Quantum field theory Exercises 4. Solutions 2005-11-21 • Exercise 4.1. • In the rest fram E = m, p = 0. Performing a boost with rapidity η (along x axis for example) we have E = mcoshη, p

Quantum field theory Exercises 4. Solutions

2005-11-21

• Exercise 4.1.

• In the rest fram E = m, p = 0. Performing a boost with rapidity η (along x axis for example) we have E = m cosh η, p _x = m sinh η. Thus, (E + p)/(E − p) = e ^2η .

• Performing a second boost with rapidity η ⁰ along x we get E → E cosh η ⁰ + p sinh η ⁰ and p → E sinh η ⁰ + p cosh η ⁰ , so

e ^2η → (E cosh η ⁰ + p sinh η ⁰ ) + (E sinh η ⁰ + p cosh η ⁰ ) (E cosh η ⁰ + p sinh η ⁰ ) − (E sinh η ⁰ + p cosh η ⁰ )

= e ^2η

E + p

E − p = e ^2(η+η

⁾ .

• Exercise 4.2.

Let v ⁰ = ξ _R ^† ψ _R and v ⁱ = ξ _R ^† σ ⁱ ψ _R . Using the rules for spinor transformations under infinites- imal boosts and rotations

ψ _R → e ^(−iθ

^+η

⁾

i

2

ψ _R = (1 + (−iθ _i + η _i ) σ ⁱ

2 + . . . )ψ _R , we get for v ⁰

v ⁰ → ξ _R ^† (1 + (+iθ _i + η _i ) σ ⁱ

2 )(1 + (−iθ _i + η _i ) σ ⁱ

2 )η _R = ξ _R ^† (1 + η _i σ ⁱ + . . . )η _R

= v ⁰ + η _i v ⁱ + . . .

and for v ⁱ

v ⁱ → ξ _R ^† (1 + (iθ _j + η _j ) σ ^j

2 )σ ⁱ (1 + (−iθ _k + η _k ) σ ^k 2 )η _R

= ξ _R ^† (σ ⁱ + i

2 θ _j [σ ^j , σ ⁱ ] + 1

2 η _j {σ ^j ,σ ⁱ } + . . .)η _R = ξ _R ^† (σ ⁱ − θ _j ε ^jik σ ^k + η _j δ ^ji + . . . )η _R

= v ⁱ − θ _j ε ^jik v ^k + η _j δ ^ji v ⁰ + . . . .

This is exactly transformation properties of a (contravariant) four-vector under infinitesimal boost η i (boost of rapidity |η| in the direction of η ⁱ /|η|). We also checked that it has correct properties under usual rotations θ ⁱ .

Calculations for ξ _L ^† σ ¯ ^µ η _L are completely analogous.

You can also check everything explicitly for finite boosts, using the formula e ^η

^σ

= coshη + η _i

η σ ⁱ sinh η , where η = p

η ⁱ η ⁱ is rapidity of the boost and η _i /η — its direction.

• Exercise 4.3.

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Transformations of the double-contravariant tensor are F ^µ ν → Λ ^µ _ρ Λ ^ν _σ F ^{ρ σ} ,

where Λ ^µ _ρ = exp{−(i/2)ω _{α β} (J ^{α β} ) ^µ _ρ }, and

(J ^{α β} ) ^µ _ρ = i(η ^{α µ} δ _ρ ^β − η ^{β µ} δ _ρ ^α ) . Then, to the first order

δ F ^{µ ν} = ω ^µ _ρ F ^{ρ ν} − ω ^ν _ρ F ^{ρ µ} .

Using the explicit form ω _i0 = −ω ⁱ⁰ = −η ⁱ and ω _{i j} = ω ^{i j} = ε ^{i jk} θ ^k one gets for B and E δ E = −[η η η × B] + [θ θ θ × E] ,

δ B = +[η η η × E] + [θ θ θ × B] .

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