HW11_soln

HW11_soln - PHY 5667 Problem Set no 11 solution Problem 1(i...

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Unformatted text preview: PHY 5667 Problem Set no. 11 solution Problem 1 (i) In the basis where γ μ = parenleftbigg σ μ ¯ σ μ parenrightbigg we find γ 5 = i γ γ 1 γ 2 γ 3 = i parenleftbigg 1 1 parenrightbiggparenleftbigg σ 1 σ 1 parenrightbiggparenleftbigg σ 2 σ 2 parenrightbiggparenleftbigg σ 3 σ 3 parenrightbigg = parenleftbigg − 1 1 parenrightbigg Therefore P L ≡ 1 2 ( 1 − γ 5 ) P R ≡ 1 2 ( 1 + γ 5 ) = parenleftbigg 1 parenrightbigg = parenleftbigg 1 parenrightbigg are the correct projectors to pick the left-/righthanded component of a Dirac- Spinor. (ii) (a) ( a ) γ μ γ 5 = i γ μ γ γ 1 γ 2 γ 3 Since on the r.h.s all γ-matrices appear exactly once, γ μ meets 3different γ s, with which it anticommutes, and once the same γ , with which it commutes. γ μ γ 5 = i( − 1) 3 γ γ 1 γ 2 γ 3 γ μ = − γ 5 γ μ ⇒{ γ μ , γ 5 } = 0 (b) Tr[ γ 5 ] = iTr[ γ γ 1 γ 2 γ 3 ] = 4i( η 01 η 23 − η 02 η 13 + η 03 η 12 ) = 0 (c) Using the cyclicity of the trace: Tr[ γ μ γ 5 ] = 1 2 Tr[ { γ μ , γ 5 } ] = 0 1 Also, you could argue that this trace has to be zero since it contains an odd number of γ matrices. (d) Use that γ ρ γ ρ = c · 1 , where c = 1 if ρ = 0 and c = − 1 if ρ = 1 , 2 , 3. Therefore, cγ ρ γ ρ = 1 . Tr[ γ 5 γ μ γ ν ] = c Tr[ γ ρ γ ρ γ 5 γ μ γ ν ] Inserting identity, let’s choose ρ negationslash = μ, ν = − c Tr[ γ ρ γ 5 γ μ γ ν γ ρ ] γ ρ anticommutes with γ 5 , γ μ , γ ν = − c Tr[ γ ρ γ ρ γ 5 γ μ γ ν ] cyclicity of the trace = − Tr[ γ 5 γ μ γ ν ] ⇒ Tr[ γ 5 γ μ γ ν ] = 0 (e) Tr[ γ 5 γ μ γ ν γ ρ ] = iTr[ γ γ 1 γ 2 γ 3 γ μ γ ν γ ρ ] contains an odd number of γ-matrices, therefore it vanishes. (e) As long as you can find a matrix γ α where α negationslash = μ, ν, ρ, σ , the same trick as in part (d) works and the trace Tr[ γ 5 γ μ γ ν γ ρ γ σ ] vanishes. The trace therefore survives only if μ, ν, ρ, σ are some permutation of 0 , 1 , 2 , 3. Because all the γ in the trace then anticommute (they are all different from each other), this also implies that Tr[ γ 5 γ μ γ ν γ ρ γ σ ] is totally antisymmetric under exchange of any two indices, or Tr[ γ 5 γ μ γ ν γ ρ γ σ ] ∝ ε μνρσ . The proportionality constant can be obtained e.g. from Tr[ γ 5 γ γ 1 γ 2 γ 3 ] = − iTr[ γ 5 γ 5 ] = − iTr[ 1 ] = − 4i = − 4i ε 0123 . Therefore Tr[ γ 5 γ μ γ ν γ ρ γ σ ] = − 4i ε μνρσ (f) γ α γ μ γ ν γ ρ γ σ γ α = { γ α , γ μ } γ ν γ ρ γ σ γ α − γ μ γ α γ ν γ ρ γ σ γ α = 2( γ ν γ ρ γ σ γ μ + γ μ γ σ γ ρ γ ν ) = γ ν γ ρ { γ σ , γ μ } − γ ν γ ρ γ μ γ σ + { γ μ , γ σ } γ ρ γ ν − γ σ γ μ γ ρ γ ν + γ ν { γ ρ , γ σ } γ μ − γ ν γ σ γ ρ γ μ + γ μ { γ σ , γ ρ } γ ν − γ μ γ ρ γ σ γ ν = { γ ν , γ ρ }{ γ σ , γ μ } + { γ ν , γ μ }{ γ ρ , γ σ } −...
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HW11_soln - PHY 5667 Problem Set no 11 solution Problem 1(i...

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