# Fluctuations in Ginzburg-Landau Model for superconductivity

Given the Ginzburg-Landau Hamiltonian: $$\beta\mathcal{H}=\int d^3x\left[\frac{K}{2}(D_\mu\psi)(D_\mu\psi)^*+\frac{t}{2}|\psi|^2+u|\psi|^4+\frac{L}{2}(\nabla\times \vec{A})^2\right]$$ with $$\psi$$ a complex order parameter, I managed to find the Ginzburg-Landau equations and find the saddle-point constant solution $$\psi(x)=\bar{\psi}=\sqrt{-\frac{t}{4u}}$$ (for $$t<0$$) for $$\vec{A}=0$$.

Now I am trying to expand the hamiltonian about small fluctuations: $$\psi(x)=e^{i\theta(x)}(\bar{\psi}+\phi(x))$$ and $$\vec{A}=\vec{a}$$ with the Coloumb gauge: $$\nabla\cdot \vec{a}=0$$.

I want to expand the hamiltonian to quadratic terms in $$\psi,\theta,\vec{a}$$, so I got to: $$\beta\mathcal{H}\approx const+\int d^3x \left(\frac{K}{2}\left(i(\partial_\muθ)e^{iθ}\phi+e^{iθ}\partial_\muφ-iea_\mu e^{iθ}\phi\right)\left(i(\partial_\muθ)e^{iθ}\phi+e^{iθ}\partial_\muφ-iea_\mu e^{iθ}\phi\right)^*+\frac{t}{2}|\phi|^2+\frac{L}{2}|\nabla\times\vec{a}|\right)$$

But I am unsure how to proceed, especially regarding the $$|\nabla\times\vec{a}|^2$$ term - how does the Coloumb gauge come into play here?

So I managed to solve this, in a way: $$\beta\mathcal{H}=\int d^3x\left[\frac{K}{2}(D_\mu\psi)(D_\mu \psi)^*+\frac{t}{2}|\psi|^2+u|\psi|^4+\frac{L}{2}|\nabla\times\vec{A}|^2\right]$$ $$=\int d^3x \left[ \frac{K}{2}\left(\partial_\mu-iea_\mu\right)(e^{i\theta}(\bar{\psi}+\phi)\left((\partial_\mu-iea_\mu)(e^{i\theta}(\bar{\psi}+\phi))\right)^*+\frac{t}{2}|\phi|^2+\frac{L}{2}|\nabla\times \vec{a}|^2 \right]$$ $$=\int d^3x\biggl[ \frac{K}{2} \biggl(e^{i\theta}\left(i(\partial_\mu\theta)(\bar{\psi}+\phi)+\partial_\mu(\bar{\psi}+\phi)-iea_\mu(\bar{\psi}+\phi) \right)\biggr) \biggl(e^{-i\theta}\biggl(-i(\partial_\mu\theta)(\bar{\psi}+\phi)+$$$$\partial_\mu(\bar{\psi}+\phi)+iea_\mu(\bar{\psi}+\phi) \biggr)\biggr) +\frac{t}{2}|(\bar{\psi}+\phi)|^2+u|(\bar{\psi}+\phi)|^4+\frac{L}{2}|\nabla\times \vec{a}|^2 \biggr]$$ Now we want to keep only up to quadratic terms, meaning that terms such as $$ea_\mu(\partial_\mu\theta)\phi^2$$ are ignored, which leaves us with: $$\beta\mathcal{H}\approx \int d^3x \left[\frac{K}{2}\left( \bar{\psi}^2(\partial_\mu\theta)^2+(\partial_\mu\phi)^2-2e\bar{\psi}^2a_\mu(\partial_\mu\theta)+e^2\bar{\psi}^2a_\mu^2 \right)+\frac{t}{2}\phi^2+6u\bar{\psi}^2\phi^2+\frac{L}{2}(\nabla\times \vec{a})^2\right]$$ Integration by part and using the Coloumb gauge $$\nabla\cdot\vec{a}=0$$ gets rid of the $$2e\bar{\psi}^2a_\mu(\partial_\mu\theta)$$ term, and since repeated indeces are summed we can write $$(\partial_\mu \xi)^2=(\nabla\xi)^2$$ so: $$\beta\mathcal{H}\approx \int d^3x \left[\frac{K}{2}\left( \bar{\psi}^2(\nabla\theta)^2+(\nabla\phi)^2+e^2\bar{\psi}^2|\vec{a}|^2 \right)+\frac{t}{2}\phi^2+6u\bar{\psi}^2\phi^2+\frac{L}{2}(\nabla\times \vec{a})^2\right]$$
Now taking the fourier transform of $$\phi$$, $$\vec{a}$$ and $$\theta$$ easily gives rise to their expectation values and variances.