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Unformatted text preview: PHY 5667 Problem Set no. 11 solution Problem 1 (i) In the basis where = parenleftbigg parenrightbigg we find 5 = i 1 2 3 = i parenleftbigg 1 1 parenrightbiggparenleftbigg 1 1 parenrightbiggparenleftbigg 2 2 parenrightbiggparenleftbigg 3 3 parenrightbigg = parenleftbigg 1 1 parenrightbigg Therefore P L 1 2 ( 1 5 ) P R 1 2 ( 1 + 5 ) = parenleftbigg 1 parenrightbigg = parenleftbigg 1 parenrightbigg are the correct projectors to pick the left/righthanded component of a Dirac Spinor. (ii) (a) ( a ) 5 = i 1 2 3 Since on the r.h.s all matrices appear exactly once, meets 3different s, with which it anticommutes, and once the same , with which it commutes. 5 = i( 1) 3 1 2 3 = 5 { , 5 } = 0 (b) Tr[ 5 ] = iTr[ 1 2 3 ] = 4i( 01 23 02 13 + 03 12 ) = 0 (c) Using the cyclicity of the trace: Tr[ 5 ] = 1 2 Tr[ { , 5 } ] = 0 1 Also, you could argue that this trace has to be zero since it contains an odd number of matrices. (d) Use that = c 1 , where c = 1 if = 0 and c = 1 if = 1 , 2 , 3. Therefore, c = 1 . Tr[ 5 ] = c Tr[ 5 ] Inserting identity, lets choose negationslash = , = c Tr[ 5 ] anticommutes with 5 , , = c Tr[ 5 ] cyclicity of the trace = Tr[ 5 ] Tr[ 5 ] = 0 (e) Tr[ 5 ] = iTr[ 1 2 3 ] contains an odd number of matrices, therefore it vanishes. (e) As long as you can find a matrix where negationslash = , , , , the same trick as in part (d) works and the trace Tr[ 5 ] vanishes. The trace therefore survives only if , , , are some permutation of 0 , 1 , 2 , 3. Because all the in the trace then anticommute (they are all different from each other), this also implies that Tr[ 5 ] is totally antisymmetric under exchange of any two indices, or Tr[ 5 ] . The proportionality constant can be obtained e.g. from Tr[ 5 1 2 3 ] = iTr[ 5 5 ] = iTr[ 1 ] = 4i = 4i 0123 . Therefore Tr[ 5 ] = 4i (f) = { , } = 2( + ) = { , } + { , } + { , } + { , } = { , }{ , } + { , }{ , }...
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 Fall '10
 Okui

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