CHEM 614
MODULE 6
ANGULAR MOTION AND ANGULAR MOMENTUM
We started up this subject in Module 3 and now we need to expand on where we left it. In that Module
we examined the solution to the Schrödinger equation for motion in a circular trajectory about a
central, fixed point. The spinning of a CD typifies such motion in the classical regime. We assumed
that the potential energy of the particle was zero as it goes around its trajectory. We found we could
solve the linear, second order differential equation and arrive at an expression for the corresponding
quantized energy levels according to:
 m  Aeim   Beim 
l
(6.1)
l
l
1/ 2
 2 IE 
where  is the angular variable and ml   2 


with I = mr2 (the moment of inertia).
In order to satisfy the cyclical boundary conditions we saw that ml is restricted to the values
0,  1,  2,... The energy solutions corresponding to the functions were shown to be
Eml 
( ml ) 2
2I
(6.2)
One thing to note about this is that ml = 0 is an allowed value which leads to E0 = 0. In words this
means that the rotating disk has no zero point energy, very different from our other quantum
mechanical systems.
Angular Momentum Considerations
Up to this point the integer ml is simply a quantum number that specifies individual wavefunctions and
energy levels, but now we show that it has a wider significance.
We start from the classical concept of angular momentum.
In Figure 6.1 we depict the perspective view of the motion
under consideration. The particle of mass m rotates in the
plane that is perpendicular to the defined z-axis.
xy
The
system rotates clockwise when viewed from below with
angular velocity  (radians per second).
an
The radial
vector is defined by r and the linear momentum vector
Figure 6.1
( p ) is directed tangentially to the radial vector.
[NOTE: in order to be consistent with the vector notation in Barante and Lowe, I now show vectors in
bold face type.]
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Classical physics now tells us that such motion produces an angular momentum vector ( L ) in the
direction orthogonal to the plane of the motion. (See the gyroscope example in Lowe, chapter 4).
The classical expression is
L  r p
(6.3)
the cross product of the vectors yields a third vector that is orthogonal to the components.
In the case where the motion is in three dimensions we use the determinant
i
L x
px
j
y
py
k
z
pz
(6.4)
where the i, j, and k terms are the unit vectors in the x, y, and z directions, respectively. The
magnitude of the z-component of angular momentum can be picked out of determinant (6.4) as
Lz  xp y  yp x
(6.5)
whence the operator for the z-component of angular momentum becomes:
Lˆz  x


y
i y
i x
(6.6)
which, in polar coordinates becomes
Lˆz 

i 
(6.7)
Now let us see how our wavefunction (6.1) responds to this operator
Lˆz ml 

( Aeiml  Be  iml )
i 
(6.8)
Recall from our earlier discussion that the two exponential terms correspond to motion in the two
directions, clockwise and anti-clockwise. Let us consider a single direction by putting B = 0 then
Lˆz ml 

Aeiml  ml Aeiml  ml  ml
i 
(6.9)
Equation (6.9) is therefore an eigenvalue equation and ml is the (REAL) eigenvalue for the zcomponent of angular momentum. Thus we note that the z-component of the angular momentum is
quantized according to the quantum mechanical treatment and we see that the quantity ml is the
quantum number for the z-component of the angular momentum due to motion about a central point.
Normalization of the wavefunction yields A 
1
.
2
2
Angular momentum operators
If i, j,and k are the orthogonal unit vectors, then the position and linear momentum vectors of a
particle can be expressed as
r  xi  yj  zk
(6.10)
p  p x i  p y j  pz k
(6.11)
and
it follows that the angular momentum can be similarly expressed
L  Lx i  Ly j  Lz k
(6.12)
We saw above that angular momentum is the vector product of position and linear momentum vectors
and thus
i
L  r xp  x
px
j
y
py
k
z
pz
(6.13)
and picking out the individual components from the determinant we obtain
L  ( ypz  zp y )i  ( zpx  xpz ) j  ( xp y  yp x ) k
(6.14)
thus from (6.12) and (6.14) the three components can be identified as
Lx  ( ypz  zp y )
Ly  ( zpx  xpz )
(6.15)
Lz  ( xp y  ypx )
Note that each of these identities can be generated by cyclic permutation of its predecessor.
The magnitude (L, a scalar quantity) of the angular momentum is related to its components in the same
way that any vector magnitude can be constructed
L2  Lx 2  Ly 2  Lz 2
(6.16)
According to the principles of classical mechanics, no constraints are placed on the energy of a rotating
system, and therefore none on the magnitude of the angular momentum, or any of the components
thereof. The quantum interpretation paints a different picture.
The above definitions carry forward into the quantum realm where the q and pq in the definitions of Lq
become interpreted as operators. Recall that we construct a position operator from q by replacing it
with the multiplier q, and a momentum operator by replacing pq by ( / i ) / q . Thus
 
 
 
 
 
 
Lˆx   y  z  Lˆ y   z  x  Lˆz   x  y 
i  z
y 
i  x
z 
i  y
z 
3
(6.17)
We now establish the commutation relationships of the Lq operators. First the commutator of the
operators Lx and Ly can be found:
[ Lˆ x , Lˆ y ]  [( ypz  zp y ),( zpx  xpz )]
 [ ypz , zpx ]  [ ypz , xpz ]  [ zp y , zp x ]  [ zp y , xpz ]
 y[ pz , z ] px  0  0  p y [ z, pz ]x  i (  yp x  xp y )
 i Lˆz
In the first line of the above sequence we simply replace the operators by the magnitudes of the
observables they represent (see equation (6.15). This is a stratagem that simplifies the procedure
(compare this with the normal” approach used by Lowe in A4-3 which generates lots of mixed second
order partial derivatives). To generate line 2 we multiplied out term by term. The zeros in line 3
appear because the z-component of p commutes with both y and x, and z commutes with the x- and ycomponents of p. Writing out the commutators and rearranging generates the non-zero terms in line 3,
at the end of which we have reverted to the operator symbolism.
The other two commutators can be found in the same way, or can be simply written down using
cyclical permutation. In summary then
[ Lˆ x , Lˆ y ]  i Lˆz
[ Lˆ y , Lˆz ]  i Lˆ x
(6.18)
[ Lˆz , Lˆ x ]  i Lˆ y
The remaining operator we need is that corresponding to the square of the magnitude of the angular
momentum, L̂2 . We need to know whether it commutes with the Lq components
[ Lˆ2 , Lˆz ]  [ Lˆx 2  Lˆ y 2  Lˆz 2 , Lˆz ]  [ Lˆx 2 , Lˆz ]  [ Lˆ y 2 , Lˆz ]  [ Lˆz 2 , Lˆz ]
 [ Lˆx 2 , Lˆz ]  [ Lˆ y 2 , Lˆz ]  0
(6.19)
Now we examine each of the two non-zero terms in turn
[ Lˆx 2 , Lˆz ]  Lˆx Lˆx Lˆz  Lˆz Lˆx Lˆx
 Lˆx Lˆx Lˆz ( Lˆx Lˆz Lˆx  Lˆx Lˆz Lˆx )  Lˆz Lˆx Lˆx
 Lˆ [ Lˆ , Lˆ ]  [ Lˆ , Lˆ ]Lˆ
x
x
z
x
z
(6.20)
x
 i ( Lˆx Lˆ y  Lˆ y Lˆx )
The parenthesis placed in line 2 is the overall addition of zero to the RHS.
Similarly
[ Lˆ y 2 , Lˆz ]  i ( Lˆx Lˆ y  Lˆ y Lˆx )
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(6.21)
Then the sum of the two terms in equation (6.21) is zero and therefore Lˆ2 and Lˆz must commute.
Moreover, Lx, Ly, and Lz all occur symmetrically in L2 and therefore all must commute with L2, if any
one of them does. In other words
[ Lˆ2 , Lˆq ]  0
(6.22)
The commutation relations expressed in (6.18) and (6.22) are fundamental to the theory of angular
momentum, and are quite general. In fact we can usually say that an observable is an angular
momentum if its operators satisfy the above commutation relationships.
The Particle on a Sphere
The next step up from a particle moving in a circle (2-dimensions) is the case of a particle moving on
the surface of a sphere (3-dimensions).
This is equivalently the motion of an actual particle
constrained to move at a constant radius from a central point, or the motion of a point in a solid body
that is representative of the motion of the whole body. Although the system might seem to have little
relevance, it turns out that the results of solving the equations of motion for this system are very useful
when we come to look at the motion of electrons in spherically symmetrical Coulomb fields, and at the
energy states of rotating diatomic molecules.
As in the case of the particle undergoing motion in a circle the potential energy of the particle is
constant and we equate it to zero. Then the hamiltonian is
Hˆ  
2
2m
2
(6.23)
The most useful form of the laplacian to use is the one expressed in spherical polar coordinates
x  r sin  cos  ; y  r sin  sin  ; z  r cos 
and when these are substituted into the Cartesian laplacian it becomes
2 
1 2
1
r  2 2
2
r r
r
(6.24)
where the legendrian operator 2 is defined as
2 
1 2
1 


sin 
2
2
sin  
sin  

(6.25)
now you can see that the legendrian is the angular part of the laplacian operator-it contains no terms in
r. Since the particle is confined to a spherical surface of fixed radius we can ignore the radial
derivatives in the laplacian and we are left with the legendrian as being the operator in the Schrödinger
equation
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Hˆ  
2
2
1 2



2
2m r 2
2I
(6.26)
where I is the moment of inertia.
Thus the Schrödinger equation that we must solve is
 2 IE 
 2 ( ,  )    2  ( , )


(6.27)
where the wavefunction is a function of  and See Figure 6.2 for a diagram of the coordinate
system.
One way to solve the Schrödinger equation (6.27) is to recognize the analogy that a sphere can be
regarded as simply a stack of rings. Thus the solution should resemble that for the case of the ring
except that now the particle can travel from ring to ring always maintaining its same distance from the
z


r
y
x
Figure 6.2
center of the sphere. In this analogy we would expect equation (6.27) to be separable and of the form
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 ( , )  ( )( )
(6.28)
This is in fact the case and the  component turns out to be
 2
 constant x 
 2
(6.29)
which is the same as that for the particle on the ring. Moreover the cyclical boundary conditions are
the same, the solutions are the same, and specified by the quantum number ml. The quantum number
has the same values as before ( 0, 1, 2,... ) and it specifies the magnitude of the angular momentum
component on the z-axis.
We still have to worry about the  function, which is not so simply
dealt with. We would expect, however that just as the particle on the
ring had to obey the cyclical boundary condition which led to the
quantum number ml, now there will be a circum-polar cyclical
boundary condition to be satisfied and this will no doubt lead to
another quantum number. Figure 6.3 shows this in a schematic way.
One approach to finding the solution to equation (6.27) is simply to
recognize that it corresponds to a well-known mathematical function
Figure 6. 3
and to refer to the appropriate tables for its solution. Taking this
approach we find that the solutions are the spherical harmonics, Ylml ( ,  ) . They satisfy the
eigenvalue equation
2Ylml  l (l  1)Ylml
(6.30)
The labels l and ml are found to have the following values:
l  0,1,2,...and ml  l , l  1,...  l . Thus the solutions behave in a quantized manner and for a given
value of l there are 2l+1 values of ml..
Table 6.1 lists the spherical harmonic functions for l = 0 through 3, and their amplitudes at different
points on a spherical surface are depicted in Figure 6.4. The spherical harmonics (see Table 6.1) are
complex functions except for the cases where ml = 0. In order to envisage their shapes it is usual to
plot the boundary surfaces of  2 . The boundary surface for l = ml = 0, has spherical symmetry, but
those for higher values of these two quantum numbers exhibit nodes.
As predicted in the discussion above the Schrödinger equation (6.27) is separable which leads to the
idea that the spherical harmonics are composed of two factors
Ylml ( ,  )  lml ( ) ml ( )
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(6.31)
The  functions are the same as those for the particle on the ring as already described; the  functions
are called the associated Legendre functions.
Comparing equations (6.27) and (6.30) it is easy to see that the particle energies are confined to the
values
 2
Elml  l (l  1)  
 2I 
(6.32)
The quantum number l specifies the energy of the particle, and the energy is independent of ml.
Therefore each energy level is (2l+1)-fold degenerate.
Table 6.1
l
ml
Yl ,ml
0
0
 1 


 4 
1
0
 3 


 4 
1
 3 


 8 
1/ 2
1/ 2
cos 
1/ 2
sin  e  i
1/ 2
2
0
 5 


 16 
1
 15 


 8 
2
 15 


 32 
(3cos 2   1)
1/ 2
cos  sin  e  i
FIGURE 6.4: note that the number of nodes
1/ 2
sin 2  e 2 i
increases as l increases.
1/ 2
3
0
 7 


 16 
(5cos3   3cos  )
1/ 2
1
 21 


 64 
2
 105 


 32 
(5cos 2   1) sin  e  i
1/ 2
sin 2  cos  e 2 i
1/ 2
3
 35 


 64 
sin 3  e 3i
Angular Momentum Considerations
In the same way that the quantum number ml specified the magnitude of the z-component of angular
momentum of the particle moving in a circle, we might anticipate that the quantum number l likewise
has implications to the angular momentum of the particle on the sphere. Classical physics tells us that
the rotational energy of a spherical body of moment of inertia I and angular velocity  is given
by E  I  2 / 2 . (This is analogous to the equation for the kinetic energy of a moving particle of mass
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m and velocity v, i.e. Ekin = mv2/2). Moreover, and again analogous to the linear case, the angular
momentum of the spherical rotator is given as
L = I
From these relationships we see that energy can be expressed in terms of momentum by
E = L2/2I.
Comparing this expression with equation (6.32) leads to the conclusion that the magnitude of angular
momentum is given by
L  l (l  1)
1/ 2
(6.33)
Thus we see that the magnitude of the angular momentum is restricted by l to certain values and l is
called the angular momentum quantum number.
It turns out that the spherical harmonics are eigenfunctions of
Lˆ z , the operator for the z-component of the angular
momentum, and the eigenvalues are ml . Thus ml specifies
the component of the angular momentum that can be ascribed
to rotation around that axis. But, since ml is restricted to
certain values, the z-component of the angular momentum is
restricted to 2l+1 discreet values for any value of l.
If the angular momentum is represented as a vector of length
that is proportional to its magnitude (of length {l(l+1)}1/2
units), then the vector must be oriented with respect to the
(arbitrary) z-axis such that its projection thereon is equal to a
length of ml units.
This result means that the plane of
rotation of the particle on the sphere can assume only a
Figure 6.5
discrete set of orientations (Figure 6.5 which shows the
permitted orientations for l = 2). Thus, contrary to what
classical physics tells us, quantum mechanics requires that the orientation of a rotating body is subject
to quantum restrictions. This is referred to as space quantization. Figure 6.5 is useful to visualize the
idea of space quantization, but it is deficient in that it implies that the components Lx and Ly are
determined, as is Lz. However, the three components of angular momentum do not commute with one
another (see above). Therefore, if Lz is precisely defined, the other components are highly uncertain.
For the same reasons Figure 6.6a, which is a summarized version of the vector picture of Figure 6.5, is
also misleading. This leads to the construction of Figure 6.6b in which the cones represent the
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uncertainty of the directions of the x- and y-components of the angular momentum. The vector can be
considered as being of length {l(l+1)}1/2 units and laying along the cones with its tip at the cone mouth.
Each cone has a z-projection of ml units on the z-axis. Each vector however lies at an unspecified
azimuthal angle on its cone.
Figure 6.6
From the above we see that quantization of the angular momentum forbids the vector to be exactly
parallel to the specified z-axis. The maximum value of the magnitude of the z-component occurs when
ml = l, so the maximum co-latitude angle (max) will be given by
cos max 
(ml ) max
l

1/ 2
{l (l  1)}
{l (l  1)}1/ 2
for low values of l, cosmax < 1or max> 0.
Only when l is very large (in the classical limit) is {l (l  1)}1/ 2  l and then the co-latitude will become
zero and the axis of rotation will lie along the z-axis. This corresponds to classical behavior.
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