1

Angular Momentum Theory.

Marcel Nooijen, University of Waterloo

In these lecture notes I will discuss the operator form of angular momentum theory.

Angular momentum theory is used in a large number of applications in chemical physics.

Some examples are: atomic orbital theory, rotational spectra, many electron theory of

atoms, NMR and ESR spectroscopy, spin-orbit coupling. All of these topics can be

treated in a unified and elegant way using the theory of angular momentum. The topic of

angular momentum theory will also introduce us to chemical systems that are not easily

described using “single determinant” molecular orbital theory.

We will first look at an alternative solution to the solution of the time-independent

Schrödinger equation for the hydrogen atom. The hydrogen atom is the simplest atom and

can be solved exactly. It is a useful model for all other atoms. The S.E. for the hydrogen

atom can be reduced to a one-body problem in three dimensions. Even more essential the

problem has spherical symmetry, and it will therefore be advantageous to use spherical

coordinates to describe the solutions and tackle the problem. The angular part of the

problem shows up in many guises in physical chemistry and is not restricted at all to

finding atomic orbitals. We will use a very powerful way of finding solutions to this

problem that can be used in precisely the same way for such diverse problems as finding

the eigenfunctions of the rigid rotor, the description of spin eigenstates (singlets, triplets

etc.), or hyperfine splitting in atomic absorbtion spectra. The method of solving the

angular problem involves working with operators and commutators and this type of

approach is used very often in the literature nowadays. It is useful to know, and very

elegant too, I may add - I can't resist telling you about it for the sheer beauty of it. The

second part of the Hydrogen atom problem involves the radial part of the Schrödinger

equation. I will also discuss this in some detail, but it has far less general applicability

than the angular part. In the second part of these notes we will consider angular

2

momentum theory for many-particle systems. The treatment will be quite general, and the

focus is on the concepts rather than mathematical detail.

Part A: The hydrogen atom.

I. Introduction.

As always in quantum theory we analyse the classical problem first in order to derive the

quantum hamiltonian. The classical energy of a proton and an electron would consist of

the kinetic energy + Coulomb interaction

Epm

pm

er r

p

p

e

e e p

= + −−

2 2 2

02 2 4πε r r (6.1)

where the subscript p indicates the proton while e refers to the electron. As discussed

very nicely in MS (problems 5-5/5-6), any two-body problem in which the potential

energy depends only on the distance between the two bodies can be separated in a center

of mass problem and a one-body problem that involves the reduced mass. Hence if we

define

r

r r

Rm r m r

m mM m mc

p p e e

p ep e=

++

= +; (6.2a)

r r rr r rm m

m mm

mm m

me pe p

e pe

p

e pe= − =

+=

+≈; µ (6.2b)

the energy becomes

E PM

p er

r rC= + − =2 2 2

02 2 4µ πε; r (6.3)

The classical energy can be immediately translated into the quantum mechanical

Hamiltonian by making the substitutions p ixx → −∂∂

h , etc, and realizing that

p p p px y z2 2 2 2= + + . This leads to the quantum mechanical Hamiltonian

3

$HM

ertotal

C= −∇

−∇

−h h22

22 2

02 2 4µ πε (6.4)

where ∇ =∂∂

+∂∂

+∂∂

22

2

2

2

2

2x y z (it is pronounced "nabla squared"). We are interested in

finding the eigenstates of this hamiltonian. The center of mass motion separates easily if

we try the solution Φ Ψ( ) ( )r rR rc , which leads to

−∇

+ −∇

− =h

r

r r hr2

22

2 2

021

2 4C c

ctotal

RM R r

er

r EΦΦ Ψ

Ψ( )( ) ( )

[ ] ( )µ πε

(6.5)

Each of the terms on the left hand side must equal a constant. The total energy is given by

the sum of the translational energy of the center of mass and the internal energy of the

hydrogen atom. One possible solution for the center of mass wave function is a plane

wave Φ( )r r r

R eCik Rc= ⋅ with energy E k

MCM =h2 2

2. It is infinitely degenerate though (every

direction of rk with the same length corresponds to the same energy) and this translational

energy can take on any positive value. It is not quantized. Henceforth we will only

consider the internal wave function Ψ( )rr which is associated with the Hamiltonian

$H er m

ere

= −∇

− ≈ −∇

−h h22 2

0

22 2

02 4 2 4µ πε πε (6.6)

The latter Hamiltonian corresponds to a fixed proton in the origin. We could have started

from this formulation and we would have made only a very small error. Our problem

becomes $ ( , , ) ( , , )H x y z E x y zΨ Ψ=

Because the potential only depends on the distance to the origin it is convenient to us

spherical coordinates. This will allow us to seperate variables, and investigate various

problems separately. Defining

x r y r z r

r= = =≤ < ∞ ≤ ≤ ≤ ≤

sin cos ; sin sin ; cos ;; ; ;θ ϕ θ ϕ θ

θ π ϕ π0 0 0 2 (6.7)

with the corresponding inverse transformation

r x y z yx

z x y z= + + = = + +−2 2 2 1 2 2 2 2 1 2c h c h/ /

; arctan( ) ; arccos( )ϕ θ (6.8)

we can write the Schrödinger equation

4

$ ( , , ) ( ) ( , , ) ( , , )H r er

r E rΨ Ψ Ψθ ϕµ πε

θ ϕ θ ϕ= − ∇ − =h2

22

02 4 (6.9)

In order to discuss this problem further we need to obtain ∇2 in spherical coordinates.

This can be done by straightforward but very tedious manipulation using essentially the

chain rule. For example

∂∂

=∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

f rx

fr

rx

fx

fx

( , , ) * * *θ ϕθ

θϕ

ϕ for any f r( , , )θ ϕ (chain rule).

∂∂

→∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂x

rx r x x

θθ

ϕϕ

∂∂

=∂∂

+ + = + + = =−rx x

x y z x y z x xr

( ) ( ) * sin cos/ /2 2 2 1 2 2 2 2 1 212

2 θ ϕ

So the procedure would be: "Take derivatives, and express (eventually) everything in

spherical coordinates." Very tedious! Let us work out one example explicitly, as it

involves a famous operator:

$ ( ) ( ) ( ) ( )L i xy

yx

i x ry

y rx r

i xy

yx

i xy

yxz =

∂∂

−∂∂

=∂∂

−∂∂

∂∂

+∂∂

−∂∂

∂∂

+∂∂

−∂∂

∂∂

h h h hθ θ

θϕ ϕ

ϕ

x ry

y rx

xyr

yxr

∂∂

−∂∂

= − = 0

xy

yx

mess xy yx∂∂

−∂∂

= − =θ θ *( ) 0

xy

yx

x xx y

y yx y

∂∂

−∂∂

=+

−−+

=ϕ ϕ

2 2 2 2 1( )

and hence $L iz =∂∂

hϕ

.

Similar manipulations yield for − ∇h2

2

2µ:

− ∇ = −∂∂

∂∂

+h h2

22

22

2

22 21

2µ µ µr rr

rL

r( )

$ (6.10)

where $ $ $ $L L L Lx y z2 2 2 2= + + is precisely the square of the angular momentum operator

$ [sin

(sin )sin

]L2 22

2

21 1

= −∂∂

∂∂

+∂∂

hθ θ

θθ θ ϕ

(6.11)

5

Please note that h2 is part of the definition of the operator $L2 . The operator $L2 was

encountered before when we discussed the rigid rotor. It shows up very frequently in

quantum mechanics, and below we will discuss the eigenfunctions of this operator, which

are functions of the angular coordinates only. For now let us just assume that such

eigenfunctions exist and let us denote them ga ( , )θ ϕ , where

$ ( , ) ( , )L g aga a2 2θ ϕ θ ϕ= h , (6.12)

such that the eigenvalue is h2a . Assuming this, we try solutions for the Schrödinger

equation of the form

Ψ( , , ) ( )* ( , )r f r gaθ ϕ θ ϕ= (6.13)

and substituting this form in the Schrödinger equation

$ ( ) ( , ) ( ) ( , )Hf r g Ef r ga aθ ϕ θ ϕ= (6.14)

we obtain

−∂∂

∂∂

+ − =h h2

22

2

2

2

021

2 4µ µ πεr rr f

rar

er

f r Ef r( ) ( ) ( ) ( ) (6.15)

This is a one-dimensional differential equation for f r( ), that depends on the eigenvalue

a of the angular part of the wave function. We are hence left with two subproblems:

Problem a: Find the precise eigenvalues and eigenfunctions of the $L2 operator (the so-

called angular problem).

Problem b: Find solutions of the radial equation for fixed value of a .

Intermezzo: Working with commutators.

In the following we will make extensive use of commutation relations. The following

rules come in very handy:

1.

cA B A cB c A B$ , $ $ , $ $ , $= =

6

In the above c is a number, but it could also be an operator that commutes with both $A

and $B ! If an operator commutes with everything else we can always simply put it in

front of everything.

2.

$ $, $ $ $, $ $ , $ $AB C A B C A C B= +

proof:

$ $ $ $ $ $ ( $ $ $ $ $ $ ) ( $ $ $ $ $ $ ) $ $, $ $ , $ $ABC CAB ABC ACB ACB CAB A B C A C B− = − + − = +

This rule allows us to express unknown commutators in terms of elementary

commutators and operator products. It is often a very quick way to simplify matters.

Examples abound below!

3.

$ , $ $ $( $ $ ) ( $ $ ) $ $ , $ $ , $A B C A B C B C A A B A C+ = + − + = +

II. Solution of angular problem using operator algebra.

We will show that the angular eigenvalue problem can be solved using only the

commutation relations between the angular momentum operators. So everything follows

from the following 'definitions' :

$ , $ $

$ , $ $

$ , $ $

$ $ $ $

L L i L

L L i L

L L i L

L L L L

x y z

z x y

y z x

x y z

=

=

=

= + +

h

h

h

2 2 2 2

(A.1)

In fact you derived these anticommutation relations yourself (MS problem 4-17) from the

definition of $L r p= ∧r r , and the commutation relations r p ik l kl, = hδ , where k l, label the

three cartesian directions in space (x,y,z). Using the above mentioned rules this is very

easy:

7

$ , $ $ $ $$ , $$ $$

$ $ $ , $ $ , $ $ , $ $$ $, $ ( $$ $ $ ) $

L L yp zp zp xp

yp p z yx p p z p p xp z p i xp yp i L

x y z y x z

x z z z y x y z y x z

= − −

= − − + = − =2 h h

Please note how we moved commuting operators in front for immediate simplifications!

We will also use the following auxillary operators:

$ $ $

$ $ $

L L iL

L L iLx y

x y

+

−

= +

= − (A.2)

Using the commutation relations you can derive:

$ , $ $ , $ $ , $L L L L L Lx y z2 2 2 0= = = (A.3)

$ $ $ $

$ $ $

L L L L L

L L L Lz z

z z

2 2

2

= + +

= + −− +

+ −

h

h (A.4)

$ , $ $ , $L L L L2 2 0+ −= = (A.5)

$ , $ $L L Lz + += h (A.6)

$ , $ $L L Lz − −= −h (A.7)

examples:

$ , $ $ , $ $ , $ $ , $

$ $ , $ $ , $ $ $ $ , $ $ , $ $

( $ $ $ $ ) ( $ $ $ $ )

L L L L L L L L

L L L L L L L L L L L L

i L L L L i L L L L

x x x y x z x

y y x y x y z z x z x z

y z z y z y y z

2 2 2 2

0

0

= + +

= + + + +

= − + + + =h h

(A.8)

$ , $ $ , $ $ $ , $ $ , $

( ) ( $ $ ) $

L L L L iL L L i L L

i L i i L L iL L

z z x y z x z y

y x x y

+

+

= + = +

= + − = + =h h h h (A.9)

Let us now continue and derive the eigenvalues of $L2 just by using the commutation

relations. In fact we will use that since $ , $L Lz2 0= these operators must have common

eigenfunctions. We could have used just as easily $Lx or $Ly , but is is the standard

8

convention to use the pair $L2 and $Lz . Let us call these common eigenfucntions ga b, ( , )θ ϕ ,

where a is the eigenvalue of $L2 while b is the eigenvalue of Lz . They are arbitrary (real)

numbers at the moment.

$, ,L g aga b a b

2 = (A.10)

$, ,L g bgz a b a b= (A.11)

However because

$L L L Lx y z2 2 2 2= + + (A.12)

we immediately deduce that

a b≥ ≥2 0 (A.13)

Next, if ga b, is a common eigenfunction of $ , $L Lz2 then $ ,L ga b+ , a new function, is also an

eigenfunction of both these operators. Proof:

$ ( $ ) $ $ ( $ ), , ,L L g L L g a L ga b a b a b2 2

+ + += = (A.15)

$ ( $ ) $ $ $ ( )( $ ), , , ,L L g L L g L g b L gz a b z a b a b a b+ + + += + = +h h (A.16)

hence $ ,L ga b+ is an eigenfunction with eigenvalues ( , )a b + h , or $ ,L ga b+ is zero..... It is seen

that acting with L+ keeps the eigenvalue of $L2 the same but it increases the eigenvalue of

$Lz by the amount h . This clearly indicates that the eigenvalue a of $L2 is degenerate: in

general there is more than one eigenfunction with the same eigenvalue!

Similarly

$ ( $ ) $ $ ( $ ), , ,L L g L L g a L ga b a b a b2 2

− − −= = (A.17)

$ ( $ ) $ $ $ ( )( $ ), , , ,L L g L L g L g b L gz a b z a b a b a b− − − −= − = −h h (A.18)

$,L ga b− is a common eigenfunction of $ , $L Lz

2 with eigenvalues ( , )a b − h (or it is zero.....).

$L+ and $L− are called ladder operators. They define eigenfunctions having adjacent

eigenvalue of $Lz (shift b by ±h) but they leave the eigenvalue of $L2 unchanged.

9

However the ladder operators cannot act indefinitely, since b a2 ≤ . Let us call the

maximum eigenvalue b lmax = h . This means that the next higher function generated by $L+

has to vanish!

$,L ga l+ =h 0 (A.19)

In the sequel I will suppress h in the subscribt: g ga l a l, ,h → . It will turn out that the natural

unit of angular momentum is h , and the formulas take a simpler form if we write

b lmax = h . Since l is arbitrary, this is not a limitation, just a convenience. Interestingly

enough we can immediately find the eigenvalue of $L2 if b lmax = h . We use the specific

form for $L2 that acts with $L+ first (see Eqn. A.4), because we know $ ,L ga l+ = 0.

$ ( $ $ $ $ ) ( ) ( ), , , ,L g L L L L g l l g l l ga l z z a l a l a l2 2 2 2 2 2 1= + + = + = +− + h h h h (A.20)

Similarly acting by $L− the laddering down process must end. Let us call −kh the

minimum value, such that $ ,L ga k− − = 0. The corresponding eigenvalue of $L2 is given by

(now we use the form of $L2 in which $L− acts first (eqn. A.4)):

2 2 2 2 2 2, , , ,

ˆ ˆ ˆ ˆ ˆ( ) ( ) ( 1)a k z z a k a l a kL g L L L L g k k g k k g− + − − − −= + − = + = +h h h h (A.21)

Since the eigenvalue a is the same we must have k l= .

We can summarize the above result by saying we get the complete set of eigenvalues:

a l l= +( )1 2h , eigenvalues of $L2 , l ≥ 0 (A.22)

Corresponding values of $Lz : m m l l l lh, , ,....., ,= − − + −1 1 (A.23)

And we can use as the defining equations for the (unnormalized) eigenfunctions:

$ ; $, , ,L g g L gl l l m l m+ − −= =0 1 (unnormalized, laddering down) (A.24)

or

$ ; $, , ,L g g L gl l l m l m− − + += =0 1 (unnormalized, laddering up) (A.25)

10

What are allowed values of l , (which up to now was only restricted to be ≥ 0)? By

raising −l by 1 each time we must end up at +l and this means that there are only two

types of possibilities

A. l is integer

B. l is half integer.

The first possibility can describe functions gl m, ( , )θ ϕ (see below). With the second type

of eigenvalue we cannot associate a well defined eigenfunction in the angular coordinates

however. Still, they turn out to have a physical meaning. They turn up when we describe

the spin of particles! Each value of l describes a set of eigenfunctions

g m l l l ll m, , , ,..., ,= − − + −1 1 . They are said to form a multiplet of dimension 2 1l + .

In the table below I have listed the lowest types of multiplets. With each value of l we

have 2 1l + values of m, that range from − − + −l l l l, ,...., ,1 1 , as shown

$ ( )L2 2h $ ( )Lz h degeneracy $L2 spatial spin

l l( )+1 m "name" "name"

l = 0 0 0 1 s singlet

l =1 2 -1,0,1 3 p triplet

l = 2 6 -2,-1,0,1,2 5 d quintet

l = 3 12 -3,-2,-1,0,1,2,3 7 f septet

l = 12

34

−12

12

, 2 - doublet

l = 32

154

− −32

12

12

32

, , , 4 - quartet

l = 52

354

− − −52

32

12

12

32

52

, , , , , 6 - sextet

11

At this point we have shown the general structure of the solutions, which we derived

using only the commutation relations between the operators. The first four relations in

this section is the only thing we needed, and all of the rest follows or can be derived.

Presently, in quantum mechanics the commutation relations are taken as the definition of

angular momentum. This is for example why spin is considered angular momentum: the

spin operators simply satisfy the same commutation relations!

If we assume the standard definition for angular momentum we can do a little more and

also derive the corresponding eigenfunctions in spherical coordinates. The operators $ , $ , $L L Lz + − can be expressed in spherical coordinates, just like we did for Lz . They would

take the form:

$

$ [ cossin

]

$ [ cossin

]

L i

L e i

L e i

z

i

i

=∂∂

=∂∂

+∂∂

=∂∂

−∂∂

+

−−

h

h

h

ϕ

θθθ ϕ

θθθ ϕ

ϕ

ϕ

(A.26)

It is easy to find solutions that are eigenfuctions of $Lz

−∂∂

= → =i f m f f eimh hϕ

ϕ ϕ ϕ ϕ( ) ( ) ( ) (A.27)

Boundary condition: f f( ) ( )ϕ π ϕ+ =2 → m is integer (only integer values allowed for

m!) This is the reason that the half-integer (spin) functions cannot be expressed in θ ϕ,

coordinates. They would not be single-valued functions in 3d-space!

Next we can solve for the θ -part. In spherical coordinates the eigenfunctions gl m, are

conventionally denoted as Ylm( , )θ ϕ . They are called the spherical harmonics. The ϕ -

dependent part of these functions is determined above, and the Ylm( , )θ ϕ can be written as

Y P elm

lm im( , ) ( )θ ϕ θ ϕ= , (A.28)

12

where the Plm( )θ are so-called associated Legendre polynomials. They can be easily

generated using the ladder operators. If we take the function with m l= it has to satisfy

L Yll

+ =( , )θ ϕ 0; Y P elm

ll il( , )θ ϕ θ ϕ= b g (A.29)

And using the spherical coordinate form for the $L+ operator we find

e e l Pi illlϕ ϕ

θθθ

θ[ cossin

] ( )∂∂

− = 0 (A.30)

It is easily verified that the solution is

Pll l l( ) (sin ) sinθ θ θ= =

as

l ll lsin cos cossin

sin− − =1 0θ θ θθ

θ

The highest m-valued function in a multiplet, Yll θ ϕ,b g hence has the simple form

Y ell l il( , ) sinθ ϕ θ ϕ= . (A.31)

All of the other functions in the multiplet can be found by acting with $L− . There is one

further simplification in that we only need to generate functions up to m = 0. One can

show that

Y P elm

lm im( , ) ( )θ ϕ θ ϕ=

Hence the θ -part is the same for +m and −m . This follows from the form of $L2

$sin

(sin )sin

L22

2

21 1

=∂∂

∂∂

+∂∂θ θ

θθ θ ϕ

and $ ( , )L P elm im2 θ ϕ ϕ and $ ( )L P el

m im2 − −θ ϕ yields the same differential equation for Plm ( )θ

$ (sin

(sin )sin

) ( ) ( ) ( )L m P l l Plm

lm2

2

21 1=

∂∂

∂∂

− = +θ θ

θθ θ

θ θ

Let us look at the non-trivial example of l = 2 (d-functions).

Y e i22 2 2( , ) sinθ ϕ θ ϕ= (general formula)

$ ( , ) ~ sin cosL Y ei− → 2

1 θ ϕ θ θ ϕ

13

$ ( , ) ~ ( sin cos cos ) cosL Y− → − + + = −20 2 2 2 23 1θ ϕ θ θ θ θ

$ ( , ) ~ sin cosL Y e i−

− −→ 21 θ ϕ θ θ ϕ

$ ( , ) ~ ( sin cos cos ) ~ sinL Y e ei i−

− − −→ − + −22 2 2 2 2 2 2θ ϕ θ θ θ θϕ ϕ

$ ( sin cos cossin

sin )L− → − =2 2 02θ θ θθ

θ

It is seen that the form of the functions is generated quite easily. The normalization

factors are less important, although even they can be obtained very generally from the

commutation relations!

To solve the Schrödinger equation for the Hydrogen atom we actually only require

angular eigenfunctions of $L2 , not of $Lz . We develop the above formalism because of its

elegance and generality. In practice it is easier to think of real eigenfunctions. We can

therefore make linear combinations of degenerate eigenfunctions of $L2 . In particular we

can combine P elm im( )θ ϕ and P el

m im( )θ ϕ− into P mlm ( )cos( )θ ϕ and P ml

m ( )sin( )θ ϕ . For

the above d-functions this leads to the familiar cartesian forms of the d-orbitals

sin cos sin (cos sin ) ~

sin sin sin sin cos ~

2 2 2 22 2

2

2 22

2

2 2

θ ϕ θ ϕ ϕ

θ ϕ θ ϕ ϕ

= −−

=

x yr

xyr

sin cos cos ~

sin cos sin ~

θ θ ϕ

θ θ ϕ

xzrxyr

2

2

3 1 3 122

2cos ~−−z

r

14

All angular functions can easily be generated this way: Start from sin l ileθ ϕ , act with $L−

sequentially, and combine e im± ϕ into cos( ), sin( )m mϕ ϕ . Voila!

III. The radial equation for the Hydrogen atom and its solutions.

Above we discussed the angular part of the equations that determine the atomic orbitals

for the Hydrogen atom in great detail. Here we will discuss how the full set of solutions

can be obtained. The Hamiltonian in spherical coordinates was given by

$ ( )$

Hr r

rr

Lr

er

= −∂∂

∂∂

+ −h2

22

2

2

2

021

2 4µ µ πε (r.1)

And we try the function Ψ( , , ) ( ) ( , )r f r Ylmθ ϕ θ ϕ= →

[ ( ) ( ( ( ) ) ( )] ( , ) ( ) ( , )−∂∂

∂∂

++

− =h h2

22

2

2

2

021 1

2 4µ µ πεθ ϕ θ ϕ

r rr f

rl l

re

rf r Y Ef r Yl

mlm (r.2)

Let us simplify the notation somewhat and multiply through with 22µh

, and define

22µ εEh

= . Let us also use 24

24

22

02

2

02

0

µπε πε

e m ea

e

h h≈ = , where a0 is the Bohr radius. This

yields the radial equation

−∂∂

∂∂

++

− =1 1 2

22

20r r

r fr

l lr

f ra r

f r f r( ) ( ) ( ) ( ) ( )ε (r.3)

or better yet,

−∂∂

−∂∂

++

− =2 1 22

2 20r

fr

fr

l lr

f ra r

f r f r( ) ( ) ( ) ( )ε (r.4)

One more substitution to make. Try f r p r e r( ) ( )= −α , where p r( ) will be a polynomial in

r , p r a br cr( ) ...= + + +2 , hence

∂∂

=∂∂

−

∂∂

=∂∂

−∂∂

+

− −

−

fr

pr

e p r e

fr

pr

pr

p r e

r r

r

α α

α

α

α α

( )

[ ( )]2

2

2

222

e r−α is multiplied throughout, so we can cancel this. We end up with an equation for p r( )

15

− + − + − ++

− =2 2 2 1 22

22

20r

dpdr r

p r d pdr

dpdr

p r l lr

p ra r

p r p rα α α ε( ) ( ) ( ) ( ) ( ) ( ) (r.5)

Let us first examine some of the lower degree equations before discussing the general

solution. Remember that l = 0 for s-orbitals, l =1 would yield p −orbitals, and so forth.

Let me just list some solutions:

l p r dpdr

d pdr

= = = =0 1 02

2, ( ) : , substitute in (r.5)

2 2

10

2

2

0

α α ε

ε α α

r a rr

a

− − = ∀

→ = − =; ; (r.6)

and therefore

f r e Ea

r a( ) ;/= = −− 0

2

022

h

µ (r.7)

Another example:

l p r r c dpdr

d pdr

= = − → = =0 1 02

2, ( ) ,

− + − + − − − − = − ∀2 2 2 2

0

2

r rr c

a rr c r c r c rα α α ε( ) ( ) ( ) ( ) (r.8)

in such an equation the terms must match for each power in r, hence

order r

order unity rr

r ra

orderr

c c a

:

: /

: /

− =

+ − =

− − + =

α εα α

α

2

0

0

2 2 2 0

1 2 2 2 0

(r.9)

and we obtain the solutions:

ε α α= − = =20 01 2 2, / ,a c a (r.10)

or

f r r a e Ea a

r a( ) ( ) ;( )

/= − = − = −−22 2

14 20

22

02

2

02

0h h

µ µ (r.11)

16

Let us take the simplest example of a p function

l p r r dpdr

d pdr

= = = =1 1 02

2, ( ) , , :

− + + + − − =2 2 2 2 2

0

2

r r ar rα α α ε

α ε= = −1 2 1 40 02/ , /a a f r re E

ar a( ) ,/= = −− 2

2

02

014 2

h

µ

General characteristics:

Suppose highest power in polynomial is rm. Then, substituting in (r.5):

Terms of order m: − =α ε2 (always!)

Terms of order m−1: 2 2 2 0 2 1 2

0 0

α α α+ − = → + =ma

ma

( )

αµ

=+

= −+

=11 2

11

0 1 20

2

02 2( )

,( )

, , , ,...m a

Ea m

mh

Independent of l ! Usually we put n m= +1 and call it the principle quantum number:

r e Em a n

n r na

e

− − = −12

02 2

0

21/ , h (n −1: highest power in r )

Call s the smallest exponent in p r( ), p r r ar brs( ) ( ...)= + + +1 2 .

In equation (r.5) we will then obtain terms starting from rs−2: − − − + + = → =2 1 1 0s s s l l s l( ) ( )

The first radial solution corresponding to Ylm θ ϕ,b g (no radial nodes) starts with rl !

Depends on quantum number l.

17

Summary of general solutions:

Ψn l m n l

r nal mr p r e Y

E na n

, , ,/

,( , , ) ( ) ( , );

( )

θ ϕ θ ϕ

µ

=

= −

− 0

2

02 22

h

where p r ar br rn ln n l

, ( ) ....= + + +− −1 2 is a polynomial in r having n l− −1 radial nodes.

Possible energy levels and their degeneracies:

nl nm l l l l

== −= − − + −

1 2 30 1 2 3 1

1 1

, , ,...., , , ,...,

, ,...., ,

Degeneracy En : ( ) ( )2 1 2 2 12

10

1

1

12l n l n n n n

l

n

l

n

+ = + = + ⋅ − ==

−

=

−

∑ ∑

We note that for the hydrogen atom the energy only depends on the principal quantum

number n . Hence the 2 2s p, orbitals are degenerate as are 3 3 3s p d, , and so forth. This is

only true for one-electron atoms (H, Ne7+, etc), but not for many-electron atoms in

general.

18

Part B: Many-Particle Angular Momentum Operators.

The commutation relations determine the properties of the angular momentum and spin

operators. They are completely analogous:

$ , $ $ , .

$ $ $

$ $ $ $ $

$ $ $ $

L L i L etc

L L iL

L L L L L

L L L L

x y z

x y

z z

z z

=

= ±

= + −

= + +

±

+ −

− +

h

h

h

2 2

2

$ , $ $ , .

$ $ $

$ $ $ $ $

$ $ $ $

S S i S etc

S S iS

S S S S S

S S S S

x y z

x y

z z

z z

=

= ±

= + −

= + +

±

+ −

− +

h

h

h

2 2

2

The one-electron eigenfunctions for the $ , $L Lz2 operators are the spherical harmonics

2 2

1 1 1

1 1 1

ˆ ( , ) ( 1) ( , )ˆ ( , ) ( , )ˆ ( , ) ~ ( , ) ( 1) ( 1) ( , ) ( , ) ( , )ˆ ( , ) ~ ( , ) ( 1) ( 1) ( , ) ( , ) ( , )

m ml l

m mz l l

m m m ml l l l

m m m ml l l l

L Y l l Y

L Y m Y

L Y Y l l m m Y C l m Y

L Y Y l l m m Y C l m Y

θ ϕ θ ϕ

θ ϕ θ ϕ

θ ϕ θ ϕ θ ϕ θ ϕ

θ ϕ θ ϕ θ ϕ θ ϕ

+ + ++ +

− − −− −

= +

=

= + − + ≡

= + − − ≡

h

h

h

h

Here I included the precise proportionality constants. For later convenience they are

abbreviated as ( , )C l m± . The one-electron spin eigenfunctions are denoted as

Y Y1 21 2

1 21 2

//

//;= =−α β . Explicitly the various equations read

$ ( )

$ , $ , $

S

S S Sz

2 2 212

12

1 34

12

0

α α α

α α α α β

= + =

= = =+ −

h h

h h

$ ( )

$ , $ , $

S

S S Sz

2 2 212

12

1 34

12

0

β β β

β β β α β

= + =

= − = =+ −

h h

h h

A single electron (a so-called spin 1/2 particle) is always described by the spin-functions

α and β . Higher than spin 1/2 functions show up in many-electron wave functions.

Nuclear spin operators (indicated I ) also satisfy precisely the same commutation

relations ˆ ˆ ˆ,x y zI I i I = h and cyclic permutations. Nuclei on the other hand can have

higher spins (they consist of protons and neutrons that individually are spin 1/2 particles).

19

This means that the nuclear spin functions might for example be a triplet (I=1), or a

quartet (I=3/2). The mathematics underlying all of these different physical phenomena is

precisely the same, and we will focus on the case of angular momentum for an atom

consisting of a certain number of electrons.

The many-electron operators are defined in an analogous fashion

$ $ ( ) $ ( ) ..., .L L L etcxtotal

x x= + +1 2 $ $ ( ) $ ( ) ..., .S S S etcxtotal

x x= + +1 2

The operators $ $ $ $ $ $ $,S S S S S S Stotalxtotal

xtotal

ytotal

ytotal

ztotal

ztotal2 = + + , (and similarly $ ,L total2 ) are

complicated (two-electron) operators that contain mixed terms like $ ( ) $ ( )S Sx x1 2 . We can

always express things in term of products of $ $S Stotal total+ − etc, so we do not need to use $S 2

directly. We can work with the sum-operators on a product of functions. As examples

consider $ ( ( ) ( )) [ $ ( ) $ ( )]( ( ) ( )) [( $ ( ) ( )) ( ) ( ) $ ( ) ( )]

( ) ( ) ( ) ( ) ( ) ( )

S S S S Sztotal

z z z zα α α α α α α α

α α α α α α

1 2 1 2 1 2 1 1 2 1 2 2

21 2

21 2 1 2

= + = +

= + =h h

h

and $ ( ( ) ( )) [ $ ( ) $ ( )]( ( ) ( )) [( $ ( ) ( )) ( ) ( ) $ ( ) ( )]

( ) ( ) ( ) ( )S S S S Stotal− − − − −= + = +

= +α α α α α α α α

β α α β1 2 1 2 1 2 1 1 2 1 2 2

1 2 1 2h h

hence we can think of acting with the one-electron operator on each one-electron function

separately and summing the result.

The procedure works in the same way if we use the $Ltotal− operator on a product of

spherical harmonics. Let us take the p functions and abbreviate

p Y p Y p Y1 11

0 10

1 11= = =−−; ; . Then

$ ( ( ) ( )) ( ( ) ( ) ( ( ) ( ))L p p p p p ptotal+ − −= +0 1 1 1 0 01 2 2 1 2 1 2h

$ ( ( ) ( )) ( ) ( ) ( ) ( ) ( )L p p p p p pztotal

0 1 0 1 0 11 2 0 1 1 2 1 2− − −= − = −h h

For p-functions (l =1) the factor h hl l m m l m( ) ( ) ( , , )+ − + = = = −1 1 2 1 0 1 .

We also wish to examine what happens if we act on an antisymmetric Slater determinant

of spin-orbitals. Also here we can just act with the one-electron operator on each of the

20

product functions in the determinant and sum the result. The reason is that the sum

operator is symmetric and commutes with any permutation of electron labels, e.g.

[ $ ( ) $ ( )][ ( , ) ( ) ( )] [ $ ( ) $ ( )] ( ) ( )

( , )[( $ ( ) $ ( )) ( ) ( )] [ $ ( ) $ ( )] ( ) ( )

S S P S S

P S S S Sz z z z

z z z z

1 2 1 2 1 2 1 2 2 1

1 2 1 2 1 2 2 1 2 1

+ = +

= + = +

α β α β

α β α β

This argument is clearly general and so ˆ ˆ.... ( ....)total totalz a b z a bS Sϕ ϕ ϕ ϕ=

Let us consider the spin orbitals in a p-manifold: p p p p p p1 0 1 1 0 1, , , , ,− − . In this notation

p p p p1 1 1 1= =α β; , etc. Below I will give a set of examples of operations of spin and

angular momentum operations. You can verify the results, and see that the basic rules are

not very difficult.

$ , $

$ ( !), $

L p p L p p p p

S p p p p antisymmetry S p pz

z

+

−

= =

= = =

1 1 1 1 1 1

1 1 1 1 1 1

0 2

0 0

h

The above relations suffice to show that p p1 1 is the ml = 2 component of the 1 D

multiplet. You would need to use the form 2 2ˆ ˆ ˆ ˆ ˆz zL L L L L− += + + h . All other functions in the

multiplet can be generated by acting successively by $L− . For example

$ ( )$ ( ) ( )

L p p p p p p

L p p p p p p p pz

− = +

+ = +

1 1 0 1 1 0

0 1 1 0 0 1 1 0

2h

h

Importantly, this eigenstate with eigenvalue lm = h is not a determinant but a linear

combination of determinants. In this state you cannot say that these orbitals are occupied

and the rest empty. The wave function is more complicated. You can act with the 2L

operator on this new function to show that the eigenvalue has not changed: 2(2+1) 2h is

the answer you should find. But it is more work to show this explicitly (give it a try). You

can keep applying the $L− operator until you find the state 1 1p p− − ( 2lm = − h ), and acting

with $L− once more yields zero.

21

As another example $ , $

$ , $

$ ( )$ ( )

L p p p p L p p p p

S p p S p p p p

L p p p p p p p p

S p p p p p p

z

z

+

+

− − −

−

= = =

= =

= + =

= +

1 0 1 1 1 0 1 0

1 0 1 0 1 0

1 0 0 0 1 1 1 1

1 0 1 0 1 0

2 0

0

2 2

h h

h

h h

h

The first two equations establish that p p1 0 is the m ml s= =1 1, component of the 3 P

multiplet. Please verify. All other 9 states can be obtained by successive application of $L−

and $S− as illustrated by the last two examples given.

We will do some exercises with the angular momentum operators elaborating on some

examples as illustrated above. There is a vast literature on the topic and the above is a

very brief summary. Let me re-emphasize that angular momentum theory underlies NMR

spectra (nuclear spin functions). Electron spin resonance can be treated analogously. This

will be illustrated in class. To treat these phenomena we act on product of nuclear spin

functions. The principles are similar, and spin functions are in fact a bit easier because

the proportionality constants for spin 12

particles are unity rather than 2 . Let me also

mention that we can use a similar treatment to find eigenfunctions of the total angular

momentum operator ˆˆ ˆJ L S= + , which is particularly relevant for atomic term symbols.

We will go through some examples in class and in the problem set.

III. General decomposition of a product basis of angular momentum eigenfunctions

into eigenfunctions of the total angular momentum operators.

In this section we consider the construction of eigenfunctions of the angular momentum

operators for a composite particle (in fact it applies to any product function, for example

spin-orbitals 1pα ). The structure is very general. We will consider the products for two

particles, but from this you can construct products for an arbitrary number of particles

22

using the same general principles. You have seen examples in NMR before we get to this

part of the lecture notes.

Consider the individual multiplets 1 1 1 1 1, , ...l m m l l= − and 2 2 2 2 2, , ...l m m l l= − which are

eigenfunctions of the angular momentum operators 2ˆ ˆ(1), (1)zL L and 2ˆ ˆ(2), (2)zL L

respectively. Here we are using the Dirac notation for states: 1 1,l m indicates an

eigenfunction characterized by the quantum numbers in the so-called ket. More on this

notation shortly. The full set of product functions 1 1 2 2, ,l m l m spans a space of

dimension 1 2(2 1)(2 1)l l+ + . We want to decompose this product basis into a set of new

basis functions that are eigenfunctions of the composite angular momentum operators 2

,ˆ ˆ,total z totalL L , where ˆ ˆ ˆ(1) (2)totalL L L= + . Let us indicate these eigenfunctions as

, , ...L M M L L= − . Such a set of functions with a given value of L is called a multiplet,

and the dimension of the multiplet is 2L+1. We will see below that the possible

eigenvalues L range from 1 2 1 2 1 2, 1,...,L l l l l l l= + + − − , or, assuming that 1 2l l≥ , we can

also say that the multiplets that are occurring are 1 2 2, 0,..,2L l l k k l= + − = .

The total dimension of the space spanned by these multiplets is

2 2

1 2 2 1 20,2 0,2

2 1 2 2 2 2

2 1

{2( ) 1} (2 1)(2( ) 1) 2

1(2 1)(2 1) 2 (2 1) 2 (2 (2 1))2

(2 1)(2 1)

k l k ll l k l l l k

l l l l l l

l l

= =

+ − + = + + + −

= + + + + − +

= + +

∑ ∑

This is consistent with the total number of product functions, as should be the case.

Let us rationalize this result further by an explicit construction of the eigenfunctions.

Each product function itself is an eigenfunction of ,ˆ

z totL :

, 1 1 2 2 1 2 1 1 2 2ˆ , , ( ) , ,z totL l m l m m m l m l m= +h . Hence we can easily arrange the product

functions in a table, such that along a row, they all have the same value of M.

23

Table: Arrangement of product functions according to M-values (eigenvalue of ,ˆ

z totL )

M= 1 2l l+ 1 1 2 2, ,l l l l …1

M= 1 2 1l l+ − 1 1 2 2, 1 ,l l l l− 1 1 2 2, , 1l l l l − … 2

M= 1 2 2l l+ − 1 1 2 2, 2 ,l l l l− 1 1 2 2, 1 , 1l l l l− − 1 1 2 2, , 2l l l l − … 3

………… ……………… ……………….. …………………

M= 1 2 2l l− − + 1 1 2 2, 2 ,l l l l− + − 1 1 2 2, 1 , 1l l l l− + − + 1 1 2 2, , 2l l l l− − + …3

M= 1 2 1l l− − + 1 1 2 2, 1 ,l l l l− + − 1 1 2 2, , 1l l l l− − + …2

M= 1 2l l− − 1 1 2 2, ,l l l l− − …1

# functions 1 22( ) 1l l+ + 1 22( 1) 1l l+ − + 1 22( 2) 1l l+ − +

The overall multiplet structure can be discerned now. In the second column, top row, we

find the function with the maximum value 1 2M L l l= = + . This function is also an

eigenfunction of 2ˆtotalL . This follows because , 1 1 2 2

ˆ , ,totalL l l l l+ =0. You can complete the

argument: why is it an eigenfunction of 2ˆtotalL ? The other functions in this multiplet are

generated by acting with ,ˆ ˆ ˆ(1) (2)totalL L L− − −= + , hence

, 1 1 2 2 1 1 1 1 2 2 2 2 1 1 2 2ˆ , , ( ( , ) , 1 , ( , ) , , 1 )totalL l l l l C l l l l l l C l l l l l l− − −= − + −

and so forth. It is seen that in general we find a linear combination of products states,

except for the highest and lowest possible M-value in the space. This process continues

as you ladder down and decrease the value of M , until 1 2M L l l= − = − − . The number of

functions is precisely the number of functions in the first column, 1 22( ) 1l l+ + . The next

multiplet is constructed by starting from the eigenfunction that has 1 2 1M l l= + − , and is

orthogonal to the linear combination function already found. Alternatively we might try

to find that linear combination such that acting with ,ˆ

totalL+ on this combination yields 0.

Then it will be an eigenfunction of both 2ˆtotalL and ,

ˆz totalL . For example

24

, 2 2 1 1 2 2 1 1 1 1 2 2

2 2 1 1 1 1 2 2 1 1 2 2

ˆ ( ( , 1) , 1 , ( , 1) , , 1 )

( , 1) ( , 1)[ , , , , ] 0totalL C l l l l l l C l l l l l l

C l l C l l l l l l l l l l+ + +

+ +

− − − − −

= − − − =.

This means it will be an eigenfunction of 2ˆtotalL with 1 2 1L l l= + − .

Even though we require a linear combination of the product states to construct the true

eigenfunctions, the number of functions in this second multiplet is precisely the same as

in the second column of the table, namely 1 22( 1) 1l l+ − + . The scheme now repeats itself.

To start constructing the next multiplet find the remaining highest M-state that is

orthogonal to the states already found and ladder down. Alternatively, and this is usually

easier, find that linear combination of functions of specific M-values such that acting

with ,ˆ

totalL+ yields precisely 0.

The above scheme is completely general, and applies to a variety of problems in physics.

For example we can construct eigenfunctions of the spin-operators in this fashion, and in

this way we find the proper combinations of singlet, doublet, triplet spin functions. The

same strategy applies to nuclear spin functions. These problems are very instructive:

They give exact results because the basis set is complete, while the algebra is often not so

tedious. We can also couple different angular momentum operators for example to get

eigenfunctions of the total angular momentum ˆˆ ˆJ L S= + . These problems are always

completely analogous if you work on constructing many-particle functions. Essentially

this is true because the angular momentum operators for different particles commute. It is

also possible to use angular momentum theory to construct for example integrals and

orbitals of higher l-value. As an example you can evaluate integrals like ˆi jp x d . The

selection rules are easily evaluated (meaning you can know when an integral is zero). The

precise evaluation of non-zero integrals requires more work.

To help you digest this material there are some exercises!