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Unformatted text preview: PHY 5667 Problem Set no. 12 solution Problem 1 L = ( )( ) m 2 + i L L i T L 2 L + i L 2 L (i) The U(1) charge of L has to be 1 2 . The kinetic term of the lefthanded spinor remains unchanged with under any U(1)charge (because it contains L and L , the first contributes a factor e i q and the second e i q , and the two cancel out for any q ), but the interaction terms require q = 1 2 to be invariant: i T L 2 L i e i e i q T L 2 e i q L = e i (1+2 q ) i T L 2 L q = 1 2 i L 2 L i e i e i q L 2 e i q L = e i (1+2 q ) i L 2 L q = 1 2 (ii) When we promote ( x ), extra terms in the Lagrangian can stem only from derivatives. In particular ( )( ) [ ( e i ( x ) )][ ( e i ( x ) )] = ( )( ) i( ( x ))[ ] i L L i e i ( x ) / 2 L ( e i ( x ) / 2 L ) = i L L i i 2 ( ( x )) L L So L [( e i ( x ) ) , e i ( x ) / 2 L ] L [ , L ] = i( ( x )) [ + i 2 L L ] bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright J Thus, the Noether current is J = + i 2 L L (iii) Lets first get the equations of motion from the variation of the action. By the usual manipulations (in particular integration by parts and reshuffling of spinors, T L 2 L = T L 2 L and L 2 L = L 2 L ) we obtain S = 0 = integraldisplay d 4 x parenleftBig [ 2 m 2 + i L 2 L ] + [ 2 m 2 i T L 2 L ] + [ i( L ) 2i T L 2 ] L + L [i L + 2i 2 L ] parenrightBig 1 So the equations of motion are 2 m 2 + i L 2 L = 0 2 m 2 i T L 2 L = 0 i( L ) 2i T L 2 = 0 i L + 2i 2 L = 0 Now J = [ + i 2 L L ] = 2 2 + i 2 ( L ) L + i 2 L ( L ) = [ m 2 + i L 2 L ] [ m 2 i T L 2 L ] i T L 2 L i L 2 L = 0 (iv) We need the interaction terms, which contain...
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 Fall '10
 Okui
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