Gauge Fields

This is the story of the fields that induce interactions when acting as preservers of a local phase transformation.

If a local phase transformation linked to the $U(1)$ group

$$\phi(x) \rightarrow \phi'(x) = e^{i \alpha(x)} \phi(x)$$ (165)

is performed in the Lagrangian

$${\cal L} = \partial_{\mu}\phi^{\ast} \partial^{\mu}\phi - m^{2} \phi^{\ast} \phi$$ (166)

the mass term presents no difficulties, it is trivially invariant in front of the transformation (167). But what happens with the term involving derivatives? In this case one has

$$\partial_{\mu}\phi(x) \rightarrow \partial_{\mu}(e^{i \alph... ...ha(x)} [\partial_{\mu} + i \partial_{\mu}\alpha(x)] \phi(x)$$ (167)

so that

$$\partial\phi^{\ast} \partial^{\mu}\phi \rightarrow (\partia... ...i^{\ast} (\partial^{\mu} + i \partial^{\mu} \alpha(x)) \phi$$ (168)

and therefore the Lagrangian (168) is not invariant under second class (local) gauge transformations due to the contribution of the gradient of $\alpha(x)$.

How could one turn ${\cal L}$ to be invariant?

Remember that the electromagnetic field $A_{\mu}$ is defined modulus a gradient as a consequence of the gauge invariance of electromagnetism. Then it is possible that the invariance of ${\cal L}$ in front of local phase transformations can be restored taking advantage of the freedom of election of $A_{\mu}$. How?, adding this electromagnetic field as a compensating term in the derivative

$$\partial_{\mu}\phi \rightarrow (\partial_{\mu} - i q A_{\mu}) \phi \equiv D_{\mu}\phi$$ (169)

which defines the so called gauge covariant derivative $D_{\mu}$. It is clear that when $\phi(x)$ changes due to a transformation according to (167), simultaneously the field $A_{\mu}$ is required to change as

$$A_{\mu}(x) \rightarrow A'_{\mu}(x) = A_{\mu}(x) - \frac{1}{q} \partial_{\mu} \alpha(x)$$ (170)

Let us notice that $D_{\mu}\phi$ transforms in the same way as the field $\phi(x)$, that is

$$D_{\mu}\phi(x) \rightarrow D'_{\mu}\phi'(x) = e^{i \alpha(x)} D_{\mu}\phi(x)$$ (171)

It is then concluded that the Lagrangian

$${\cal L} = (\partial_{\mu} + i q A_{\mu}(x)) \phi'(x)\;(\... ... - i q A^{\mu}(x)) \phi(x) - m^{2} \phi^{\ast}(x) \phi(x)$$ (172)

is local phase transformation invariant. The field $A_{\mu}$ is called gauge field. It is a compensating field to restore the symmetry and a comparative field because its presence allows to physically distinguish the particles of different charges, that is to say it gives physical meaning to the charge attribute.

The immediate consequence of the requirement of invariance under local phase transformations is the natural appearance of the minimal interaction between the charged particles (charged fields) and the electromagnetic or gauge field $A_{\mu}$.

If we start now from a Lagrangian for fermions with mass $m$ - electrons for example -

$${\cal L}_{f} = i \bar{\psi} \gamma_{\mu} \partial^{\mu}\psi - m \bar{\psi} \psi$$ (173)

the invariant Lagrangian in front of gauge transformations of second class is

$${\cal L}_{f-A} = i \bar{\psi} \gamma^{\mu} (\partial_{\mu} - i e A_{\mu}) \psi - m \bar{\psi} \psi$$ (174)

or equivalently

$${\cal L}_{f-A} = i \bar{\psi} \gamma^{\mu}\partial_{\mu}\p... ...,\bar{\psi} \psi + e \bar{\psi} \gamma^{\mu} \psi A_{\mu}$$ (175)

where the appearance of the minimal interaction, as a consequence of the requirement of a local phase symmetry or gauge symmetry is again clear.