Transcript Presentation #8

MODULE 6
ANGULAR MOTION AND ANGULAR MOMENTUM
In Module 3 we solved the Schrödinger equation for motion in a
circular trajectory about a central, fixed point.
The wavefunction and energy solutions were shown to be
iml
 m  Ae
l
 Be
iml
(ml ) 2
Eml 
2I
the integer ml is a quantum number that specifies individual
wavefunctions and energy levels
but now we show that it has a wider significance.
MODULE 6
Angular Momentum
Considerations
The particle of mass m rotates
in the xy plane that is
perpendicular to the defined
z-axis.
The system rotates clockwise
when viewed from below with
an angular velocity w
(radians per second).
The radial vector is defined by r
the linear momentum vector (p) is directed tangentially to the
radial vector.
MODULE 6
Classical physics tells us that such motion produces an angular
momentum vector (L) in the direction orthogonal to the plane of
the motion.
L  rp
The cross (vector) product of the vectors yields a third vector that
is orthogonal to the components.
when the motion is in three dimensions we use the determinant
i, j, k, are the
unit vectors in
x, y, z directions
i
L x
px
j
y
py
k
z
pz
MODULE 6
The magnitude of the z-component of angular momentum can be
picked out of the determinant
Lz  xp y  ypx
whence the operator for the z-component of angular momentum
becomes:
Lˆz  x


y
i y
i x
in polar coordinates this becomes
Lˆz 

i 
MODULE 6
Operating on the wavefunction for rotary motion with this operator
Lˆz ml 

( Aeiml  Beiml )
i 
consider motion in one direction by putting B = 0
Lˆz ml 

Aeiml  ml Aeiml  ml  ml
i 
This is an eigenvalue equation
the (REAL) eigenvalue for the z-component of AM is
ml
MODULE 6
. Thus we note that the z-component of the angular momentum is
quantized
and we see that 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
1
A
2
MODULE 6
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
And similarly
p  px i  p y j  pz k
L  Lx i  Ly j  Lz k
i
L  r xp  x
px
j
y
py
k
z
pz
MODULE 6
L  ( ypz  zp y )i  ( zpx  xpz ) j  ( xp y  ypx ) k
Lx  ( ypz  zp y )
Ly  ( zpx  xpz )
Lz  ( xp y  ypx )
Note that each of these identities can be generated by cyclic
permutation of its predecessor.
Also recognize that vectors are in bold and scalars (e.g. the
magnitudes of the vectors) are in normal type.
MODULE 6
The magnitude (L, a scalar quantity) of the angular momentum is
related to its components:
L  Lx  Ly  Lz
2
2
2
2
In classical mechanics, no constraints are placed on the energy of
a rotating system
Nor on L, or any of its components.
What does the quantum treatment give?
We need first to construct the operators.
MODULE 6
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
Lx  ( ypz  zp y )
Ly  ( zpx  xpz )
Lz  ( xp y  ypx )
 
 
y z 
i  z
y 
 
 
ˆ
Ly   z  x 
i  x
z 
Lˆx 
Lˆz 
 
 
x  y 
i  y
z 
MODULE 6
Commutation relationships of the AM operators
[ Lˆx , Lˆ y ]  [( ypz  zp y ),( zpx  xpz )]
 [ ypz , zpx ]  [ ypz , xpz ]  [ zp y , zpx ]  [ zp y , xpz ]
 y[ pz , z ] px  0  0  p y [ z , pz ]x  i ( ypx  xp y )
 i Lˆz
In the first line of the above sequence we replaced the operators
by the magnitudes of the observables they represent
To generate line 2 we multiplied out term by term.
The zeros in line 3 appear because pz commutes with both y and
x, and z commutes with px and py
MODULE 6
The other two commutators can be found in the same way, or can
be simply written down using cyclical permutation.
[ Lˆx , Lˆ y ]  i Lˆz
[ Lˆ y , Lˆz ]  i Lˆx
[ Lˆz , Lˆx ]  i Lˆ y
See Module 6 to see how to find the commutation relationships for
the operator for the square of the magnitude of the AM
2 ˆ
ˆ
[ L , Lq ]  0
MODULE 6
The Particle on a Sphere
a particle moving in a circle (2-dimensions)
a particle moving on the surface of a sphere (3-dimensions).
This is equivalent to the motion of a particle constrained to move
at a constant radius from a central point, or the motion of a
point in a solid body that represents the motion of the whole
body.
the results of solving the equations of motion for this seemingly
irrelevant 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.
MODULE 6
We define the potential energy of the particle on the sphere to be
constant and we equate it to zero.
Then the hamiltonian is
Hˆ  
2
2m

2
To convert the Laplacian to spherical polar coordinates we use
x  r sin  cos  ; y  r sin  sin  ; z  r cos 
2
1

1 2
2
 
r 2 L
2
r r
r
L2 is the Legendrian operator
2
1

1 

2
L  2

sin 
2
sin   sin  

MODULE 6
The legendrian contains no terms in r
It is the angular part of the Laplacian operator.
Since the particle is confined to a spherical surface of fixed radius
we can ignore the radial derivatives in the Laplacian.
Thus the legendrian becomes the operator we need
2
2
1
2
2
ˆ
H 
L  L
2
2m r
2I
 2 IE 
L  ( ,  )    2  ( ,  )


2
MODULE 6
z


r
y
x
the wavefunction is a function of  and  only.
MODULE 6
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 center of the sphere.
In this analogy we would expect the ,  equation to be separable
and of the form
 ( ,  )  ( )( )
This is so and the  component is the same as that for the particle
on the ring.
 2
 constant .
2

MODULE 6
The cyclical boundary conditions are the same, the solutions are
the same, and are specified by the quantum number ml.
The quantum number has the same values as before.
It specifies the magnitude of the AM component on the z-axis.
We still have to worry about the  function, which is not so
simple.
The particle on the ring had to obey cyclical boundary conditions
which led to the quantum number ml, now there will be a
circum-polar cyclical boundary condition also to be satisfied
This will no doubt lead to another quantum number.
MODULE 6
the solutions to the  equation depend on a set of functions called
the spherical harmonics,
Ylml ( ,  )
L2Ylml  l (l  1)Ylml
MODULE 6
The spherical harmonics are a set of functions that satisfy the EV
equation
L Ylml  l (l  1)Ylml
2
The labels l and ml have integral values:
l = 0, 1, 2, …. And ml = l, l-1, l-2, … -l
Thus the solutions behave in a quantized manner and for a given
value of l there are 2l+1 values of ml.
l
0
ml
Yl , ml
MODULE 6
0
1/ 2
1
0
 3 


 4 
cos 
1/ 2
1
 3 


 8 
1/ 2
2
0
 5 


 16 
1
 15 


 8 
2
 15 


 32 
1/ 2
1/ 2
1/ 2
3
0
 7 


 16 
1/ 2
1
 21 


 64 
2
 105 


 32 
1/ 2
sin  e  i
(3 cos 2   1)
cos  sin  e  i
sin 2  e 2 i
(5 cos 3   3 cos  )
(5 cos 2   1) sin  e  i
sin 2  cos  e 2 i
1/ 2
3
1/ 2
 1 


 4 
 35 


 64 
sin 3  e 3i
MODULE 6
The spherical harmonics are complex functions except for the
cases where ml = 0.
2

To envisage their shapes we plot the boundary surfaces of
The boundary surface for l = ml = 0, has spherical symmetry
The surfaces for l, ml > 0 have nodes.
MODULE 6
Note that the number of nodes increases as l increases
MODULE 6
Comparing the following
 2 IE 
L  ( ,  )    2  ( ,  )


2
L Ylml  l (l  1)Ylml
2
We arrive at the set of energy states for the particle
Elml
 2
 l (l  1)  
 2I 
The quantum number l specifies the energy of the particle
(independent of ml).
Each energy level is (2l+1)-fold degenerate.
MODULE 6
Angular Momentum Considerations
The quantum number ml specified the magnitude of the zcomponent of AM 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 w is given by
E = Iw2/2.
This is analogous to the equation for the kinetic energy of a
moving particle of mass m and velocity v, i.e. Ekin = mv2/2.
MODULE 6
the angular momentum of the spherical rotator is given as
L = Iw (cf p = mv)
E = L2/2I.
Comparing this with
Elml
 2
 l (l  1)  
 2I 
L  l (l  1)
1/ 2
Thus the magnitude of the AM is restricted by l – it is quantized
l is the angular momentum quantum number.
MODULE 6
The spherical harmonics are eigenfunctions of the operator for the
z-component of the angular momentum, Lˆ z
The eigenvalues are ml
Thus ml specifies the component of the angular momentum that
Lˆ
can be ascribed to rotation
around the z-axis.
z
But, since ml is restricted to certain values, governed by l, the zcomponent of the angular momentum is restricted to 2l+1
discreet values for any value of l.
MODULE 6
The AM can be represented as a vector of length that is
proportional to its magnitude ( {l(l+1)}1/2 units),
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 is restricted in its spatial orientation
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 called space quantization.
MODULE 6
The Lz component is defined
absolutely, but Lx and Ly do not
commute with Lz.
Therefore, if Lz is precisely defined,
the other components are highly
uncertain.
In the Figure the vector arrows are
pointing to a specific direction in
space, which implies that all 3
components of AM are precisely
defined.
Thus the Figure, while useful, is not
a true representation of the QM
situation.
MODULE 6
This often seen Figure is in
error for the same reason,
viz., it implies that all 3
components of AM are
precisely defined.
MODULE 6
This Figure is an improvement.
The cones represent the uncertainty
in the directions of the x- and ycomponents 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.
MODULE 6
Thus we see that the quantization of AM 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 (the classical limit) is {l(l+1)}1/2 ~ l
Then the co-latitude will become zero and the axis of rotation will
lie along the z-axis. This corresponds to classical behavior.