# Gauge invariance of Landau-Ginsburg model [closed]

Free energy density for Landau-Ginsburg model is given by: $$F=\frac{1}{2\mu}(\vec\nabla \times \vec A)^2+|(\vec\nabla-ie\vec A)\phi|^2+a(T-T_c)|\phi|^2+\lambda|\phi|^4$$ And I was trying to show that it is gauge invariant under: $\vec A \rightarrow\vec A'=\vec A+\frac{1}{e}\vec\nabla \Lambda \,\$ and $\,\ \phi \rightarrow \phi'=e^{i\Lambda}\phi$. I managed to show that for $\,\ \phi\,\$ and $\,\ \vec\nabla \times \vec A \,\$ by direct substitution. But then $$|(\vec\nabla-ie\vec A')\phi'|=|(\vec\nabla-ie(\vec A+\frac{1}{e}\vec\nabla \Lambda)||\phi'|=|(\vec\nabla-ie\vec A -i\vec\nabla \Lambda)||\phi|$$

So it looks as if there is $+i\vec\nabla \Lambda$ missing. Is there any mistake in that derivation?

• $|D\phi|\neq|D|\,|\phi|$. May 21, 2018 at 17:54

You forgot to differentiate the exponent in the product: $(\vec\nabla-ie\vec A')\phi'=(\vec\nabla-ie(\vec A+\frac{1}{e}\vec\nabla \Lambda)(e^{i\Lambda} \phi)=e^{i\Lambda}(\vec\nabla-ie\vec A -i\vec\nabla \Lambda+i\vec\nabla\Lambda)\phi=e^{i\Lambda}(\vec\nabla-ie\vec A )\phi$