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# generates 3 -dimensional rotations - PowerPoint PPT Presentation

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 3-dimensional rotations

and with the basis:

serve as the angular

momentum operators!

and with the simpler basis:

Us3U =

Us1U =

Us2U =

[12 - 21 ] = i3

U†[12 - 21 ]U =U†i3U

U†12U- U†21U = iU†3U

U†1UU†2U- U†2UU†1U = iU†3U

SinceU†U = UU†= I

Jx

Jx Jy-Jy 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

,

3-dimensional transformations (like rotations) are

not limited to 3-dimensional “representations”

produce other 3×3 matrix representations of this algebra

R that could ( ) = e- iJ·/ħ

^

can take many forms

The 3-dimensional representation

in the orthonormal basis

that diagonalizes z is the

“DEFINING” representation

of vector rotations

that could= 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 that could

alkali metals

(here Sodium)

all show

lots of doublets

1924:Paulisuggested electrons

posses some new, previously

un-recognized & non-classical

2-valued property

…too classical

perhaps the operator relations

Such “Commutation Rules”

are recognized by

mathematicians as

the “defining algebra”

of a non-abelian

(non-commuting) 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 literal

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

2-dimensional basis set

 1 0

 0 1

=  + 

with a 2 + b 2 = 1

obviously: literal

eigenvalues of each are alsoħ/2 but

their matrices are not diagonal

in this basis

s-

s-

= 1ħ

ħ

s-

s+

= 1ħ

ħ

obviously work on

the basis we’ve defined

obeying the same commutation rule:

but 2-dimensionally!

What if we used THESE as generators?

Should still describe literal “rotations”.

Its 3-components will still require

3 continuously variable

independent parameters:

x ,y ,z

But this is not the defining representation

and can not act on 3-dimensional space vectors.

These are operators that obviously act on the SPINORS

(the SPIN space, not the 3-dimensional wave functions.

intermediate between scalars (1-component) and vectors (3-component)

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

Rt=R-1i.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

^

and obviously operators:zzz = z

Notice: zz =

(

)

(

)

+ iz

an operator analog to: operatorsei = 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.

1-particle

equation

2-particle

equation:

mutual interaction

But in operatorsmany high energy reactions

the number of particles is not conserved!

np+e++e

n+p  n+p+3

e-+ p  e-+ p + 6 +3g

i.e.

lowered

since operates on x

and as we’ve argued before,

the starting point for a relativistic

QM equation:

then

The Klein-Gordon

Equation

or if you prefer:

But this equation has a drawback: operators

Look at Schrödinger’s Equation for a free particle

*( )

( )

-

= 0

The Continuity

Equation

probability

density

probability

current density

*( ) -

( )*= 0

not positive definite

The Klein-Gordon Equation is 2nd order in t!

much more

complicated

time evolution

Need to know initial (t=0) state as well as (0)