Mc 2 ub 0 1 ua 5 u p 1 0 ub 0 1

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Unformatted text preview: state ⎛ c( p ⋅ σ ) ⎜ 2 ⎜ E + mc ⎜ 0 ⎜ ⎝ ⎞ 0 ⎟ ⎟ u ( p) c( p ⋅ σ ) ⎟ 2⎟ E − mc ⎠ If the parBcle is massless, E = p c and we have ( c p ⋅σ E + mc [see next slide] ) = c( p ⋅ σ ) = p ⋅ σ ˆ 2 pc 2 We had (see Lecture on solutions to the Dirac equation): uA = c p ⋅ σ uB E − mc 2 ( ) uB = ⎛ 0 1 ⎞ ⎛ uA 5 γ u( p ) = ⎜ ⎟⎜ ⎝ 1 0 ⎠ ⎜ uB ⎝ ⎛ ⎞⎛ 0 1 ⎞⎜ ⎟ =⎜ ⎟ ⎝ 1 0 ⎟⎜ ⎠ ⎠ ⎝ ⎛ γ 5u( p ) = ⎜ ⎜ ⎜ ⎝ 0 c E + mc ( p ⋅σ ) 2 0 ⎞ ⎟⎛ ⎜ c ( p ⋅σ ) ⎟ ⎝ ⎟ E − mc2 ⎠ u u A B c E − mc c ( ) ( p ⋅σ ) u B ⎞ 2 E + mc2 ⎛ ⎞ =⎜ ⎟⎜ ⎠⎜ ⎝ c p ⋅ σ uA E + mc 2 ( p ⋅σ u ) c E + mc A ⎛ 1 0 ⎞⎛ ⎟ =⎜ ⎟⎜ ⎟ ⎝ 0 1 ⎠⎜ ⎠ ⎝ ( p ⋅σ ) 2 0 ⎛ If the particle is massless this becomes γ 5u( p ) = ⎜ ⎜ ⎝ ( c E + mc c ( p ⋅σ ) u A ⎞ 2 E − mc2 ( p ⋅σ u ) B ⎟ ⎟ ⎠ ⎞ ⎟ u( p ) c ( p ⋅σ ) ⎟ ⎟ 2 E − mc ⎠ 0 ˆ p ⋅σ 0 ) ⎞ ⎟ u( p ) = ˆ p ⋅σ ⎟ ⎠ 0 ( ) ( ˆ p⋅Σ u p )() 3 ⎛σ 0 ⎞ ˆ γ 5 So, in the case of a massless parBcle: u...
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