Download presentation
Presentation is loading. Please wait.
Published byRussell Perry Modified over 9 years ago
1
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
2
We classify types of transformations (matrix operator groups) as OrthogonalO(2)SO(2) O(3)SO(3) UnitaryU(2)SU(2) U(3)SU(3) groups in the algebraic sense: closed within a defined mathematical operation that observes the associative property with every element of the group having an inverse
3
O (n) set of all orthogonal U T =U 1 (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 0 Rz( ) = -sin cos 0 0 0 1 U (n) set of all n×n UNITARY matrices U † =U 1 i.e. U † U SU(n) “special” unimodular subset of the above det(U)=1 SO (n) “special” subset of the above: unimodular, i.e., det(U)=1 group of all rotations in a space of n -dimensions ALL known “external” space-time symmetries in physics new “internal” symmetries (beyond space-time)
4
SO(3) cos 3 sin 3 0 -sin 3 cos 3 0 0 0 1 cos 2 0 -sin 2 0 1 0 sin 2 0 cos 2 1 0 0 0 cos 1 sin 1 0 -sin 1 cos 1 R( 1, 2, 3 )= cos 3 cos 2 +sin 3 sin 2 cos 1 sin 3 -sin 1 sin 2 sin 3 sin 1 sin 3 -cos 1 sin 2 cos 3 -cos 2 sin 3 cos 1 cos 3 -sin 1 sin 2 sin 3 sin 1 cos 3 -cos 1 sin 2 sin 3 sin 2 -sin 1 cos 2 cos 1 cos 2 = Contains SO(2) subsets like: cos sin 0 -sin cos 0 0 0 1 R z ( ) = acting on vectors like v = vxvyvzvxvyvz in the i, j, k basis ^^ ^ do not commute do commute NOTICE: all real and orthogonal
5
Call this SO(2) cos sin - sin cos R v = vxvyvxvy Obviously “reduces” to a 2-dim representation What if we TRIED to diagonalize it further? seek a similarity transformation on the basis set: U†xU†x ^ which transforms all vectors: UvUv and all operators: URU†URU†
6
cos - sin 0 -sin cos - 0 = 0 0 0 1- An Eigenvalue Problem Eigenvalues: =1, cos + isin , cos isin (1- [1 - 2 cos + ]=0 =1 = (1- [cos 2 -2 cos + sin 2 ]=0
7
cos sin 0a a -sin cos 0b = b 0 0 1 c c To find the eigenvectors a/b = b/a ?? a=b=0 acos + b sin = a asin + b cos = b c = c a( cos ) = bsin b( cos ) = asin for =cos +isin for =cos isin for =1 b = i a, c = 0 since a*a + b*b = 1 a = b = b = i a, c = 0 since a*a + b*b = 1 a = b = acos + b sin = a(cos +isin ) asin + b cos = b(cos +isin ) c = c(cos +isin )
8
cos sin 0 0 -sin cos 0 0 0 0 1 0 1 0 With eigenvectors URU † cos +isin 0 0 = 0 1 0 0 0 sin icos
9
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: v 1 (v 1 +iv 2 ) / U v = U v 2 = v 3 v 3 (v 1 iv 2 ) /
10
SO(3) R(1,2,3)R(1,2,3) cos 3 cos 2 +sin 3 sin 2 cos 1 sin 3 -sin 1 sin 2 sin 3 sin 1 sin 3 -cos 1 sin 2 cos 3 -cos 2 sin 3 cos 1 cos 3 -sin 1 sin 2 sin 3 sin 1 cos 3 -cos 1 sin 2 sin 3 sin 2 -sin 1 cos 2 cos 1 cos 2 = Contains SO(2) subsets like: cos sin 0 -sin cos 0 0 0 1 R z ( ) = acting on vectors like v = vxvyvzvxvyvz in the i, j, k basis ^^ ^ which we just saw can be DIAGONALIZED: Rv = e +i 00 010 00 ie i
11
Block diagonal form means NO MIXING of components! Rv = e +i 00 010 00 ie i Reduces to new “1-dim” representation of the operator acting on a new “1-dim” basis: e+ie+i ie i + ii
12
R( 1 ) R( 2 )= R( 1 + 2 ) UNITARY now! (not orthogonal…) e i 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)
13
SO(2) is supposed to be the group of all ORTHOGONAL 2 2 matrices with det(U) = 1 a b c d a c b d = a 2 +b 2 ab+bd ac+bd c 2 +d 2 ac = bd a 2 + b 2 = c 2 + d 2 = and along with: det(U) = ad – bc = 1 abd – b 2 c = b a 2 c – b 2 c = b c(a 2 + b 2 ) = b c = b which means: ac = ( c)d a = d
14
a b -b a SO(2)So all matrices have the SAME form: a 2 + b 2 = 1 with i.e., the set of all rotations in the space of 2-dimensions is the complete SO(2) group!
15
det ( A ) n 1 n 2 n 3 ··· n N A n 1 1 A n 2 2 A n 3 3 … A n N N n1,n2,n3…nNn1,n2,n3…nN N some properties det( AB ) = (det A )(det B ) = (det B )(det A ) = 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 ) … ( n 1 n 2 n 3 ··· n N A n 1 1 ) A n 2 2 A n 3 3 … A n N N nNnN N n3n3 N n2n2 N n1n1 N Only A 11 term 0 only diagonal terms survive, here that’s A 22 det ( A ) 123…N A 11 A 22 A 33 … A NN = N Determinant values do not change under similarity transformations! completely antisymmetric tensor (generalized Kroenicker )
16
( A kk +B kk ) k=1 N another useful property det( A+B ) = ( A + B ) 11 ( A + B ) 22 ( A + B ) 33 … ( A + B ) NN = ( A 11 + B 11 )( A 22 + B 22 )( A 33 + B 33 ) … ( A NN + B NN ) If A and B are both diagonal* ( A k + B k ) k=1 N *or are commuting Hermitian matrices det( A+B+C+D+ … ) = so ( A kk +B kk +C kk +D kk + … ) k=1 N
17
Tr ( A ) A ii = A 11 +A 22 +A 33 … +A NN i N We define the “trace” of a matrix as the sum of its diagonal terms Tr ( AB ) = (AB) ii = A ij B ji = B ji A ij i 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 determinants are invariant under basis transformations So…IF A is HERMITIAN (which means it can always be diagonalized) Tr ( A ) = N
18
Operators like if unitary U : † † † † †† which has to equal = 1 G = G † The generators of UNITARY operators are HERMITIAN and those kind can always be diagonalized Since in general In a basis where A is diagonal, so is AA, AAA,… I is already! So U=e A is diagonal (whenever A is)! What does this digression have to do with the stuff WE’VE been dealing with??
19
If U=e i A detU=e i Tr(A) For SU(n)…unitary transformation matrices with det=1 detU = 1Tr( A )=0
20
SO(3) is a set of operators (namely rotations) on the basis 100100 010010 001001 such that: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)
21
p p + 1951 m =1115.6 MeV m p =938.27 MeV Look!
22
By 1953 + p + m =1115.6 MeV + m =1321.0 MeV p +
23
100100 010010 001001 Where a general state (particle) could be expressed where for some set of generators (we have yet to specify)
24
A model that considered ’s paired composites of these 3 eigenstates pn (n + n ) and successfully accounted for the existence, spin, and mass hierachy of +, 0, K +, K 0, K , K 0 , , ppp ppnpnnnnn unfortunately also predicted the existence of states like: ppp ppnpnnnnn
25
none the less for some set of generators (we have yet to specify) SU(3) and means U NITARYThe G i must all be HERMITIAN and det U = 1 The G i must all be TRACELESS As an example consider
26
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! U which described rotations (in Dirac space) of spinors Are these generators HERMITIAN ? TRACELESS ? Does cover ALL possible Hermitian 2×2 matrices? In other words: Are they linearly independent? Do they span the entire space?
27
What’s the most general traceless HERMITIAN 2 2 matrices? c a ib a+ib c a ib a ib c cc and check out: = a +b +c 0 1 1 0 0 -i i 0 1 0 0 -1 They do completely span the space! Are they orthogonal (independent)? You can’t make one out of any combination of the others!
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.