PHY 745 Group Theory Lecture 4: Great Orthogonality Theorem
The Great Orthogonality Theorem in PHY 745 lecture 4, covering topics such as Schur's lemmas and proving the theorem. Delve into Chapter 2 of DDJ1 and enhance your understanding of this fundamental concept in physics. Join the discussion on 1/20/2017 at Olin room 102 to deepen your knowledge of group theory.
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PHY 745 Group Theory 11-11:50 AM MWF Olin 102 Plan for Lecture 4: The great orthogonality theorem Reading: Chapter 2 in DDJ 1. Schur s lemmas 2. Prove the Great Orthgonality Theorem 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 1
1/20/2017 PHY 745 Spring 2017 -- Lecture 4 2
The great orthogonality theorem on unitary irreducible representations order of the group element of the group ( ) R Notation: h R i th representation of i R denote matrix indices imension of the representation d l i h ( ) * = i j ( ) R ( ) R ij l R i 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 3
Proof of the great orthogonality theorem Prove that all representations can be unitary matrices Prove Schur s lemma part 1 any matrix which commutes with all matrices of an irreducible representation must be a constant matrix Prove Schur s lemma part 2 Put all parts together 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 4
Proof of the great orthogonality theorem Prove that all representations can be unitary matrices Prove Schur s lemma part 1 any matrix which commutes with all matrices of an irreducible representation must be a constant matrix Prove Schur s lemma part 2 Put all parts together Schur s lemma part 1: A matrix M which commutes with all of the matrices of an irreducible representation must be a constant matrix: M=sI where s is a scalar constant and I is the identity matrix. 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 5
Schurs lemmas part 1: A matrix M which commutes with all of the matrices of an irreducible representation must be a constant matrix: M=sI where s is a scalar constant and I is the identity matrix. Proof: Suppose is a diagonal matrix , the premise becomes M = = d i i ( ) R ( ) for all elements of the gro R d up d R ( ) ( ) i i ( ) R ( ) R d d kl kl ( ) )( = ( ) = i i ( ) R ( ) R d d kk ll kl kl ) ( = i ( ) 0 for all , and for al k l l d d R R kk l l k l i if ( ) is irre R duc ble i d ll d kk 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 6
Example irreducible representation: 3 1 2 1 0 0 1 1 0 0 ( ) E ( ) A ( ) B = = = 2 3 3 3 1 3 1 2 2 3 3 3 1 2 1 2 1 2 ( ) C ( ) D ( ) F = = = 2 2 2 3 3 3 3 3 3 1 2 1 2 1 2 2 2 2 Extension of proof to more general matrix Note that i M H H + = : M = 1 2 + where H M M 1 ( ) = H i M M 2 In these terms, the premise is: ( ) ( ) + = + i i ( ) R ( ) R for all elements of the group H iH H iH R 1 2 1 2 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 7
( ) ( ) + = + i i ( ) R ( ) R for all elements of the group H iH H iH R 1 2 1 2 = = i i i i ( ) R ( ) R H and ( ) R ( ) R H H H 1 1 2 2 Hermitian matrices can be diagonalized by a unitary transformation U = i i ( ) R ( ) R H for all UH U U U R 1 1 R = = i i ( ) R ( ) UH U U U U U UH U 1 ' ( ) R 1 i i ' ( ) ' ( ) for all is an equivalent representation of the group d R d R i where R The same construction ho lds for , H 2 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 8
Proof of the great orthogonality theorem Prove that all representations can be unitary matrices Prove Schur s lemma part 1 any matrix which commutes with all matrices of an irreducible representation must be a constant matrix Prove Schur s lemma part 2 Put all parts together Schur's lemma part 2 Consider two irreducible representations of the same group Suppose there exists a re ctangular ( ) for all . It follows that either or ( ) and ( ) are equivalent. R R 1 2 ( ) with dimension R and ( ) w R ith dimensions . 1 2 matrix such tha t M 1 2 = 1 2 ( ) 0 M R R M R M 1 2 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 9
Schur's lemma part 2 Consider two irreducible representations of the same group Suppose there exists a re ctangular ( ) for all . It follows that either or ( ) and ( ) are equivalent. R R 1 2 ( ) with dimension R and ( ) w R ith dimensions . 1 2 matrix such tha t M 1 2 = 1 2 ( ) 0 M R R M R M 1 2 = 1 2 S uppose ( ) R ( ) R M for all M R ( ) ( ) = 1 2 ( ) R ( ) R M M ( ) ( ) ) for all ( ) fo R = 2 1 ( ) R ( ) R M M 1 2 1 1 ( ) = = ( R M M R R 2 1 ( ) R M r a ll M R 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 10
Schurs lemma part 2 -- continued 1 2 ( ) ( ) ( ) = ( ) ( ) M = 2 = for all ( ) for all ( ) ( ) R M M R R M R 2 1 R M M = = R R = = 2 2 1 ( ) ( ) R M M for all for all M R M MM R R MM R 2 1 1 M R M M R 2 1 M = 1 1 M M = 1 MM = 2 2 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 11
Schurs lemma part 2 -- continued 2 2 ( ) ( ) From Schur's first lemma: = = ( ) ( ) R M M for all for all MM R R MM R 1 1 M M R R = = ( ) MM s I 2 2 2 ) Consider the case where ( M M s I 1 1 1 = = , it follows that 0 s s 1 2 1 2 ( ) otherwise there is an inconsistenc y MM M M = = C For the case onsider the case where that = , it follows that * s s 1 2 1 2 0 : s 1 0: = = 1 2 1 1 2 ( ) R ( ) R M ( ) Representations are eq ( ) R for all R uivalent M R M M For th c e a se that = s 1 MM 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 0 0 M 12
Proof of the great orthogonality theorem Prove that all representations can be unitary matrices Prove Schur s lemma part 1 any matrix which commutes with all matrices of an irreducible representation must be a constant matrix Prove Schur s lemma part 2 Put all parts together h ( ) * = i j ( ) R ( ) R ij l R i 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 13
Construct a matrix 2 1 1 2 1 ( ) R X ( ) M R R = 2 1 Note that: ( ) S M ( ) where is a member of the group S S M = 2 2 2 1 1 Details: ( ) S M ( ) S ( ) R X ( ) R R = 2 1 1 ( ) ( ) SR X R R = ) ( 2 1 1 1 1 1 ( ) ( ) ( ) S SR X R S R ( ) 1 = ) ( ) 2 1 1 ( ) ( SR X SR S R 1 = ( ) S M 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 14
Proof continued: h ( ) * = i j ( ) R ( ) R ij l R i Construct a matrix 2 1 1 2 1 ( ) R X ( ) M R R = = 2 1 Since: for ( ) where is a member of the group 0 for arbitrary . Choosing particular indice X = ( ( ) R ( ) , S M M S S s: M 2 1 ( ) ( ) 2 1 1 2 1 1 ( ) R X ( ) ( ) ( ) R R R R R ) ( ) * = = 2 1 ( ) R 0 R 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 15
h ( ) matrix * = i j Proof continued: ( ) R ( ) R ij l R i Construct a 2 1 1 2 1 ( ) R X ( ) M R R = 2 1 Since: fo r ( ) where is a member of the group for arbitrary . Choosing particular indices X = ( ) articular choice of : R ( ) S M M , M S S = = 2 1 : s I 1 1 1 ( ) R ( ) X R s R = = 1 1 1 For p ( ) X R R s s 1 = = = 1 1 1 ( ) ( ) E RR h h s 1 R h = s 1/20/2017 PHY 745 Spring 2017 -- Lecture 4 16 1
