Quantum field theory Exercises 8. Solutions

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Quantum field theory Exercises 8. Solutions

2005-12-19

• Exercise 8.1.

Let us first obtain the zero momentum solution given in the problem. First, let us notice, that any solution of Dirac equation also satisfies the Klein–Gordon equation, and thus should have the form of plane waves with dispersion relation p

²

= m

²

. Really, if we act on the Dirac equation by the conjugate operator, we get

0 = (−i ∂ / − m)(i ∂ / − m)ψ = (γ

^µ

γ

^ν

∂

_µ

∂

_ν

− m

²

)ψ = (η

^{µ ν}

∂

_µ

∂

_ν

− m

²

)ψ = 0 , where we used the equation γ

^µ

γ

^ν

= η

^{µ ν}

+ iσ

^{µ ν}

, and the fact that ∂

_µ

∂

_ν

is symmetric.

So, generic solution of the Dirac equation should be the sum of the plane waves with all possible p

²

= m

²

. For each momentum it is

ψ (x) = u( p)e

^−ip^µ^x^µ

(1) for so called positive energy part or

ψ(x) = v(p)e

^ip^µ^x^µ

for negative energy part. The Dirac equation givus us further constraints on the spinors u(p) and v( p).

Let us focus on the positive frequency part. Substituting (1) into the Dirac equation one gets

( / p − m)u(p) = 0 .

In Weyl parametrisation of the gamma matrices, for p

^µ

= (m,~ 0) this just leads to the require- ment that upper two components of u(p) are equal to the lower ones, u

_L

= u

_R

. The choice of normalisation √

mξ , given in the problem, is arbitrary (for linear equation with zero right part we can get a solution only up to the overall normalisation) and is a usual convention.

Let us now perform the boost with repidity η (along the z axis, for definiteness)

u(p) = exp

− 1 2 η

σ

³

0 0 −σ

³

√ m

ξ ξ

=

cosh η 2

1 0 0 1

− sinh η 2

σ

³

0 0 −σ

³

√ m

ξ ξ

=



 e

^η/2

1−σ³ 2

+ e

^−η/2

1+σ³

2

0 0 e

^η^/2

1+σ³ 2

+ e

^−η/2

1−σ³ 2





√ m ξ

ξ

.

For rapidity one has the relations, following from Lorentz transformations, e

^η/2

= p

E + p

³

/ √

m , e

^−η/2

= p

E − p

³

/ √ m .

This leads to the following result for a particle with positive energy moving along the z axis

u(p) =







p E − p

³

0 0 p

E + p

³

! ξ

p E + p

³

0 0 p

E − p

³

! ξ





 .

15

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One can verify, that for general case it can be written in the form

u(p) =

p p

^µ

σ

_µ

ξ p p

^µ

σ ¯

_µ

ξ

,

where square root of the matrix is understood as rotating the matrix to the diagonal form, taking square root of each of the eigenvalues, and then rotating it back.

A similar calculation for the negative energy solution leads to

v(p) =

p p

^µ

σ

_µ

ξ

− p p

^µ

σ ¯

µ

ξ

,

• Exercise 8.2.

Let us proove that trace of an odd number of gamma matrices γ

^µ

is zero.

tr γ

^µ

= tr(γ

^µ

γ

⁵

γ

⁵

) = − tr(γ

⁵

γ

^µ

γ

⁵

) = − tr(γ

^µ

γ

⁵

γ

⁵

) = − tr γ

^µ

,

where we used the fact γ

⁵

γ

⁵

= 1, then anticommuted γ

^µ

with γ

⁵

, then used cyclic permutation under the trace. So, the only possibility is tr γ

^µ

= 0.

Absolutely analogous arguments works for any odd number of gamm matrices, because after odd number of commutations with γ

⁵

wi get minus sign. That is tr γ

^µ

γ

^ν

γ

^λ

= 0.

To calculate trace of two gamma matrices we do

tr γ

^µ

γ

^ν

= tr(2η

^{µ ν}

1 1 1 − γ

^ν

γ

^µ

) = 2η

^{µ ν}

tr(1 1 1) − tr(γ

^µ

γ

^ν

) so, as far as in 4 dimensions tr 1 1 1 = 1,

tr γ

^µ

γ

^ν

= 4η

^{µ ν}

. For four gamma matrices

tr γ

^µ

γ

^ν

γ

^λ

γ

^ρ

= 2η

^{µ ν}

tr γ

^λ

γ

^ρ

− tr γ

^ν

γ

^µ

γ

^λ

γ

^ρ

= 2η

^{µ ν}

tr γ

^λ

γ

^ρ

− 2η

^{µ λ}

tr γ

^ν

γ

^ρ

+ tr γ

^ν

γ

^λ

γ

^µ

γ

^ρ

= 2η

^{µ ν}

tr γ

^λ

γ

^ρ

− 2η

^{µ λ}

tr γ

^ν

γ

^ρ

+ 2η

^{µ ρ}

tr γ

^ν

γ

^λ

+ tr γ

^ν

γ

^λ

γ

^ρ

γ

^µ

= 8η

^{µ ν}

η

^{λ ρ}

− 8η

^{µ λ}

η

^{ν ρ}

+ 8η

^{µ ρ}

η

^{ν λ}

+ tr γ

^µ

γ

^ν

γ

^λ

γ

^ρ

so, finally

tr γ

^µ

γ

^ν

γ

^λ

γ

^ρ

= 4(η

^{µ ν}

η

^{λ ρ}

− η

^{µ λ}

η

^{ν ρ}

+ η

^{µ ρ}

η

^{ν λ}

) For traces with γ

⁵

we can use the definition γ

⁵

= iγ

⁰

γ

¹

γ

²

γ

³

. Then

tr γ

^µ

γ

^ν

γ

⁵

= itr γ

^µ

γ

^ν

γ

⁰

γ

¹

γ

²

γ

³

= 0 ,

because for µ = ν we just get tr γ

⁵

= 0, and for µ 6= µ we always get a trace of two different gamma matrices, which is zero.

The last nontrivial trace is

tr(γ

^µ

γ

^ν

γ

^λ

γ

^ρ

γ

⁵

) = −4iε

^{µ ν λ ρ}

,

because it is nonzero only for for different indices, antysimmetric under exchange of any two gamma matrices, and thus should be proportional to the epsilon-symbol. And the overall coef- ficient can be obtained just by calculating

i tr(γ

⁰

γ

¹

γ

²

γ

³

γ

⁰

γ

¹

γ

²

γ

³

) = −i tr(γ

¹

γ

²

γ

³

γ

¹

γ

²

γ

³

) = i tr(γ

²

γ

³

γ

²

γ

³

) = i tr(γ

³

γ

³

) = −i tr(1 1 1)

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• Exercise 8.3.

The two Lagrangians differ by a total derivative,

L

⁰

= L − (i/2)∂

_µ

( ψ γ ¯

^µ

ψ ) ,

so the lead to the same equations of motion for ψ and ¯ ψ . With L we find T

^{µ ν}

= i ψ γ ¯

^µ

∂

^ν

ψ − η

^{µ ν}

L . Note, that if the fields satisfy the equations of motion, then L = 0, so we can write T

^{µ ν}

= i ψ γ ¯

^µ

∂

^ν

ψ. With L

⁰

we get T 0µ ν = T

^{µ ν}

− (i/2)∂

^ν

j

^µ

, where j

^µ

= ψ γ ¯

^µ

ψ . Note, that L

⁰

= 0 on the equations of motion also. The conservation law ∂

_µ

T

^{µ ν}

= 0 is not spoiled by this term because ∂

_µ

j

^µ

= 0 (use Dirac equations again here).

The conserved energy-momentum P

^ν

= ^R d

³

xT

^ν⁰

changes by (−i/2) ^R d

³

x∂

^ν

j

⁰

. This is zero because, if ν is a spatial index it is a spatial derivative and the spatial integral vanishes, assuming the fields decrease sufficiently fast at infinity. For ν = 0 we use ∂

₀

j

⁰

= −∂

_i

j

ⁱ

and again get a spatial derivative. Therefore P

^ν

= P

^0ν