Feynman Diagrams For Pedestrians - Herbstschule Maria Laach

• the adjoint solution∫¯ψ(x) = ψ(x) † γ 0 = ˜dp ( ¯ψ (+) (p)e ipx + ¯ψ (−) (p)e −ipx) (49)satisfies¯ψ(x)i ←− /∂ = i∂ µ ¯ψ(x)γ µ = i∂ µ ψ(x) † γ 0 γ µ γ 0 γ 0= i∂ µ ψ(x) † γ µ† γ 0 = (−i∂ µ γ µ ψ(x)) † γ 0 = (−i/∂ψ(x)) = −m ¯ψ(x) (50)∴(¯ψ(x) i ←− )/∂ + m = 0 , (51)or¯ψ (+) (p) (/p − m) = 0¯ψ (−) (p) (/p + m) = 0(52a)(52b)• general solution of the Dirac equationψ (+) (p) =ψ (−) (p) =2∑u k (p)b k (p)k=12∑v k (p)d k (p)with four independent basis solutions u 1 (p), u 2 (p), v 1 (p), v 2 (p) satisfying(/p − m)u k (p) = 0(/p + m)v k (p) = 0and the corresponding expansion coefficients b 1 (p), b 2 (p), d 1 (p), d 2 (p).k=1(53a)(53b)(54a)(54b)• in the rest frame of the particle, i. e. for p = (m, ⃗ 0), we have /p = mγ 0 and theDirac equation simplifies to( )m(γ 0 − 1)u k ( ⃗ 0 00) =u0 −2m · 1 k ( ⃗ 0) = 0(55a)( )m(γ 0 + 1)v k ( ⃗ 2m · 1 00) =v0 0 k ( ⃗ 0) = 0(55b)• the independent solutions in the rest frame are therefore⎛ ⎞⎛ ⎞10u 1 ( ⃗ ⎜0⎟0) = ⎝0⎠ , u 2 ( ⃗ ⎜1⎟0) = ⎝0⎠00⎛ ⎞⎛ ⎞00v 1 ( ⃗ ⎜0⎟0) = ⎝0⎠ , v 2 ( ⃗ ⎜0⎟0) = ⎝1⎠10(56a)(56b)10