Making Lagrangian gauge invariant

[Maybe_Memorie]

Homework Statement

[/B]
The Lagrangian $\mathcal{L}\frac{1}{2}(\partial_\mu\phi^\nu)^2+\frac{1}{2}(\partial_\mu\phi^\mu)^2+\frac{m^2}{2}(\phi_\mu\phi^\mu)^2$ for the vector field $\phi^\mu$ is not invariant with respect to the gauge transformation $\phi^\mu\rightarrow \phi^\mu+\partial^\mu\alpha$. Introduce a new real scalar field $\sigma$ and find a new interacting Lagrangian $\mathcal{L}'(\phi^\mu,\sigma)=\mathcal{L}(\phi^\mu)+\tilde{\mathcal{L}}(\phi^\mu,\sigma)$ which is gauge invariant under the given transformation and satisfies $\mathcal{L}'(\phi^\mu,0)=\mathcal{L}(\phi^\mu)$.

Can we solve this with a $\sigma$ that has a canonical kinetic term $-\frac{1}{2}(\partial_\mu\sigma)^2$?

Homework Equations

The Attempt at a Solution

It's the part about the canonical kinetic term I don't understand. On the requirement that $\sigma \rightarrow \sigma + \alpha$ I found a $\tilde{\mathcal{L}}(\phi^\mu,\sigma)$ that satisfies the required properties. I won't write it because it's rather long. I was told the answer to whether or not there can be a canonical kinetic term is "no", but I don't know why.

Is there a particular reason why this should be? Something about having unbounded negative energy perhaps?

 

[mark726]

Yes, there is a reason why a canonical kinetic term for $\sigma$ cannot be used in this case. The Lagrangian given in the problem already has a canonical kinetic term for the vector field $\phi^\mu$, but it is not gauge invariant. This means that it is not invariant under the gauge transformation $\phi^\mu\rightarrow \phi^\mu+\partial^\mu\alpha$.

In order to make the Lagrangian gauge invariant, we need to introduce a new field $\sigma$ and a new term $\tilde{\mathcal{L}}(\phi^\mu,\sigma)$ that will cancel out the non-invariant term in the original Lagrangian. However, if we were to include a canonical kinetic term for $\sigma$, it would not cancel out the non-invariant term and the Lagrangian would still not be gauge invariant. This is because the gauge transformation for $\sigma$ is different from the one for $\phi^\mu$.

In summary, the reason why a canonical kinetic term for $\sigma$ cannot be used in this case is because it would not satisfy the requirement of making the Lagrangian gauge invariant.