Gauge invariance of Landau-Ginsburg model [closed]

Question

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?

Answer 1

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