## Quantum Field Theory

### Set 14: solutions

### Exercise 1

### Consider a massless Dirac fermion ψ = ^{ψ} _{ψ}

^{L}

R

### which has the usual free Lagrangian density:

### L = i ψ6 ¯ ∂ ψ ≡ iψ ^{†} γ ^{0} 6 ∂ ψ.

### Recalling the definition of the gamma matrices and their algebra, γ ^{µ} =

### 0 σ ^{µ}

### ¯ σ ^{µ} 0

### , σ ^{µ} = (1, σ ^{i} ), σ ¯ ^{µ} = (1, −σ ^{i} ), {γ ^{µ} , γ ^{ν} } = 2 η ^{µν} , one can express the Lagrangian in terms of the left and right Weyl spinors:

### L = i

### ψ ^{†} _{L} , ψ ^{†} _{R} 0 1 1 0

### 0 σ ^{µ}

### ¯ σ ^{µ} 0

### ∂ µ ψ L

### ∂ µ ψ R

### = iψ _{L} ^{†} σ ¯ ^{µ} ∂ µ ψ L + iψ ^{†} _{R} σ ^{µ} ∂ µ ψ R .

### Notice the presence of different σ’s in the two terms. The theory is manifestly invariant under two distinct U (1) transformations

### U (1) L :

### ψ _{L} ^{0} = e ^{iα} ψ L

### ψ ^{0} _{R} = ψ R , U (1) R :

### ψ _{L} ^{0} = ψ _{L} ψ ^{0} _{R} = e ^{iβ} ψ R ,

### which rotate separately left and right spinors while leaving unchanged the coordinates. The Noether’s currents associated to the symmetries are given by

### J _{L} ^{µ} = ∂L

### ∂(∂ µ ψ L ) ∆ ψ

L### + ∂L

### ∂(∂ _{µ} ψ ^{†} _{L} ) ∆ _{ψ}

^{†}

L

### = iψ _{L} ^{†} σ ¯ ^{µ} (iψ L ) = −ψ _{L} ^{†} σ ¯ ^{µ} ψ L , J _{R} ^{µ} = ∂L

### ∂(∂ _{µ} ψ _{R} ) ∆ _{ψ}

_{R}

### + ∂L

### ∂(∂ µ ψ ^{†} _{R} ) ∆ _{ψ}

†
R
### = iψ _{R} ^{†} σ ^{µ} (iψ _{R} ) = −ψ ^{†} _{R} σ ^{µ} ψ _{R} .

### In the derivation we have used the fact that the Lagrangian density is written in an asymmetric way with respect to ψ _{L(R)} ^{†} and ψ _{L(R)} : only the former appear in the Lagrangian with the derivative ∂ µ . However one can make the symmetry manifest simply integrating ’half’ Lagrangian density by parts: _{2} ^{i} ψ _{L} ^{†} σ ¯ ^{µ} ∂ µ ψ L −→ − _{2} ^{i} ∂ µ ψ ^{†} _{L} σ ¯ ^{µ} ψ L ; this doesn’t not modify the theory nor the present discussion.

### Considering the sum of the two Noether’s currents gives:

### J _{L} ^{µ} + J _{R} ^{µ} = −ψ _{L} ^{†} σ ¯ ^{µ} ψ _{L} − ψ _{R} ^{†} σ ^{µ} ψ _{R} = − ψγ ¯ ^{µ} ψ ≡ J _{V} ^{µ} .

### J _{V} ^{µ} can be thought of as the Noether’s current associated to a U (1) transformation acting in the same way on left-handed and on right-handed fields (it is called vectorial U(1), or U (1) _{V} , since the spatial components of its Noether’s current form a true vector which changes sign under parity, giving rise to a parity-invariant Lagrangian L when contracted with the true vector ∂ i ):

### U (1) V : ψ ^{0} = e ^{iα} ψ.

### U(1) V is the subgroup of U(1) L × U(1) R obtained by performing left and right transformations with equal param-

### eters α = β, but can also be seen as a transformation acting on the Dirac field.

### On top of considering the sum of the original Noether’s currents one can also take their difference, J _{A} ^{µ} ≡ J _{R} ^{µ} − J _{L} ^{µ} . We now show that the following transformation is a symmetry of the theory and it gives rise to the current J _{A} ^{µ} :

### U (1) _{A} : ψ ^{0} = e ^{iαγ}

^{5}

### ψ = exp

### iα

### −1 0

### 0 1

### ψ =

### e ^{−iα} 0 0 e ^{iα}

### ψ L

### ψ R

### .

### Still this transformation corresponds to a subgroup of U (1) L × U(1) R where this time the left and right parameters are taken opposite in sign. In order to implement this transformation on the Dirac field we need the matrix γ ^{5} : the action on ψ doesn’t consist in a simple phase multiplication but involves the 4×4 matrix e ^{−iαγ}

^{5}

### which however is parametrized by a single parameter α ∈ [0, 2π].

### The free Lagrangian density is invariant under this axial transformations. In fact ψ ¯ ^{0} γ ^{µ} ∂ µ ψ ^{0} =

### e ^{iαγ}

^{5}

### ψ †

### γ ^{0} γ ^{µ} e ^{iαγ}

^{5}

### ∂ µ ψ = ψ ^{†} e ^{−iαγ}

^{5}

### γ ^{0} γ ^{µ} e ^{iαγ}

^{5}

### ∂ µ ψ,

### and recalling that γ ^{5} anticommutes with all the Dirac matrices, {γ ^{5} , γ ^{µ} } = 0, and that (γ ^{5} ) ^{2} = 1, one has ψ ¯ ^{0} γ ^{µ} ∂ _{µ} ψ ^{0} = ψ ^{†} e ^{−iαγ}

^{5}

### γ ^{0} γ ^{µ} e ^{iαγ}

^{5}

### ∂ _{µ} ψ = ψ ^{†} γ ^{0} γ ^{µ} e ^{−iαγ}

^{5}

### e ^{iαγ}

^{5}

### ∂ _{µ} ψ = ¯ ψγ ^{µ} ∂ _{µ} ψ.

### In the end the Noether’s current reads:

### J _{A} ^{µ} = ∂L

### ∂(∂ _{µ} ψ) ∆ ψ = − ψγ ¯ ^{µ} γ ^{5} ψ = ψ ^{†} γ ^{0} γ ^{µ} γ ^{5} ψ = ψ ^{†} _{L} σ ¯ ^{µ} ψ L − ψ ^{†} _{R} σ ^{µ} ψ R = J _{R} ^{µ} − J _{L} ^{µ} .

### Therefore the independent symmetries U(1) L × U (1) R , acting on Weyl spinors, can be recast into two equivalent symmetries, U (1) V × U(1) A , acting on the Dirac field. The reason why the transformation with γ ^{5} is called axial is that the spatial components of its Noether’s current form an axial vector, which does not change sign under parity, as it can be checked by explicit computation.

### Let’s add a mass term for the Dirac field:

### L = i ψ6 ¯ ∂ ψ − m ψψ ¯ = iψ ^{†} _{L} ¯ σ ^{µ} ∂ µ ψ L + iψ ^{†} _{R} σ ^{µ} ∂ µ ψ R − m(ψ ^{†} _{L} ψ R + ψ _{R} ^{†} ψ L ).

### One can immediately realize that now the left-handed and right-handed Weyl spinors cannot be rotated indepen- dently as in the massless case; however the transformation U (1) _{V} is still a symmetry of the massive theory. This is not the case for the axial transformation, since

### ψ ¯ ^{0} ψ ^{0} = ψ ^{†} e ^{−iαγ}

^{5}

### γ ^{0} e ^{iαγ}

^{5}

### ψ = ¯ ψe ^{2iαγ}

^{5}