I don't understand why the Helmholtz energy is used in the Ginzburg-Landau theory of superconductivity: $\mathcal{F} = \mathcal{F}_n + \alpha\left|\Psi\right|^2 + \frac{\beta}{2}\left|\Psi\right|^4 + \frac{1}{2m^*}\left|\left(-i\overrightarrow{\nabla}-e^*\overrightarrow{A}\right)\Psi\right|^2 + \frac{\left|\overrightarrow{\nabla}\times\overrightarrow{A}\right|^2}{2}$

I would say that this is the energy of the system, not the Helmholtz free energy, I may be losing something.


1 Answer 1


Ginzburg-Landau theory is really a theory of phase transition, grounded in the Landau theory of phase transitions, with the superconducting wave function, $\Psi$ serving as the order parameters. Thus, this is neither Helmholtz energy, nor the Hamiltonian, but the Landau function. In fact, the first three terms are a usual Landau function for describing a second-order phase transition, whereas the remaining terms are added to describe the variation of the order parameters in space (and sometimes in time) - this implies that this variation is adiabatic, i.e., the potentials are changing slowly on the scales of the cooper pair size.

What gives misleading impression that this is energy is the Ginzburg-Landau equation, which resembles the Schrödinger equation, which makes one associate the Ginzburg-Landau function with a Hamiltonian.

  • $\begingroup$ I don't know much about phase transition theory. A question arises: why the integral over volume of this free energy (Landau function) is the energy? So then we can minimize it and obtain GL equations. $\endgroup$ Commented May 19, 2021 at 8:11
  • $\begingroup$ After some reading on Landau phase transition theory, I found that the equation corresponds to Helmholtz energy, so I'll say that the answer is wrong. Reference: Lectures on Landau Theory of Phase Transitions, Peter D Olmsted (Department of Physics, Georgetown University) $\endgroup$ Commented May 19, 2021 at 18:03
  • $\begingroup$ @AbelGutiérrez Landau function is not free energy, although it is quite acommon error. Free energy cannot describe phase transitions, since it is averaged over all states. I recommend the goldenfeld's Lectures on phase transitions. $\endgroup$
    – Roger V.
    Commented May 19, 2021 at 18:21
  • $\begingroup$ Read "On the theory of superconductivity" at "Collected Papers of L. D. Landau", where Landau and Ginzburg say "Thus near $T_C$ we may write for the free energy..." and then they write equation (6) on the paper, where you have the three first terms of the energy in question. $\endgroup$ Commented May 20, 2021 at 6:16
  • 1
    $\begingroup$ @AbelGutiérrez People call it "free energy" by analogy, because its minima correspond to the state of the system. But the true free energy has only one minimum. $\endgroup$
    – Roger V.
    Commented May 20, 2021 at 6:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.