Loading in 3 Seconds
generates 3 dimensional rotations. and with the basis:. serve as the angular momentum operators!. And of course we already had. and with the simpler basis:. U s 3 U =. †. U s 1 U =. †. U s 2 U =. †. [ 1 2  2 1 ] = i 3. U † [ 1 2  2 1 ] U = U † i 3 U.
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generates 3dimensional rotations
and with the basis:
serve as the angular
momentum operators!
And of course we already had
and with the simpler basis:
Us3U =
†
Us1U =
†
Us2U =
†
[12  21 ] = i3
U†[12  21 ]U =U†i3U
U†12U U†21U = iU†3U
U†1UU†2U U†2UU†1U = iU†3U
SinceU†U = UU†= I
Jx
Jx JyJy Jx=iJz
This 6×6 matrix also satisfies the same algebra:
a
b
c
x
y
z
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
,
3dimensional transformations (like rotations) are
not limited to 3dimensional “representations”
Besides the infinite number of similarity transformations that could
produce other 3×3 matrix representations of this algebra
R ( ) = e iJ·/ħ
^
can take many forms
The 3dimensional representation
in the orthonormal basis
that diagonalizes z is the
“DEFINING” representation
of vector rotations
ℓ = 2
mℓ= 2, 1, 0, 1, 2
ℓ = 1
mℓ= 1, 0, 1
2
1
0
1
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:Paulisuggested electrons
posses some new, previously
unrecognized & nonclassical
2valued property
Perhaps our working definition of angular momentum was too literal
…too classical
perhaps the operator relations
Such “Commutation Rules”
are recognized by
mathematicians as
the “defining algebra”
of a nonabelian
(noncommuting) group
may be the more fundamental definition
[ Group Theory; Matrix Theory ]
Reserving L to represent orbital angular momentum, introducing the
more generic operator J to represent any or all angular momentum
study this as an
algebraic group
Uhlenbeck & Goudsmitfind actually J=0, ½, 1, 3/2, 2, … are all allowed!
quarks
3
2
1
2
1
2
1
2
1
2
1
2
leptons
spin :
p, n,e, , , e , , ,u, d, c, s, t, b
the fundamental constituents of all matter!
spin “up”
spin “down”
s = ħ = 0.866 ħ
ms = ±
sz = ħ
( )
1
0
 n l m >  > = nlm
“spinor”
( )
( )
( )
the most general state is
a linear expansion in this
2dimensional basis set
1 0
0 1
= +
with a 2 + b 2 = 1
obviously:
eigenvalues of each are alsoħ/2 but
their matrices are not diagonal
in this basis
How about the operatorssx, sy ?
s
s
= 1ħ
ħ
s
s+
= 1ħ
ħ
obviously work on
the basis we’ve defined
You already know these as the Pauli matrices
obeying the same commutation rule:
but 2dimensionally!
What if we used THESE as generators?
Should still describe “rotations”.
Its 3components will still require
3 continuously variable
independent parameters:
x ,y ,z
But this is not the defining representation
and can not act on 3dimensional space vectors.
These are operators that obviously act on the SPINORS
(the SPIN space, not the 3dimensional wave functions.
Spinors are 2component objects
intermediate between scalars (1component) and vectors (3component)
When we rotate the “coordinate system” scalars are unchanged.
Vector components are mixed by the prescriptions we’ve outlined.
What happens to SPINORS?
actually can only act on the spinor part
The rotations on 3dim vector space involved ORTHOGONAL operators
Rt=R1i.e.RtR=RRt= 1
These carry complex elements,
cannot be unitary!
As we will see later this
is a UNITARY MATRIX
of determinate = 1
Let’s stay with the simplified case of rotation = z
(about the zaxis)
^
and obviously:zzz = z
Notice: zz =
(
)
(
)
+ iz
an operator analog to: ei = cos + isin
( )
( )
1 0
0 1
1 0
0 1
Let’s look at a rotation of2(360o)
= 1 + 0
This means
A 360o rotation does not bring a spinor “full circle”.
Its phase is changed by the rotation.
Limitations of Schrödinger’s Equation
1particle
equation
2particle
equation:
mutual interaction
But in many high energy reactions
the number of particles is not conserved!
np+e++e
n+p n+p+3
e+ p e+ p + 6 +3g
Let’s expand the DEL operator from 3 to 4dimensions
i.e.
lowered
since operates on x
and as we’ve argued before,
the starting point for a relativistic
QM equation:
then
The KleinGordon
Equation
or if you prefer:
But this equation has a drawback:
Look at Schrödinger’s Equation for a free particle
*( )
( )

= 0
The Continuity
Equation
probability
density
probability
current density
starting from the KleinGordon Equation
*( ) 
( )*= 0
not positive definite
The KleinGordon Equation is 2nd order in t!
much more
complicated
time evolution
Need to know initial (t=0) state as well as (0)