The Schr¶dinger Equation in Three Dimensions

The Schrödinger Equation in Three Dimensions

Introduction

In three dimension the Schrödinger equation can be expressed by the equation

� f Ð<ß >Ñ � Z < Ð<ß >Ñ œ 3h Ð<ß >Ñh `

#7 `>t t t t

##G G G� � (10.1)

Just as we showed in one dimension, the wave function can be factored to giveGÐ<ß >Ñt

G <Ð<ß >Ñ œ Ð<Ñ/t t �3 >= (10.2)

where is the solution of the three-dimensional time-independent Schrödinger<Ð<Ñtequation

� f Ð<Ñ � Z < Ð<Ñ œ I Ð<Ñh

#7t t t t

##< < <� � (10.3)

We can also express this equation in operator form, by defining the three-dimensional

momentum operator (consistent with the one-dimensional definition) by the equation

: œ �3Äss s shf œ �3h / � / � /

Ä ` ` `

`B `C `DŒ B C D (10.4)

and the position operator by<Äs

< / B � / C � / DÄss s sœ B C D (10.5)

With these definitions the kinetic energy operator is given by

X œ f † f œ � fh h

#7 #7

Ä Äs

s s: † :Ä Ä

#7œ �

# ## (10.6)

The Free Particle in Three Dimensions

The three-dimensional free particle Schrödinger equation becomes

� f Ð<Ñ œ I Ð<Ñh

#7t t

f Ð<Ñ œ � I Ð<Ñ œ �5 Ð<Ñt t t#7

h

##

# ##

< <

< < <

(10.7)

To solve this last equation, we assume that each dimension is independent so the wave

function is factorable into three independent components

< <Ð<Ñ œ ÐBß Cß DÑ œ \ B ] C ^ Dt � � � � � � (10.8)

Three Dimensional Motion and Angular Momentum 2

The Schrödinger equation then becomes

Œ � � � � � � � � � � � �` ` `

`B `C `D� � \ B ] C ^ D œ �5 \ B ] C ^ D

2 2 2

# # ## (10.9)

or, dividing left and right sides by we obtain\ B ] C ^ D� � � � � �" ` \ B " ` ] C " ` ^ D

\ B `B ] C `C ^ D `D� � œ �5� � � � � �� 2 2 2

# # ## (10.10)

If we let we can write5 œ 5 � 5 � 5# # # #B C D

” • ” • ” •� � � � � �� " ` \ B " ` ] C " ` ^ D

\ B `B ] C `C ^ D `D� 5 � � 5 � � 5 œ !

2 2 2

# # #B C D# # # (10.11)

which can be valid for all values of , , and only if each of these brackets areB C Dindependently zero! This leads to three independent equations of the form

" ` \ B

\ B `B� 5 œ !� � � �2

# B# (10.12)

with a solution of the form

\ B œ E/� � 35 BB (10.13)

Since a free particle cannot be reflected, the solution is valid for any possible value of ,5Beither positive or negative – but not both at the same time. The general three-dimensional

solution, then, has the form

< T TÐ<Ñ œ \ B ] C ^ D œ / œ /t (10.14)� � � � � � 3 5 B�5 C�5 D 35†<t t� �B C D

where is the wave vector pointed in the direction of propogation of the wave with a5Ä

magnitude given by

l5l œ 5 � 5 � 5 œt #7I

h# # # #

B C D #(10.15)

The fact that the energy of the system can be written in terms of three differentIindependent quantities

I œ 5 � 5 � 5h

#7

#

B C D# # #ˆ ‰ (10.16)

means that there is in this system. The degerneracy same value of the energy can be

obtained with different combination of quantum numbers associated with the eigenvalues

5 5 5B C D, , and .

From our one-dimensional treatment, we know that , so we can write: œ h5B B

: œ h5Ä Ä(10.17)


which means that we can write the solution to the free particle in terms of the momentum

vector as

< TÐ<Ñ œ /t (10.18)3:†<Îht t

The time-dependent solution, then, is given by

< TÐ<Ñ œ / /t (10.19)3:†<Îh �3 >t t =

where .= œ IÎh œ : Î#7h#

Just as in the one-dimensional case, this wave function is not strictly allowed,

since it is not normalizable. We can get around this is some cases by looking as the

probability current density, which is now defined in relation to the individual components

of the momentum. In general, however, we must form an acceptible solution by utilizing

Fourier transforms to obtain

< T

1

Ð<ß >Ñ œ : / / .:t t t"

# hŠ ‹È ( � �$3:†<Îh �3 >t t = (10.20)

The Three-Dimensional Infinite Potential Well

As another simple example of a three-dimensional problem, we will consider the

infinite potential well. This problem is similar to the infinite square well. The potential

energy within a region defined by

! Ÿ B Ÿ P

! Ÿ C Ÿ P

! Ÿ D Ÿ P

B

C

D

(10.21)

is zero, while the potential energy outside this region is infinite. This confines a particle

to the volume of space described above, so the wave function must go to zero at the

boundaries.

Inside the “well” Schrödinger's equation is identical to that of the free particle and

reduces to

” • ” • ” •� � � � � �� " ` \ B " ` ] C " ` ^ D

\ B `B ] C `C ^ D `D� 5 � � 5 � � 5 œ !

2 2 2

# # #B C D# # # (10.22)

or more precisely to three independent equations of the form

" ` \ B

\ B `B� 5 œ !� � � �2

# B# (10.23)

The only difference between the free particle and the infinite “well” problem is the

boundary conditions. The wave function must go to zero at the boundaries of the well.

This means that reflection is possible so that the most general solution to the equation\above is given by

\ B œ E/ � F/� � 35 B �35 BB B (10.24)


The boundary condition

\ ! œ !� � (10.25)

requires that

E œ �F (10.26)

so that is given by\ÐBÑ

\ B œ =38 5 B� � � �T B (10.27)

where is complex in general. The additional boundary condition applied atT œ EÎ#3B œ PB gives

\ P œ =38Ð5 P Ñ œ !� �B B BT (10.28)

implying that

5 P œ 8 Ê 5 œ 8 ÎPB B B B B B1 1 (10.29)

which finally gives

\ B œ =38 8 BÎP� � � �T 1B B (10.30)

The solution has the same form for each independent coordinate, giving for <Ð<Ñt

< U 1 1 1Ð<Ñ œ =38 8 BÎP =38 8 CÎP =38 8 DÎPt � � � � � �B B C C D D (10.31)

with

5 œ 5 � 5 � 5 œ 8 ÎP � 8 ÎP � 8 ÎP œ #7IÎh# # # # # # # # # # # #B C D B B C C D D1 ‘ (10.32)

or

I œ 8 ÎP � 8 ÎP � 8 ÎPh

#78 8 8

# #

B B C C D D# # # # # #

B C D

1 ‘ (10.33)

Thus, for the three-dimensional infinite potential well there are three quantum numbers –

one associated with each of the degrees of freedom of the system. Notice that the

quantum numbers , , and ! If any of these quantum numbers were8 8 8B C D cannot be zero

zero, the wave function would be zero implying zero probablilty of finding a particle in

the well. This means that the ground state energy cannot be zero! Notice also that the

values of the quantum numbers could be positive or negative, but we obtain the same

result for and for whether the numbers are positive or negative. Thus, we will<Ð<Ñ Ittake the quantum numbers to run from 1 to positive infinity.

One new feature that arises in situations with more than one dimension is the

feature of degeneracy. When more than one quantum state has the same energy, those

states are said to be degenerate. Consider the case where our three-dimensional infinite


potential well is a cube with . The equation for the energy givesP œ P œ P œ PB C D

I œ 8 � 8 � 8 œ 8 � 8 � 8h

#7P8 8 8

# #

# B C D B C D# # # # # #

B C D

1% ‘ ‘ (10.34)

Table 9.1 lists the first 10 energy levels of the three-dimensional infinite potential well

along with the corresponding quantum numbers. Each unique set of quantum numbers

gives a unique spacial wave function, corresponding to a unique probability of locating a

particle at some point within three-dimensional space. However, several of these unique

wave functions have the same total energy. The number of different unique wave

functions all having the same total energy is called the of that particulardegeneracy

energy .level

A plot of the first 10 energy levels for the three-dimensional infinitecubic

potential well is shown in Figure 9.1, along with the degeneracy of each level.

Degeneracy is usually associated with symmetry within the system. This particular

energy level diagram and its degeneracies arise from the fact that we assumed , , andP PB C

PD were all the same. If this were the case, the degeneracy of these individual levelsnot

disappears (or at least some of the degeneracy disappears). For example, if , , andP PB C

PD are all slightly different, the energy levels are only slightly split – there wouldgenerally be three closely spaced energy levels wherever four where etc.1 œ $ß 1 œ %ß________________________________________________________________________

Problem 10.1

Use Excel to plot a bar graph of the energies of the 3D infinite potential well, were you

can vary , , and . This will give you some idea about how the degeneracy works.P P PB C D

________________________________________________________________________

Figure 10.1

0

5

10

15

20

25

30

Energy Levels

Energy Level Diagram for the 3D Infinite Potential Well

g = 1

g = 3

g = 3

g = 3g = 1

g = 6

g = 3g = 3

g = 6

g = 4


Table 10.1 Energy Values for Different Quantum Numbers

n n nB C D IÎ

" " " $# " " '" # " '" " # '" # # *# " # *# # " *$ " " """ $ " """ " $ ""# # # "#" # $ "%# " $ "%" $ # "%# $ " "%$ " # "%$ # " "%$ # # "(# $ # "(# # $ "(% " " ")" % " ")" " % ")% " # #"% # " #""

%

% # #"# % " #"" # % #"# " % #"$ $ $ #(& " " #(" & " #(" " & #(


Angular Momentum

Introduction

In three dimensions, we often encounter forces and potentials which are a function

only of the distance from the force center. To take advantage of this fact, the Schrödinger

equation for these so-called central forces can be written in spherical coordinates. The

three-dimensional Schrödinger equation

� f Ð<Ñ � Z Ð<Ñ Ð<Ñ œ I Ð<Ñh

#7t t t t

##< < < (10.35)

can be expressed in terms of , , and by writing the Laplacian in terms of sperical< ) 9coordinates:

f œ � � �` # ` " " ` ` " `

`< < `< < ` ` `

f œ f � f"

<

## #

# # # #

# # #< ß#

Œ ” •Œ sin sin

sin (10.36)) ) ) ) 9

)

) 9

where, in this last equation, we have expressed the Laplacian in terms of radial and

angular operators. The Schrödinger equation, therefore, becomes

� f � f Ð<Ñ � Z Ð<Ñ Ð<Ñ œ I Ð<Ñh "

#7 <t t t t

#

< ß# #

#” •) 9 < < < (10.37)

which we can express as

– —Œ � h f

#7 #7<� Ð<Ñ � Z Ð<Ñ Ð<Ñ œ I Ð<Ñ

� h ft t t t

# #<

# #ß

#

) 9< < < (10.38)

or

– — : Ps

#7 #7<� Ð<Ñ � Z Ð<Ñ Ð<Ñ œ I Ð<Ñ

st t t t<

# #

#< < < (10.39)

where the first term is the kinetic energy due to radial momentum, and the second term is

the kinetic energy due to angular momentum. For a system in which the radius is ,fixed

the radial term is zero and the only remaining term is the one dependent upon the angular

momentum. Such would be the case for a frictionless particle moving on a circular wire

of fixed radius, or for a rigid rotator (an example is a diatomic molecule or fixed radius).

Since the Laplacian can be separated into a radial part and an angular part, and

since the potential energy function is only a function of the radius, we should be able to

separate the wave function into a radial part times an angular part, such as

< ) 9Ð<Ñ œ V < ] ßt � � � � (10.40)


Substituting this into the last equation, and dividing by the wave function gives

– — � � � �� " : V < " P ] ß

V < #7 ] ß #7<

s� � Z Ð<Ñ œ I

st<

# #

#) 9

) 9(10.41)

If we multiply through by and move terms that are only a function of the radius to#7<#

one side of the equation, we have

P ] ß < : V <s

] ß V <œ � � #7< I � Z <

st

# #<#

#� � � �� c d� �) 9

) 9(10.42)

where

: œ � h fs

P œ � h fs< <# # #

# # #ß) 9

(10.43)

Since the Schrödinger equation must be valid for all possible values of , , and < ) 9the right-hand-side and the left-hand-side of this equation must always be equal to the

same constant which we will call . Thus, we are left with two differential equations-which we must solve

P ] ßs

] ßœ Ê P ] ß œ ] ßs

##� �� ) 9

) 9- ) 9 - ) 9 (10.44)

an eigenvalue equation for the square of the angular momentum operator, and

� � #7< I � Z < œ< : V <s

V <

#<#

#� �� c d� � - (10.45)

or

Ÿ” •� � � � � � Ÿ” •� � � � � �

:s

#7 #7<� � Z < V < œ I V <

:s

#7 #7<� � Z < V < œ I V <

<#

#

<#

#

-

j#

(10.46)

which is the Schrödinger equation expressed in terms of a rotating coordinate system

where is the radial distance from the origin and where is the square of the< œ j- #

magnitude of the angular momentum vector.

Analytical Approach

One way to make further progress in the treatment of angular momentum is to

follow the analytic approach where we need to solve the eigenfunction equation for the


square of the angular momentum

P ] ß œ ] ßs

� h f ] ß œ ] ß

� h � ] ß œ ] ß" ` ` " `

` ` `

#

# #ß

## #

#

� � � �� ” •Œ � � � �

) 9 - ) 9

) 9 - ) 9

) ) ) ) 9) ) 9 - ) 9

) 9

sin sinsin

(10.47)

Now this last equation is a function of the two angles and , ) 9 and is separable into two

independent equations. We can demonstrate this by assuming the function is] ß� �) 9separable so that , and then multiplying through by sin , and] ß œ� � � � � �) 9 @ ) F 9 )#

dividing by to obtain@ ) F 9� � � �� h ` ` � h `

` ` `� œ

# # #

##

@ ) ) ) F 9 9) ) - )

@ ) F 9� � � �” •Œ � � � �sin sin sin (10.48)

Moving all the terms that are functions only of to one side of the equation gives9

� h ` � h ` `

` ` `œ �

# # #

##

F 9 9 @ ) ) )

F 9 @ )- ) ) )� � � �� ” •Œ sin sin sin (10.49)

Now, one can show that the -component of the angular momentum vector, expressed inDterms of spherical coordinates is

P œ � 3hs `

`D

9(10.50)

so that this last equation can be written

P � h ` `sœ �

` `D

#

##F 9 @ )

F 9 @ ) ) )- ) ) )

� � � �� ” •Œ sin sin sin (10.51)

Again, if this equation is to be valid for all possible combinations of and , then both) 9sides must be set equal to the same constant, . Thus, we are left with two differential.equations which we must solve

Psœ Ê P œsD

#

D

#F 9

F 9. F 9 .F 9

� �� (10.52)

an eigenvalue equation for the square of the -component of the angular momentumDoperator, and

- ) ) ) .@ ) ) )

@ )sin sin sin (10.53)#

#

� œ� h ` `

` `� � ” •Œ � �


Dividing this last equation by sin , multiplying by , and regrouping, we obtain the#� � � �) @ )differential equation for the function:@ )� �

� h � œ" ` `

` `#

#” •Œ � � � � � �sin sin

sin (10.54)) ) ) )

) @ ) -@ )@ ) .

We first solve the eigenvalue equation for square of the -component of theDangular momentum

P œs

� 3h œ`

`

`

` hœ �

D

#

#

#

# #

F 9 .F 9

9F 9 .F 9

F 9 .

9F 9

� � � �Œ � � � �

� � � �

(10.55)

The function must be of the formF 9� �F 9� � œ E/α9 (10.56)

from which we obtain

α.##

E/ œ � E/h

α9 α9 (10.57)

or

α α. .

##

œ � Ê œ „ 3 œ „ 37h h

È(10.58)

where . This gives, for the solution7 œ ÎhÈ � �. F 9

F 9� � œ E/ � F/�37 �379 9 (10.59)

Consider the time-dependence of this solution, which would be of the form

F 9� �ß > œ E/ � F/�3 7 � > �3 7 � >� � � �9 = 9 = (10.60)

The terms on the right-hand-side represent wave functions with a phase velocity of

„ Î7= , respectively. A single particle can not move in both directions at the time,same

so each of these solutions must be eigensolutions for the problem. The superposition

simply indicates the uncertainty of knowing the direction of motion.

We require the eigenfunctions to be single valued, so that , orF 9 1 F 9� � � �„ # œ

E/ œ E/ œ E/ /�37 �37 „# �37 „37#9 9 1 9 1� � (10.61)

which is valid as long as

/ œ "„37#1 (10.62)

which requires that be an intiger. In fact, we can absorb the into the constant so that7 „


7 must be a positive or a negative integer and the eigenfunction be expressed as

F 9� � œ E/379 (10.63)

where

7 œ Îh 7 œ !ß „ "ß „ #ßâÈ. where (10.64)

or

. œ h 7# # (10.65)

The eigenfunction equation for then becomesPs#

D

P œ h 7sD

# # #F 9 F 9� � � � (10.66)

or

P œ h7 7 œ !ß „ "ß âsD F 9 F 9� � � � where (10.67)

The function now must satisfy the equation@ )� �� h � œ

" ` ` h 7

` `#

# #

#” •Œ � � � � � �sin sin

sin (10.68)) ) ) )

) @ ) -@ )@ )

or

� � � œ !" ` ` 7

` ` h” •Œ � � � � � �sin sin

sin (10.69)) ) ) )

) @ ) @ )@ ) -

# #

#

The solution to this differential equation can be determined using a power series solution

if we make use of the change of variable

' )œ cos (10.70)

since

`0 `0 ` `0 " `0 `0

` ` ` ` ` `œ † œ Ê œ

) ) ) ) ) ) )

))� � � � � ��

cos cos sin cos

cos sin (10.71)

Writing the differential equation in the form

� � � œ !" ` ` 7

` ` h” •Œ � � � � � �sin sin sin

sin(10.72)

) ) ) ) )

) @ ) -@ ) @ )

# #

# #

we make the change of variables to obtain

” • Œ Œ ˆ ‰ � � � � � �` `T 7

` ` h " �" � � � T œ !

' ' '' '

' -## #

#

(10.73)


The solution to this differential equation must be continuous, square-integrable, and

single-valued throughout the domain of , which is from to . We will make the' �" �"further simplification at this point of setting , and attempt a solution to the7 œ !equation

” •� � � � � � � �` T `T ` T

` ` ` h� # � � T œ !

# #

# # ##' ' ' -

' ' '' ' ' (10.74)

of the form

T œ -� � �' '8œ!

88 (10.75)

where

` T

`œ 8 8 � " - œ 8 � # 8 � " -

� # œ � # 8 - œ � # 8 -`T

`

� œ � 8 8 � " - œ � 8` T

`

#

#8œ!

8 8 �#8�# w w 8

8 œ!

8œ! 8œ!

8 88�" 8

# # 8�##

#8œ! 8œ!

8

� � � ��

'

'' '

' ' ' ''

'

' ' ''

'

w

ww

� �8 � " -88'

(10.76)

giving

�œ � �� ” •8œ!

8�# 8#88 � # 8 � " - � #8 � 8 8 � " � - œ !

h

-' (10.77)

In order for this equation to be valid for all possible values of , the coefficients of each'power must be zero, giving

� �� ” •8 � # 8 � " - � #8 � 8 8 � " � - œ !h

8�# 8#

-(10.78)

or

- œ - œ -# 8 � 8 8 � " � Îh 8 8 � " � Îh

8 � # 8 � " 8 � " 8 � #8�# 8 8

# #� � � � � � � �� - -(10.79)

Notice that as , so that the infinite sum would blow up. This means8p∞ - Î- p "8�# 8

that we must terminate the infinite series at some point. Thus, for a particular value of ,8say , we must have that , orj - œ !j�2

j j � " � Îh œ !� � ˆ ‰- # (10.80)


giving

- œ h j j � "# � � (10.81)

Development of Lagendre Polynomials

The Associated Lagendre Polynomials

Operator Method

We showed above that the Schrödinger equation can be written as

– — : Ps

#7 #7<� Ð<Ñ � Z Ð<Ñ Ð<Ñ œ I Ð<Ñ

st t t t<

# #

#< < < (10.82)

For the central force problem, the potential energy term is a function only of | and the<lttorque is zero, which means that the angular momentum is a constant of the motion. We

know that for any quantum mechanical operator,

.

.>ßØ > E > Ù œ Ø > E > Ù � Ø > L E > Ù

` 3

`> h< < < < < <� � � � � � � � � � � �’ “| | | | | | (10.83)s s ss

This equation tells us that for the expectation value of the angular momentum operator to

be zero, that operator must explicitly depend upon the time, and commute withnot must

the Hamiltonian. We therefore need to define the angular momentum operator and varify

that it is time independent and that it commutes with the Hamiltoninan.

We begin with the classical definition of the angular momentum

P œ < ‚ :Ä

t t (10.84)

The three components of the corresponding to this vector we expect to be of theoperator

form

P œ C: � D:

P œ D: � B:

P œ B: � C:

s ss ss

s ss ss

s ss ss

B D C

C B D

D C B

(10.85)

In one dimension, we found that the commutator

c dBß :s sB œ 3h (10.86)

Here we have products of position operators in one dimension with momentum operators

in another dimension. How do these commute? To determine this we examine the


commutator in position representation as follows:

c d � � � �Œ Œ Œ

C 0ÐBß Cß DÑ œ C �3h 0 Bß Cß D � �3h C0 Bß Cß D` `

`D `D

œ �3h C � C œ !`0 `0

`D `D

s sß :D (10.87)

so we find, in general,

c d ‘ ‘B B

: :

B ß :

s s

s s

s s

3 4

3 4

3 344

ß œ !

ß œ !

œ 3h$

(10.88)

Since the different components of the position operator commute, we designate a three-

dimensional state vector for position with the notation

l<Ù œ lBß Cß DÙ œ lB ß B ß B Ùt " # $ (10.89)

because all of these position measurements are . Likewise we can designatecompatible

the three-dimensional state vector for momentum by the equation

l:Ù œ l: ß : ß : Ù œ l: ß : ß : Ùt B C D " # $ (10.90)

When we examine the commutator relationships of the components of with each other,Ps

with and with the Hamiltonian, we find the following relationshipsPs#

’ “P ßP œ 3hP BCDs s sB C D cyclic (10.91)

’ “P ßP œ ! 3 œ Bß Cß Ds s#3 or (10.92)

and

” •Pß œ !t Ls s (10.93)

where we define

P œ P � P � Ps s s s# # # #

B C D (10.94)

Since both and any of the other components commute, we know that we can define anPs#

eigenvector of which is simultaneously an eigenvector of one of the other componentsPs#

of . It is conventional to choose the -component and write the eigenvalue equationsP Ds

P l ß7Ù œ l ß7Ùs

P l ß7Ù œ h7 l ß7Ùs

#

D

- - -

- -

(10.95)

where and are to be determined.- 7


Since the angular momentum operator commutes with the Hamiltonian operator,

we can also write this last equation in the form

P lIß ß7Ù œ lIß ß7Ùs

P lIß ß7Ù œ h7 lIß ß7Ùs

#

D

- - -

- -

(10.96)

In what follows, however, we will include the notation for the total energy to keep thenot

notation simpler.

________________________________________________________________________

Problem 10.2

Beginning with the defining relationships for the components of the angular momentum

operator, show that

’ “P ßP œ 3hP BCDs s sB C D cyclic

________________________________________________________________________

Problem 10.3

Show that

’ “P ßP œ ! 3 œ Bß Cß Ds s#3 or

________________________________________________________________________

Problem 10.4

Show that

” •Pß œ !t Ls s

________________________________________________________________________

As we have already seen, quantum mechanical commutator relationships are very

fundamental, since these relationships are independent of a particular representation

(position, momentum, etc.). Although the commutator relationships we developed above

were based upon the classically defined angular momentum operator, there are ,other

similar operators which obey these same commutator relationships. These other

operators must, therefore, correspond to different types of angular momentum operators.

It is conventional to designate whichany arbitrary angular momentum operator

satisfies the commutator relationships above by the notation and to reserve theNs

designation of for the angular momentum operator. Thus, we rewrite thePs orbital

commutator relationships in terms of the generic angular momentum operator Ns

’ “N ß N œ 3hN BCDs s sB C D cyclic (10.97)

’ “N ß N œ ! 3 œ Bß Cß Ds s#3 or (10.98)


where we define

N œ N � N � Ns s s s# # # #

B C D (10.99)

Again, since both and any of the other components commute, we know that we canNs#

define an eigenvector of which is simultaneously an eigenvector of one of the otherNs#

components of . We choose the -component and write the eigenvalue equationsN Ds

N l ß7Ù œ l ß7Ùs

N l ß7Ù œ h7 l ß7Ùs

#

D

- - -

- -

(10.100)

where and are to be determined.- 7

Step-up and Step-down Operators

In order to determine the eigenvalues of these equations we will make use of the

non-Hermetian operators

N œ N � 3N

N œ N � 3N

s s s

s s s

� B C

� B C

(10.101)

We find that the commutator relationships between these operators and with are givenNsDby

’ “’ “’ “

N ß N œ � hN

N ß N œ � hN

N ß N œ �#hN

s s s

s s s

s s s

D � �

D � �

� � D

(10.102)

In addition, we notice that

N N œ N � 3N N � 3N œ N � N � 3 N ß N

œ N � N � hN

s s s s s s s s s s

s s s

� � B C B C C BB C

# #

# #

D D

Š ‹Š ‹ ’ “ (10.103)

and, similarly,

N N œ N � N � hNs s s s s� � D

# #

D (10.104)

We will now use these raising and lowering operators to determine the

eigenvalues of the eigenvalue equations for and . First, from the commutationN Ns s#

D

relationship between and , we findN Ns sD �

’ “N ß N œ � hNs s sD � � (10.105)


or

N N � N N œ hN Ê N N œ N N � hs s s s s s s s sD � � D � D � � DŠ ‹ (10.106)

Operating on an eigenvector with this operator gives

N N l ß7Ù œ N N � h l ß7Ù

N h7 � h l ß7Ù œ h 7� " N l ß7Ù

s s s s

s s

D � � D

� �

- -

- -

Š ‹� � � �

(10.107)

which demonstrates that must be an eigenvector whose -value is one greaterN l ß7Ù 7s� -

than the eigenvector ! Similarly,l ß7Ù-

N N l ß7Ù œ h 7� " N l ß7Ùs s sD � �- -� � (10.108)

which likewise demonstrates that must be an eigenvector whose -value is oneN l ß7Ù 7s� -

less than the eigenvector ! This means that the step-up and step-down operatorsl ß7Ù-have the property

N l ß7Ù œ - l ß7 � "Ù

N l ß7Ù œ - l ß7 � "Ù

s

s

� �

� �

- -

- -

(10.109)

where are constants as yet to be determined.-„________________________________________________________________________

Problem 10.5

Using the definition of and and the commutator relationships for angularN Ns s� �

momentum, show that

’ “’ “’ “

N ß N œ � hN

N ß N œ � hN

N ß N œ �#hN

s s s

s s s

s s s

D � �

D � �

� � D

________________________________________________________________________

Determination of the Eigenvalues of and N Ns s#

D

But now that we know the properties of the and operators, we can use themN Ns s� �

to determine the eigenvalues and . In addition, we know that the -component of the- 7 Dangular momentum can never be larger than the magnitude of the angular momentum.

This means that

ØN Ù Ÿ ØN Ùs s# #

D (10.110)


or, in terms of the eigenvectors

Ø ß7l N � N l ß7Ù !- -Š ‹s s# #

D (10.111)

But the result of this equation is obvious, since are eigenvectors of both andl ß7Ù N- s#

NsD . It is just

- � h 7 !# # (10.112)

or

h 7 Ÿ# # - (10.113)

This means that may be positive or negative, but cannot be greater in magnitude than a7certain value determined by . Thus, has a maximum value, and a minimum- 7 7 ß7+B

value, . Since the step-up operator will always increase the value of the quantum7 7738

number by one, if we operate on the ket with the step-up operator, we cannotl ß7 Ù- 7+B

get another eigenvector, so we must have

N l ß7 Ù œ !s� 7+B- (10.114)

Likewise, the operator operating on the ket must also giveN l ß7 Ùs� 738-

N l ß7 Ù œ !s� 738- (10.115)

Now we can use a cleaver “trick” to determine the value of and . We7 77+B 738

operate on the maximum and minimum kets with the combined operators andN Ns s� �

N Ns s� �. Since

N l ß7 Ù œ !s� 7+B- (10.116)

we also have

N N l ß7 Ù œ !s s� � 7+B- (10.117)

but

N N l ß7 Ù œ N � N � hN l ß7 Ù œ !

œ � h 7 � h 7 l ß7 Ù œ !

œ � h 7 7 � " l ß7 Ù œ !

s s s s s� � 7+B D 7+B

# #

D

# # #7+B 7+B 7+B

#7+B 7+B 7+B

- -

- -

- -

Š ‹ˆ ‰ ‘� �

(10.118)

Now the ket cannot be zero, so that the coefficients out front must satisfy the equation

- � h 7 7 � " œ !#7+B 7+B� � (10.119)

Similarly, for the operator operating on the minimum ket, we haveNs�

N l ß7 Ù œ !s� 738- (10.120)


Again we operate on this vector with the operator to obtainNs�

N N l ß7 Ù œ !s s� � 738- (10.121)

but

N N l ß7 Ù œ N � N � hN l ß7 Ù œ !

œ � h 7 � h 7 l ß7 Ù œ !

œ � h 7 7 � " l ß7 Ù œ !

s s s s s� � 738 D 738

# #

D

# # #738 738 738

#738 738 738

- -

- -

- -

Š ‹ˆ ‰ ‘� �

(10.122)

Again, the ket cannot be zero, so that the coefficients out front must satisfy the equation

- � h 7 7 � " œ !#738 738� � (10.123)

The two equations relating and to together give7 7738 7+B -

7 7 � " œ 7 7 � "738 738 7+B 7+B� � � � (10.124)

This can be written as

7 �7 � 7 �7 œ !

7 �7 � 7 �7 œ !

7 �7 7 �7 � 7 �7 œ !

7 �7 � " 7 �7 œ !

# #738 7+B738 7+B

# #738 7+B 738 7+B

738 7+B 738 7+B 738 7+B

738 7+B 738 7+B

ˆ ‰ˆ ‰ � ��

which can have two possible solutions:

7 œ 7 � "

7 œ �7738 7+B

738 7+B

(10.125)

the first of which obviously leads to a contradiction. Thus, we now know that the

maximum and minimum values of the quantum number are equal in magnitude. Let's7call this number . Thus, and , and either of our equations for 4 7 œ 4 7 œ � 47+B 738 -yield

- � h 4 4 � " œ !# � � (10.126)

from which we obtain

- œ h 4 4 � "# � � (10.127)

where . We therefore have a relationship between and the maximum and4 œ 77+B -minimum values of and . From this point on, we will now write our kets in7 4 � 4 Ð� �the form rather than , since we know the value of . But we still don'tl4ß7Ù l ß7Ù Ñ- -know just what the values of are. To determine the possibilities, consider operating on7

the minimum ket multiple times with the operator. Each time we operate,l4ß7 Ù N738 �s

the value of increases by one unit (in terms of ). This means we have7 h

7 � " � " � " �â œ 7738 7+B (10.128)


or

7 �7 œ 87+B 738 (10.129)

where is an integer or zero. But and so that this last equation8 ß 7 œ 4 7 œ � 47+B 738

gives

4 � � 4 œ 8

#4 œ 8

4 œ 8Î#

� � (10.130)

which means that is either an integer half-integer or zero! Thus, we can have the4 ß ßfollowing situations:

4 œ ! 7 œ !

4 œ 7 œ � ß�" " "

# # #4 œ " 7 œ �"ß !ß�"

4 œ 7 œ � ß� ß� ß�$ $ " " $

# # # # #ã

(10.131)

The Significance of Half-Integer Spins

When we examine analytically the solutions for the -component of the D orbital

angular momentum, we find that the acceptable values for the quantum numberP ß 7D

are integers (or zero). The half-integer solutions, then, do not correspond to orbital

angular momentum. These solutions correspond to the quantum mechanical angularspin

momentum (and actually arise when one attempts to solve the Schrödinger equation

relativistically). The spin of a system is, therefore, not classical, although there are some

classical analogues (for example the relationship between angular momentum and theany

magnetic moment). We find in nature that different elementary particles (and

combinations of elementary particles) have different values of ,intrensic spin

characteristic of the particle. The intrensic spin of the particle does not change and is not

a function of the coordinates of space. We will examine the special case of half-integer

electron spin in the next section.

Evaluating -„ We are now in a position to determine the coefficients , arising from the action-„of the and operators. To do this we examine the expectation value of the N N N Ns s s s

� � � �

operator . Notice that is the adjoint of , or . ThisØ4ß 7lN N l4ß 7Ù N N N œ Ns s s s s s� � � � � �

†

means we can write

Ø4ß 7lN N l4ß 7Ù œ Ø4ß 7lN N l4ß7Ù

œ Ø4ß7lN l4 ß7ÙØ4 ß7lN l4ß7Ù

s s s s

s s

� � ��

4�

w w�

†

†�w

(10.132)


Now, using the definition of the adjoint operation, we can write this last equation as

Ø4ß 7lN N l4ß 7Ù œ Ø4 ß7lN l4ß7Ù Ø4 ß7lN l4ß7Ù

œ - Ø4 ß7l4ß7 � "Ù - Ø4 ß7l4ß7 � "Ù

œ l- l Ø4ß7 � "l4 ß7ÙØ4 ß7l4ß7 � "Ù

œ l- l

s s s s� � � �

4

w w

4�

w w�

4

�# w w

�#

��

w

w

w

*

* *

(10.133)

But we can evaluate the expectation value of easily, just as we did earlier, toN Ns s� �

obtain

l- l œ Ø4ß 7lN N l4ß 7Ù

œ Ø4ß7l N � N � hN l4ß7Ù

œ h 4 4 � " � h 7 � h 7

œ h 4 4 � " �7 7� "

� � �#

# #

D D

# # # #

#

s s

s s sŠ ‹� �c d� � � �

(10.134)

Similarly, one can show that

l- l œ Ø4ß 7lN N l4ß 7Ù œ h 4 4 � " �7 7� "� � �# #s s c d� � � � (10.135)

This gives, assuming that and are ,- -� � real

N l4ß7Ù œ h 4 4 � " �7 7� " l4ß7 � "Ù

N l4ß7Ù œ h 4 4 � " �7 7� " l4ß7 � "Ù

s

s

�

�

È � � � �È � � � �(10.136)

We have now developed all the equations we need to completely characterize the angular

momentum of a system. We now wish to examine half-integer spin systems.

The Schr¶dinger Equation in Three Dimensions

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