chapter 14. angular momentum in quantum me- chanics

QM 14. Angular Momentum, June 15, 2005 1

Chapter 14. Angular momentum in Quantum Me-

chanics

14.1. Introduction

x 1. Angular momentum appeared naturally when we analyzed classical

mechanics of a two-particle system (see Chapter 13), where it is connected

to the rotational energy and the angular velocity. To understand the rota-

tional motion of a molecule, we need to know how Quantum Mechanics treats

angular momentum.

The usefulness of the angular momentum is not limited to systems that

have two particles. Any molecule that is not subjected to an external force

will rotate without impediment and its angular momentum must be con-

served. This happens no matter how many atoms the molecule has.

Why would anyone be interested in the rotational motion of a molecule?

It turns out that rotational energy is quantized and a rotating molecule can

absorb or emit photons, by undergoing transitions between rotational states.

The magnitude of the rotational energies is such that the frequency of these

photons corresponds to microwave radiation. Microwave spectroscopy has de-

veloped complex instruments and procedures for determining the rotational


energies of various molecules and for using these data to extract informa-

tion about molecular structure. The bond lengths and the shapes of many

molecules have been determined very accurately by this method.

You may heard at a seminar or read in a magazine that various mole-

cules have been detected in interstellar space. The detection was not made

by sending a graduate student to a remote galaxy to do some analytical

chemistry there. It was done by measuring the radiation emitted by these

molecules.

To emit radiation a molecule must be in an excited state. This excitation

occurs when molecules in interstellar space collide with each other. Because

the temperature of interstellar space is very low, the molecules located there

have very low kinetic energy (when you study Statistical Mechanics you will

learn why). When two of them collide, neither can gain or lose much energy

(a collision between two slow-moving cars does not cause much damage).

There is enough energy to excite the rotational motion, but not enough to

excite the vibrational motion or the motion of the electrons. Because of

these collisions, some of the molecules are in a excited rotational state and

can radiate photons. A microwave spectroscopist on earth, or a spectrometer

on a space station, can detect this radiation and determine from its frequency


the type of molecule emitting it.

Angular momentum also plays a role in explaining the properties of the

electron in the hydrogen atom (the other two-body system studied here). The

s, p, and d orbitals that you learned about in General Chemistry describe

states in which the electron has di®erent energies and angular momenta.

It is very likely that you have learned that electrons and nuclei have a

mysterious property called `spin'. This is a misleading name, inherited from

the time when it was thought that spin existed because the electron was a

spinning sphere. It turns out that spin is an intrinsic property of a particle,

like mass and charge. A variety of experiments have shown that the spin

states of electrons and nuclei behave as if they are the eigenstates of an

angular-momentum operator. Nowadays we say that the spin is an intrinsic

angular momentum of the particle.

The existence of the nuclear spin made possible the development of nu-

clear magnetic resonance (NMR) spectroscopy. This technique has revolu-

tionized chemistry and medicine, and one cannot imagine a decent chem-

istry laboratory or hospital without a good NMR facility. NMR is based on

changing the spin state of the nuclei by absorption of very low frequency

electromagnetic radiation. To understand the properties of these spin states


one must understand angular momentum.

In this Chapter I give a brief introduction to orbital angular momentum.

As with other dynamical quantities, we need to ¯nd the operators correspond-

ing to the classical angular momentum and to calculate their eigenvalues and

eigenstates. Knowing these will tell us what values the angular momentum

can take and how to calculate the transitions produced by photon absorption.

Since the angular momentum L is a vector, it is fully described when

we know its three components Lx, Ly and Lz. We expect therefore that in

Quantum Mechanics the angular momentum is an operator L , and its three

components Lx, Ly and Lz are all operators. Of these operators, two play an

important role: L2 ´ L2x+ L2y+ L2z and Lz. The reason for this is peculiar to

quantum mechanics. It turns out that we can only measure simultaneously

the values of L2 and one of its components. It is conventional to work with

Lz. Since Lz does not commute with either Lx or Ly, these components of L

cannot be measured simultaneously with Lz. I will tell you later what that

means to a person performing experiments. For now, just be prepared to

hear some weird, but perfectly true and exhaustively veri¯ed, statements.

There is a theorem in quantum mechanics, which I have mentioned in

Chapter 3, that says that two operators that commute (such as L2 and Lz)


can have the same eigenfunctions but di®erent eigenvalues. The joint eigen-

functions of L2 and Lz are called spherical harmonics. They were discovered

before the invention of quantum mechanics and appear frequently in other

¯elds of physics.

The spin is a special kind of angular momentum, not connected to a

rotational motion. The theory of spin is based on a generalization of the

results obtained here for the orbital angular momentum. We postulate that

the spin angular momentum is represented by three operators, analogous to

Lx, Ly, and Lz, which have the same commutation relations as Lx, Ly, and

Lz, but share no other property. Therefore, all orbital angular momentum

properties that can be derived from the commutation relations are also valid

for spin. Most of the concepts required for understanding spin later are

introduced in this chapter.

Finally, at the end of the chapter we show that the two-particle Hamil-

tonian derived in the previous chapter can be written in terms of the angular

momentum squared, as suggested in Chapter 13, x15.


14.2. The operators representing the angular momentum in Quan-

tum Mechanics

x 2. Angular momentum in classical mechanics. We have already encoun-

tered the angular momentum operator, very brie°y, in Chapter 2, x15. Since

we plan to study it extensively in this chapter, I will remind you a few im-

portant properties.

In classical mechanics the angular momentum of a particle is de¯ned by

L = r£ p (14.1)

Here r is the position of the particle and p is its momentum. The symbol

£ appearing in Eq. 14.1 denotes the vector product (or cross product) of r

with p. The components of the vector L are

Lx = y pz ¡ z py (14.2)

Ly = z px ¡ x pz (14.3)

Lz = x py ¡ y px (14.4)

where x; y; z are the components of the vector r, and px, py, pz are those of

the vector p.

Eq. 14.1 is the general de¯nition of the angular momentum of a particle

at position r, having momentum p. You should, however, keep in mind that


we are interested here in a two-particle system; in that case r is the vector

pointing from particle 1 to particle 2, whose length is equal to the distance

between the particles, and p is the vector with components ¹dxdt, ¹dy

dt, ¹dz

dt,

where ¹ is the reduced mass (see Chapter 13).

x 3. The angular momentum operator in Quantum Mechanics. As you have

learned in Chapter 2, all classical dynamical quantities are represented in

Quantum Mechanics by operators. The operators x , y and z, corresponding

to the coordinates x, y and z are de¯ned by:

x Ã(x; y; z) ´ xÃ(x; y; z) (14.5)

y Ã(x; y; z) ´ y Ã(x; y; z) (14.6)

z Ã(x; y; z) ´ z Ã(x; y; z) (14.7)

When applied to a function, these operators multiply the function by the

corresponding coordinate.

The operators corresponding to the momentum components are:

px Ã(x; y; z) ´ ¹h

i

@

@xÃ(x; y; z) (14.8)

py Ã(x; y; z) ´ ¹h

i

@

@yÃ(x; y; z) (14.9)

pz Ã(x; y; z) ´ ¹h

i

@

@zÃ(x; y; z) (14.10)

with i =p¡1. Here we show how the operators act on an arbitrary function


Ã(x; y; z).

Replacing x, y, z, px, py, pz in Eqs. 14.2{14.4 with the operators de¯ned

by Eqs. 14.5-14.7 and 14.8-14.10 gives us the operators for the components

of the angular momentum:

LxÃ =¹h

i

Ãy@Ã

@z¡ z @Ã

@y

!(14.11)

LyÃ =¹h

i

Ãz@Ã

@x¡ x@Ã

@z

!(14.12)

LzÃ =¹h

i

Ãx@Ã

@y¡ y@Ã

@x

!(14.13)

x 4. Angular momentum in spherical coordinates. Angular momentum is

connected to the rotational motion of a particle. As you have seen in the

previous chapter, this motion is best described by using the spherical coordi-

nates fr; µ; Ág (see Chapter 13, Figure 13.2). This means that in Eqs. 14.11{

14.13 we must replace x, y, and z with expressions containing r, µ, and Á

and also replace @@x, @@y, @@zwith expressions containing only functions of r,

µ, and Á and the partial derivatives @@r, @@µ, @@Á

The result is:

Lx = i¹h

ÃsinÁ

@

@µ+ cot µ cosÁ

@

@Á

!(14.14)

Ly = i¹h

Ã¡ cosÁ @

@µ+ cot µ sinÁ

@

@Á

!(14.15)

Lz = ¡i¹h @@Á

(14.16)


The algebra needed for performing this change of variables is straight-

forward but rather tedious. In Supplement 14.1 I show, brie°y, how such a

change of coordinates is performed. The procedure uses the equations con-

necting the spherical coordinates to the Cartesian ones (Chapter 13, x11,

Eqs. 10.29{10.31) and the chain rule of calculus.

If you like performing lengthy and tedious calculations, the derivation of

Eqs. 14.14{14.16 from Eqs. 14.11{14.13 is fun. If your de¯nition of fun di®ers

from mine, do not panic. Going back and forth between various coordinate

systems is so common in all branches of physics (electrodynamics, quantum

mechanics, °uid mechanics, theory of elasticity) that it is well covered in

many books. A classic is P. M. Morse and H. Feshbach, Methods of Theo-

retical Physics (McGraw Hill Book Company, New York, 1953), where the

change of coordinates is explained in Chapter I. A practicing physical chemist

is most likely to look up in that book the necessary equations, rather than

spending time doing the transformation.

x 5. The operator L2. In the classical analysis of rotational motion of a two-

particle system (see Chapter 13), the rotational energy is proportional to the

angular momentum squared. It is reasonable to expect that this quantity is

an important player in the quantum theory of this system. In preparation


for this, we determine here the expression for the operator L2.

The angular momentum is a vector and its length is given by

L2 = L2x + L2y + L

2z (14.17)

To obtain L2 in spherical coordinates, I insert in Eq. 14.17 the expressions

from Eqs. 14.14{14.16 and simplify the result. This is another tedious (but

straightforward) calculation that I will not do in detail. You can test that

the result is:

L2Ã = ¡¹h2"1

sin µ

@

@µsin µ

@Ã

@µ+

1

sin2 µ

@2Ã

@Á2

#(14.18)

We will have an opportunity to see this operator at work shortly.

Exercise 14.1 Convince yourself that the procedure described above leads

to Eq. 14.18.

14.3. The commutation relations between L2 and Lx, Ly, Lz

x 6. Introduction. When we studied the eigenvalue problem in Chapter 3,

I mentioned two general theorems. Two operators commute if and only if

they have essentially the same eigenfunctions. Moreover, I mentioned that if

two operators commute then the magnitude of the quantities represented by

these operators can both be measured without one measurement interfering


with the other. I can measure L2 alone, or Lz alone, or ¯rst L2 and then

Lz, and will get the same results in any of these measurements. We say that

the two quantities can be measured simultaneously. If two operators do not

commute, then measuring the magnitude of one quantity alters the value of

the other one. For example, Lz and Lx do not commute. This means that if

I prepare two identical systems and on one I measure Lz and then I measure

Lx, and on the other I measure Lx directly, I can get di®erent results for Lx.

We say that Lz and Lx cannot be measured simultaneously.

These theorems make clear that it is important to know whether the

operators L2, Lx, Ly, Lz commute with each other.

There is however a deeper reason to study the commutation relations.

We have introduced the angular momentum by replacing r and p, in the

classical de¯nition L = r £ p of the orbital angular momentum, with the

appropriate operators. This gives us the operators L2, Lx, Ly and Lz. Using

these expressions we can calculate the commutators of each pair of operators

(e.g. [L2; Lx]). It turns out that these commutation relations provide a more

general de¯nition of the angular momentum than L = r£p. This means that

we can derive, from the commutation relations, expressions for all physically

relevant quantities that we can derive from L = r £ p. In addition the


de¯nition based on the commutation relations can be applied equally well

for the spin, for which L = r£ p makes no sense.

x 7. The commutation relations. I remind you the commutator [A; B] of two

operators A and B is

[A; B]Ã ´ ABÃ ¡ BAÃ (14.19)

The commutator is an operator and I show it acting on an arbitrary function

Ã. Two operators commute when their commutator satis¯es

[A; B]Ã = 0

when applied to any arbitrary function Ã.

It is tedious, but not di±cult, to show that (use the de¯nitions Eq. 14.14{

14.16 and the de¯nition of the commutator Eq. 14.19)

hLx; Ly

i= i¹hLz (14.20)

hLy; Lz

i= i¹hLx (14.21)

hLz; Lx

i= i¹hLy (14.22)

The components of the angular momentum do not commute with each other.

Therefore, we cannot measure them simultaneously.

It is again tedious but not di±cult, to use the de¯nition (Eq. 14.18) of


L2 and the de¯nition (Eqs. 14.14{14.16) of Lx, Ly, Lz, to show that

hL2; Lx

iÃ(r; µ; Á) = 0 (14.23)

hL2; Ly

iÃ(r; µ; Á) = 0 (14.24)

hL2; Lz

iÃ(r; µ; Á) = 0 (14.25)

These equations show that the angular momentum squared (which is pro-

portional to the rotational energy) commutes with each component of the

angular momentum. This means that we can measure simultaneously the

value of L2 and that of one component. It is customary to choose this com-

ponent to be Lz, although it makes no di®erence which component is used.

These results tell us that it is impossible to place the system in a state

in which we can measure simultaneously the values of Lx, Ly, and Lz. We

can only place the system in a state in which we know L2 and one of the

components (here we choose Lz). This means that it is impossible in Quan-

tum Mechanics to create a state in which we can determine the direction of

the angular momentum vector. We will discuss this in more detail when we

examine the rotational motion of a diatomic molecule and the rotation of the

electron in a hydrogen atom.

Exercise 14.2 Use a symbolic manipulation program to test that the com-


mutation relations written above are correct, regardless of what function the

commutator is applied to.

14.4. The eigenvalue equations for L2 and Lz

x 8. As you have learned in Chapter 4, if we want to know what values a

certain dynamical variable can take in an experiment, we need to ¯nd the

eigenvalues of the corresponding operator.

The eigenvalue equation for an arbitrary operator O is (see Chapter 3) is

OÃ(µ; Á) = ¸Ã(µ; Á) (14.26)

In writing this equation, I assumed that the operator acts on functions that

depend on µ and Á, which is the case for the operators related to the angular

momentum.

As explained in Chapter 3, there may be many pairs f¸i; Ãi(µ; Á)g, i =

0, 1, 2, . . . that satisfy this equation. The functions Ãi(µ; Á) are called the

eigenfunctions of the operator O and the numbers ¸i are the eigenvalues of

O corresponding to the eigenfunctions Ãi(µ; Á).

The eigenfunctions and the eigenvalues contain all the information needed

for understanding the phenomena that involve the quantity O.


x 9. The eigenvalue problem for L2. The general formula Eq. 14.26 applied

to the operator eigenvalue problem for L2 is (replace O with L2 in Eq. 14.26

and then use Eq. 14.18 for L2)

L2Ã(µ; Á) = ¡¹h2"1

sin µ

@

@µsin µ

@Ã(µ; Á)

@µ+

1

sin2 µ

@2Ã(µ; Á)

@Á2

#

= ¸Ã(µ; Á) (14.27)

This equation appears in many branches of physics and was solved in the

nineteenth century. The eigenfunctions are called spherical harmonics and

are denoted by Y m` (µ; Á). The eigenvalues are

¸` = ¹h2`(`+ 1); ` = 0; 1; 2; : : : (14.28)

This means that the spherical harmonics Y m` (µ; Á) and the eigenvalues ¸`

satisfy the equation

L2 Y m` (µ; Á) = ¹h2`(`+ 1)Y m` (µ; Á) (14.29)

where ` = 0; 1; 2; : : : (14.30)

and m = ¡`;¡`+ 1; : : : ; `¡ 1; ` (14.31)

The eigenfunctions Y m` (µ; Á) are labeled by two indices, ` and m. It is

not unusual to have more than one index labeling an eigenfunction or even

an eigenvalue. In Chapter 8 you have seen that the eigenfunctions and the

eigenvalues of a particle in a box were labeled by three indices.


The values the indices ` and m can take are prescribed by Eqs. 14.30 and

14.31. For every value of ` ,m can take 2`+1 values. Since the eigenvalues are

independent of m, there are 2`+1 eigenfunctions Y m` (namely, Y ¡`` , Y ¡(`¡1)` ,

. . . , Y `¡1` , Y `` ) for each eigenvalue ¹h2`(` + 1). We say that the eigenvalue

¹h2`(`+ 1) is degenerate and its degeneracy is equal to 2`+ 1.

We will discuss shortly the physical meaning of this result. But before

we do this, let us look at the eigenvalue problem for the operator Lz.

x 10. The eigenvalue problem for Lz. Since Lz and L2 commute, they must

have the same eigenfunctions. Indeed, one can show that the equation

Lz Ym` (µ; Á) = ¹hmY

m` (µ; Á) (14.32)

is satis¯ed for any values of ` and m allowed by Eqs. 14.30 and 14.31. Thus,

the eigenvalues of Lz are

¹hm with m = ¡`;¡`¡ 1; : : : ; `¡ 1; ` (14.33)

For example, if ` = 1, Y ¡11 , Y 01 , Y11 are eigenfunctions of Lz with the eigen-

values ¡¹h, 0 and ¹h, respectively.

x 11. Spherical harmonics. The spherical harmonics Y m` (µ; Á) have been

extensively studied in many branches of physics. They have very interest-

ing properties, but we don't have time to examine them here. The general


formula is

Y m` (µ; Á) =(¡1)`+m2``!

vuut2`+ 14¼

(`¡m)!(` +m)!

eimÁ(sin µ)md`+m

d(cos µ)`+m

³1¡ cos2 µ

´`(14.34)

I give below the functional form of Y m` for a few values of ` and m.

Y 00 (µ; Á) =1

2p¼

(14.35)

Y ¡11 (µ; Á) =1

2e¡iÁ

s3

2¼sin µ (14.36)

Y 01 (µ; Á) =1

2

s3

¼cos µ (14.37)

Y 11 (µ; Á) = ¡12eiÁ

s3

2¼sin µ (14.38)

Y ¡22 (µ; Á) =1

4e¡2iÁ

s15

2¼sin2 µ (14.39)

Y ¡12 (µ; Á) =1

2e¡iÁ

s15

2¼cos µ sin 2µ (14.40)

Y 02 (µ; Á) =1

4

s5

¼(¡1 + 3 cos2 µ) (14.41)

Y 12 (µ; Á) = ¡12eiÁ

s15

2¼cos µ sin µ (14.42)

Y 22 (µ; Á) =1

4e2iÁ

s15

2¼sin2 µ (14.43)

Y ¡33 (µ; Á) =1

8e¡3iÁ

s35

¼sin3 µ (14.44)

Y ¡23 (µ; Á) =1

4e¡2iÁ

s105

2¼cos µ sin2 µ (14.45)

Y ¡13 (µ; Á) =1

8e¡iÁ

s21

¼(¡1 + 5 cos2 µ) sin µ (14.46)


Y 03 (µ; Á) =1

4

s7

¼(¡3 cos µ + 5 cos3 µ) (14.47)

Y 13 (µ; Á) = ¡18eiÁ

s21

¼(¡1 + 5 cos2 µ) sin µ (14.48)

Y 23 (µ; Á) =1

4e2iÁ

s105

2¼cos µ sin2 µ (14.49)

Y 33 (µ; Á) = ¡18e3iÁ

s35

¼sin3 µ (14.50)

Y ¡44 (µ; Á) =3

16e¡4iÁ

s35

2¼sin4 µ (14.51)

Y ¡34 (µ; Á) =3

8e¡3iÁ

s35

¼cos µ sin3 µ (14.52)

Y ¡24 (µ; Á) =3

8e¡2iÁ

s5

2¼(¡1 + 7 cos2 µ) sin2 µ (14.53)

Y ¡14 (µ; Á) =3

8e¡iÁ

s5

¼(¡3 + 7 cos2 µ) cos µ sin µ (14.54)

Y 04 (µ; Á) =3

16p¼(3¡ 30 cos2 µ + 35 cos4 µ) (14.55)

Y 14 (µ; Á) = ¡38e¡iÁ

s5

¼(¡3 + 7 cos2 µ) cos µ sin µ (14.56)

Y 24 (µ; Á) =3

8e2iÁ

s5

2¼(¡1 + 7 cos2 µ) sin2 µ (14.57)

Y 34 (µ; Á) = ¡38e3iÁ

s35

¼cos µ sin3 µ (14.58)

Y 44 (µ; Á) =3

16e4iÁ

s35

2¼sin4 µ (14.59)

These expressions were generated in WorkBookQM.14.2 by using the

function SphericalHarmonicY[`;m; µ; Á] provided by Mathematica, which

returns the expression for function Y m` (µ; Á). In WorkBookQM.14.3, I tested


that the general expression (Eq. 14.34) gives the same result.

If you use other sources to obtain expressions for the spherical harmonics,

you should be aware that di®erent authors sometimes use slightly di®erent

de¯nitions. They are all eigenfunctions of L2 and Lz, because they di®er

only by multiplicative factors.

x 12. The physical interpretation of these eigenstates. As you have seen in

the previous chapters, the eigenstates of an operator can be used to study

a variety of physical properties of the quantity that the operator represents.

I postpone studying the physical properties of the angular momentum until

we examine the properties of a diatomic molecule and those of a hydrogen

atom. In those systems the angular momentum properties are connected

to the rotation of the diatomic or the rotation of the electron around the

nucleus. The angular momentum properties are entangled with the radial

motion and are easier to understand in the context of speci¯c systems.

Exercise 14.3 Show that if ªm` (µ; Á) is an eigenfunction of an arbitrary

operator, then ®ªm` (µ; Á) , where ® is an arbitrary complex number, is also

an eigenfunction, corresponding to the same eigenvalue as ªm` (µ; Á).

Exercise 14.4 Take several values of ` and m and verify Eqs. 14.29 and

14.32 for Y m` given above. Some examples are given in WorkBookQM.14.2.


Exercise 14.5 1. Show that the function

Xm=¡`

cm(r)Ym` (µ; Á) (14.60)

is an eigenfunction of L2 with the eigenvalue ¹h2`(`+ 1).

2. Show that the function de¯ned by Eq. 14.60 is not an eigenfunction of

Lz.

3. Choose the coe±cients cm in Eq. 14.60 to construct 2` + 1 real (as

opposed to imaginary or complex) eigenfunctions ´`(r; µ; Á) of L2. For

example, for ` = 1 you should construct three real eigenfunctions ´1,

´2, ´3 corresponding to three di®erent choices of c¡1, c0, and c1. All

three should be eigenfunctions of L2 with the eigenvalue 1(1+1)¹h2. For

` = 2 there are ¯ve such real eigenfunctions, etc. Going beyond ` = 2

becomes complicated, but perhaps you see a pattern that will help you

out. The real functions constructed in this way give the angular part of

the famous s, p, d orbitals you have learned about in general chemistry.

We will construct and use them when we study the hydrogen atom and

the chemical bond.


Supplement 14.1. A brief explanation of the procedure for changing

coordinates

x 13. The change of variables from fx; y; zg to fr; µ; Ág is motivated by

physics. However, the calculation is a tedious exercise in calculus. If you are

interested in such manipulations, you will ¯nd here a hint of how they are

done. However, you can skip this material without irreparable damage to

your education.

To go from LxÃ given by Eq. 14.11 to LxÃ given by Eq. 14.14, I express

x, y, and z in terms of r, µ, and Á by using Eq. 13.29. This is the easy part.

It is harder to express (@Ã=@z)x;y and (@Ã=@y)x;z in terms of (@Ã=@µ)r;Á,

(@Ã=@Á)r;µ, and (@Ã=@r)µ;Á. The subscripts in these formulae are added to

remind me which variables are held constant when the derivative is taken.

The chain rule gives

Ã@Ã

@x

!y;z

=

Ã@Ã

@r

!µ;Á

Ã@r

@x

!y;z

+

Ã@Ã

@µ

!r;Á

Ã@µ

@x

!y;z

+

Ã@Ã

@Á

!r;µ

Ã@Á

@x

!y;z

(14.61)

I will calculate derivatives (@r=@x)y;z, (@µ=@x)y;z, and (@Á=@x)y;z from Eq. 13.31,

which expresses r, µ, and Á as functions of x, y, and z. From r =px2 + y2 + z2,


I get Ã@r

@x

!y;z

=@

@x

qx2 + y2 + z2 =

2x

2px2 + y2 + z2

=x

r=r sin µ cosÁ

r= sin µ cosÁ (14.62)

In this calculation, I used x = r sin µ cosÁ (Eq. 13.29).

To calculate (@µ=@x)y;z, I use cos µ = z=px2 + y2 + z2 (Eq. 13.31). Tak-

ing the derivative givesÃ@ cos µ

@x

!y;z

=@

@x

Ãzp

x2 + y2 + z2

!y;z

(14.63)

which leads to

¡ sin µÃ@µ

@x

!y;z

= ¡z 1

(px2 + y2 + z2)2

2x

2px2 + y2 + z2

(14.64)

Usingpx2 + y2 + z2 = r, x = r sin µ cosÁ, and z = r cos µ (Eq. 13.29), this

becomes Ã@µ

@x

!y;z

= +(r sin µ cosÁ)(r cos µ)

r3 sin µ=cos µ cosÁ

r(14.65)

To calculate (@Á=@x)y;z, I use Eq. 13.31:Ã@ tanÁ

@x

!y;z

=@

@x

µy

x

¶y;z

(14.66)

Using (@ tanÁ=@x)y;z = sec2 Á (@Á=@x)y;z and

@@x(y=x)y;z = ¡y=x2 to perform

the derivatives gives

sec2 Á

Ã@Á

@x

!y;z

= ¡ yx2

(14.67)


Using y = r sin µ sinÁ, x = r sin µ cosÁ, and sec2 Á = 1= cos2 Á in this equa-

tion gives Ã@Á

@x

!y;z

= ¡ sinÁ

r sin µ(14.68)

Let us see what we have achieved. I want to calculate (@Ã=@x)y;z by

using Eq. 14.61. The derivatives (@r=@x)y;z, (@µ=@x)y;z, and (@Á=@x)y;z that

appear in Eq. 14.61 are given by Eqs. 14.62, 14.65, and 14.68. Putting these

in Eq. 14.61 gives

Ã@Ã

@x

!y;z

= sin µ cosÁ

Ã@Ã

@r

!µ;Á

+cos µ cosÁ

r

Ã@Ã

@µ

!r;Á

¡ sinÁ

r sin µ

Ã@Ã

@Á

!r;µ

(14.69)

You can perform the same kind of calculations to get (@Ã=@y)x;z and

(@Ã=@z)x;y. This is a tedious exercise in calculus. You should try to do it to

see if you can obtain Eqs. 14.14{14.16. The intermediate results are

Ã@r

@y

!x;z

= sin µ sinÁ (14.70)Ã@Á

@y

!x;z

=cosÁ

r sin µ(14.71)Ã

@µ

@y

!x;z

=cos µ sinÁ

r(14.72)Ã

@r

@z

!x;y

= cos µ (14.73)Ã@Á

@z

!x;y

= 0 (14.74)


Ã@µ

@z

!x;y

= ¡sin µr

(14.75)

Since

@Ã

@y=

Ã@Ã

@r

!Ã@r

@y

!x;z

+

Ã@Ã

@µ

!Ã@µ

@y

!x;z

+

Ã@Ã

@Á

!Ã@Á

@y

!x;z

; (14.76)

I use Eqs. 14.70{14.72 to obtainÃ@Ã

@y

!x;z

= sin µ sinÁ@Ã

@r+cos µ sinÁ

r

@Ã

@µ+cosÁ

r sin µ

@Ã

@Á(14.77)

Similarly, I get Ã@Ã

@z

!x;y

= cos µ@Ã

@r¡ sin µ

r

@Ã

@µ(14.78)

Next, use Eq. 13.29 for x; y; z and Eqs. 14.69, 14.77, and 14.78 for the

derivatives (@Ã=@x)y;z, (@Ã=@y)x;z, and (@Ã=@z)x;y in the de¯nitions (Eqs. 14.11{

14.13) of LxÃ, LyÃ, and LzÃ, and you will obtain Eqs. 14.14{14.16. This

is what textbook writers describe as a trivial but tedious calculation. In

most cases, they mean: we know that this is right, but we are unwilling to

prove it. If you go on and learn more about Physical Chemistry, you will

¯nd out that there are specialized books that give anything you need to go

from one coordinate system to another. You don't need to do this sort of

thing yourself. It is more important to know when it is advantageous to use

a particular coordinate system. Here the force has spherical symmetry and

this suggests using spherical coordinates.

chapter 14. angular momentum in quantum me- chanics

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