5) for the Laguerre polynomials and, in fact, when j 0, equation (F.15) reduces to (F.5).

Differential equation

The differential equation satis®ed by the associated Laguerre polynomials may be obtained by repeatedly differentiating equations (F.8) j times

r	d3 Lk	(2 ÿ r)	d2 Lk	(k ÿ 1)	dLk	0
r	dr3	(2 ÿ r)	dr2	(k ÿ 1)	dr	0
r	d4 Lk	(3 ÿ r)	d3 Lk	(k ÿ 2)	d2 Lk		0
r	dr4	(3 ÿ r)	dr3	(k ÿ 2)	dr2		0
.
.
.

dj 2 Lk

(j 1 ÿ r)

d j 1 Lk

d j Lk

(k ÿ j)

drj 2

drj 1

drj

When the polynomials Lkj (r) are introduced with equation (F.9), the differential

equation is

d2 Lkj

(j 1 ÿ r)

dLkj

(r) 0

(k ÿ j)Lk

(F:16)

dr2

Integral relations

In order to obtain the orthogonality and normalization relations of the associate Laguerre polynomials, we make use of the generating function (F.10). We multiply together g(r, s; j), g(r, t; j), and the factor rj íeÿr and then integrate over r to give an integral that we abbreviate with the symbol I

1	X X		1
	1 1 sá tâ
I …0	rj íeÿr g(r, s; j)g(r, t; j) dr á j â j	á!â!	…0	rj íeÿr Láj (r)Lâj (r) dr

(F:17)

Laguerre and associated Laguerre polynomials

315

To evaluate the left-hand integral, we substitute the analytical forms of the generating functions from equation (F.10) to give

(st) j

…0

rj íeÿar dr

(F:18)

where

(1 ÿ s)j 1(1 ÿ t)j 1

1 ÿ st

1 ÿ s

1 ÿ t

(1 ÿ s)(1 ÿ t)

The integral in equation (F.18) is just the gamma function (A.26), so that

Ã(j

í 1)

( j

í)!

…0

rj íeÿar dr

j í . 0

a j í 1

aj í 1

where we have restricted í to integer values. Thus, I in equation (F.18) is

(j í)!(st)j(1 ÿ s)í(1 ÿ t)í

st) j

Applying the expansion formula (A.3), we have

st)ÿ( j í 1)

( j í i)!

(st)i

i 0 (j í)!i!

If we replace the dummy index i by á, where á i j, then this expression becomes

and I takes the form

I (1 ÿ s)í(1 ÿ t)í X1 (á í)! (st)á á j (á ÿ j)!

Combining this result with equation (F.17), we have

1 1 sá tâ

j íeÿr L j (

)L j

(

) d

s)í(1

t)í

(á í)!

(st)á

(F:19)

…0

á j â j á!â!

á j (á ÿ j)!

We now equate coef®cients of like powers of s and t on each side of this equation. Since the integer í appears as an exponent of both s and t on the right-hand side, the effect of equating coef®cients depends on the value of í. Accordingly, we shall ®rst have to select a value for í.

For í 0, equation (F.19) becomes

1 1 sá tâ	1	rjeÿr Láj (r)Lâj	1	á!	(st)á
á j â j á!â!	…0	rjeÿr Láj (r)Lâj	(r) dr á j	(á ÿ j)!	(st)á
XX			X

Since the exponent of s on the right-hand side is always the same as the exponent of t, the coef®cients of sá tâ…for1 á 6â on the left-hand side must vanish, i.e.

r	jeÿr L j (	r	)L j	(	r	) d	r	0;	á â	(F:20)
r	á	r	â		r		r		6

Thus, the associated Laguerre polynomials form an orthogonal set over the range

0 < r < 1 with a weighting factor rjeÿr. For the case where s and t on the left-hand side have the same exponent, we pick out the term â á in the summation over â, giving

316

Appendix F

á!

1 (st)á

…0

rjeÿr[Láj (r)]2 dr á j

(st)á

á j (á!)2

(á ÿ j)!

Equating coef®cients of (st)á on each side yields

(á!)3

…0

rjeÿr[Láj (r)]2 dr

(F:21)

(á ÿ j)!

Equations (F.20) and (F.21) may be combined into a single expression

(á!)3

…0

rjeÿr Láj (r)Lâj (r) dr

äáâ

(F:22)

(á ÿ j)!

For í 1, equation (F.19) becomes

1 1 sá tâ

á j â j á!â!

…0

rj 1eÿr Láj (r)Lâj (r) dr

(á 1)!

[(st)á

(st)á 1

sá 1 tá

sá tá 1]

á j (á ÿ j)!

Equating coef®cients of like powers of s and t on both sides of this equation, we see

that

…1

j 1eÿr L j

(

)L j (

) d

â á, á

(F:23)

and that

…0

á 1

r ÿ

(á ÿ j)!

j 1eÿr L j (

)L j

(

) d

á![(á 1)!]2

(F:24)

…0

(á ÿ j)!

(á ÿ 1 ÿ j)!

(á ÿ j)!

j 1eÿr[L j (

)]2 d

(á!)2

(á

1)!

á!

(2á ÿ j 1)(á!)3

(F:25)

The term in which â á ÿ 1 is equivalent to the term in which â á 1 after the dummy indices á and â are interchanged. Equations (F.23), (F.24), and (F.25) are pertinent to the wave functions for the hydrogen atom.

Completeness

We de®ne the set of functions ÷kj(r) by the relation

÷kj(r)	(k j)!	rjeÿr	1=2
			j
	ÿ		Lk	(r)	(F:26)
	(k!)3

According to equation (F.22), the functions ÷kj(r) constitute an orthonormal set. We now show1 that this set is complete.

Substitution of equation (F.15) into (F.26) gives

	ÿ jer	1=2 dkÿ j
÷kj(r) (ÿ1) j	r			(rk eÿr)	(F:27)
	k!(k ÿ j)!		drkÿ j

If we apply equation (A.11), we may express the derivative in (F.27) as

1D. Park, personal communication. This method parallels the procedure used to demonstrate the completeness of the set of functions in equation (D.15).

Laguerre and associated Laguerre polynomials

dkÿ j

k! dkÿ j

eirs

ikÿ j k! 1

eirs

(rk eÿr)

…ÿ1

skÿ j

drkÿj

2ð

drkÿ j

(1 is)k 1

2ð

(1 is)k 1

so that ÷kj(r) in integral form is

…ÿ1

1=2

1) jik

jer

eirs

÷kj(r)

(ÿ

!rÿ

s kÿ j ds

2ð

j)!

is)k 1

To demonstrate that the set ÷kj(r) is complete, we need to evaluate the sum

÷kj(r)÷kj(r9)

k j

317

(F:28)

Expressing (F.28) in terms of the dummy variable of integration s for r and in terms of t for r9, we obtain for the summation

÷kj(r)÷kj(r9)

k j

1 ( 1)k

j k

(st)k

(rr9)ÿ j=2e(r r9)=2

1 1 ei(rs r9t)

"k j

# ds dt

4ð2

j)!

i(s

st]k 1

…ÿ1…ÿ1

By letting á k ÿ j, we may express the sum on the right-hand side as

(ÿ1)á(á j)!

i(s

st]ÿ(á j 1)(st)á

j![1

i(s

t)]ÿ( j 1)

á 0

á!

where we have applied equation (A.3) to evaluate the sum over á. We now have

1 1

ei(rs r9t)

÷kj(r)÷kj(r9)

(rr9)ÿ j=2e(r r9)=2

…ÿ1…ÿ1

ds dt

(F:29)

4ð2

[1 i(s t)] j 1

To evaluate the double integral, we introduce the variables u and v

s t

s ÿ t

ds dt 2 du dv

The double integral then factors into

1 ei(r r9)u

du 1 ei(rÿr9)v dv

2ð

r r9

jeÿ(r r9)=2[2ðä(

r ÿ r

9)]

…ÿ1 (1 2iu) j 1

…ÿ1

where the ®rst integral is evaluated by equation (A.11) and the second by (C.6).

Equation (F.29) becomes

1 ÷

( )÷

(

r r9

jä(

2(rr9)1=2

ÿ r

By applying equation (C.5e), we obtain the completeness relation

÷kj(r)÷kj(r9) ä(r ÿ r9)

(F:30)

k j

demonstrating according to equation (3.31) that the set ÷kj(r) is complete.

Appendix G

Series solutions of differential equations

General procedure

The application of the time-independent SchroÈdinger equation to a system of chemical interest requires the solution of a linear second-order homogeneous differential equation of the general form

p(x)	d2 u(x)		q(x)	du(x)		r(x)u(x) 0	(G:1)
	dx2			dx

where p(x), q(x), and r(x) are polynomials in x and where p(x) does not vanish in some interval which contains the point x 0. Equation (G.1) is linear because each term contains u or a derivative of u to the ®rst power only. The order of the highest derivative determines that equation (G.1) is second-order. In a homogeneous differential equation, every term contains u or one of its derivatives.

The Frobenius or series solution method for solving equation (G.1) assumes that the solution may be expressed as a power series in x

1
u X ak x k s a0 xs a1 xs 1	(G:2)
k 0

where ak (k 0, 1, 2, . . .) and s are constants to be determined. The constant s is chosen such that a0 is not equal to zero. The ®rst and second derivatives of u are then given by

	du			X
	du			1
		dx	u9	ak (k s)x k sÿ1 a0 sxsÿ1 a1(s 1)xs	(G:3)
d2 u		X		k 0
		X
	1
	u0 ak (k s)(k s ÿ 1)x k sÿ2 a0 s(s ÿ 1)xsÿ2 a1(s 1)sxsÿ1
dx2	u0 ak (k s)(k s ÿ 1)x k sÿ2 a0 s(s ÿ 1)xsÿ2 a1(s 1)sxsÿ1
		k 0

(G:4)

A second-order differential equation has two solutions of the form of equation (G.2), each with a different set of values for the constant s and the coef®cients ak .

Not all differential equations of the general form (G.1) possess solutions which can be expressed as a power series (equation (G.2)).1 However, the differential equations encountered in quantum mechanics can be treated in this manner. Moreover, the power

1For a thorough treatment see F. B. Hildebrand (1949) Advanced Calculus for Engineers (Prentice-Hall, New York).

318

Series solutions of differential equations

319

series expansion of u is valid for many differential equations in which p(x), q(x), and/ or r(x) are functions other than polynomials,2 but such differential equations do not occur in quantum-mechanical applications.

The Frobenius procedure consists of the following steps.

1.Equations (G.2), (G.3), and (G.4) are substituted into the differential equation (G.1) to obtain a series of the form

ak [(k s)(k s ÿ 1) p(x)x k sÿ2 (k s)q(x)x k sÿ1 r(x)x k s] 0

ck x k sÿ2 0	(G:5)
k á
k 0
2. The terms are arranged in order of ascending powers of x to obtain
1
X

where the coef®cients ck are combinations of the constant s, the coef®cients ak , and the coef®cients in the polynomials p(x), q(x), and r(x). The lower limit á of the summation is selected such that the coef®cients ck for k , á are identically zero, but cá is not.

3.Since the right-hand side of equation (G.5) is zero, the left-hand side must also equal zero for all values of x in an interval that includes x 0. The only way to

meet this condition is to set each of the coef®cients ck equal to zero, i.e., ck 0 for k á, á 1, . . .

4.The coef®cient cá of the lowest power of x in equation (G.5) always has the form cá a0 f (s), where f (s) is quadratic in s because the differential equation is

	second-order. The expression cá a0 f (s) 0 is called the indicial equation and
	has two roots, s1 and s2, assuming that a0 60.
5.	For each of the two values of s, the remaining expressions ck 0 for k á 1,
	á 2, . . . determine successively a1, a2, . . . in terms of a0. Each value of s yields
	a different set of values for ak ; one set is denoted here as ak , the other as a9k .
6.	The two mathematical solutions of the differential equation are u1 and u2
	u1 a0 xs1 [1 (a1=a0)x (a2=a0)x2 ]
	u2 a90 xs2 [1 (a91=a90)x (a92=a90)x2 ]

where a0 and a90 are arbitrary constants. Physical solutions are obtained by applying boundary and normalization conditions to u1 and u2.

7.For some differential equations, the two roots s1 and s2 of the indicial equation differ by an integer. Under this circumstance, there are two possible outcomes: (a) steps 1 to 6 lead to two independent solutions, or (b) for the larger root s1, steps 1 to 6 give a solution u1, but for the root s2 the recursion relation gives in®nite values for the coef®cients ak beyond some speci®c value of k and therefore these steps fail to provide a second solution. For some other differential equations, the two roots of

2See for example E. T. Whittaker and G. N. Watson (1927) A Course of Modern Analysis, 4th edition (Cambridge University Press, Cambridge), pp. 194±8; see also the reference in footnote 1 of this Appendix.

320	Appendix G

the indicial equation are the same (s1 s2) and therefore only one solution u1 is obtained. In those cases where steps 1 to 6 give only one solution u1, a second solution u2 may be obtained3 by a slightly more complex procedure. This second solution has the form

u2 cu1 ln x c bk x k s2

k 0

where c is an arbitrary constant and the coef®cients bk are related to the coef®cients ak . However, a solution containing ln x is not well-behaved and the arbitrary constant c is set equal to zero in quantum-mechanical applications.

8. The interval of convergence for each of the series solutions u1 and u2 may be determined by applying the ratio test. For convergence, the condition


	ak 1
lim	ak	jxj	,	1
k!1	ak	jxj		1

must be satis®ed. Thus, a series converges for values of x in the range

1			1
ÿ		, x	,
ÿ	R	, x	,	R
where R is de®ned by


	lim		ak 1
R k!1			ak

For R equal to zero, the corresponding series converges for ÿ1 , x , 1. If R equals unity, the corresponding series converges for ÿ1 , x , 1.

Applications

In Chapters 4, 5, and 6 the SchroÈdinger equation is applied to three systems: the harmonic oscillator, the orbital angular momentum, and the hydrogen atom, respectively. The ladder operator technique is used in each case to solve the resulting differential equation. We present here the solutions of these differential equations using the Frobenius method.

Harmonic oscillator

The SchroÈdinger equation for the linear harmonic oscillator leads to the differential equation (4.17)

	d2ö(î)	î2ö(î)		2E
ÿ					ö(î)	(G:6)
ÿ	dî2			"ù	ö(î)	(G:6)
If we de®ne ë by the relation
	2ë 1		2E			(G:7)

			"ù
and introduce this expression into equation (G.6), we obtain
	ö0 (2ë 1 ÿ î2)ö 0					(G:8)

3 See footnote 1 of this Appendix.

Series solutions of differential equations

321

We ®rst investigate the asymptotic behavior of ö(î). For large values of î, the

constant 2ë 1 may be neglected in comparison with î2 and equation (G.8) becomes

ö0 î2ö

The approximate solutions of this differential equation are

because we have

ö ce î2=2

The function e

î2

ö0 (î2 1)ö î2ö

for large î

is not a satisfactory solution because it becomes in®nite as

î ! 1, but the function eÿî2=2 is well-behaved. This asymptotic behavior of ö(î)

suggests that a satisfactory solution of equation (G.8) has the form

ö(î) u(î)eÿî2=2

(G:9)

where u(î) is a function to be determined.

Substitution of equation (G.9) into (G.8) gives

u0 ÿ 2îu9 2ëu 0

(G:10)

We solve this differential equation by the series solution method. Applying equations

(G.2), (G.3), and (G.4), we obtain

ak (k s)(k s ÿ 1)îk sÿ2

k 0

ak [ÿ2(k s) 2ë]îk s 0

(G:11)

k 0

The coef®cient of îsÿ2 gives the indicial equation

a0 s(s ÿ 1) 0

îsÿ1 gives

(G:12)

with two solutions, s

0 and s

1. The coef®cient of

a1(s 1)s 0

(G:13)

For the case s 0, the coef®cient a1 has an arbitrary value; for s 1, we have a1 0.

If we omit the ®rst two terms (they vanish according to equations (G.12) and

(G.13)) in the ®rst summation on the left-hand side of (G.11) and replace the dummy

index k by k 2 in that summation, we obtain

fak 2(k s 2)(k s 1) ak [ÿ2(k s) 2ë]gîk s 0

(G:14)

k 0

Setting the coef®cient of each power of î equal to zero gives the recursion formula

k 2

2(k s ÿ ë)

(G:15)

2)(k

For the case s 0, the constants a0 and a1 are arbitrary and we have the following

two sets of expansion constants

2(1 ÿ

ë)

ëa

2(2 ÿ ë)

ë(2 ÿ ë)

2(3 ÿ ë)

22(1 ÿ ë)(3 ÿ ë)

4 . 3

5 . 4

2(4 ÿ ë)

ë(2 ÿ ë)(4 ÿ ë)

2(5 ÿ ë)

23(1 ÿ ë)(3 ÿ ë)(5 ÿ ë)

6 . 5

7 . 6

Thus, the two solutions of the second-order differential equation (G.10) are

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