Transcript Outline of section 5

Angular momentum in
quantum mechanics
• Classical definition of angular momentum
• Linear Hermitian Operators for angular momentum
Commutation relations
Physical consequences
• Simultaneous eigenfunctions of total angular
momentum and the z-component
Vector model
• Spherical harmonics
Orthonormality and completeness
Classical angular momentum
For a classical particle, the angular momentum is
defined by
L
L  r p
p
 Lx i  Ly j  Lz k
r
In components
Lx  ypz  zp y
F
Ly  zpx  xpz
Lz  xp y  ypx
Angular momentum is very important in problems
involving a central force (one that is always
directed towards or away from a central point)
because in that case it is conserved
Same origin for r and F
dL d
dr
dp
 (r  p ) 
p r 
dt dt
dt
dt
p
 (  p)  (r  F)  0.
m
Hermitian operators for quantum
angular momentum
In quantum mechanics we get linear
Hermitian angular momentum
operators from the classical expressions
using the postulates
r  L x  r  r , p  p  i 
L  r  p  L  i r  
Lx  ypz  zp y
Lx  y pz  z p y
Ly  zpx  xpz
Ly  z px  x pz
Lz  xp y  ypx
Lz  x p y  y px
 

ˆ
Lx  i  y  z 
y 
 z

 
ˆ
Ly  i  z  x 
z 
 x
 

ˆ
Lz  i  x  y 
x 
 y
Commutation relations
The different components of angular momentum
do not commute with one another, e.g.
Proof:  L x , L y   L x L y  L y L x


Lx , L y   i Lz


Lx  y pz  z p y
Ly  z px  x pz
Lz  x p y  y px
[ x, p x ]  i
[ y, p y ]  i
[ z , pz ]  i
Similar arguments give
the cyclic permutations
L y , Lz   i Lx


Summarize these as
Lz , Lx   i L y


 Li , L j   i L k


where i, j, k obey a
cyclic (x, y, z) relation
Commutation relations (2)
The different components of L do not commute with
each another, but they do commute with the squared
magnitude of the angular momentum vector:
Proof:
2
2
x
2
y
L L L L
 L x , L2  


Similar proofs for the other components
 L x , L 2    L y , Lˆ2    L z , L 2   0
 

 

2
z
Commutation relations (3)
The different components of angular momentum do not commute
• Lx, Ly and Lz are not compatible observables
• They do not have simultaneous eigenfunctions (except when L = 0)
• We can not have perfect knowledge of any pair at the same time
BUT, the different components all commute with L2
• L2 and each component are compatible observables
• We can find simultaneous eigenfunctions of L2 and one component
CONCLUSION
We can find simultaneous eigenfunctions of one
component of angular momentum and L2 .
Conventionally we chose the z component. Next step is
to find these eigenfunctions and study their properties.
L z   
2
L    2
What determines the direction of the z-axis?
In an experiment we usually have one or more privileged directions (e.g. the
direction of an external electric or magnetic field) which gives a natural z axis.
If not, this direction is purely arbitrary and no physical consequences depend
on what choice we make.
Angular momentum in
spherical polar coordinates
z
Spherical polar coordinates are the natural coordinate system
in which to describe angular momentum: x, y, z  r , ,
θ
r
The angular momentum operators only depend on the angles
θ and φ and not on the radial coordinate r.

 cos   
Lˆx  i  sin 


 tan   


 sin   
Lˆ y  i   cos 




tan   

 

Lˆx  i  y  z 
y 
 z

 
Lˆ y  i  z  x 
z 
 x
 

Lˆz  i  x  y 
x 
 y
Lˆz  i
Lˆ  Lˆ  Lˆ  Lˆ  
2
2
x
2
y
2
z
φ
x
x  r sin  cos 
y  r sin  sin 
z  r cos 


2
y
 1  
 
1 2 
 sin    sin     sin 2   2 




Note: The angular momentum operators commute with any operator which only depends on r.
L2 is closely related to the angular part of the Laplacian
1     L2
2 
r

2
r r  r 
2
2 2
r
Lz in spherical polars
Proof that
x  r sin  cos 
y  r sin  sin 
z  r cos 

Lz  i

 x  y  z 



  x  y  z
x  r sin  cos  
y  r sin  sin  
z  r cos 

 

Lˆz  i  x  y 
x 
 y
x
 r sin  sin    y

y
 r sin  cos   x

z
0




i ˆ
  y  x  Lz

x
y
Eigenfunctions of Lz
Look for simultaneous eigenfunctions of L2 and Lz
Lˆz        
First find the eigenvalues and
eigenfunctions of Lz. Can only
depend on the angle φ

i
          

     A exp  i 
2
Normalize
solution
 d      1
2
0
    A* exp  i  A exp  i   A
2
2
A
2
 d  2 A  1 
2
0
A
2
1
2
1
    
exp  i 
2
Lz  i


Eigenfunctions of Lz (2)
Boundary condition: wave-function must be single-valued
 (  2 )  ( )
    
1
exp  i 
2
 exp  i   exp  i  exp  2 i 
 exp  2 i   1
   m  0, 1, 2, 3
The angular momentum about the z-axis is
quantized in units of hbar (compare Bohr model).
The possible results of a measurement of Lz are
Lz  m
m  integer
So the eigenvalue equation and eigenfunction solution for Lz are
Lˆz m    m m  
1
 m   
exp  im 
2
Orthonormality and
completeness
Lz is a Hermitian operator. Its eigenfunctions are
orthonormal and complete for all functions of the angle
φ that are periodic when φ increases by 2π.
   2    
Orthonormality
2
*

 m   n  d 
0
Completeness
   
1
2
2
am 
 a   
m
m
*

 m    d
0
mn
0

m 
2
 exp i  n  m d  
1
 m   
exp  im 
2
Example
A particle has the angular wavefunction
Find, by inspection or otherwise, the
coefficients am in the expansion
 m   
1
exp  im 
2
1
   
1  i cos3  , 0    2
3

   
 a   
m 
Hence confirm that the wavefunction is normalized.
What are the possible results of a measurement of Lz
and their corresponding probabilities?
Hence find the expectation value of Lz for many such
measurements on identical particles.
m
m
Eigenfunctions of L2
Now look for eigenfunctions of L2
Lˆ2 f  ,    
2
f  ,  
 

2
2  1
ˆ
L  
sin


sin






1 2 

 2
2
 sin   
Try a separated solution of the form
1
f ( ,  )  ( )( ) 
exp(im )( )
2
(this ensures the solutions remain eigenfunctions of Lz)
Eigenvalue equation is
Lˆ2 exp(im )( )  
2
exp(im )( )
We get the equation for Θβm(θ) which depends on both β and m
 1  
 
1

2
sin


m


 m    0

 sin   

2
  sin 



Eigenfunctions of L2 (2)
Make the substitution
  cos  
 d 


  sin 
 d 

 1  
 
1

2
sin


m


 m    0

 sin   

2
  sin 



This gives the Legendre equation, solved in 2B72 by the Frobenius method.
d  m     
d 
m2 
2
  m     0
1   
   
2
 

1    
 
We need solutions that are finite at μ = ±1 (i.e. at θ = 0 and θ = π since μ =
cosθ). This is only possible if β satisfies
  l (l 1) where l  0,1,2,
and l  m
This is like the SHO where we found restrictions on the energy
eigenvalue in order to produce normalizable solutions.
Eigenfunctions of L2 (3)
Label solutions to the Legendre equation by the values of l and m
  m      lm   

d 
m2 
2 d lm    
 lm     0
1   
  l  l  1 
2
 
  
1



 

  l (l  1)
For m = 0 the finite solutions are the Legendre polynomials
l ,m0     Pl     Pl  cos 
P0     1
P1     
P2    
1
3 2  1

2
For non-zero m the solutions are the associated Legendre polynomials
lm     Pl
m
    Pl  cos 
m
m
Pl m (  )  (1   2 )
m /2
 d 

 Pl (  )
 d 
Note that these only depend on the size of m not on its sign
Eigenvalues of L2
So the eigenvalues of L2 for physically allowed solutions are

2
 l (l 1) 2 , where l  0,1,2,
and l  m
The possible results for the measurement of the squared magnitude of the
2
2
2
angular momentum are
L 
 l (l  1)
The possible results for a measurement of the magnitude of the angular
momentum are
L  l (l  1)
From l  m we get  l  m  l
For each l there are 2l+1 possible integer values of m
The restriction on the possible values of m is reasonable. The z-component
of angular momentum can not be greater than the total!
In fact, unless l = 0, the z-component is always less than the total and can
never be equal to it. Why?
Summary
The simultaneous eigenfunctions
of Lz and L2 are
f ( , )  Pl m  cos  exp im 
Eigenvalues of Lˆ2 are l (l 1) 2 , with l  0,1, 2,
The integer l is known as the principal angular momentum quantum
number. It determines the magnitude of the angular momentum
Eigenvalues of Lˆz are m , with  l  m  l
(i.e. m  l , l  1 , 1, 0,1,
l  1, l )
The integer m is known as the magnetic quantum number. It
determines the z-component of angular momentum. For each
value of l there are 2l+1 possible values of m.
The simultaneous eigenfunctions of L2 and Lz do not correspond to definite values
of Lx and Ly, because these operators do not commute with Lz. We can show,
however, that the expectation value of Lx and Ly is zero for the functions f(θ,φ).
The vector model
This is a useful semi-classical model of the quantum results.
Imagine L precesses around the z-axis. Hence the magnitude of L and
the z-component Lz are constant while the x and y components can take a
range of values and average to zero, just like the quantum eigenfunctions.
A given quantum number l determines the
magnitude of the vector L via
L2  l (l  1)
z
2
L  l (l  1)
L
The z-component can have the 2l+1 values
corresponding to
Lz  m , l  m  l
In the vector model this means that only
particular special angles between the angular
momentum vector and the z-axis are allowed
θ
The vector model (2)
Example: l=2
Lz
Magnitude of the angular momentum is
L  l (l  1)
2
2
6
2
Ly
L  l (l  1)  6
Component of angular momentum
in z- direction can be
Lx
l  m  l  Lz  2 ,  , 0, , 2
Quantum eigenfunctions correspond to a cone of solutions for L in the vector model
Spherical harmonics
The simultaneous eigenfunctions of
L2 and Lz are usually written in terms
of the spherical harmonics
f ( ,  )  Ylm  ,  
 Nlm Pl m  cos   exp  im 
Proportionality constant Nlm is
chosen to ensure normalization
NB. Some books write the
spherical harmonics as
LˆzYlm  ,    m Ylm  ,  
Lˆ2Ylm  ,    l  l  1 2Ylm  ,  
First few examples (see 2B72):
Y00 ( ,  ) 
Y11 ( ,  )  
Yl m ( ,  )
Remember
x  r sin  cos 
y  r sin  sin 
z  r cos 
1
4
3
3 ( x  iy )
sin  exp(i )  
8
8
r
Y10 ( ,  ) 
3
3 z
cos  
4
4 r
Y11 ( ,  ) 
3
3 ( x  iy )
sin  exp(i ) 
8
8
r
Y20  ,   
5
3cos 2   1

16
Shapes of the spherical harmonics
Y00
Y11
Y10
z
y
x
Re[Y11 ]
l  1, m  0
l  0, m  0
1
Y00 
4
3
cos 
4
Y10 
l  1, m  1
Y11  
Imaginary
3
sin  exp(i )
8
To read plots: distance from origin corresponds to magnitude
(modulus) of plotted quantity; colour corresponds to phase (argument).
(Images from http://odin.math.nau.edu/~jws/dpgraph/Yellm.html)
Real
Shapes of spherical harmonics (2)
Y22
Y21
Y20
Re[Y22 ]
Re[Y21 ]
z
y
x
l  2, m  0
Y20 
l  2, m  2
Y22 
15
sin 2  exp(2i )
32
5
(3cos 2   1)
16
l  2, m  1
Y21  
Imaginary
15
sin  cos  exp(i )
8
Real
To read plots: distance from origin corresponds to magnitude (modulus)
of plotted quantity; colour corresponds to phase (argument).
(Images from http://odin.math.nau.edu/~jws/dpgraph/Yellm.html)
Orthonormality of spherical harmonics
The spherical harmonics are eigenfunctions of Hermitian operators.
Solutions for different eigenvalues are therefore automatically orthogonal
when integrated over all angles (i.e. over the surface of the unit sphere).
They are also normalized so they are orthonormal.
Integration is over the solid angle
d   sin  d d
2

0
0
d 3r  r 2 drd 
which
comes from
 r 2 dr sin  d d
*
d

d

sin

Y
lm (   )Yl ' m ' (   )  1 if l  l ' and m  m '
 
 0 otherwise
Convenient shorthand
*
d

Y
lm (   )Yl ' m ' (   )   ll '  mm '

Compare: 1D Cartesian version of orthonormality



n*  x m  x  dx   mn
Completeness of spherical harmonics
The spherical harmonics are a complete, orthonormal set for
functions of two angles. Any function of the two angles θ and φ can
be written as a linear superposition of the spherical harmonics.

l
f  ,      almYlm (   )
l  m  l
l  0 m  l
Using orthonormality we can show that the expansion coefficients are
alm   d  Ylm* (   ) f (   )

Compare: 1D version
2

0
0
*
d

d

sin

Y
lm (   ) f (   )
 
 ( x)   ann ( x)
n

an   n*  x   x  dx

Examples

l
f  ,      almYlm (   )
l  0 m  l
1) A particle has the un-normalized angular wavefunction
  ,  
4
3
2
6
Y00  Y11  Y10  Y11  Y21
5
5
5
5
a) Normalize this wavefunction.
b) What are the possible results of a measurement of
Lz and their corresponding probabilities? What is the
expectation value of many such measurements?
c) What are the possible results of a measurement of
L2 and their corresponding probabilities? What is the
expectation value of many such measurements?
Examples (2)
1) A particle has normalized angular wavefunction
  ,  
15
cos 2 sin 
14
Find the probability of measuring

You can use the result
L2  2
2
2
cos
2

sin
 d   / 4

0
Y11 ( ,  )  
3
sin  exp(i )
8
Y10 ( ,  ) 
3
cos 
4
Y11 ( ,  ) 
3
sin  exp( i )
8
Summary
The simultaneous eigenfunctions of Lz and
L2 are the spherical harmonics Ylm  ,  
Eigenvalues of Lˆ2 are l (l 1) 2 , with l  0,1, 2,
LˆzYlm  ,    m Ylm  ,  
Lˆ2Ylm  ,    l  l  1 2Ylm  ,  
l = principal angular momentum quantum number.
Determines the magnitude of the angular momentum.
Eigenvalues of Lˆz are m , with  l  m  l
m = magnetic quantum number.
Determines the z-component of angular momentum.
The spherical harmonics are a complete orthonormal set
for functions of two angles
 d Y
*
lm
(   )Yl ' m ' (   )   ll '  mm '
2

0
0
 d    d  d sin 

l
f  ,      almYlm (   )
l  0 m  l
alm   d  Ylm* (   ) f (   )