we’ve found an exponential expression for operators

/

/1 Gi

Gji

eeH

n

jj

We’ve found an exponential expression for operators

n number of dimensions ofthe continuous parameter Generator G

The order (dimensions) of G is the same as H

We classify types of transformations (matrix operator groups) as

Orthogonal O(2) SO(2)O(3) SO(3)

Unitary U(2) SU(2)U(3) SU(3)

groups in the algebraic sense: closed within a defined mathematical operation

that observes the associative propertywith every element of the group having an inverse

O(n) set of all orthogonal UT=U1 (therefore real) matrices of dimension n×n

Rotations in 3-dim space SO(3)4-dim space-time Lorentz transformations SO(4)

Orbital angular momentum rotations SO(ℓ ) (mixing of quantum mechanical states)

cos sin 0Rz() = -sin cos 0

0 0 1

U(n) set of all n×n UNITARY matrices U†=U1

i.e. U†USU(n) “special” unimodular subset of the above det(U)=1

SO(n) “special” subset of the above: unimodular, i.e., det(U)=1group of all rotations in a space of n-dimensions

ALL known“external”space-timesymmetriesin physics

new“internal”

symmetries(beyond

space-time)

SO(3) cos3 sin3 0

-sin3 cos3 0

0 0 1

cos2 0 -sin2

0 1 0

sin2 0 cos2

1 0 0

0 cos1 sin1

0 -sin1 cos1

R(1,2,3)=

cos3cos2+sin3sin2 cos1sin3-sin1sin2sin3 sin1sin3-cos1sin2cos3

-cos2sin3 cos1cos3-sin1sin2sin3 sin1cos3-cos1sin2sin3

sin2 -sin1cos2 cos1cos2

=

Contains SO(2) subsets like:

cos sin 0

-sin cos 0

0 0 1

Rz( ) =

acting on vectors like

v =vx

vy

vzin the i, j, k basis^ ^ ^

do notcommute

do commute

NOTICE:all real

and orthogonal

Call this SO(2)

cos sin -sin cos

R v = vx

vy

Obviously “reduces” to a 2-dim representation

What if we TRIED to diagonalize it further?

seek a similarity transformation on the basis set: U†x

which transforms all vectors: Uv

and all operators: URU†

cos - sin 0 -sin cos - 0 = 00 0 1-

An Eigenvalue Problem

2

4cos4cos2 2

1coscos 2

2cos1 cos i

sin cos i

Eigenvalues: =1, cos + isin , cos isin

(1-[1 - 2cos + ]=0

=1

= (1-[cos2-2cos+sin2]=0

cos sin 0 a a-sin cos 0 b = b 0 0 1 c c

To find the eigenvectors

2

1

2

1

2

i

2

i

a/b = b/a ?? a=b=0

acos + b sin = aasin + b cos = b

c = c

a(cos ) = bsin b(cos ) = asin

for =cos+isin

for =cosisin

for =1

b = i a, c = 0since a*a + b*b = 1

a = b =

b = i a, c = 0since a*a + b*b = 1

a = b =

acos + b sin = a(cos+isin)asin + b cos = b(cos+isin)

c = c(cos+isin)

cos sin 0 0

-sin cos 0 0

0 0 1 0 1 0

With < v | R | v >

eigenvectorsURU†

cos +isin 0 0

= 0 1 0

0 0 sin icos

2

1

2

1

2

i2

i

cos +isin 0 0

= 0 1 0

0 0 sin icos

and under a transformation to this basis(where the rotation operator is diagonalized)

vectors change to:

v1 (v1+iv2)/Uv = U v2 = v3

v3 (v1iv2)/

2

2

< v | R | v >

SO(3)

R(1,2,3)cos3cos2+sin3sin2 cos1sin3-sin1sin2sin3 sin1sin3-cos1sin2cos3

-cos2sin3 cos1cos3-sin1sin2sin3 sin1cos3-cos1sin2sin3

sin2 -sin1cos2 cos1cos2

=

Contains SO(2) subsets like:

cos sin 0

-sin cos 0

0 0 1

Rz( ) =

acting on vectors like

v =vx

vy

vzin the i, j, k basis^ ^ ^

which we just saw can be DIAGONALIZED:

Rv = e+i 0 0 0 1 0 0 0 iei

)(2

1yx ivv

)(2

1yx ivv

zv

Block diagonal form means NO MIXING of components!

Rv = e+i 0 0 0 1 0 0 0 iei

)(2

1yx ivv

)(2

1yx ivv

zv

Reduces to new “1-dim” representation of the operatoracting on a new “1-dim” basis:

e+i )(2

1yx ivv

)cossin(

)sincos(

))(sin(cos

2

1

2

1

yx

yx

yx

vvi

vv

ivvi

iei )(2

1yx ivv

)cossin(

)sincos(2

1

yx

yx

vvi

vv

+

i

)( 2121 iii eee321321 )()( iiiiii eeeeee

1 ii ee

R(1) R(2)= R(1+2)

UNITARY now!(not orthogonal…)

ei is the entire set of all 1-dim

UNITARY matrices, U(1)

obeying exactly the same algebra as SO(2)

SO(2) is ISOMORPHIC to U(1)

SO(2) is supposed to be the group of all ORTHOGONAL 22 matriceswith det(U) = 1

a bc d

a cb d

= a2+b2 ab+bdac+bd c2+d2

ac = bd a2 + b2 = c2 + d2 = and

along with: det(U) = ad – bc = 1 abd – b2c = b a2c – b2c = b c(a2 + b2) = b c = b

which means:ac = (c)d a = d

a b-b a

SO(2)So all matrices have the SAME form:

a2 + b2 = 1with

i.e., the set of all rotations in the space of 2-dimensionsis the complete SO(2) group!

det(A) n1 n2 n3···nN An11 An22 An33 … AnNN

n1,n2,n3…nN

N

some properties

det(AB) = (detA)(detB) = (detB)(detA) = det(BA)

since these are just numbers

which means

det(UAU†) = det(AU†U) = det(A)

So if A is HERMITIAN it can be diagonalized by a similarity transformation (and if diagonal)

det(A) …(n1 n2 n3···nN An11)An22 An33 … AnNN nN

N

n3

N

n2

N

n1

N

Only A11 term0 only diagonal termssurvive, here that’s A22

det(A) 123…N A11 A22 A33 … ANN

= N

Determinant values do not changeunder similarity transformations!

completely antisymmetric tensor (generalized Kroenicker )

(Akk+Bkk ) k=1

N

another useful property

det(A+B) = (A+B)11 (A+B)22 (A+B)33 … (A+B)NN

= (A11+B11)(A22+B22)(A33+B33)…(ANN+BNN)

If A and B are both diagonal*

(Ak+B

k ) k=1

N

*or are commuting Hermitian matrices

det(A+B+C+D+…) =

so

(Akk+Bkk+Ckk+Dkk+ …) k=1

N

Tr(A) Aii = A11+A22 +A33…+ANNi

N

We define the “trace” of a matrix as the sum of its diagonal terms

Tr(AB) = (AB)ii = AijBji = BjiAiji

N

i

N

j

N

j

N

i

N

= (BA)jj = Tr(BA)

Notice:

Tr(UAU†) = Tr(AU†U) = Tr(A) Which automatically implies: Traces, like

determinantsare invariantunder basis

transformations

So…IF A is HERMITIAN (which means it can always be diagonalized)

Tr(A) = N

Operators like GieU

if unitary U: GiGi eeU

†† †

) ( GGiGiGi eeeUU

† † † which has to equal

= 1

G = G†

The generators of UNITARY operators are HERMITIAN

and those kind can always be diagonalized

Since in general AAAAAAIeU A

!3

1

!2

1

In a basis where A is diagonal, so is AA, AAA,… I is already!So U=eA is diagonal (whenever A is)!

What does this digression have to do with the stuff WE’VE been dealing with??

)!3

1

!2

1Idet()det(det AAAAAAeU A

N

k

iN

k

kkk e11

32 )!3

1

!2

11(

)(Tr3214321 ANN eeeeeee

If U=eiA detU=eiTr(A)

For SU(n)…unitary transformation matrices with det=1

detU = 1 Tr(A)=0

SO(3) is a set of operators (namely rotations) on the basis

100

x010

y001

z

zvyvxvvzyxˆˆˆ

1

0

0

0

1

0

0

0

1

zyxvvv

such that:

z

y

x

z

y

x

v

v

v

v

v

v

R preserves LENGTHS and DISTANCES

SU(3) NEW operators (not EXACTLY “rotations”, but DO scramble components) which also act on a 3-dim basis (just not 3-dim space vectors)

p

p +

1951

m=1115.6 MeVmp=938.27 MeV

Look!

By 1953

+ p + m=1115.6 MeV

+ m=1321.0 MeV p +

100

p010

n001

Where a general state (particle) could be expressed

1

0

0

0

1

0

0

0

1

n pn

p

N

where NNN GieU

for some set of generators (we have yet to specify)

A model that considered ’s paired composites of these 3 eigenstates

pn pn (n+ n)2

1

and successfully accounted for the existence, spin, and mass hierachy of

+, 0,

K+, K0, K, K0

, ,

ppp ppn pnn nnnunfortunately also predicted the existence of states like:

ppp ppn pnn nnn

none the less NNN GieU

for some set of generators (we have yet to specify)

SU(3)and

means UNITARY The Gi must all be HERMITIAN

and det U = 1 The Gi must all be TRACELESS

As an example consider

SU(2) set of all unitary 2×2 matrices with determinant equal to 1.

I claim this set is built with the Pauli matrices as generators!

2/

ieUU which described rotations (in Dirac space) of spinors

01

10x

0

0

i

iy

10

01z

Are these generators HERMITIAN? TRACELESS?

10

01

0

0

01

10

i

iDoes

cover ALL possibleHermitian

2×2 matrices?

In other words: Are they linearly independent? Do they span the entire space?

What’s the most general traceless HERMITIAN 22 matrices?

c aiba+ib c

aibaib

cc

and check out:

= a +b +c 0 11 0

0 -ii 0

1 00 -1

They do completely span the space!Are they orthogonal (independent)?

You can’t make one out of any combination of the others!