جاهعة الانبار-كلية العلوم – قسن الفيسياء

Quantum Mechanics

The Time- Independent Schrödinger Equation ( TISE) and its solutions:

Three Dimensional Problems

أ.د. نبيل ابراهين فواز

Ref.: Quantum Mechanics

Concepts and Applications

Second Edition

Nouredine Zettili

Jacksonville State University, Jacksonville, USA

The Hydrogen Atom:

Hydrogen atom considered as a system of two interacting point particles ( nucleus &

electron). The charge on the nucleus (Ze) and electron charge (-e). The P.E. of the system is (-

Ze2/r) where r is the distance between the nucleus and the electron. Since the electron moving

around the nucleus in spherical path, so we need the spherical polar coordinates to solve the

equ. of motion.

The Schrödinger wave equ. in 3-D written as:

)1(0][2

2

2 VEm

In spherical polar coordinates, we have

cossinsincossin rzryrx

)sin

1

tan

1(

122

2

22

2

22

22

rrrr

The Schrodinger equ. becoms

)2()(]sin

1)(sin

sin

1)(

1[

2 2

2

222

2

2

2

ErV

rrrr

rrm

Equ.(2) can be separated by writing )3(........),()(),,( YrRr

)4.......(0),()(])([2

),(sin

1)(

),()(sinsin

1)()()(

1),(

0),()(])([2

)],()([sin

1

),()()(sinsin

1),()()(

1

22

2

22

2

2

2

22

2

22

2

2

2

YrRErVm

Yr

rR

Yr

rRrRr

rrr

Y

YrRErVm

YrRr

YrRr

YrRr

rrr

Dividing the equation by ),()( YrR , multiplying by 2r , and rearranging terms, this

becomes

)5(......0]),(sin),(

1),()(sin

sin),(

1[

]})([2

)()()(

1{

2

2

2

2

22

YY

YY

ErVmr

rRr

rrrR

The two terms in the curly braces ( { }) depend only on r, and the two terms in the square

brackets ( [ ]) depend only upon angles.

Both have been chosen to equal the constant )1( , where is the angular momentum

quantum number. Then

)6(........)1(])([2

)()()(

12

22

ErV

mrrR

rr

rrR

Which is call the radial equation, and

)7(.....)1(),(sin),(

1),()(sin

sin),(

12

2

2

Y

YY

Y

Called the angular equation. Notice that [ )1( +(- )1( =0]

The angular equation:

The solutions to equ. ( 7 ) are the spherical harmonic functions ),(,

m

Y

, where m,

are the orbital and magnetic quantum numbers. The spherical harmonic are also separable

)8(.......)()(,

),(,

m

gm

fm

Y

Using equ. ( 8 ) in equ. ( 7 )

)1()(sin)(

1)()(sin

sin)(

1

)1()()(sin)()(

1)()()(sin

sin)()(

1

2

2

2

2

2

2

gg

ff

gfgf

gfgf

Multiplying by 2sin and rearranging

)9(.......0)()(

1sin)1()()(sin

)(

sin2

22

g

gf

f

The first two terms depend only on θ, and the last term depends only on ϕ. Again using same

constant as a separation constant

)11(........)()(

1

)10(........sin)1()()(sin)(

sin

2

2

2

22

mgg

mff

)12().(exp02

2

2

imAgissolutionThegmd

gd

The constant A may be evaluated by normalizing Φ.

)13().(exp2

1

2

111

2

0

2

2

0

imgAdAdgg

The single valued ness of the function Φ demands that it should have the same value at φ=0

and φ= 2π.

)14(,...2,1,0).(exp2

1

12sin2cos)2.(exp

)2.(exp)0.(exp

mwhereimg

mimmior

miAAAg

m = ± correspond to two solutions, while for m = 0, we have only one solution. The quantity

m is called magnetic quantum number which obviously possesses the integral nature.

For real solution ...,2,1cos2

1sin

2

1 mformormg

And 2

1g for m = 0.

However, the general solution for angular equation can be given by sperical harmonic

function as:

]:[2

1sin2/1]

2

!)12([

!2

1),(

,

]:[2

1sin2/1]

2

!)12([

!2

1),(

,

,.......,&,........3,2,1,0)()(,

),(,

ee mii

NOTEe im

Yand

ee mii

NOTEe im

Y

mwherem

gm

fm

Y

And the general solution for radial equation can be given as:

)

0

(1210

/1)

0

(1

)(

)

0

2(12

10/2/3)

0

2(

2/1}

3)]1([2

!)1({

an

r

nLe anr

an

r

rr

nRaswrittenbecanAlso

annLe anrr

annn

n

nR

,......)4,3,2,1(][!

)()(

,)2

(12

1 0

0

nenxndx

nd

n

eexL

n

andpolynomialLaguerretheisna

ZrL

n

radiusBohraWhere

Laguerre polynomials of any order can be calculated using the generating function

][)(j

ej

xj

dx

jdj

exLj

Associated Laguerre polynomials can be calculated from Laguerre polynomials using the

generating function

)()1()( xkj

Lkdx

kdkxLk

j

The complete solution of Schrödinger equation for hydrogen atom is written after

multiplying the functions R(r) , Θ(θ) and Φ(φ) and we have

)(cos)

0

2(

12

1)

0

2()

0

.(exp21

]3})!{(2

)!1(3)

0

2(

)!(4

)!)(12([

mPim

ena

ZrL

nna

Zr

na

Zr

nn

n

na

Z

m

m

mn

And ,.....3,2,12 22

42

nn

emZEE nn

The total degeneracy is 2

1

0 2

)1(2)12( nn

n nn

The W.F. for the ground state n=1, ℓ=0, m=0 is: )

0

.(exp30

1

100 a

r

a

The probability density is : ).

0

2.(exp

30

1

110110 a

r

aP

The probability of finding the electron in the volume dV, dddrrdV sin2( ) is given by:

110110

dddrr

a

r

adddrr sin2)

0

2.(exp

30

1sin2

For the whole surface of the sphere :

drra

r

a

dddrra

r

adrrP

2

0

3

0

2

00

2

0

3

0

)2

.(exp4

sin)2

.(exp1

)(

Which it is the probability that the electron lies between the distance r+dr from the nucleus.

The prob. Is maximum when dP(r)/dr = 0.

0

0

2

0

2

0

3

0

000

2

3

0

)(

02

20]2

2[)2

.(exp4

0])2

.(exp2)2

.(exp)2

(.[4

ara

rr

a

rr

a

r

a

a

rr

a

r

ar

adr

dP r

Thus, in the normal state the max. prob. is at distance equal to the radius of 1st Bohr orbit.

NOTE:

2|&21)1(0coscos|cossin 2

0

2

00

0 dd

Radiative Transition:

We know according to Bohr theory, that the frequency of radiation emitted from an atom, is

given by nmh

EE nm

.

To prove this according to the quantum theory, the W.F. describing an electron in a state,

can be written as:

nnnnn tiE

andtiE

[).(exp).(exp

function of position only]

The mean ( expectation) value of the position of the electron is given by:

dxx

dxtiE

tiE

xdxxx

nn

nn

])()(.[exp

Since nn and is function of position only ( independent of time), so the electron in one state

(pure) Ψ do not radiate ( do not oscillate).

For electron transition between two states from state (Em) to another state (En), so the

electron can be in any states (n) or (m). The W.F. which describe electron mixed between two

states is given by:

)(1

..

..

conditionionnormalizatbbaaand

mstateinparticlethefindingofprobtheisbb

nstateinparticlethefindingofprobtheisaa

ba mn

At t=0: a=1 and b=0

At t ≠0: i) a=0 & b=1 – electron excited from state n to state m.

ii) a=1 & b=0 – electron return to state n from state m.

iii) a≠0 & b≠0 – electron can be in any states m & n with different

probabilities and the atom will radiate electromagnetic

radiation.

To find the expectation value of position

dxxbdxtiE

tiE

xba

dxtiE

tiE

xabdxxa

dxbbaabax

dxbabaxx

mmm

mn

n

nn

mmnn

mmnmnmnn

mnmn

2

2

22

]).[(exp]).[(exp

]).[(exp]).[(exp

)(

)()(

In special case for nmmnandabba

Then, the 2nd

and the 3rd

terms can be added as

tth

EEt

EE

dxxtEE

ba

eegU

dxtEE

itEE

ixba

nmnm

mnnm

ii

nmnmmn

2cos)(2cos)(cos

)(cos2

cos2sin

]})(.[exp])(.[{exp

So, the electron oscillating between two states in time t will have frequency of h

EE nm

and its expectation value for its position is given by:

dxxtbadxxbdxxax mnmmnn

2cos222

When, the electron in state n, so b=0, while when the electron in state m, a = 0. In both cases

the expectation value of electron position will be fixed and dose not change with time. But,

when the electron oscillate between these two states (n) and (m), then its oscillation between

these two states will be with a frequency h

EE nm , same as the frequency of photon (

radiation) given by Bohr theory.

Ex.1: Prove that the most likely distance from the origin of an electron in the n=2, ℓ=1 state is

4a0. The radial probability density is ).(exp24

1

0

2

0

2

3

0

2

)(a

r

a

r

arP r

.

Sol.:

To find its maximum ( most likely) we take the 1st derivative

.4.max

0]4[24

1

0])1

(4[24

1

)(24

1

0

0

43

5

0

0

43

5

0

4

5

0

)(

0

00

0

arisayieldsthatsolutiononlyThe

a

rre

a

ea

rera

erdr

d

adr

dP

ar

ar

ar

ar

r

2

2

2

2

2

2

2

2

2

2

2

2

2

22

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

22

4

3]

4

411[

8][

8

4

3]

4

4164[

8]

4

221[

8][

8

xxxxz

z

y

y

x

x

xxxxxz

z

y

y

x

x

Lm

h

LLLm

h

L

n

L

n

L

n

m

hE

Lm

h

Lm

h

LLLm

h

L

n

L

n

L

n

m

hE

2/12/3

32

11

11

}!)1(2

!)({

!)12([),22,1(

112/

),22,1(11

2/)(

),,(11

2/1)(

......!3)2()1(

)2)(1(

!2)1(

)1(

!11),,(

nn

nNnFeN

FeNR

caFeu

x

ccc

aaax

cc

aax

c

axcaF

functiontrichypergeomeconfluenttheisF

2/]

4

11[)(

...]04

11[),4,1(

11,1,3)(

),22,1(11

2/)(

31

31

eNrR

FnrR

nFeNrRn

3

2/3329

]2

16

1

2

5

0

4[23

2]2

16

1

21[

0

23

22

0

2|)(31

|1

NN

deN

deN

drrrRionNormalizat

oo a

Zrr

na

Zr

e

oa

ZrR

eerR

3

22

3/]

6

2[2/3)(

627

8)(

31

3/]

61[.2/3

9

22/]

41[.

3

2/3)(

31

2/]

6

11[2)(

42

]!6

1(...0!16

11[),6,1(

11,2,4)(

42

eNrR

FnrR