From Special Relativity to Feynman Diagrams.pdf

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

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

View Full Document Right Arrow Icon
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
Image of page 174

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

View Full Document Right Arrow Icon
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 .
Image of page 175
Image of page 176
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern