HW11_soln

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

This preview shows pages 1–3. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 } −...
View Full Document

{[ snackBarMessage ]}

### Page1 / 11

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online