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

252	Approximation methods

assume in the continuing presentation that all the roots are indeed different from each other.

`Correct' zero-order eigenfunctions

The determination of the coef®cients cáã is not necessary for ®nding the ®rstorder perturbation corrections to the eigenvalues, but is required to obtain the `correct' zero-order eigenfunctions and their ®rst-order corrections. The coef®- cients cáã for each value of á (á 1, 2, . . . , gn) are obtained by substituting the value found for E(1)ná from the secular equation (9.65) into the set of simultaneous equations (9.64) and solving for the coef®cients cá2, . . . , cá g n in terms of cá1. The normalization condition (9.57) is then used to determine cá1. This procedure uniquely determines the complete set of coef®cients cáã (á,

ã1, 2, . . . , gn) because we have assumed that all the roots E(1)ná are different. If by accident or by clever choice, the initial set of unperturbed eigenfunc-

tions

(0)

is actually the `correct' set, i.e., if in the limit

ë !

0 the perturbed

øná

(0)

eigenfunction øná reduces to øná

for all values of á, then the coef®cients cáã

are given by cáã äáã and the secular determinant is diagonal

H(1)n1,n1 ÿ E(1)ná

0

0

0

H(1)n2,n2

ÿ

E(1)ná

0

0

(1)

(1)

0

0

H ng

n

,ng

n ÿ

Ená

corrections to the eigenvalues are then given by

The ®rst-order

(1)

^ (1)

á 1, 2, . . . , gn

(9:66)

Ená H ná,ná,

It is obviously a great advantage to select the `correct' set of unperturbed eigenfunctions as the initial set, so that the simpler equation (9.66) may be used. A general procedure for achieving this goal is to ®nd a hermitian operator

^

^

(0)

and

^ (1)

and has eigenfunctions և with non-

A that commutes with both H

H

degenerate eigenvalues ìá, so that

^

^

(0)

]

^

^

(1)

] 0

(9:67)

and

[A,

H

[A,

H

^

^

^

(0)

A÷á ìá÷á

commute, they have simultaneous eigenfunctions. Therefore,

Since A and

H

we may select ÷1, ÷2, . . . , ÷g n

as the initial set of unperturbed eigenfunctions

ø(0)ná ÷á,

á 1, 2, . . . ,

gn

^

^

(1)

]j÷ái (â 6á), which of course vanishes

We next form the integral h։j[A,

H

according to equation (9.67),

9.5

Degenerate perturbation theory

253

^

^ (1)

^ ^ (1)

^ (1)

^

h։j[A,

H

]jևi h։jA H

j÷ái ÿ h÷âjH

Ajևi

^

^ (1)

^

(1)

jևi

hA։jH

j÷ái ÿ ìáh÷âjH

(0) ^

(1)

(0)

(ìâ ÿ ìá)hønâjH

jønái

^ (1)

(ìâ ÿ ìá)H nâ,ná

0

^ (1)

Since

ìâ

ì , the off-diagonal elements

equal zero and the set of

H

6 (0)á

nâ,ná

^

functions

øná ÷á

is the `correct' set. The parity operator Ð discussed in

Section 3.8 can often be used in this context for selecting `correct' unperturbed eigenfunctions.

First-order corrections to the eigenfunctions

To obtain the ®rst-order corrections ø(1) to the eigenfunctions øná, we multiply

ø(0) 6 ná

equation (9.62) by â for k n and integrate over all space

k

(0) ^ (0)	(0)	(1)	(0) ^ (1)	(0)	(1) (0) (0)
høkâjH	ÿ En	jønái ÿhøkâjH		jönái Enáhøkâjönái

Applying the hermitian property of	^ (0)	(0)
	H	and noting that økâ

all eigenfunctions belonging to the eigenvalue E(0), we have
				n
(0)	(0)	(0)	(1)	(0) ^ (1)	(0)
(Ek	ÿ En	)høkâjønái ÿhøkâjH			jönái

is orthogonal to

(9:68)

We next expand the ®rst-order correction ø(1)ná in terms of the complete set of unperturbed eigenfunctions

	gj
ø(1)ná	X X
	aná, jãø(0)jã	(9:69)

j(6n) ã 1

where the terms with j n are omitted for the same reason that they are omitted in equation (9.30). Substitution of equations (9.54) and (9.69) into (9.68) gives

		X X		X
		gj		gn
(0)	(0)		(0) (0)	^ (1)
(Ek	ÿ En	)	aná, jãhøkâjøjã i ÿ	cáã H kâ,nã
		j(6n) ã 1		ã 1

In view of the orthonormality relations, the summation on the left-hand side may be simpli®ed as follows

gj	gj
X X	X X
aná, jãhø(0)kâjø(0)jã i	aná, jãäkjäâã aná,kâ
j(6n) ã 1	j(6n) ã 1

Therefore, we have

254		Approximation methods
				gn	^ (1)
			ÿ
				cáã H kâ,nã
				ã 1
	a			X		(9:70)
		ná,kâ		(E(0)k
					ÿ E(0)n )

The eigenfunctions øná for the perturbed system to ®rst order are obtained by combining equations (9.61), (9.69), and (9.70)

					gn	^ (1)
				X X	cáã H kâ,nã
			(0)		ã 1	ÿ			(1)
				gk
ø	ná	ö	ná ÿ	ë	X			ø	kâ	(9:71)
				k(6n) â 1	(E(0)		E(0))
					k		n

Example: hydrogen atom in an electric ®eld

As an illustration of the application of degenerate perturbation theory, we consider the in¯uence, known as the Stark effect, of an externally applied electric ®eld E on the energy levels of a hydrogen atom. The unperturbed

Hamiltonian operator	^	(0)	for the hydrogen atom is given by equation (6.14),
	H

and its eigenfunctions and eigenvalues are given by equations (6.56) and (6.57), respectively. In this example, we label the eigenfunctions and eigenva-

lues of	^	(0)	with an index starting at 1 rather than at 0 to correspond to the
	H
principal quantum number n. The perturbation				^
				H9 is the potential energy for

the interaction between the atomic electron with charge ÿe and an electric ®eld E directed along the positive z-axis

^	^	(1)	eE z eE r cos è	(9:72)
H9	H

If spin effects are neglected, the ground-state unperturbed energy level E(0)1 is non-degenerate and its ®rst-order perturbation correction E(1)1 is given by equation (9.24) as

E(1)1 eE h1sjzj1si 0

This integral vanishes because the unperturbed ground state of the hydrogen atom, the 1s state, has even parity and z has odd parity.

The next lowest unperturbed energy level E(0)2 , however, is four-fold degenerate and, consequently, degenerate perturbation theory must be used to

determine its	perturbation corrections. For simplicity of notation, in the
(0)	(0)	^ (1)	we drop the index n, which has the value
quantities øná,	öná, and	H ná,nâ

n 2 throughout. As the initial set of eigenfunctions for the unperturbed system, we select the 2s, 2p0, 2p1, and 2pÿ1 atomic orbitals as given by equations (6.59) and (6.60), so that

9.5

Degenerate perturbation theory

255

ø1(0) j2si,

ø2(0) j2p0i

(9:73)

ø3(0) j2p1i,

ø4(0) j2 pÿ1i

The `correct' set of unperturbed eigenfunction ö(0)

are, then

X

á

öá(0)

4

â 1

cáâø(0)â

cá1j2si cá2j2p0i cá3j2p1i cá4j2pÿ1i,

á 1, 2, 3, 4

(9:74)

^

(1)

The matrix elements H

áâ in this example are

^

(1)

(0)

(0)

(0)

(0)

i

H

áâ

eE høá

jzjøâ i eE høá

jr cos

èjøâ

eE …

2ð

…

ð

…

1

øá øâ

r cos èr

sin è dr dè dj

(9:75)

(0)

(0)

2

0

0

0

These

matrix

elements vanish

unless

Ä

m

0 and

Ä

l

1. Thus,

only

the

^ (1)

^ (1)

matrix element

H

12 , which equals H21

, is non-zero.

^ (1)

To evaluate the matrix element

H12 , we substitute the 2s wave function from

equation (6.59) and the 2p0 wave function from equation (6.60a) into (9.75)

(1)

(1)

1

r

ð

2ð

H^ 12 H^ 21 eE [ð(2a0)4]ÿ1

…0

r4 1 ÿ

eÿr=a0 dr…0

cos2è sin è dè…0

dj

2a0

ÿ3eE a0

where equations (A.26) and (A.28) are used. The secular determinant (9.65) is

ÿ			E(1)	ÿ			(1)
	ÿ		2	ÿ	(1)	ÿ
		3eE a0			E2		0
			0		0		E2
			0		0		0
which expands to

	0				0
	0
	0
	E	2
ÿ
		(1)

		[(E2(1))2 ÿ (3eE a0)2](E2(1))2 0
The four roots are E2(1) ÿ3eE a0, 3eE a0,							0, 0, so that to ®rst order the
perturbed energy levels are
E			ÿe92	ÿ	3e	a ,	E			ÿe92
	21		8a0			E 0		23		8a0
										(9:76)
E			ÿe92		3e	a ,	E			ÿe92
	22		8a0			E 0		24		8a0

The four linear homogeneous simultaneous equations (9.64) are

ÿ1=2

ÿ1=2

256	Approximation methods
		(1)	^ (1)	0
	ÿcá1 Eá		cá2 H12
		^ (1)	(1)	0
	cá1 H12		ÿ cá2 Eá			(9:77)
			ÿcá3 Eá(1) 0
			ÿcá4 Eá(1) 0
To ®nd the `correct' set of unperturbed eigenfunctions ö(0)					, we substitute ®rst
				á
E2(1) ÿ3eE a0, then successively E2(1) 3eE a0, 0, 0 into the set of equations
(9.77). The results are as follows
(1)	^ (1)		c1 c2; c3 c4 0
for E2	H12 ÿ3eE a0	:
(1)	^ (1)		c1 ÿc2; c3 c4 0
for E2	ÿH12 3eE a0	:
for E2(1) 0:			c1 c2 0; c3 and c4			undetermined

Thus, the `correct' unperturbed eigenfunctions are ö(0)1 2 (j2si j2p0i)

ö(0)2 2 (j2si ÿ j2p0i)

(9:78)

ö(0)3 j2p1i ö(0)4 j2pÿ1i

The factor 2ÿ1=2 is needed to normalize the `correct' eigenfunctions.

9.6 Ground state of the helium atom

In this section we examine the ground-state energy of the helium atom by means of both perturbation theory and the variation method. We may then compare the accuracy of the two procedures.

The potential energy V for a system consisting of two electrons, each with mass me and charge ÿe, and a nucleus with atomic number Z and charge Ze

is
V ÿ	Ze92	ÿ	Ze92		e92
	r1		r2		r12

where r1 and r2 are the distances of electrons 1 and 2 from the nucleus, r12 is the distance between the two electrons, and e9 e for CGS units or e9 e=(4ðå0)1=2 for SI units. If we assume that the nucleus is ®xed in space, then the Hamiltonian operator for the two electrons is

^	"2	2	2	Ze92		Ze92		e92
H ÿ 2me		(=1	=2) ÿ r1		ÿ r2		r12		(9:79)

	9.6 Ground state of the helium atom								257
The operator H^	applies to He for	Z		2, Li for	Z		3, Be2 for Z		4, and

so forth.

Perturbation theory treatment

We regard the term e92=r12 in the Hamiltonian operator as a perturbation, so that

^	^	(1)		e92
H9	H		r12		(9:80)

In reality, this term is not small in comparison with the other terms so we should not expect the perturbation technique to give accurate results. With this choice for the perturbation, the SchroÈdinger equation for the unperturbed Hamiltonian operator may be solved exactly.

The unperturbed Hamiltonian operator is the sum of two hydrogen-like Hamiltonian operators, one for each electron

	^	(0)		^	(0)			^	(0)
where	H			H	1		H2
^	(0)			"2			2			Ze92
H	1	ÿ		2me		=1		ÿ		r1
^	(0)			"2			2			Ze92
H	2	ÿ		2me		=2		ÿ		r2
If the unperturbed wave function ø(0) is written as the product

ø(0)(r1, r2) ø(0)1 (r1)ø(0)2 (r2)

and the unperturbed energy E(0) is written as the sum

E(0) E(0)1 E(0)2

then the SchroÈdinger equation for the two-electron unperturbed system

^ (0)		(0)	(r1, r2) Eø	(0)	(r1	, r2)
H	ø
separates into two independent equations,
H(0)i	ø(0)i E(0)i ø(0)i ,				i 1, 2

which are identical except that one refers to electron 1 and the other to electron 2. The solutions are those of the hydrogen-like atom, as discussed in Chapter 6. The ground-state energy for the unperturbed two-electron system is, then, twice the ground-state energy of a hydrogen-like atom

E(0) ÿ2	2 e92	ÿ	Z2 e92
	Z			(9:81)
	2a0		a0

The ground-state wave function for the unperturbed two-electron system is the product of two 1s hydrogen-like atomic orbitals

258 Approximation methods

Table 9.1. Ground-state energy of the helium atom

Method			Energy (eV)	% error
Exact			ÿ79.0
Perturbation theory:
(0)			ÿ108.8	ÿ37.7
E(0)	E	(1)
E			ÿ74.8	5.3
Variation theorem (E )			ÿ77.5	1.9

1

Z

3

1

Z

3

ø(0)(r1, r2)

eÿ Zr1=a0 eÿ Zr2=a0

eÿ(r1 r2)=2

(9:82)

ð

a0

ð

a0

where we have de®ned

ri

2Zri

,

i 1, 2

(9:83)

a0

The ®rst-order perturbation correction E(1) to the ground-state energy is obtained by evaluating equation (9.24) with (9.80) as the perturbation and

(9.82) as the unperturbed eigenfunction

E(1)

*ø(0)

e92

ø(0)

2Z

e92

ø(0)

+

Ze92

I

(9:84)

where

r12

and

where

jr2 ÿ r1j

eÿ(r1 r2)

I … …

r12r22 sin è1 sin è2 dr1 dè1 dj1 dr2 dè2 dj2

(9:85)

r12

This six-fold integral I is evaluated in Appendix J and is equal to 20ð2. Thus, we have

5Ze92			5
E(1)		ÿ		E(0)					(9:86)
	8a0		8Z
The ground-state energy of the perturbed system to ®rst order is, then
E E(0) E(1) ÿ Z2 ÿ					Z		e92
					5				(9:87)
					8		a0
Numerical values of E(0) and E(0) E(1)			for the helium atom (Z 2) are

given in Table 9.1 along with the exact value. The unperturbed energy value

E(0) has a 37.7% error when compared with	the exact value. This large
inaccuracy is expected because the perturbation	^
	H9 in equation (9.80) is not

small. When the ®rst-order perturbation correction is included, the calculated energy has a 5.3% error, which is still large.

9.6 Ground state of the helium atom

259

Variation method treatment

As a normalized trial function ö for the determination of the ground-state energy by the variation method, we select the unperturbed eigenfunction ø(0)(r1, r2) of the perturbation treatment, except that we replace the atomic number Z by a parameter Z9

ö ö1ö2

	1		Z9		3=2
ö1					eÿ Z9r1=a0	(9:88)
	ð1=2		a0
	1		Z9		3=2
ö2					eÿ Z9r2=a0
	ð1=2		a0

The parameter Z9 is an effective atomic number whose value is determined by the minimization of E in equation (9.2). Since the hydrogen-like wave functions ö1 and ö2 are normalized, we have

hö1jö1i hö2jö2i 1

(9:89)

The quantity E is obtained by combining equations (9.2), (9.79), (9.88), and (9.89) to give

E *ö1

ÿ 2me =12 ÿ r1

ö1+

*ö2

ÿ 2me

=22 ÿ r2

ö2+

"2

Ze92

"2

Ze92

e92

*ö1ö2

ö1ö2+

(9:90)

r12

Note that while the

trial

function ö ö1ö2 depends on the parameter Z9, the

Hamiltonian operator contains the true atomic number Z. Therefore, we rewrite equation (9.90) in the form

E *ö1

ÿ

"2

ÿ

Z9e92

ö1+

*ö1

(Z9

Z)e92

ö1+

2me =12

r1

ÿr1

ÿ

"2

=22

ÿ

Z9

e92

ö2+

( Z9

Z)

e92

ö2

+

*ö2

2me

r2

*ö2

ÿr2

e92

ö1ö2

+

(9:91)

*ö1ö2

r12

The ®rst term on the right-hand side is just the energy of a hydrogen-like atom with nuclear charge Z9, namely, ÿZ92 e92=2a0. The third term has the same value as the ®rst. The second term is evaluated as follows

260 Approximation methods

*ö1

(Z9

Z)e92

+

1 Z9 3 1

ÿ

ö1

( Z9 ÿ Z)e92

…0

r1ÿ1 eÿ2 Z9r1=a0 4ðr12 dr1

r1

ð

a0

( Z9 ÿ Z)e9

2 Z9

a0

where equations (A.26) and (A.28) have been used. The fourth term equals the second. The ®fth term is identical to E(1) of the perturbation treatment given by equation (9.86) except that Z is replaced by Z9 and therefore this term equals

5Z9e92=8a0. Thus, the quantity E in equation (9.91) is

E

2

Z92 e92

2

Z9( Z9 ÿ Z)e92

5Z9e92

[Z92

ÿ

2(Z

ÿ

5

)Z9]

e92

ÿ

2a0

a0

8a0

16

a0

(9:92)

We next minimize E with respect to the parameter Z9

dE

2[Z9 ÿ (Z ÿ

5

)]

e92

0

d Z9

16

a0

so that

Z9 Z ÿ

5

16

Substituting this result into equation (9.92) gives

E

ÿ(Z ÿ

5

)

2 e92

> E0

(9:93)

16

a0

as an upper bound for the ground-state energy E0.

When applied to the helium atom (Z 2), this upper bound is

27 2 e92

e92

E ÿ

ÿ2:85

(9:94)

16

a0

a0

The numerical value of E is listed in Table 9.1. The simple variation function (9.88) gives an upper bound to the energy with a 1.9% error in comparison with the exact value. Thus, the variation theorem leads to a more accurate result than the perturbation treatment. Moreover, a more complex trial function with more parameters should be expected to give an even more accurate estimate.

Problems

9.1The Hamiltonian operator for a hydrogen atom in a uniform external electric ®eld E along the z-coordinate axis is

	H^		ÿ"2	=2	ÿ	e92	ÿ	eEz
			2ì			r

Problems

261

Use the variation trial function ö ø1s (1 ëz), where ë is the variation parameter, to estimate the ground-state energy for this system.

9.2 Apply the gaussian function

ö eÿër2=a20

where ë is a parameter, as the variation trial function to estimate the energy of the ground state of the hydrogen atom. What is the percent error?

9.3Apply the variation trial function ö(x) x(a ÿ x)(a ÿ 2x) to estimate the energy of a particle in a box with V (x) 0 for 0 < x < a, V (x) 1 for x , 0, x . a. To which energy level does this estimate apply?

9.4Consider a particle in a one-dimensional potential well such that

V (x) (b"2=ma4)x(x ÿ a),	0 < x < a
1,	x , 0, x . a

where b is a dimensionless parameter. Using the particle in a box with V (x) 0 for 0 < x < a, V (x) 1 for x , 0, x . a as the unperturbed system, calculate the ®rst-order perturbation correction to the energy levels. (See Appendix A for the evaluation of the resulting integrals.)

9.5 Consider a particle in a one-dimensional potential well such that

V (x) (b"2=ma3)x,	0 < x < a
1,	x , 0, x . a

where b is a dimensionless parameter. Using the particle in a box with V (x) 0 for 0 < x < a, V (x) 1 for x , 0, x . a as the unperturbed system, calculate the ®rst-order perturbation correction to the energy levels. (See Appendix A for the evaluation of the resulting integral.)

9.6Calculate the second-order perturbation correction to the ground-state energy for the system in problem 9.5. (Use integration by parts and see Appendix A for the evaluation of the resulting integral.)

9.7Apply the linear variation function

ö c1(2=a)1=2 sin(ðx=a) c2(2=a)1=2 sin(2ðx=a)

for 0 < x < a to the system in problem 9.5. Set the parameter b in the potential equal to ð2=8. Solve the secular equation to obtain estimates for the energies E1 and E2 of the ground state and ®rst-excited state. Compare this estimate for E1 with the ground-state energies obtained by ®rst-order and second-order perturbation theory. Then determine the variation functions ö1 and ö2 that correspond to E1 and E2.

9.8 Consider a particle in a one-dimensional champagne bottle2 for which

V (x) (ð2"2=8ma2) sin(ðx=a),	0 < x < a
1,	x , 0, x . a

2 G. R. Miller (1979) J. Chem. Educ. 56, 709.