Transcript Angular Momentum

Quantum
Physics II
Angular Momentum
Particle on a Sphere
Recommended Reading:
Harris, Chapter 6
Angular Momentum
Angular Momentum: L = m r  v = r  mv = r  p
O
r
L=r  p
m
v
p
r
Note that the angular momentum vector is perpendicular to both the
vectors r and p
Some Math:
The vectors r and p can be expressed in terms of their components
r  xi  y j  zk
p  px i  p y j  p z k
Then the angular momentum is defined by the cross product
i
L  rp  x
px
j
y
py
k
z  i ypz  zpy  j xpz  zpx   k xpy - ypx
pz
(11)




General expression for the angular momentum in Cartesian coordinates
Angular Momentum
If the particle is confined to the x-y plane, so the only non-zero
component of its angular momentum is the z-component LZ.
Lz = (r  p)Z = xpy - ypx
x and y are the coordinates of the particle, px and py are the x and y
components of the momentum of the particle. We can now find the
quantum mechanical operator corresponding to the angular
momentum by using the operator definition for the momenta p
L̂ Z  x i 


  
 y i    i   (12)
y
x
 φ 
Operate on the wavefunctions found above with this angular
momentum operator

L̂ Z ψ  L̂ Z A expimL φ   i A expimL φ 
(13)
φ
 mL A expimL φ   mL φ
Eigenvalue equation. The wavefunction Aexp(imLx) corresponds to a
state with with angular momentum mL 
Z component of Angular momentum is quantised in units of mL 
Rotation in three dimensions
In this section we consider a particle that is free to move in three
dimensions but always at a constant distance r from a fixed point O. In
this way the motion of the particle is confined to the surface of a
sphere.
The potential V is constant and can be set equal to zero
This model can be applied to the
rotation of rigid molecules, and
also to describe electron and
nuclear spin (see later), but here
we we introduce it as a first step
towards obtaining a solution for the
3_D Schrodinger equation for the
Hydrogen atom.
O
r
Cartesian Coordinates
If we consider the proton to be at the origin then the position of
the electron relative to the proton is specified by three
coordinates (x,y,z)
z
r  x 2  y2  z 2
Because of the form of r
solving the Schrodinger
equation
in
Cartesian
coordinates is very difficult.
So we change to another
coordinate
system
to
specify the position of the
electron relative to the
proton.
Spherical
coordinates are the best for
this problem.
(x,y,z)
r
x
(x,y)
y
Spherical Coordinates (r,,)
Instead of the three
coordinates (x,y,z), we can
specify the position of the
electron by three Spherical
coordinates (r,,).
z
r

range 0  
r = radius,
 = polar angle range 0  
 = azimuthal angle
range 0  2
y

y = r sin  sin 
x
x = r sin  cos 
Transformation from Spherical (r,,) to
Cartesian (x,y,z) Coordinates
We can transform form spherical to cartesian coordinates and
vice-versa using the following equations
(x,y,z)  (r,,)
(r,,)  (x,y,z)
x  r sinθ cos φ
y  r sinθ sinφ
z  r cosθ
r  x2  y2  z 2



 2
2  z2 
x

y


z

1
θ  cos    cos1
r
y
φ  tan1 
x
and the volume element dV, is given by
dV  dxdydz  r 2 sinθdrdθdφ
z
Schrodinger Equation in Spherical Coordinates
The time independent Schrodinger equation in three-dimensions is
2 2

 ψr, θ, φ   Uψr, θ, φ   Eψr, θ, φ 
2m
We must also express the Laplacian operator 2 in polar coordinates
(see additional notes) the result is that
1   2  
1  
 
1 2 
2
 

sinθ  
 r



2
2
2
θ  sin θ φ
r r  r  sinθ θ 


therefore the Schrodinger equation in 3D polar coordinates is
1   2  
1  
 
1  2  2m
 r
  2 E  Uψ  0

 sinθ  
2
2
2
θ  sin θ φ  
r  r  r  sinθ θ 
this looks terrible but it in not much different from a particle in a 3d
box or the SHO equation.
Particle confined to a Sphere
We now return to the particle confined to move on the surface of a
sphere. This is really a 2-dimensional problem, since we can specify the
position of the particle with just two coordinates, (, ). Since r is
constant
1    2  
 r  r r    0
2

r 
and the Schrodinger equation becomes
2
 1  

1 2 

   Eψ
 sinθ  
2
2
2
θ  sin θ φ 
2mr  sinθ θ 
2 2
Which we can write as 
L̂   Eψ
2I
 
 
1 2 

 sinθ  
2
2
θ  sin θ φ 
 sinθ θ 
 1
2
L̂  
where
is the square of the angular
momentum operator
and I = mr2 is the Moment of Inertia of the particle about the origin.
We now want to find the eigenvalues and eigenvectors of the L2 operator.
This will give us the allowed values of the angular momentum.
Particle confined to a Sphere
The Schrodinger equation is
with
2 2

L̂   Eψ
2I
(1)
 
 
1 2 

 sinθ  
2
2
θ  sin θ φ 
 sinθ θ 
 1
2
L̂  
Yet again we use the method of separation of variables. We try a solution
of the form
 ,        
where () is a function of  only and () is a function of  only, I.e.
these are one dimensional functions. When we substitute into equation
(1) above and rearrange, we get
   
 
   2    2IE
  0


 sinθ

2
2
2


θ  sin θ  φ   
 sinθ θ 

Multiplication by sin2 and division across by  gives:
sinθ  
  1   2   2IE
2
 sinθ
   2   2 sin   0
 θ 
θ    φ  
The two variables  and  can now be separated by moving the term in 
to the right hand side of the equation
sinθ  
  2IE
1   2  
2
 sinθ
  2 sin   -  2 
 θ 
θ  
  φ 
(2)
only depends on 
only depends on 
Since  and  can be varied independently, this equation can be satisfied
only when both sides of the equation are equal to a constant. Set this
constant equal to ml. Then
sinθ  
  2IE
1   2  
2
2
sin   ml   sinθ

 θ 
θ   2
  φ 2 
We have already seen the equation on the right-hand side, (particle on a
ring).
1   2  
 2
2
 ml 
 m2
(3)
l 0
2
  φ 2 
φ
1


expiml  with ml = 0, ±1, ±2...
solutions (eigenfunctions) are
2
ml is known as the azimuthal quantum number.
The equation on the left hand side for  becomes after rearranging,
m2
1
 
  2IE
l
0
 sinθ
 2 
2
 sinθ θ 
θ  
sin 
putting
2 IE
2

and multiplying across by  gives
2
1  
  ml
   0
 sinθ

sinθ θ 
θ  sin2 
(4)
we now have another differential equation to solve. This one is a little
more complicated than those encountered previously. Solutions can be
found for all values of  but most of these diverge and therefore are nt
physically acceptable solutions.
It can be shown that non-divergent (physically acceptable) solutions
only exist when
  l (l  1) wherel  0, 1, 2, 3,...
and
ml  0,  1,  2,  3,... l
so if l = 0 then ml = 0;
l =1 then ml = -1, 0, +1;
l =2 then ml = -2, -1, 0, +1, +2; l =3 then ml = -3, -2, -1, 0, +1, +2, +3 etc.
inserting
  l (l  1) and
1   2  
2
ml    φ 2 
into eqn (4) gives
1 d
d 
  1 d2  
  l l  1
 sinθ

2
2


sinθ dθ 
dθ  sin    d 
multiply across by 
 d
d 
  d2  
  l l  1
 sinθ

2
2
sinθ dθ 
dθ  sin   d 
(5)
Recall that (, ) = ()() then we can write equation (5) as
L̂2  ,    l l  1 , 
(6)
This is an eigenvalue equation for the square of the angular momentum
operator, L2. From this equation we see that, since l is an integer the
the angular momentum is quantised. So we have found another quantum
number l.
What about the solutions to the differential equation for  (equation 4) ?
2
1  
  ml
   0
 sinθ

2
sinθ θ 
θ  sin 
Associated Legendre Polynomials
This is a difficult equation to solve, it is called the The Associated
Legendre Equation. We use the same technique that we used for the
Simple Harmonic Oscillator Equation, that is we look for a Series
Solutions
The solutions are a called the
Associated Legendre Functions, Pl,mL(cos ) they are polynomials that
depend on the angle  and the two quantum numbers l and mL. For each
allowed set of quantum numbers (l, mL) there is a solution Pl,ml(). The
first few are given in the following table
l
0
1
1
2
2
2
ml
Pl,mL(cos )
0
1
0
cos 
1
sin 
0 (3cos2 - 1)/2
 1 3cos sin 
2
3sin2
l
3
3
3
3
What do these functions look like ?
ml
0
1
2
3
Pl,mL(cos )
(5cos3 - 3cos)/2
3sin(5cos2 -1)/2
15cossin2
15sin3
Polar Plots to visualise the () wavefunctions
P1,0
l = 1, mL = 0
z

x-y plane
P1,1
l = 1, mL =  1
P2,0
l = 2, mL = 0
P2,1
l = 2, mL =  2
P2,1
l = 2, mL =  1
P3,0
P3, 2
l = 3, mL = 0
l = 3, mL =  2
P3,1
P3, 3
l = 3, mL =  1
l = 3, mL =  3
We now have the solutions to both parts of the wave function,
 ,        
The solutions to the  - dependent part () are given by

1
expiml 
2
with ml = 0, ±1, ±2...
While the  - dependent solutions () are the associated Legendre
polynomials Pl,ml(cos ).
Putting this all together we see that the wavefunctions for a particle
confined to move on the surface of a sphere are
1
 ,     
expiml  Pl,m cosθ
l
2
The product of these two functions are usually combined into a single
function called a Spherical Harmonic, denoted Yl,ml(, ), where l and ml
are the two quantum numbers that determine the particular form that
the wavefunction takes. Note that the Spherical Harmonics depend on
two variables  and . With this notation the eigenvalue equation for L2
becomes
L̂2 Yl ,ml ,   l l  1Yl ,ml , 
The first few Spherical Harmonics Yl,ml
l ml
0
0
2l  1 l -m!
Pl,ml cos  eiml 
4  l  m!
Yl ,ml ,  
Definition
Yl ,ml ,  
1
4
l ml
2
0
2 1
l ml
1
0
1 1
Yl ,ml ,  
3
cos 
4
3
sin e i
8
2 2
Yl ,ml ,  


5
3 cos2   1
16 
15
cos  sin e i
8
15
sin2 e  2i
32 
These can be visualised using the
applets on the following websites
http://www.falstad.com/qmrotator/
http://www.bpreid.com/applets/poasDemo.html
http://www.uniovi.es/qcg/harmonics/harmonics.html
These images show the
Real part of the first
few
Spherical
Harmonics
l = 1: ml = -1,
l = 2: ml = -2,
-1,
l=0
ml = 0
0,
0,
+1
+1,
+2
Energies of the Wavefunctions
Going back to equation 1 we have
2 2

L̂   Eψ
2I
But we know that the wavefunctions for a particle confined to a sphere
are the spherical harmonics so this equation becomes:
2 2
2

L̂ Yl ,m 
l l  1Yl ,m  EYl ,m
2I
2I
This is an eigenvalue equation and we see that the allowed energies of
a particle confined to the surface of a sphere and rotating about the
originare quantised according to the equation
2
2
E
l l  1 
l l  1
2
2I
2mr
where l = 0, 1, 2, 3, ...
Note that the energy is independent of the value of ml, so there will be
2 l + 1 states with the same energy, that is, each energy level is 2 l + 1
degenerate.
Examples: l = 0, degeneracy = 1:
l = 1, degeneracy = 3
l = 2, degeneracy = 5 ...
Angular Momentum and Spatial Quantisation
The equation relating rotational energy and angular momentum is
1 2
L2
L2
E  I 

2
2
2I
2 mr
We have also seen that the angular momentum is quantised and its
eigenvalues are given by
L̂2 Yl ,ml ,   l l  1Yl ,ml , 
from this we see that the magnitude of the angular momentum is limited
to the following values
L̂  L̂2  l l  1 
where l = 0, 1, 2, 3, ...
We have also seen that for a particle confined to rotate in the xy-plane its
z-component of angular momentum is also quantised according to the
equation
L̂ z Yl ,ml ,   ml Yl ,ml , 
So that component of the angular momentum along the z-axis s
quantised according to the equation
L z  ml 
where ml = 0, ±1, ±2, ± 3, ... ± l
l=2
Putting this together we see that the
quantised angular momentum for a given
value of l can be represented by a vector
of length proportional to l l  1 
and orientated so that the
projection of the vector on
the z-axis is equal to ml 
z
ml = 2
z
ml = 1
l = 1, ml = 1
ml = 0
l = 1, ml = 0
l = 0, ml = 0
l = 1, ml = -1
ml = -1
ml = -2
for a given value of the quantum number l there are 2l +1 directions in
which the angular momentum vector can point.
If all directions in space are equivalent, then the direction we define for
the z-axis is arbitrary and the 2l +1 orientations that the rotating particle
can adopt all have the same energy, that is, they are degenerate.
In the presence of an externally applied electric or magnetic field ,
however, the z-axis is determined by the direction of this field and the
orientation of the angular momentum with respect to this axis will affect
the energy, that is, the external field can remove the degeneracy
(Zeeman efect, see later)
So far we have only discussed the component of angular momentum
along the z-axis, what about the components in the x and y directions?
The Uncertainty Principle forbids complete knowledge of the orientation
of the angular momentum vector. So if the component in the z-direction
is known then the other two components remain undetermined. This
situation is represented by the following diagram:
one end of the angular momentum vector is considered to be at the apex
of the appropriate cone and the other end then lies anywhere along the
circular cross section at the top of the cone, so its projection can lie
anywhere in the xy-plane
Figure depicts this for l = 2
c
ml = 2h/2
c
ml = h/2c
ml = 0
ml = -h/2
ml = -2h/2
Application: Rotational Spectroscopy of
Diatomic Molecules
r
m2
r
m1

Centre of mass
The motion of the two masses m1 and m2 is mathematically equivalent
to the rotation of a single particle of mass  about a fixed point with
the distance between the particle and fixed point being equal to the
bond length r. It is therefore possible to treat the rotation of a diatomic
molecule as the motion of a particle of mass  confined to the surface
of a sphere
m1m2
reduced mass   
m1  m2
In rotational spectroscopy it is customary to use J as the angular
quantum number instead of l. With this modification the allowed
rotational energy levels are
2
2
EJ 
J(J  1) 
J(J  1)
2
2
I
2 r
Where I = r2 is the moment of inertia of the molecule.
For a given rotational energy there will be (2J + 1) possible orientations
of the molecule in space, so each rotational state is
(2J +1) -fold
degenerate .
The molecule can absorb or emit a photon and make a transition to a
higher or lower rotational state. However there is a selection rule that
states that J can only change by 1 in a transition. So the change in
rotational energy if the molecule absorbs a photon is given by
2
2
(J  1)(J  2)  J(J  1)  (J  1)
E  E J1  E J  
2I
I
Example
In the absorption spectrum of 12C16O the rotational transition from the
J = 12 to the J = 13 rotational state causes absorption of radiation at a
wavenumber of 50.2 cm-1. (a) Calculate the energy change involved in
the transition (b) the moment of inertia and (c) the bond length of CO
(a)
E(photon)  h 
hc
 hc~
  9.98  1022 J

(b) This energy is equal to the difference in rotational energy between
the J = 12 and J = 13 states
2
E  E13  E12   13
I
So the moment of inertia is
2
I  13
 1.45x10- 46 kg.m2  r 2
E
The reduced mass is  
Which gives a bond length
12x16
amu  1.14x10-26 kg
12  16
r
I
 1.13x10-10 m

rendered images of the Spherical Harmonics
websites
on the following
http://www.uniovi.es/qcg/harmonics/harmonics.html
http://odin.math.nau.edu/~jws/dpgraph/Yellm.html