Phy489_Lecture21-22_2013 - Phy489 Lecture 2122 0 Chiral...

Info icon This preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Phy489’Lecture’21-22’ 0’
Image of page 1

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

View Full Document Right Arrow Icon
Chiral’Fermion’States’&’Electroweak’UnifcaBon’ QuesBon:’how’can’we’contemplate’uni±ying’two’±orces’that’appear’to’have’couplings’ that’are’very’di²erent’in’±orm’(not’just’in’“apparent’magnitudes”’since’it’was’already’ suspected’that’the’“weakness”’o±’the’charged’weak’interacBon’could’be’aPributed’to’ the’mass’o±’the’exchanged’parBcle)?’ Compare’the’couplings’(vertex’±actors’in’Feynman’rules)’ ig e γ μ i g w 2 2 1 5 ( ) i g z 2 c V f c A f 5 ( ) QED’ Charged’Weak’ Neutral’Weak’ Note’that’QED’is’all’“neutral’current”.’We’will’see’that’the’neutral’weak’and’EM’ currents’“mix”,’hence’the’and’terms’instead’o±’the’pure’ V-A ’o±’the’charged’ weak’interacBon.’ c V f c A f vector pure vector-axial vector (V-A) mix of vector and axial vector 1’
Image of page 2
Chiral’Fermions’ How’do’we’deal’with’the’“structural’diferences”?’( e.g. ’the’diferent’vertex’±actors)’ Difficulty’is’associated’with’the’±actors’o±’.’This’can’be’dealt’with’by’“absorbing”’ a’±actor’o±’into’the’de²niBon’o±’the’parBcle’spinors:’ (1 γ 5 ) 5 ) /2 u L p ( ) = 1 − γ 5 2 u p ( ) we’call’this’a’le]’handed’spinor,’though’in’ general’it’is’not’a’helicity’eigenstate’ Look’at’the’term’ 5 u p ( ) = c p σ ( ) E + mc 2 0 0 c p ( ) E mc 2 u p ( ) I±’the’parBcle’is’massless,’and’we’have’ E = p c c p ( ) E + mc 2 = c p ( ) p c = ˆ p [see next slide] 2’
Image of page 3

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

View Full Document Right Arrow Icon
We had (see Lecture on solutions to the Dirac equation): u A = c E mc 2 p σ ( ) u B u B = c E + mc 2 p ( ) u A γ 5 u ( p ) = 0 1 1 0 u A u B = 0 1 1 0 c E mc 2 p ( ) u B c E + mc 2 p ( ) u A = 1 0 0 1 c E + mc 2 p ( ) u A c E mc 2 p ( ) u B 5 u ( p ) = c E + mc 2 p ( ) 0 0 c E mc 2 p ( ) u A u B = c E + mc 2 p ( ) 0 0 c E mc 2 p ( ) u ( p ) If the particle is massless this becomes 5 u ( p ) = ˆ p ( ) 0 0 ˆ p ( ) u ( p ) = ˆ p Σ ( ) u p ( ) 3’
Image of page 4
So,’in’the’case’of’a’massless’parBcle:’where’ γ 5 u p ( ) = ˆ p Σ ( ) u p ( ) Σ σ 0 0 Recall’that’is’the’spin’matrix’for’a’Dirac’parBcle’(see’secBon’7.2),’so’ 2 Σ ˆ p Σ ( ) represents’the’helicity,’with’eigenvalues’of’±1.’
Image of page 5

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

View Full Document Right Arrow Icon
Image of page 6
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