1

Lecture Notes PH 411/511 ECE 598 A. La Rosa

INTRODUCTION TO QUANTUM MECHANICS ________________________________________________________________________

CHAPTER-11 The SCHRODINGER EQUATION in 3D

Description of the motion of two interacting particles

11.1 General case of an arbitrary interaction potential

11.2 Case when the potential depends only on the relative position

of the particles U )( 21 rr

U

Decoupling the Center of Mass motion and the Relative Motion

The CM variable and relative position variable

Solution by separation of variables

Equation of Motion for the Center of Mass

Equation governing the Relative Motion

11.3 Central Potentials )( rUU

11.3A Separation of variables method ),()( YR r

11.3B The angular equation

The Legendre Eq.

11.3C The radial equation, using the Coulomb Potential

Radial wavefunctions of the bound states (E’ < 0 )

References:

"Introduction to Quantum Mechanics" by David Griffiths; Chapter 4

2

CHAPTER-11

SCHRODINGER EQUATION in 3D Description of two interacting particles’ motion One particle motion

In the case in which a particle of mass m moves in 1-D and inside

a potential V(x,t) , the Schrodinger Eq. is,

),( 2

2

22

txVxmt

i

, where = (x, t) (1)

When the particle moves in the 3-D space, the equation adopts the form,

),(2

22

tVmt

i r

, where = ( r

, t) (2)

here 2 stands for the Laplacian operator

_________________________

In Cartesian coordinates:

2

2

2

2

2

22

zyx

In spherical coordinates:

2

2

2 2

2

2

1

1

12

2sinsin

sin rθ

θrrr

rr

X

Z

Y

r

(x, y, z, t)

3

11.1 General case of an arbitrary interaction potential

Two particles motion

X

Z

Y

r1

1r

=(x1, y1, z1, t)

r2

2r

=(x2, y2, z2, t)

m1 m2

The mutual interaction between the two particles is specified through

the potential ),,( tV 21 rr

and the corresponding Schrodinger Eq. is

),,(22

22 22

tVmmt

i2111

21 rrrr

, (3)

where = ( ,, 21 rr

t)

In the expression above the following notation is being used,

)(

,

, 1111 zyx

r ;

222

2

2

1 11 zyx1

r ; etc

11.2 Case when the potential depends only on the relative position of the particles

Here we consider that the potential ),,( tV 21 rr

depends only on

the relative position,

),,( tV 21 rr

= ),( tU 21 rr

(4)

Decoupling the Center of Mass motion and the Relative Motion

For this particular case it can be shown that the motion of the two particles can be decoupled into

the motion of the center of mass, and

the motion relative to each other.

The CM and the relative-position variables

Such decoupling can be achieved through the following change of

variables: )( ),( r ,Rrr

21 , where

4

rr

R21

2211

mm

mm

, 21 rr r

(5)

( ,, 21 rr

t) = ( ,, rR

t)

Notice, by calling

),,( zyx RRRR

and ) , ,( zyx r

the definitions in (5) can be re-written more explicitly as,

M

xmxmR 2211

x

,

M

ymymR 2211

y

,

M

zmzmR 2211

z

)( 21 xxx , ) ( 21 yyy , ) ( 21 zzy (6)

( ,, 21 rr

t) = ( ,, rR

t)

where we are using 21 mmM

_________________________

You may want to jump to expression (10), which gives the form in which Eq. (3) is transformed upon the application of the change of variables defined in (5).

( ,, 21 rr

t) = ( ,, rR

t) implies,

xx

x M

m

Rxx

R

Rx x

x

x 111

1

11x1x1xx

x

xxx

xR

xR

m

x

x

Rx

xR

xRRx Mx1

)

)()(()(

2

2

1

)()((

2

2

2

2

)

xM

m

xR

m

RM

m

xRxx

Mx

11 1

2

22

2

2

)

( 2)(

xxRM

m

M

m

xR x

11

2

22

2

2

2

2

)

( 2)(

yRyM

m

M

m

Ry yy1

11

5

2

22

2

2

2

2

)

( 2)(

zRzM

m

M

m

R zz1z

11

Since 2

2

2

2

2

2

2

111 zyx

1r , adding the last three results gives,

)(

2 )( 22

zyx RzRyRxM

m

M

m

11 2rr R

1

RR rr

M

m

M

m 11 22 2 )( 2

where we have used the notation

)

,, zyx RRR

R ,

2

2

2

2

2

22

zyx RRR

R and

)

,, zyx

r ,

222

2222

zyx

r

Very conveniently, the previous results is expressed as,

RR rrr

MM

m

mm

12

2

222 1

11

11

1 (7)

Similarly

xx

x M

m

Rxx

R

Rx x

x

x 222

2

22x2x2xx

x

xxx

xR

xR

m

x

x

Rx

xR

xRRx Mx2

)

)()(()(

2

2

2

)()((

2

2

2

2

)

xM

m

xR

m

RM

m

xRxx

Mx

22 2

2

22

2

2

)

( 2)(

xxRM

m

M

m

xR x

22

6

Analogous results for 2

2

2y

and 2

2

2z

lead to,

RR rrr

MM

m

mm

12

2

222 2

22

11

2 (8)

Replacing (7) and (8) in (3) one obtains,

),()(22

22 11

22

2tU

mm

mm

ti

21Mr

rR

21

In terms the reduced mass

21 mm

11

1 (9)

one obtains the equation

),(

22

22 22

tUt

iM

rrR

(10)

where = ( ,, rR

t)

Solution by separation of variables

We will specialize in the case where the potential U does not depend

explicitly on the time t. For such a case, we will be looking for

solutions to Eq. (10) in the form,

( ,,rR

t) ≡ tEE

CMi

eF )(/(

)()()

rR (11)

tt EE ii

ee FCM

))) /()(

(/()(

rR [[

Replacing (11) in (10)

)()()()()(2

)()(2

)()( )(222 2

rrrrr RRRRrR

FUFFF

MEECM

7

Dividing by )()( rR

F

)()()(2

)()(2

)(222 2

rrr r

RR

R

UF

FEE

MCM

This expression indicates that the terms on the right, although

intrinsically depending on the variables r

and R

, they add up to a constant value. Selectively we choose,

)()(2

22

RR

R

M

CME , (12)

and

)()()(2

22

rrr r

UE F

F

(13)

Equation of Motion for the Center of Mass

The first expression leads to,

)(

)(2

22

RRR

CMEM

(14)

which admit solutions of the form

RnR

)( CM 2)( /

Ei M

e (15)

where n

is an arbitrary constant unit vector. (That means there are many solutions of this type, one for each selected unit vector n

.)

In terms of the wavevector nk

2

CM

CMME

, the previous expression

is written as,

)(R

=R

CM

k

(

.)ie (15)’

In short,

if a single vector CMk

were used in the separation of variables

given in (11), then the motion of the CM would be described by a

plane-wave that propagates in the direction of CMk

,

/ (/(

)(][ )(

CMCM

.)) t-t EiEee

i

RkR

CM (16)

8

Accordingly, the CM is likely to be anywhere in the space but has a very definite orientation of its momentum (that given by

CMk

).

If a more precise localization of the CM were desired, then we can

select a bunch of wavevectors nk

CM

2CM

EM within a range k

and define a wavepacket

] )/( [ )( CMCM

tEit

-e,

Rk

kCM

R

. (17)

That way the CM will be more localized (because )( t,R

is more

spatially localized) but, at the same time, there will be a corresponding uncertainty in its momentum.

It is worth mentioning that we have not been looking for a solution

)(tRR

, as we would have been doing in a classical mechanics

approach; rather we have found an amplitude probability )(R

.

Equation governing the Relative Motion

Back to Eq. (13). This leads to,

)()()()(2

22

rrrrr

FFF EU

(18)

In the next section we will specialize for the case for central potentials; that is, when the potential depends only on the magnitude

of the particles separation: )()( rUU r

11.3 Central Potentials )()( rUU r

)()()()(2

22

rrrr

FFF ErU

(19)

This is an equation for )(r

F where the energy parameter E is still to

be determined. When working with central potentials, it is more convenient to use spherical coordinates to solve (19).

At the beginning of this Chapter the Laplacian operator was given in spherical coordinates. Using that expression in (19) one obtains,

9

)(

1

1

2

1

2 2

2

22

22

r F

2sinrθsin

θsinrrr

rr

)()()( rr

FFU Er (20)

11.3A Separation of variables method

We will look for solutions of the form

),()( )( YRF rr

(21)

This leads to,

)(),(

2

1

22

2

r

rr

rrRY

),()(

2

2

22

2

1

1

2

YR2sinr

θsinθsinr

r

),()( ),()( )( YRYRU rErr

Dividing each term by ),()( YR r

)(

1

)(

1

22

2

2

r

rr

rrrRR

),(

2

2

22

2

1

1

),(

1

2

Y2sinr

θsinθsinrY

)( ErU

)(

)(

1

22

2

r

rr

rrRR

ErUr )(2 +

),(

2

22

1

1

),(

1

2

Y2sin

θsinθsinY

0

10

)( )(

1 2

r

rr

rrRR E)(

2

2

2

rUr

+

+ ),(2

2

1

1

),(

1

Y2sin

θsinθsinY

0

The first two terms depend only on r, while the third term only on the

angular variables. This means that the radial part and the angular part are each constant.

),(2

2

1

1

),(

1

Y2sin

θsinθsinY

-

)( )(

1 2

r

rr

rrRR

)(

2 2

2ErUr

Rearranging terms, one obtains

),(),(2

2

1

1

YY2sin

θsinθsin

Angular Eq. (22)

and

)( )( 2

)(

1

2

2

2

2

rr RR ErUrr

rrr

Radial Eq. (23)

In summary, we are looking for solutions to Eq. (20) in the form

),()( )(

YRF rE

r

11.3B The angular Equation

To solve the angular equation

),(),(2

2

1

1

YY2sin

θsinθsin

we will look again for solutions of the form of separated variables,

)( ))( (, Y (24)

11

)( )(

1

θsin

θsin + )(

2

2

1

2sin)( = )( )(

Dividing by )( )( and multiplying by sin2()

[)

1

()

(

θsinθsin + sin

2() ] +

)(

1

2

2

)( = 0

This means that each of the two term must be constant

[)

1

()(

θsinθsin + sin

2() ] = m2

(25)

)(

1

2

2

)( = - m2

(26)

The value of the constant m2 is determined by physics-based

considerations.

Indeed, from the second Eq. above one obtains

2

2

d

d)( + m2

)( = 0

(where the partial derivative symbol has been dropped, since the function depends only on one variable,) which has solutions of the

form )( =ei

m

. But notice that the wavefunction should take the

same value whenever is incremented by 2; that is,

)( = )( (physics-based requirement)

This implies ei

m(

= ei

m

, or

ei

m

= 1

This condition can be satisfied only if m is an integer number. Thus,

the dependence of the wavefunction on the variable is given by,

)(m

=2

1 ei

m

; m = … , -2, -1, 0, 1, 2, … (27)

12

where the factor ( 1/ 2 ) has been introduced to ensure the ortho-

normality condition of these functions in the range 20 ,

mmmm '

2

0'

)()(*

(28)

We will see below some restrictions on the allowed range of values for m (see expression (42).)

X

Z

Y

r

(x, y, z)

The other angular Equation (25) has the form,

0)(

1)(

2

2

msin

mθsin

θsin dd

dd

(29)

where it has been emphasized that the solution depends on the

parameters and m. The task is to find a solution over the range 0 ≤

≤ .

It is convenient introduce the new variable

w = cos(), for 0 ≤ ≤.

and the new function

Gm (w) = )(

m

It can be shown that the Eq. for Gm is,

0

1

2

)-(1 2

2

2

2

2

m

mm Gw-

m

ww

w

Gw

d

dG

d

d

(30)

13

Case m=0

We will see later that the functions Gm corresponding to m≠0 can be

built upon G0 . The latter satisfies the Eq.

0 2 )-(1 0

0

2

0

2

2

Gd

dG

d

Gd

ww

ww Legendre Eq. (31)

Power series solution

G0 (w) = k

k

0k

wc

(32)

From (31) one obtains,

0k

kwk

ckwd

dG1

0 ,

0k

kwk

ckwd

dGw 2

2 0

, which gives,

0

0

2

Gw

Gw

0k

kwk

ck 2 + k

k0k

wc

=

0k

k

kwck )2( (33)

Also,

0k

kwk

ckkwd

Gd2

2

2

)1(

0 ,

0k

kwk

ckkw

Gw )1(

2

2

2 0

)-(1 2

0

2

2 wd

Gdw

0k

kwk

ckk 2 )1( -

0k

kwk

ckk )1(

=

2k

k

kwckk 2

)1( -

0k

k

kwckk )1(

Using k - 2 = k’

=

0k

k

kwckk

2 )1)(2( -

0k

k

kwckk )1(

where we have replaced k’ for k again (to avoid writing too many variables)

14

= kwckkckk

0kkk] )1()1)(2[(

2

(34)

Replacing (33) and (34) in (31), one obtains

kwckkckk

0kkk] )1()1)(2[(

2

+

0k

k

kwck )2( = 0

kwk

ckkckk

0kk

] )1([ )1)(2[( 2

= 0

This requires

kk ckk

kkc

)2)(1(

)1( 2

recursion relation (35)

Thus, the solution (32) can be written more explicitly as,

G0 (w) =

...

)2)(1(

)1)(0(

)4)(3(

)3)(2(

)2)(1(

)1)(0(1 42

0wwc (36)

+

...

)3)(2(

)2)(1(

)5)(4(

)4)(3(

)3)(2(

)2)(1(1 53

1wwc

where 0

c and 1

c are arbitrary constants. Expression (31) provides two

independent solutions. Notice, each subsequent coefficient 2k

c

depends on the value of the previous one k

c (for example, if one

coefficient were equal to zero, then all the other subsequent coefficients will be null as well.)

Divergence at w =1. For large values of k, this series varies as

)2/( / 2 kkcc kk , a behavior which is similar to the divergent

series

0k

k)/1( . This indicates that (32) diverges for w=1, thus

constituting an unacceptable solution, since the wavefunction has

to have a definite value at =90o.

Polynomial solutions. It is still possible to obtain a satisfactory

solution out of (32), if were chosen of the form = l (l +1), with l

being an integer value, then the coefficient series (32) would

15

become a polynomial of degree l (with the corresponding 0

c or 1

c

conveniently chosen equal to zero, depending on whether l is odd

or even, respectively.)

Summary:

The Legendre Equation

0

2 )-(1 0

0

2

0

2

2

Gwd

dGw

wd

Gdw Legendre Eq.

admits physically acceptable solutions provided that has

the form of = l (l +1), with l being an integer. For a

given l, the resulting solutions is a polynomial of degree l.

G0 (w) = )(wwc Pkk

0kl

l

Legendre Polynomial .

where kk ckk

kkc

)2)(1(

) 1 ( )1( 2

ll

Since an arbitrary multiplicative constant, 0

c or 1

c ,

accompany to these solutions, the Legendre polynomials are defined such that

Pl (1) =1

From expression (36) one obtains,

l = 0: 1)(0

wP (37)

l = 1: wwPl

)(

l = 2: )31()( 2

2 2

1wwP

l = 3: )3

5(

2

3)( 3

3wwwP

l = 4: )3

35101(

8

3)10

6

7101()( 44

04wwwwcwP

…

16

Since Pl is a polynomial of degree l,

Pl (-w) = (-1)l

Pl (w) (38)

Without proof we state that the Legendre polynomials satisfy,

'

1

1 12

2)()( lll'l l

dwww PP (39)

The general case: m≠ 0

We are looking for solutions to Eq. (29), which is independent of the

sign of m.

0

1

2

)-(1 2

2

2

2

2

m

mm Gw-

m

ww

w

Gw

d

dG

d

d

For that purpose, let’s define the Associated Legendre functions

)(

)-(1)(

2/2 wmw

mmwwm PP

d

dll II

IIIIII

(40)

That is, the polynomials )( wmP II

lare obtained by taking derivatives of

the already known Legendre polynomials )(

wPl

given in (37). It turns

out these Associated Legendre functions satisfy Eq. (29),

0 2

2

2

22

1)1(

2

)-(1

II

IIIIm

w-

m

w

m

ww

m

w PPP

d

d

d

d

lllll (41)

Notice that, since )(

wPl

is a polynomial of degree l , )(wmP II

l will be

identically equal to zero for lIIm . Accordingly, for a given l there

will be (2 l + 1) associated Legendre polynomials

m = - l , - l +1, … , l -1, l (42)

Using the Legendre polynomials given in (37) one obtains,

17

l = 0, m= 0: 1)()( 00

0 ww PP (43)

l = 1, m= 0: www PP )()( 10

1

m= 1: 2/121

2/1211 )-(1)()-(1)(

wwww PP

dwd

l = 2, m= 0: )31(2

1)()( 2

20

2 www PP

m= 1: wwwww PPdwd

3)-(1)()-(1)( 2/122

2/121

2

m= 2: 3)-(1)()-(1)( 222

22/222

2

wwww PP

dw

d

Since )(

wmw

mP

d

dlII

IIis a polynomial of degree ( l - IIm ) and 2/2 )-(1 IImw ,

is an even function, then

)( )(-1)(- wmmwm PP IIIIII -ll

l (44)

Without proof we state that the associated Legendre polynomials satisfy,

'

1

1 )! (

)! (

)12(

2)()(

lll

l

ll'l

II

IIIIII

m

mdwww mm PP

(45)

This expression will be useful for properly normalizing the wavefunction.

Summary:

The angular equation (22)

),(),(2

2

1

1

YY2sin

θsinθsin

admits solutions of the form:

)( )(),(

mm

Y l

18

( ml

= cosmwm PP (( IIII

ll , l = 0, 1, 2, 3, …

)(m

= 2

1 ei

m

; m = - l, … , -1, 0, 1, … , l

A proper constant in front of )( )( is selected so that the solutions

are normalized. The resultant functions are called the spherical harmonics.

Spherical Harmonics

,i

2

1)os(

)!(

)!(

2

)12()1(),( e

2/1

,

mcm

m

mm PYm ll l

ll

for 0m

(46)

0 ,),( )1(),( *, ,

m formm

YY m ll

The factor (-1)m

is chosen for convenience (other authors’ notation

differ by a factor of (i)l .)

The spherical harmonics ),( ,

ml

Y constitute an orthonormal set of

functions,

''

),(),(

mm

spaceangularmm'

YYd llll'

*

Using sin ddd ,

' ),(),(

'sin

0

2

0mmmm'

YYdd

llll'

* (47)

The following website is very helpful for visualizing the spherical harmonic functions:

http://www.vis.uni-stuttgart.de/~kraus/LiveGraphics3D/java_script/SphericalHarmonics.html

l = 0, m=0

19

2/10,0)4(

1),(

Y

l=0, m=0 l=0, m=0

Absolute value of real part Absolute value,

colored by complex phase

____________________________________

l = 1, m=0

cos2/1

0,1 4

3),(

Y

l = 1, m= 0

Absolute value Absolute value,

The two different colors indicate the different sign taken by the wavefunction

____________________________________

l = 1, m = ± 1

ieY

8

3),( sin

2/1

1,1

20

l =1, m=1 l =1, m=1

Absolute value of real part Absolute value,

cossin 8

32/1

colored by complex phase

l =1, m= -1

Absolute value of complex part

sinsin 8

32/1

____________________________________

)1cos3( 22/1

0,2 16

5),(

Y

ieY

cossin

2/1

1,2 8

5),(

2

32

15),( cossin 2

2/1

2,2ieY

21

l = 2, m= 0 l =2, m=1 l =2, m=2

Absolute value of real part (illuminated by light.)

l =2, m=0 l =2, m=1 l =2, m=2

Absolute value, colored by complex phase

____________________________________

The spherical harmonic functions, given in (46) constitute a complete set of solution of the angular equation (22). We are now left with the task of solving the radial equation (23).

11.3B The Radial Equation

We have found that in Eq. (23) that the radial equation must satisfy

)( )( 2

)(

1

2

2

2

2

rE

rE

RR ErUrr

rrr

ll l l

)( 1

22

where l takes integer values l = 0, 1,2, … (this condition comes

from the solutions required for the angular component of the wavefunction), and the values for E need to be determined.

For the case of a Coulomb field

rrU

)(

where 0

2

4

Ze , Z is the atomic number (number of

protons in the nucleus) and e is the elementary

charge,

we have,

)( )( 2

2

22

22

rr RR Errrrr

)( 1 l l (48)

Using unit-less variables r’, E’

When considering the case of a Coulomb field, it is convenient to use

2

32

,,

eee

mmm

(49)

as the units of mass, length and time, respectively. Notice that the

unit of energy is 2

2

e

m . For our purpose, instead of e

m we will use

the reduced mass instead. Thus, since

2 has units of distance,

and 2

2

has units of energy, the equation above can be expressed

in terms of unit-less parameters r’ and E’ instead of the variables r

and E, where

r ≡ r’

2 , E ≡ E’

2

2

. (50)

23

Indeed, using in addition R(r)= R( r’ ),

emr

RrrR 2

r'r'

R, … etc.,

one obtains,

0R '

1 ' 2

'

''

2

' ) (

22

2

r'

rr

)(

rrrE

1 l l (51)

Let

R(r’) = '

) (

r

'ru (52)

The purpose of this change of variable is to obtain an equation the resembles the one dimensional case (as we will verify below.)

ur'r'

u

r'r'

R2

11

u'rr

u

r'r'r' 32

R

222

r'r'r'u

r'u

r'

u

r'

u

'rr'2322

R

2

2

111 2

2

u

r'

u

r'

u

r' 32 'r' r

22

2

2

1

Adding the last two expressions,

2

2

1

22

r'u

r'r'r'r'

RR

2

0 '

'

1 ' 2

'

'

1

' '

1

) (

)(

22

2

rrrr

)(

rr

r'r' uuuE

ll

0 )( '

1 ' 2 )(

'

1

'

22

2

'ru

r'ru

r

)(

r

uE

ll

0 '

1

'

1 ' 2

' ) (

22

2

'

2r

rr

)(

r

uuE

ll

This Eq. can be expressed as,

24

0 ' 2 '

) ( ) (

eff2

2

'r'r uE

r

uV (53)

where an effective potential has been defined as,

1

) (

2 eff

'r'r

1)('r

2

llV

r’

Veff

l =0

l ≠ 0

Notice, Eq. (53) resembles the case for a one dimensional problem. However,

It has significance only for r >0, and

Must be supplemented by boundary conditions at r =0.

We will require that the radial function )(

rE

Rl

remain finite at

the origin. Since r'

)u(r')(

r

ER

l, we must require that

0(0)u (54)

Notice, in the region 'r the equation (53) becomes,

0 ' 2

) ( '

2

2

'r

ruE

u

For E’ >0, the behavior of u(r’) in this region would be oscillatory.

' '2) ( rie E'ru

25

If our interest is to find bound states, we can ensure that the wavefuntion vanishes when 'r if we require that E’<0

' '2)( rEe'ru

Looking for bound states (E’ < 0 )

Bound states will be obtained with negative values for E ’.

Let’s define,

r'

where is a parameter to be chosen conveniently later.

u(r’)=v() ,

vv

r'r'

u,

2

22

2

v2r'

u

2

2

2

v0

1 ' 2 )(

2

2

v

2

)(E

ll

2

2

v0

2

1

2

' )(

2

v)(2E ll-

-

Choosing '22 E

2

2

v0

'2

1

4

1 )(

1

2

v

E

)( ll-

Let, '2

1

Ek

2

2

v0

1

4

1 )(

2

v

k)( ll-

Summary

26

)( )( 2

2

22

22

rE

rE

RR Errrrr ll

ll

)( 1

where 0

2

4

Ze

R(r) = R(r’) = '

) (

r

'ru, with r ≡ r’

2 and E ≡ E’

2

2

, gives

0 '

1

'

1 ' 2

' ) (

22

2

'

2r

rr

)(

r

uuE

ll

u(r’)=v() , r' '22 E and '2

1

Ek

gives,

k 2

2

lv

0

1

4

1

2

k

klv

)( ll- (55)

with the condition

0(0)k

lv

The function k l

v and the values of the parameter k (associated to

the energy E) need to be determined.

Let’s figure out the behavior of )(

klv near the origin .

Assume

0

)(

j

j

jcs

klv = ... 1 1 ss cco (56)

The power s is unknown (i.e. we do not know yet has fast )(

klv

tends to zero when 0 .)

The value of s is determined by replacing the expansion (56) in (55)

and analyzing the terms of lowest power, that is s . In effect, such

procedure gives,

27

0 ... 4

1

... 1

... 1 2

... 1 2

0

10

1

1

kk

)1()1(

))(1()1(

s

ss

ss

ss

c

cc

cc

sscssc

o

o

llll

For a given value of l , this expression above imposes the following

condition on s,

0)1()1( llss (condition on s ) (57)

The possible choices for s are s = l +1 and s = - l. The second

option is discarded on the basis that such selection would not satisfy the condition 0(0)

k

lv . Thus,

s = l +1 (58)

0

)(

klv 1l

Let’s figure out the behavior of )(

klv when .

In this region Eq. (55) takes the form,

k 2

2

lv

0

4

1

klv- ,

which has solutions of the form

2

1

0)(

ekl

v (59)

Full solution of Eq, (55) The results in (58) and (59) suggest looking for solutions of the form

)(1

)(

2

1

kk llge

l

v , (60)

where

0

)(

j

j

jcg

kl and 00 c

28

gg eeeg

2

1

2

1

2

1 11

2

1)1()(

lll

l

d

d

d

dv

kl

ggge 121 )1( 1

2

1 llll

dd

gg

ggg

ggg

e

e

)1( 2

1 1

2

1

1)1( )1( 2 2

2 1

1 )1(21 1

21)(

41

2

1

2

1

2

2

ll

l

llllll

llll

dd

dd

d

d

dd

d

dv

kl

Thus

ggg

gge

11

4

1

1

-)1( )1(

] )1(2[ 1)( 2

2

2

2

2

1

lll

ll

l

llll

dd

dd

d

dv

kl

ggg --e- lll ll

ll

2

1

1 1 1

1

41

4

1 2

kk

k)(

)(lv

The addition of these last two expressions should be equal to zero,

0 )]1( [] )1(2[ 1

1

2

2

g

gg ll

lll

l

-k

dd

dd

0

)]1( [ ] 22[

2

2

k

kk -kl

ll ggg

ll

d

d

d

d (61)

According to (60), we expand the function g in a power series,

0

)(

j

j

jcg

kl , where 00 c (62)

29

0'

'

1'

1 making

1

1 )1'(

j

j

j

j-j'

j

j

jcjjc

g

dd

kl

0'

'

2'

2 making

2

2

2

2

)1')(2'( )1(

j

j

j

j-j'

j

j

jcjjcjj

g

d

dkl

1'

'

1'

1 making

2

1

2

2

)')(1'( )1(

j

j

j

j-j'

j

j

jcjjcjj

g

d

dkl

0'

'

1' )')(1'(

j

j

jcjj

0'

'

'1

'

j

j

j

or

j

j

jcjjc

g

dd

kl

Replacing the expression on the right side in (61), and renaming j’

back to j, we obtain

0 )1](22[ )]1( [ ))(1( 11 jj cccc jjjj jj

ll -k

0 )1](22[ )]1( [ 1 jcc jj jj ll -k

We obtain the following recurrence formula

jccjj

jj

)1() 2(2

)1( 1

l

l k- (63)

For large j, 1

1

1

jjc

c j. This convergence is similar to the expansion

of the function j

jj

e

!

1, which is not acceptable in this case, since

the wavefunction should tend to zero in the region .

One way to obtain a useful solution is to find the conditions under which the series (62) gets truncated into a polynomial.

30

Expression (63) indicates that that is possible if k (the term that determines the energy of the system) were equal to an integer number. Indeed,

For a given l, by selecting

k = n (with 1 ln ), (64)

lkg becomes a polynomial lng of order (n – l – 1)

For a given n, the possible values for l are

l = 0, 1, 2, 3, …, (n-1); (65)

That is, different solutions will be associated to the same energy level.

From (55), '2

1

Ek

. An integer value for k= n implies that the

allowed values for the energy are discrete and,

22'

1

nnE

From (50), E ≡ E’2

2

where

0

2

4

Ze . Thus,

22

2

22

2 1

24

1

2)(

0

2

-

n

Ze

nnE

, (66)

where n = 1, 2, 3, …

In the usual spectroscopy notation, the energy levels are specified by two symbols:

The first gives the value of the principal quantum number n.

The second is a code letter (s, p, d, …) that indicates the value

of the orbital angular momentum l (0, 1, 2, ... respectively.)

31

-3.4

-13.6

-1.51

1s

2s

3s 3p

2p

3d

E(eV)

l=0 l=1 l=2 l=3 l=4

0

Radial wavefunctions of the bound states (E’ < 0 )

R(r) = R(r’) = '

) (

r

'ru=

r'k

r'

)( )(1

2

1

l

ge

lv

where n

'2

1

Ek , r'r'

2'22

n E ,

))

(2(

1

2

1

2

1

ll

nn g

ge

e

nn

ll

/2)(

)(r'R

where r' 2

n

, r ≡ r’

2 and

0

2

4

Ze ;

The expression for is further expressed as

)2 2

0/(4

0

2

2ar

n

ZZe

nr

where ma

e100 10529.0

4

2

2

0 x

That is,

32

)(

2

1

ll nngR er

l

)( )(

)/(

ln

go

an

Z re

l

(67)

where )2

0/( arn

Z and

2

2

40

0 ea

0

)(

j

j

jc

ng

l, with 00 c and jcc

jj

j nj

)1)( 2(2

)1( 1

l

l -

We have omitted the coefficient in front of the exponential term in

Rnl , since it will be absorbed by the coefficient co from the

polynomial gnl.

Case n=1

According to (65), the only possible value for the orbital angular

momentum is l = 0.

Also, from (67),

)1)( 2(

)1)( 2(

)1(

)1)( 2(2

)1( 1 jjj cccc

jj

j

jj

j

jj

j nj

1--

l

l

For j=0 the recurrence formula gives c1 =0. Hence cj =0 for any j>0.

Thus g10() = co.

Expression (67) gives, 0000 1

24

0

2

1

2

2 ccr

cRrZe

eeer

)( ;

0 /

00 1

arcR

Zr e)( (68)

where the result has been expressed in terms of the Bohr’s radius

mae

100 10529.0

4

2

2

0 x

(69)

33

The condition of normalization requires, 12

00 1

drrRr

r2

)( . Using

(68) one obtains, 12

0

22)( 0

/

0

rc

r

rZ ae ; 13)/(

4

12)( 00 Zac ;

3)2)( 00 /(4 ac Z ; and finally 2/3)0/(20 ac Z . Thus, (68) becomes,

0

2/3

0

/

20 1

ar

aR

ZeZr

)( (70)

The radial distribution function Dnl,

20 110

rRD rr2

)()( (71)

gives the probability per unit length that the electron is to be found at

a distance r from the nucleus.

R10 r2R10

a0 r r

This radial solution (71) is complemented with the spherical

harmonic corresponding to l = 0 and m=0 (given in expression (46),

2/10,0)4(

1),(

Y . Thus, the solution to the Schrodinger Eq. (19) is

given by, 2/10,0

)4(

1 /),( , 02/3 )(2

00 100 1

,

ar

YRFZ

a

Z err

)()( .

02/3 /1 ][

0

,,00 1

arrF

ZeaZ

)( (72)

34

Case n=2

According to (65), the only possible value for the orbital angular

momentum is l = 0, 1.