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/ / 1 G i Gj i e e H n j j We’ve found an exponential expression for operators n number of dimensions of the continuous parameter Generator G The order (dimensions) of G is the same as H

We’ve found an exponential expression for operators

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We’ve found an exponential expression for operators. n number of dimensions of the 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) - PowerPoint PPT Presentation

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Page 1: 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

Page 2: We’ve found an exponential expression for operators

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

Page 3: We’ve found an exponential expression for operators

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)

Page 4: We’ve found an exponential expression for operators

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

Page 5: We’ve found an exponential expression for operators

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†

Page 6: We’ve found an exponential expression for operators

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

Page 7: We’ve found an exponential expression for operators

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)

Page 8: We’ve found an exponential expression for operators

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

Page 9: We’ve found an exponential expression for operators

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 >

Page 10: We’ve found an exponential expression for operators

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

Page 11: We’ve found an exponential expression for operators

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

Page 12: We’ve found an exponential expression for operators

)( 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)

Page 13: We’ve found an exponential expression for operators

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

Page 14: We’ve found an exponential expression for operators

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!

Page 15: We’ve found an exponential expression for operators

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 )

Page 16: We’ve found an exponential expression for operators

(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

Page 17: We’ve found an exponential expression for operators

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

Page 18: We’ve found an exponential expression for operators

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??

Page 19: We’ve found an exponential expression for operators

)!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

Page 20: We’ve found an exponential expression for operators

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)

Page 21: We’ve found an exponential expression for operators

p

p +

1951

m=1115.6 MeVmp=938.27 MeV

Look!

Page 22: We’ve found an exponential expression for operators

By 1953

+ p + m=1115.6 MeV

+ m=1321.0 MeV p +

Page 23: We’ve found an exponential expression for operators

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)

Page 24: We’ve found an exponential expression for operators

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

Page 25: We’ve found an exponential expression for operators

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

Page 26: We’ve found an exponential expression for operators

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?

Page 27: We’ve found an exponential expression for operators

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!