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

122

Harmonic oscillator

hn9jxjni

2mù

1=2

(hn9j^ayjni hn9j^ajni)

"

2mù

1=2

(pn 1än9,n 1 pnän9,nÿ1)

so that

"

•••••••••••

•••

h

j j

i r•••••••••••••••••

n

1 x n

(n 1)"

(4:45a)

h

ÿ

j j

2mù

i r••••••••••

n

1 x n

n"

(4:45b)

2mù

hn9jxjni 0

for n9 6 n 1, n ÿ 1

(4:45c)

If we replace n by n ÿ 1 in equation (4.45a), we obtain

h j j

ÿ i r••••••••••

n x n

1

n"

2mù

From equation (4.45b) we see that

hn ÿ 1jxjni hnjxjn ÿ 1i

Likewise, we can show that

hn 1jxjni hnjxjn 1i

In general, then, we have

hn9jxjni hnjxjn9i

To ®nd the matrix element hn9j^pjni, we use equations (4.43b) and (4.44) to

give

ù

hn9j^ay ÿ ^ajni

hn9j^pjni i m2

1=2

"

i m2

ù

1=2

(pn 1än9,n 1 ÿ pnän9,nÿ1)

so that

"

•••••••••••

•••

h

j

j

i

r•••••••••••••••••••••••••

n

1

^p n

i

(n 1)m"ù

(4:46a)

h

ÿ

j

j

2

i ÿ r•••••••••••••

n

1 ^p n

i

nm"ù

(4:46b)

2

hn9j^pjni 0

for n9 6 n 1, n ÿ 1

(4:46c)

We can easily show that

4.4 Matrix elements

123

hn9j^pjni ÿhnj^pjn9i

The matrix element hn9jx2jni is

hn9jx2jni

"

hn9j(^ay

^a)2jni

"

hn9j(^ay)2 ^ay^a ^a^ay ^a2jni

2mù

2mù

From equation (4.34) we have

(n 1)(n 2)jn 2i

(^ay)2

n

pn 1^ayjn 1i

j

i

•••••••••••

p•••••••••••••••••••••••••••••

a

pna

n

ÿ

1

n n

^y^ajni p•••^yj

i

j i

(4:47)

^a^ay n

n

1^a n 1

(n 1) n

2j

i

j

i

j

i

•••••••••••

^a

p

n(n ÿ 1)jn ÿ 2i

n

n^a n

1

so that

j

i

•••

j

ÿ

i p•••••••••••••••••

n9

"

[

(n 1)(n 2)än9,n 2 (2n 1)än9n

x2jni 2mù

h

j

p•••••••••••••••••••••••••••••

p•••••••••••••••••

n(n ÿ 1)än9,nÿ2]

We conclude that						p•••••••••••••••••••••••••••••
h	j		"		"
n	2 x2jni hnjx2jn 2i				2mù	(n 1)(n 2)	(4:48a)
	hnjx2jni			(n 21)			(4:48b)
			mù			p•••••••••••••••••
		2			"
hn ÿ 2jx2jni hnjx2jn ÿ 2i					2mù	n(n ÿ 1)	(4:48c)
	hn9jx jni 0,			n9 6 n 2, n, n ÿ 2			(4:48d)

The matrix element hn9j^p2jni is obtained from equations (4.43b) and (4.47)

hn9j^p2jni ÿ m2		ù hn9j(^ay ÿ ^a)2jni ÿ m2ù hn9j(^ay)2			ÿ ^ay^a
"		"
ÿ ^a^ay ^a2jni
"		ù	p•••••••••••••••••••••••••••••
ÿ	m		[ (n 1)(n 2)än9,n 2 ÿ (2n 1)än9n
	2

p•••••••••••••••••

n(n ÿ 1)än9,nÿ2]

so that

124

Harmonic oscillator

hnj^p2jni m"ù(n 21)

ù

p•••••••••••••••••••••••••••••

(4:49b)

"

hn 2j^p2jni hnj^p2jn

2i ÿ

m

(n 1)(n 2)

(4:49a)

2

2

"

ù

p•••••••••••••••••

hn ÿ 2j^p2jni hnj^p2jn ÿ

2i ÿ

m

n(n ÿ 1)

(4:49c)

2

hn9j^p jni 0,

n9 6 n 2, n, n ÿ 2

(4:49d)

k

k

^a)k , we can ®nd the

Following this same procedure using the operators (^ay

matrix elements of x and of ^p

for any positive integral power k. In Chapters

9 and 10, we need the matrix elements of x3 and x4. The matrix elements hn9jx3jni are as follows:

"

3=2

p•••••••••••••••••••••••••••••••••••••••••••3=2

hn

3jx3jni hnjx3jn

3i

(n 1)(n 2)(n 3)

(4:50a)

2mù

h

j

j

i h

j

j

i

2mù

n

1

x3

n

n

x3

n

1

3

(n 1)"

(4:50b)

hn ÿ 1jx3jni hnjx3jn ÿ 1i

3 2mù

(4:50c)

n"

3=2

hn9jx3jni 0,

3=2

p••••••••••••••••••••••••••••••••

(4:50e)

n9 6 n 1, n 3

"

hn ÿ

3jx3jni hnjx3jn ÿ

3i

2mù

n(n ÿ 1)(n ÿ 2)

(4:50d)

The matrix elements hn9jx4jni are as follows

hn 4jx4jni hnjx4jn 4i 2mù

2

(n 1)(n 2)(n 3)(n 4)

"

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

(4:51a)

"

1

"

2

p•••••••••••••••••••••••••••••

3

2

hn

2jx4jni hnjx4jn

2i

2

mù

(2n 3) (n 1)(n 2)

(4:51b)

1

"

2

hnjx4jni

2

mù

n2 n 21

p•••••••••••••••••

(4:51c)

hn ÿ 2jx4jni hnjx4jn ÿ 2i

2

2

mù

(2n ÿ 1) n(n ÿ 1)

(4:51d)

4

"

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

hn ÿ

4jx4jni hnjx4jn ÿ

4i

n(n ÿ 1)(n ÿ 2)(n ÿ 3)

(4:51e)

2mù

hn9jx jni 0,

n9 6 n, n 2, n 4

(4:51f)

(n 12)1=2

(n 12)1=2

4.6 Three-dimensional harmonic oscillator

125

4.5 Heisenberg uncertainty relation

Using the results of Section 4.4, we may easily verify for the harmonic oscillator the Heisenberg uncertainty relation as discussed in Section 3.11. Speci®cally, we wish to show for the harmonic oscillator that

ÄxÄ p > 12"

where

(Äx)2 h(x ÿ hxi)2i

(Ä p)2 h(^p ÿ hpi)2i

The expectation values of x and of ^p for a harmonic oscillator in eigenstate jni are just the matrix elements hnjxjni and hnj^pjni, respectively. These matrix elements are given in equations (4.45c) and (4.46c). We see that both vanish, so that (Äx)2 reduces to the expectation value of x2 or hnjx2jni and (Ä p)2 reduces to the expectation value of ^p2 or hnj^p2jni. These matrix elements are given in equations (4.48b) and (4.49b). Therefore, we have

" 1=2

Äx mù

Ä p (m"ù)1=2

and the product ÄxÄ p is

ÄxÄ p (n 12)"

For the ground state (n 0), we see that the product ÄxÄ p equals the minimum allowed value "=2. This result is consistent with the form (equation (3.85)) of the state function for minimum uncertainty. When the ground-state harmonic-oscillator values of kxl, k pl, and ë are substituted into equation (3.85), the ground-state eigenvector j0i in equation (4.31) is obtained. For excited states of the harmonic oscillator, the product ÄxÄ p is greater than the minimum allowed value.

4.6 Three-dimensional harmonic oscillator

The harmonic oscillator may be generalized to three dimensions, in which case the particle is displaced from the origin in a general direction in cartesian space. The force constant is not necessarily the same in each of the three dimensions, so that the potential energy is

V 12kxx2 12kyy2 12kzz2 12m(ù2x x2 ù2y y2 ù2z z2)

126 Harmonic oscillator

where kx, ky,			kz are the respective force constants and ùx, ùy, ùz are the
respective classical angular frequencies of vibration.
The SchroÈdinger equation for this three-dimensional harmonic oscillator is
"2			@2ø @2ø @2ø
ÿ								! 21 m(ù2x x2 ù2y y2 ù2z z2)ø Eø
	2m		@x2		@ y2		@z2

where ø(x, y, z) is the wave function. To solve this partial differential equation of three variables, we separate variables by making the substitution

ø(x, y, z) X (x)Y (y)Z(z)

(4:52)

where X (x) is a function only of the variable x, Y (y) only of y, and Z(z) only of z. After division by ÿø(x, y, z), the SchroÈdinger equation takes the form

	"2 d2 X			ÿ 21mù2x x2		"2 d2 Y			ÿ 21mù2y y2!
	2mX		dx2			2mY d y2
						"2 d2 Z			ÿ 21 mù2z z2 E
						2mZ	dz2

The ®rst term on the left-hand side is a function only of the variable x and remains constant when y and z change but x does not. Similarly, the second term is a function only of y and does not change in value when x and z change but y does not. The third term depends only on z and keeps a constant value when only x and y change. However, the sum of these three terms is always equal to the constant energy E for all choices of x, y, z. Thus, each of the three independent terms must be equal to a constant

		"2 d2 X							ÿ 21mù2x x2 Ex
	2mX dx2
		"2 d2 Y							ÿ 21mù2y y2 Ey
	2mY dy2
		"2 d2 Z							ÿ 21mù2z z2 Ez
		2mZ dz2
where the three separation constants Ex, Ey, Ez satisfy the relation
		Ex Ey Ez E								(4:53)

The differential equation for X (x) is exactly of the form given by (4.13) for a one-dimensional harmonic oscillator. Thus, the eigenvalues Ex are given by

equation (4.30)
Enx (nx 21)"ùx,	nx 0, 1, 2, . . .
and the eigenfunctions are given by (4.41)

4.6 Three-dimensional harmonic oscillator							127
				mù	1=4	2
Xnx (x) (2nx nx!)ÿ1=2				x		Hnx (î)eÿî	=2
				ð"
	mù		1=2
î	x		x
	"

Similarly, the eigenvalues for the differential equations for Y (y) and Z(z) are, respectively

Eny (ny 21)"ùy,							ny 0, 1, 2, . . .
Enz (nz 21)"ùz,							nz 0, 1, 2, . . .
and the corresponding eigenfunctions are
							mù			1=4	2
Yny (y) (2ny ny!)ÿ1=2							y				Hny (ç)eÿç =2
							ð"
ç	mù				1=2
	y				y
	"
							mù			1=4	2
Znz (z) (2nz nz!)ÿ1=2							z			Hnz (æ)eÿæ =2
							ð"
æ	mù			1=2
	z				z
	"

The energy levels for the three-dimensional harmonic oscillator are, then,

given by the sum (equation (4.53))
Enx,n y,nz (nx 21)"ùx (ny 21)"ùy (nz 21)"ùz				(4:54)
The total wave functions are given by equation (4.52)
	m		3=4
ønx,n y,nz (x, y, z) (2nx n y nz nx!ny!nz!)ÿ1=2			(ùxùyùz)1=4
	ð"
3 Hnx (î) Hny (ç)Hnz (æ)eÿ(î2 ç2 æ2)=2				(4:55)

An isotropic oscillator is one for which the restoring force is independent of the direction of the displacement and depends only on its magnitude. For such an oscillator, the directional force constants are equal to one another

kx k y kz k

and, as a result, the angular frequencies are all the same

ùx ùy ùz ù

In this case, the total energies are

128	Harmonic oscillator
Enx,n y,nz	(nx ny nz 23)"ù (n 23)"ù	(4:56)

where n is called the total quantum number. All the energy levels for the isotropic three-dimensional harmonic oscillator, except for the lowest level, are degenerate. The degeneracy of the energy level En is (n 1)(n 2)=2.

Problems

4.1Consider a classical particle of mass m in a parabolic potential well. At time t the displacement x of the particle from the origin is given by

x a sin(ùt b)

where a is a constant and ù is the angular frequency of the vibration. From this expression ®nd the kinetic and potential energies as functions of time and show that the total energy remains constant throughout the motion.

4.2 Evaluate the constant c in equation (4.10). (To evaluate the integral, let y cos è.)

4.3Show that ^a and ^ay in equations (4.18) are not hermitian and that ^ay is the adjoint of ^a.

4.4	The operator N^ ^ay^a is hermitian. Is the operator ^^aay hermitian?
4.5	Evaluate the commutators [ H^ , ^a] and [ H^ , ^ay].

4.6Calculate the expectation value of x6 for the harmonic oscillator in the n 1 state.

4.7Consider a particle of mass m in a parabolic potential well. Calculate the probability of ®nding the particle in the classically allowed region when the particle is in its ground state.

4.8Consider a particle of mass m in a one-dimensional potential well such that

V (x) 21mù2 x2,	x > 0
1,	x , 0
What are the eigenfunctions and eigenvalues?

4.9What is the probability density as a function of the momentum p of an oscillating particle in its ground state in a parabolic potential well? (First ®nd the momentum-space wave function.)

4.10Show that the wave functions An( ã) in momentum space corresponding to ön(î) in equation (4.40) for a linear…harmonic1 oscillator are

An( ã) (2ð)ÿ1=2 ön(î)eÿi ãî dî

ÿ1

iÿn(2n n!ð1=2)ÿ1=2eÿ ã2=2 Hn( ã)

where î (mù=")1=2 x and ã (m"ù)ÿ1=2 p. (Use the generating function (D.1) to evaluate the Fourier integral.)

Problems

129

4.11 Using only equation (4.43b) and the fact that ^ay is the adjoint of ^a, prove that hn9j^pjni ÿhnj^pjn9i

4.12Derive the relations (4.50) for the matrix elements hn9jx3jni.

4.13Derive the relations (4.51) for the matrix elements hn9jx4jni.

4.14Derive the result that the degeneracy of the energy level En for an isotropic three-dimensional harmonic oscillator is (n 1)(n 2)=2.

5

Angular momentum

Angular momentum plays an important role in both classical and quantum mechanics. In isolated classical systems the total angular momentum is a constant of motion. In quantum systems the angular momentum is important in studies of atomic, molecular, and nuclear structure and spectra and in studies of spin in elementary particles and in magnetism.

5.1 Orbital angular momentum

We ®rst consider a particle of mass m moving according to the laws of classical mechanics. The angular momentum L of the particle with respect to the origin

of the coordinate system is de®ned by the relation
L r 3 p	(5:1)

where r is the position vector given by equation (2.60) and p is the linear momentum given by equation (2.61). When expressed as a determinant, the angular momentum L is

L			x	y	z
			px	py	pz
			i	j	k
The components Lx, Ly, Lz of the vector				L are

Lx ypz ÿ zpy
Ly zpx ÿ xpz	(5:2)
Lz xpy ÿ ypx

The square of	the magnitude of the vector L is given	in terms of these
components by
	L2 L : L L2x L2y L2z	(5:3)

130

5.1 Orbital angular momentum

131

If a force F acts on the particle, then the torque T on the particle is de®ned as

dp
T r 3 F r 3 dt	(5:4)

where Newton's second law that the force equals the rate of change of linear momentum, F dp=dt, has been introduced. If we take the time derivative of

equation (5.1), we obtain

dt

3 p r 3 dt

r 3 dt

dt

(5:5)

dL

dr

dp

dp

since

dr

3 p

dr

3 m

dr

0

dt

dt

dt

Combining equations (5.4) and (5.5), we ®nd that

T

dL

(5:6)

dt

If there is no force acting on the particle, the torque is zero. Consequently, the rate of change of the angular momentum is zero and the angular momentum is conserved.

The quantum-mechanical operators for the components of the orbital angular

momentum are obtained by replacing px, py,

pz in the classical expressions

(5.2) by their corresponding quantum operators,

"

@

@

^Lx y^pz ÿ z^py

y

ÿ z

(5:7a)

i

@z

@ y

"

@

@

^Ly z^px ÿ x^pz

z

ÿ x

(5:7b)

i

@x

@z

^Lz x^py ÿ y^px

"

x

@

ÿ y

@

(5:7c)

i

@ y

@x

Since y commutes with ^pz

and z commutes with ^py, there is no ambiguity

regarding the order of y and ^pz and of z and ^py in constructing

^

Lx. Similar

^

^

remarks apply to Ly and Lz. The quantum-mechanical operator for L is

^

^

^

^

(5:8)

and for L2 is

L iLx jLy kLz

^2

^

: ^

^2

^2

^2

(5:9)

L

L

L Lx

L y

Lz

^

^

^

The operators Lx, Ly, Lz can easily be shown to be hermitian with respect to a

^

^2

set of functions of x, y, z that vanish at 1. As a consequence, L and L are also hermitian.