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CHAPTER 1: GENERAL PROPERTIES OF ANGULAR MOMENTUM IN

QUANTUM MECHANICS

(From Cohen-Tannoudji et al., Volume I, Chapter VI)

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A. INTRODUCTION: THE IMPORTANCE OF ANGULAR MOMENTUM

Quantum theory of angular momentum, which will be developed here, is important in

many areas of physics, for example:

atomic, molecular and nuclear physics: classification of spectra;

particle and high energy physics: spin of elementary particles;

condensed matter physics: magnetism;

and many more.

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The angular momentum plays a very important role in mechanics: classically, the

total angular momentum of an isolated physical system is a constant of motion:

d

dt

L = 0 (1.1)

and this is true also when a particle moves in a central potential. This last fact will

become relevant in development of quantum theory of the hydrogen atom.

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B. COMMUTATION RELATIONS CHARACTERISTIC OF ANGULAR MOMENTUM

1. Orbital angular momentum

Let us start with x-component of the classical angular momentum:

Lx = ypz zpy (1.2)

The corresponding quantum operator is obtained by substituting the classical posi-tionsy and z by the position operators Yand Zrespectively, and by substituting the

classical momenta pyand pzby the operators Pyand Pz, respectively:

Lx = YPz ZPy (1.3)The other components of the orbital angular momentum can be constructed similarly

allowing us to write the vector of orbital angular momentum operators asL = R P (1.4)

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The components of the orbital angular momentum satisfy important commutation

relations. To find these, we first note that the angular momentum operators are

expressed using the position and momentum operators which satisfy the canonical

commutation relations:

[X, Px] =[Y, Py] =[Z, Pz] =i (1.5)

All the other possible commutation relations between the operators of various com-

ponents of the position and momentum are zero. The desired commutation relationsfor the angular momentum operators are then calculated as follows:

Lx, Ly =

YPz ZPy, ZPx XPz

(1.6)

=

YPz, ZPx

+

ZPy, XPz

(1.7)

= YPz, Z Px + XZ, Pz Py (1.8)= iYPx + iXPy (1.9)= i Lz (1.10)

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Similar calculations can be performed (do it as a homework!) to obtain all the com-

mutation relations between the components of the orbital angular momentum:Lx, Ly

= i Lz (1.11)

Ly, Lz = i Lx (1.12) Lz, Lx = i Ly (1.13)which can be shortened using the antisymmetric tensor i jkas follows

Li, Lj = i jki Lk (1.14)

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For the system of N particles, the total angular momentum is a sum of the angular

momenta of the individual particles

L =

N

i=1Li (1.15)

whereLiis given as follows:

Li = Ri Pi (1.16)

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1. Generalization: definition of an angular momentum

We will now define an angular momentum Jas any set of observables Jx, Jy and Jz

which satisfy

Jx, Jy =iJzJy, Jz

=iJx

Jz, Jx=iJy

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We now introduce the operator

J2 = J2x + J2y +

J2z (1.17)

that represents the scalar square of the angular momentum. It has two notable

properties:

it is a self-adjoint operator, and thus an observable, and

it commutes with the components of the angular momentum:

J2, J = 0

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J2, Jx

=

J2x + J

2y +

J2z , Jx

(1.18)

=

J2y , Jx

+

J2z , Jx

=0 (1.19)

(1.20)

J2y , Jx = J2y Jx Jy Jx Jy + Jy Jx Jy

Jx J

2y (1.21)

= Jy Jy, Jx + Jy, Jx Jy (1.22)

= iJy Jz iJz Jy (1.23)J2z , Jx

= Jz

Jz, Jx

+

Jz, Jx

Jz (1.24)

= iJz Jy + iJy Jz (1.25)

Complete set of commuting observables (C.S.C.O.) relevant to the quantum theory

of angular momentum is given by the operators J2 and Jz.

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C. GENERAL THEORY OF ANGULAR MOMENTUM

1. Definitions and notations

a. THE J+AND JOPERATORS

J+ = Jx + iJyJ

= Jx

iJy

which satisfy the following commutation relationsJz, J+

= J+ (1.26)

Jz, J = J (1.27)

J+, J = 2Jz (1.28)J2, J+

=

J2, J

=

J2, Jz

=0 (1.29)

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They also satisfy the following useful relations:

J+ J =

Jx + iJy

Jx i Jy

(1.30)

= J2x + J2y i

Jx, Jy

(1.31)

= J2x + J2y + Jz (1.32)

= J2

J2z + Jz (1.33)

J J+ = Jx iJy Jx + i Jy (1.34)

= J2x + J2y + i

Jx, Jy

(1.35)

= J2x + J2y Jz (1.36)

= J2 J2z Jz (1.37)J

2=

1

2 J+ J + J J+ + J2z (1.38)

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b. NOTATION FOR THE EIGENVALUES OF J2 AND JZ

J2 is the sum of the squares of three self-adjoint operators, i.e. for any ket |, thematrix element

|J2|

is non-negative:

|J2| = |J2x | + |J2y | + |J2z | (1.39)=

Jx|2 + Jy|2 + Jz|2 0 (1.40)That is all the eigenvalues of J2 are positive or zero.

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Let|be an eigenvector of J2

, then

|J2| = | 0 (1.41)We shall write the eigenvalues of J2 in the form

j (j + 1) 2 = (1.42)

with the convention that

j 0 (1.43)The eigenvalues of Jzare traditionally written as

m (1.44)

wherem is a dimensionless number.

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c. EIGENVALUE EQUATIONS FOR J2 AND Jz

We can now summarize the eigenvalue equations for both operators relevant to

quantum theory of angular momentum J2 and Jz:

J2|k, j,m = j (j + 1) 2|k, j,m (1.45)Jz|k, j,m = m|k, j,m (1.46)

(1.47)

wherek is included for completeness to represent an additional commuting observ-

ables from C.S.C.O.

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2. Eigenvalues of J2 andJz

a. LEMMAS

. Lemma I (Properties of the eigenvalues of J2 and Jz)

If j (j + 1) 2

and m are the eigenvalues of J2

and Jz associated with the sameeigenvector |k, j,m, then jand m satisfy the inequality

j m j (1.48)

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Substituting these expressions to the inequalities above we get

j (j + 1) m (m + 1) =(j m) (j +m + 1) 0 (1.55)j (j + 1) m (m 1) =(j m + 1) (j +m) 0 (1.56)

and consequently

(j + 1) m j (1.57)j m j + 1 (1.58)

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. Lemma II (Properties of the vectorJ|k, j,m)Let|k, j,mbe an eigenvector of J2 and Jzwith the eigenvalues j (j + 1) 2 andm

1. Ifm =

j, J

|k, j,

j

=0.

2. Ifm >j, J|k, j,m is a non-null eigenvector of J2 and Jz, with the eigenvaluesj (j + 1) 2 and(m 1).

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Proof: (1) the square norm of J

|k, j,mis equal to

2

[j (j+

1) m(m 1)]and thusgoes to zero for m =j. Since the norm of a vector goes to zero iff the vector is thenull vector, we conclude that all vectors J|k, j,jare null:

m =j J|k, j,j =0 (1.59)We can also establish the converse:

J|k, l,m =0 m =j (1.60)by letting J+act on both sides of the equation above we get the relation

2j (j + 1) m2 +m

|k, j,m (1.61)

= 2 (j +m) (j

m + 1) |k, j,m

= 0 (1.62)

whose only solution is obtained for m =j.

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(2) Assuming that m >j, the vector J|k, j,m is now a non-null vector since thesquare of its norm is different from zero. We will now establish that it is an eigenvector

of both J2 and Jz. We already know that J and J2 commute, that is:J2, J

|k, j,m = 0 (1.63)

which we can write as followsJ2 J|k, j,m = J J2|k, j,m (1.64)

= j (j + 1) 2 J|k, j,m (1.65)This expression shows that J|k, j,m is an eigenvector of J2 with the eigenvaluej (j + 1) 2.

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. Lemma III (Properties of the vectorJ+|k, j,m)

Let|k, j,mbe an eigenvector of J2 and Jzwith the eigenvalues j (j + 1) 2 andm.

1. Ifm = j, J+|k, j, j =0.

2. If m < j, J+|k, j,m is a non-null eigenvector of J2 and Jz , with the eigenvaluesj (j + 1) 2 and(m + 1).

Proof: (1) The proof is based on similar argument as the proof of Lemma II. The

square of the norm of J+|k, j,m

is zero ifm = j, therefore:

m = j J+|k, j, j =0 (1.70)

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b. DETERMINATION OF THE SPECTRUM OF J2

AND Jz

Let |k, j,m be a non-null eigenvector of J2 and Jz with the eigenvalues j (j + 1) 2and m. According to Lemma I:j m j. That is, there exists a positive or zerointeger ps.t.

j m p

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Now let J

act on ( J

)p|k, j,m

and assume that the eigenvalue (mp)of J

zasso-

ciated with the vector( J)p|k, j,mis greater thanj, i.e. m p>j.

By point (2) of Lemma II, J(J)p|k, j,m is then non-null and corresponds to theeigenvalues j (j + 1) 2 and(mp1). However, this is in contradiction with LemmaI, sincem

p

1

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