From Special Relativity to Feynman Diagrams.pdf

# N 1 n a θ r n i j where we have used the short hand

• 549

This preview shows pages 174–176. Sign up to view the full content.

n = 0 1 n ! A r ) n i j , where we have used the short-hand notation D r ) D ( g r )) . A r ) is referred to as the infinitesimal generator of the transformation D (g) . As the parameters θ r are varied A r ) i j = θ r ( L r ) i j describes a vector space A of parameters θ r with respect to the basis of infinitesimal generators ( L r ) i j . In particular the higher order terms in the expansion ( 7.28 ) are written in terms of powers of A r ) . For example, at second order the Taylor expansion (7.28) of D r ) reads: D r ) = 1 + θ r L r + 1 2 θ r θ s L r L s + O 3 ). (7.32) From ( 7.32 ) we compute, at the same order, the inverse transformation: D r ) 1 = 1 θ r L r + 1 2 θ r θ s L r L s + O 3 ). (7.33) Consider the matrix representation D 1 ), D 2 ) of two group elements, g 1 = g 1 ), g 2 = g 2 ) , where, for the sake of simplicity, we write θ for the set of n para- meters { θ 1 , . . ., θ n } , the lower index in θ 1 , θ 2 referring to two different elements. We define the commutator of D 1 ), D 2 ) as the matrix D 1 1 ) D 1 2 ) D 1 ) D 2 ) . This matrix must be a representation D 3 ) of some group element g 3 = g 3 ) g 1 1 · g 1 2 · g 1 · g 2 . Using ( 7.32 ) and ( 7.33 ) a simple computation shows that the terms linear in the θ parameters cancel against each other so that the expansion of the group commutator becomes D 1 1 ) D 1 2 ) D 1 ) D 2 ) = 1 + θ r 1 θ s 2 [ L r , L s ] + · · · (7.34) where [ L r , L s ] is the algebra commutator defined as L r L s L s L r . On the other hand from the group composition law we also have

This preview has intentionally blurred sections. Sign up to view the full version.

192 7 Group Representations and Lie Algebras D 3 ) D 1 1 ) D 1 2 ) D 1 ) D 2 ) = 1 + θ m 3 L m + · · · . (7.35) Since ( 7.34 ) and ( 7.35 ) must coincide, we deduce θ k 1 θ l 2 [ L k , L l ] = θ m 3 L m , (7.36) that is [ L k , L l ] = C kl m L m , (7.37) where we have set C kl m θ [ k 1 θ l ] 2 = θ m 3 . (7.38) The set of constants C kl m are referred to as the structure constants of the Lie group . From ( 7.37 ) we see that the structure constants are antisymmetric in their lower indices. We can easily verify that the infinitesimal generators L r satisfy the identity [ L k , [ L l , L m ]] + [ L l , [ L m , L k ]] + [ L m , [ L k , L l ]] = 0 (7.39) called Jacobi identity . As a consequence, by use of the ( 7.37 ) and ( 7.39 ), we obtain that the structure constants must satisfy the identity C kl n C mn p + C lm n C kn p + C mk n C ln p = 0 , (7.40) or, equivalently: C [ kl n C m ] p n = 0 (7.41) where the complete antisymmetrization in three indices has been defined after ( 5.17 ) of Chap.5 . A vector space of matrices A which is closed under commuta- tion, namely such that the commutator of any two of its elements is still in A , is an example of a Lie algebra . Its algebraic structure is defined by the commuta- tion relations between its basis elements, as in ( 5.37 ), i.e. by its structure constants C mn p . The Lie algebra, as we have seen, describes exhaustively the structure 7 of the abstract group G in the neighborhood of the identity of G . It follows that the structure constants C rs p , do not depend on the particular representation D of G .
This is the end of the preview. Sign up to access the rest of the document.
• Fall '17
• Chris Odonovan

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern