# How to check for invariance in Lagrangian after gauge transformation?

If I have the Lagrangian density:

$$\mathcal{L}=\left(\partial_{\mu} \phi^{*}\right)\left(\partial^{\mu} \phi\right)-m^{2} \phi^{*} \phi$$

How can I show it is invariant under the following gauge transforms (do I just sub them in)?

$$\phi^{\prime}=e^{i \theta} \phi$$

$$\phi^{*^{\prime}}=e^{-i \theta} \phi^{*}$$

And then from this find the conserved Noether current?

• After that you find exactly the same Lagrangian... – Turgon Jun 12 at 6:45
• A covariant derivative with a gauge field has to be introduced. – p6majo Jun 12 at 7:06
• Hi desirooo: Are you talking about global gauge transformations? – Qmechanic Jun 12 at 7:18

In Field Theory, a Lagrangian is said to be invariant if you only introduce a quadridivergence of a field which only depends on space-time I.E.:

Be $$\mathcal{L}=\mathcal{L}[\phi(x),\partial_\mu\phi(x),x]$$ a Lagrangian of scalar field $$\phi$$ where $$x=(t,\vec{x})$$ is a space time coordinate. Then for an infinitesimal transformation $$\phi\rightarrow\phi+\alpha\delta\phi$$, if the Lagrangian becomes $$\mathcal{L}\rightarrow\mathcal{L}+\alpha\delta\mathcal{L}=\mathcal{L}+\alpha\partial_\mu J^\mu(x)$$ where $$J^\mu$$ is any function of space-time, then the Lagrangian is said to be invariant.

With Noether's theorem, the conserved current is given by $$j^\mu\equiv J^\mu(x)-\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi$$.

In your example you have $$\mathcal{L}=\partial_\mu\phi^*\partial^\mu\phi - m^2\phi^*\phi$$ and your infinitesimal translation is $$\phi\rightarrow\phi+i\theta\phi$$ and $$\phi^*\rightarrow\phi^*-i\theta\phi^*$$. By plugging into the Lagrangian we get

$$\mathcal{L}\rightarrow\partial_\mu(\phi^*-i\theta\phi^*)\partial^\mu(\phi+i\theta\phi)-m^2(\phi^*-i\theta\phi^*)(\phi+i\theta\phi)$$ $$=\partial_\mu\phi^*\partial^\mu\phi-i\theta\partial_\mu\phi^*\partial^\mu\phi+i\theta\partial_\mu\phi^*\partial^\mu\phi+\theta^2\partial_\mu\phi^*\partial^\mu\phi-m^2(\phi^*\phi-i\theta\phi^*\phi+i\theta\phi^*\phi -\theta^2\phi^*\phi)$$ We then neglect terms of order $$O(\theta^2)$$ taking $$\theta$$ as an infinitesimal parameter. Thus we have:

$$\mathcal{L}\rightarrow\mathcal{L}+0$$

Because all other terms cancels each other! Thus the Lagrangian is invariant under this transformation and the function $$J^\mu=$$constant which we can choose to be $$0$$. The conserved current is (one for each field):

$$j^\mu_1=-\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi^*)}\delta\phi^*$$ $$=-(\partial^\mu\phi)(-i\phi^*)=i(\partial^\mu\phi)\phi^*$$

Similarly: $$j_2^\mu=-i(\partial^\mu\phi^*)\phi$$