Download Lect8_MatrixRepresentations

So as an exercise in using this notation let’s look at



 


(
x
)
g
(
y
)
( x) ( y )  
  x g 




  
   g   x y
 
y

The indices indicate very specific matrix or vector components/elements.
These are not matrices themselves, but just numbers, which we can reorder as
we wish. We still have to respect the summations over repeated indices!


g
 

 
 g
?? x y
i.e.


( x) ( y )   x y
  
x y


And remember
we just showed
(g) = g

 x y  y x
All dot products are INVARIANT
under Lorentz transformations.
even for ROTATIONS

0 0
as an example, consider
rotations about the z-axis
1
2
1
2
( x) ( y)   x y  ( x cos  x sin  )( y cos  y sin  )
 ( x cos  x sin  )( y cos  y sin  )  x y
2
1
2
1
3 3
 x1 y1 cos2   x1 y 2 sin  cos  x 2 y1 sin  cos  x 2 y 2 sin 2    3 3
 x y  2 2 2
x y
1 2
2 1
1 1
2
 x y cos   x y sin  cos  x y sin  cos  x y sin  
0 0
0 0
1 1
2 2
3 3
x y x x x x x y

 x y
The relativistic transformations:
E
px   ( p x   )
c
E
E
E    ( E  cp x )
  (  p x )
c
c
suggest a 4-vector
E 

 

that
also
p  ( ; p)
( p)   p
transforms by
c
so

p p
should be an invariant!
2
E
2
 2 p
c
In the particle’s rest frame:
E = ?mc2 pp = ?m2c2
px = ?0
In the “lab” frame:
E
px   ( p x   )   (0  mc) = mv
c
E
E
  (   p x ) =  E = mc
c
c
c
so

2
2 2
2
2
2 2
( p) ( p)    m c    m c
  (1   )m c
2
2 2
m c
2
2 2
Limitations of Schrödinger’s Equation
1-particle
equation

2 2
i  ( x , t )  
 ( x, t )  V ( x )  ( x, t )
2
t
2m x1

 
 
2-particle
i  ( x1 , x2 , t )  

(
x
,
x
,
t
)

 ( x1 , x2 , t )
1
2
2
2
equation:
t
2m2 x1
2m2 x2
2
2
2
2
 V ( x1 )  V ( x2 )  V ( x1 , x2 ) ( x1 , x2 , t )
mutual interaction
np+e++e
But in many
high energy reactions
the number of particles
is not conserved!
n+p  n+p+3
e+ p  e+ p + 6 + 3
Sturm-Liouville Equations
a class of differential equations that include:
Legendre's equation
i.e. a class
of differential eq's
to which
Schrodinger's
equations
all belong!
the associated Legendre equation
Bessel's equation
the quantum mechanical harmonic oscillator
whose solutions satisfy:
(*n  m )  n* mdr3  0
for different
eigenfunctions,
If we adopt the following as a definition of the "inner product"
     * dr3
*
v
compare this directly to the vector "dot product"  n un
n
then notice we have automatically
    *
n
Recall:
any linear combination of simple solutions to a differential equation
is also a solution,
and, from previous slide:
 m  n  m n  mn
eigenvalues are REAL
and
different eigenfunctions
are "orthogonal"
Thus the set of all possible eigenfunctions (basic solutions)
provide an "orthonormal" basis set and any general solution
to the differential equation becomes expressible as
   an  n
where
an   n    n *  dr3
n
any general solution will be a function in the "space" of all possible
solutions (the solution set) sometimes called a Hilbert Space
(as opposed to the 3-dimensional space of geometric points.
What does it mean to have a matrix representation of an operator?
of Schrödinger’s equation?
H n  En n
H n  En n
where n represents all distinguishing quantum numbers
(e.g. n, m, ℓ, s, …)
m H n  En m n   mn En
Hmn
m H  Em m
H n  En n
since
†
H H
H=
E1 0
0 E2
0 0
0 0
:.
0 0
0 0
E3 0
0 E4
0
0
0
0
...
...
...
...
with the “basis set”:
1
0
0
0
1
0
0
0
1 ...
:
:
:
·
,
·
,
·
,
This is not general at all (different electrons, different atoms require different matrices)
Awkward because it provides no finite-dimensional representation
That’s why its desirable to abstract the formalism
32
Hydrogen Wave Functions
1  z   zr / a
 100 
  e
 a
32
1  Z   Zr   Zr / 2 a
 200 
  1   e
  2a   2a 
52
1 Z
 Zr / 2 a
 i
 211 
re
sin

e
 
8  a 
52
1 Z 
 Zr / 2 a
 210 
cos 
  re
  2a 
0
1
0
0
0
::
0
0
1
0
0
::
52
1 Z
 Zr / 2 a
 211 
sin  ei
  re
8  a
1 Z 
Zr
Z 2 r 2   Zr / 3a
 300 
   27  18  2 2 e
a
a 
81 3  a  
32
1
0
0
0
0
::
0
0
0
1
0
::
0
0
0
0
1
::
0
0
0
0
But the sub-space of angular momentum
(described by just a subset of the quantum numbers)
doesn’t suffer this complication.
Angular Momentum
|lmsms…>
l = 0, 1, 2, 3, ...
Lz|lm> = mh|lm>
for m =  l,  l+1, … l1, l
L2|lm> = l(l+1)h2|lm>
Sz|lm> = msh|sms>
for ms = s, s+1, … s1, s
S2|lm> = s(s+1)h2|sms>
Of course |nℓm> is  dimensional again
can measure all the spatial(x,y,z) components


(and thus L itself) of L  r  mv
not even possible in principal !
 
  

L  r  i , ,   ir  
 x y x 
So, for
example
 
 

Lz  i x  y   i
x 

 y
azimuthal
angle in
polar
coordinates
nlm …
Angular Momentum
l
Measuring Lx alters Ly (the operators change the quantum states).
The best you can hope to do is measure:
l = 0, 1, 2, 3, ...
L2lm(,)R(r)= l(l+1)ħ2lm(,)R(r)
Lz lm(,)R(r) = mħ lm(,)R(r)
for m = l, l+1, … l1, l
States ARE simultaneously eigenfunctions of BOTH of THESE operators!
We can UNAMBIGUOULSY label states with BOTH quantum numbers
ℓ=2
mℓ = 2, 1, 0, 1, 2
2
ℓ=1
mℓ = 1, 0, 1
1
1
0
0
L2 = 1(2) = 2
|L| = 2 = 1.4142
L2 = 2(3) = 6
|L| = 6 = 2.4495
Note the always odd number
of possible orientations:
A “degeneracy” in otherwise identical states!
Spectra of the
alkali metals
(here Sodium)
all show
lots of doublets
1924: Pauli suggested electrons
posses some new, previously
un-recognized & non-classical
2-valued property
Perhaps our working definition of angular momentum was too literal
…too classical
perhaps the operator relations
Lx Ly  Ly Lx  iLz
Ly Lz  Lz Ly  iLx
Lz Lx  Lx Lz  iLy
may be the more fundamental definition
Such “Commutation Rules”
are recognized by
mathematicians as
the “defining algebra”
of a non-abelian
(non-commuting) group
[ Group Theory; Matrix Theory ]
Reserving L to represent orbital angular momentum, introducing the
more generic operator J to represent any or all angular momentum
J x J y  J y J x  iJ z
J y J z  J z J y  iJ x
study this as an
algebraic group
J z J x  J x J z  iJ y
Uhlenbeck & Goudsmit find actually J=0, ½, 1, 3/2, 2, … are all allowed!
quarks
leptons
spin 12 p,
: n, e, , , e ,  ,  , u, d, c, s, t, b
the fundamental constituents of all matter!
ms = ±
1
2
spin “up”
spin “down”
s = 32 ħ = 0.866 ħ
sz =
|nlm>|
1
2
1
2
>=
1
2
ħ
( )nlm
1
0
“spinor”
the most general state is
a linear expansion in this
2-dimensional basis set


()
() ()
=  1 +  0
0
1
with  2 +  2 = 1
SPIN
ORBITAL ANGULAR
MOMENTUM
fundamental property
of an individual component
relative motion
between objects
Earth:
orbital angular momentum: rmv
plus
“spin” angular momentum: I
in fact ALSO
“spin” angular momentum: Isunsun
but particle spin especially that of truly fundamental particles
of no determinable size (electrons, quarks)
or even mass (neutrinos, photons)
must be an “intrinsic” property of the particle itself
Total Angular Momentum
nlm sm j…
l = 0, 1, 2, 3, ...
l
s
Lz|lm> = mħ|lm>
for m = l, l+1, … l1, l
L2|lm> = l(l+1)ħ2|lm>
Sz|lm> = msħ|sms>
for ms = s, s+1, … s1, s
S2|lm> = s(s+1)ħ2|sms>
In any coupling between L and S it is the
TOTAL J = L + s that is conserved.
Example
J/ particle: 2 (spin-1/2) quarks bound in a ground
(orbital angular momentum=0) state
Example
Either
spin-1/2 electron in an l=2 orbital. Total J ? 3/2 or 5/2
possible
BOSONS
spin 1

FERMIONS
spin ½ p, n, e, 
Nuclei (combinations of p,n) can have
J = 1/2, 1, 3/2, 2, 5/2, …
BOSONS
FERMIONS
mesons
spin 0 “psuedo-scalar”
spin ½


0
  0
quarks and leptons
e, , , u, d, c, s, t, b, 
Baryon “octet”
p, n, 
mediators
spin 1 Force
spin 3/2
“vector”bosons: ,W,Z
Baryon “decupltet”
D, S, X, W
 ,  ,  , K ,K ,K
“vector” mesons
r, , , J/, 
spin 2
:
spin 5/2
:
Combining any pair of individual states |j1m1> and |j2m2>
forms the final “product state”
|j1m1>|j2m2>
What final state angular momenta are possible?
What is the probability of any single one of them?
Involves “measuring” or calculating OVERLAPS
(ADMIXTURE contributions)
or forming the DECOMPOSITION
into a new basis set of eigenvectors.
S
j1j2
|j1m1>|j2m2> =
z
j j1 j2;
m m1 m2
| jm>
j=| j1j2 |
Clebsch-Gordon coefficients
Matrix Representation
for a selected j
J2|jm> = j(j+1)h2| j m >
Jz|jm> = m h| j m > for m = j, j+1, … j1, j
J±|jm> = j(j +1)m(m±1) h | j, m1 >
The raising/lowering operators through which we identify
the 2j+1 degenerate energy states sharing the same j.
adding
J+ = Jx + iJy
J = Jx  iJy
subtracting
2Jx = J+ + J
Jx = (J+ + J )/2
2iJy = J+  J
Jy = i(J  J+)/2
The most common representation of angular
momentum diagonalizes the Jz operator:
<jn| Jz |jm> = mmn
1 0 0
= 0 0 0
0 0 -1
(j=1)
Jz
(j=2)
Jz =
2
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
-1
0
0
0
0
0
-2
J±|jm> = j(j +1)m(m±1) h | j, m1 >
< 1 0 | J | 1 1 > =
2|10>
< 1 -1 | J | 1 0 > =
2 | 1 -1 >
0
J =
0 0
2 0 0
0 2 0
J =
0
0
0
J | 1 -1 > = 0
< 1 0 | J | 1 -1 > =
2|10>
< 1 1 | J | 1 0 > =
2|11>
J | 1 1 > =
0
2
0
0 2
0 0
For J=1 states
Jx 
a matrix representation of the angular momentum operators
 0
1
 2
2
 0

 0
1
J y  i 2
2
 0

Jz 
2
0
2
i 2
0
i 2
 1 0 0


 0 1 0
 0 0 1


0 

2

0 
0
i
0

 0
1 2
0 


1 2
0
1 2


1 2
0 
 0
  0
 
2   i 2
 
  0
i
2
0
i
2


i 2

0 
0
Which you can show conform to the
COMMUTATOR relationship
you demonstrated in quantum mechanics
for the differential operators
of angular momentum
[Jx, Jy] = iJz
i 0
2
Jx J y  Jy Jx =  0 0
 i
0
2
i 
2
 i
2
0 0

i 
 i
2  2
0
0
0
i 
2
1 0 0 


0   i 0 0 0 
i 
  0 0  1
2
= iJz
J
J 1 / 2
z

0 
  1/ 2


 1/ 2
 0
J J 1z 
 1 0 0


 0 1 0
 0 0 1


J J 3 / 2 z 
0
0
0 
  3/ 2


 1/ 2
0
0 
 0
 0
0
 1/ 2
0 


0
0
 3/ 2
 0
J J 2 z 
 2 0

 0 1
 0
0

0
 0
 0
0

0
0
0
0
0
0 

0
0 
0
0 

1 0 
0  2 
0
R(1,2,3) =
1
0
0 


 0 cos1 sin 1 
 0  sin  cos 

1
1
z
z′
1
y′
1
y
x =x′
R(1,2,3) =
z′
z′′ 2 1
z
 cos 2

 0
 sin 

2
0
0 
0  sin  2   1


1
0   0 cos1 sin 1 


0 cos 2   0  sin 1 cos1 
2
y′ =y′′
1
y
x =x′
2 x′′
R(1,2,3) =
z
z′
z′′′ = z′′ 2 1
 cos 3 sin  3 0   cos 2



sin

cos

0

 0
3
3
 0
0
1   sin  2

0
0 
0  sin  2   1


1
0   0 cos1 sin 1 


0 cos 2   0  sin 1 cos1 
3
y′′′
3
2
y′ =y′′
1
y
x =x′
2 x′′
3
x′′′
R(1,2,3) =
about
about
x-axis
about
y′-axis
z′′-axis
1st
 cos 3 sin  3 0   cos 2



sin

cos

0

 0
3
3
 0
0
1   sin  2

0
0 
0  sin  2   1


1
0   0 cos1 sin 1 


0 cos 2   0  sin 1 cos1 
2nd 3rd
These operators DO NOT COMMUTE!
but as nn
Recall: the “generators” of rotations
are angular momentum operators
and they don’t commute!
Infinitesimal rotations DO commute!!
 1
3

  3 1
 
0
 2
  2 

0 
1 
0  2  1
3 0   1
3



1
0   3 1 0     3 1
 
0
1  0
0 1 
0
 2
  2 

0 
1 
3 0  1
 1




1
0

 0
3
 0
0 1  2

 1

 0
 
 2
0  2 

1
0 
0
1 