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Why when applying the Ginzburg-Landau theory for superconductors and expanding the free energy in terms the order parameter $\psi$ one has to consider only the even powers of $|\psi|$?

I suppose it has to do with some symmetry under $\psi\rightarrow -\psi$ of the system (similar to the ferromagnetic case) but I still have not figured out which.

Diracology
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    $\Psi$ being a complex order parameter, a change $\Psi$ to $-\Psi$ is just a shift of the phase (i.e. $\varphi$ when writing $\Psi=\left|\Psi\right|e^{\mathbf{i}\varphi}$ for instance) by $\pi$. If you interpret $\Psi$ as a macroscopic wave function (this picture has some limit), then only $\left|\Psi\right|^{2}$ appears as an observable. This is confirmed microscopically from the BCS theory. The presence of a magnetic field makes the Mexican-hat picture associated with second order phase transition asymmetric, according to the picture of the Higgs mechanism. – FraSchelle Nov 28 '16 at 18:54
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    Some Landau functionals contain a third order term in the order parameter, especially in liquid crystal. I'm not aware of anything of this sort in superconductors, and it's unlikely to occur, since the order parameter is complex and the Ginzburg-Landau functional represents a fictitious energy. Perhaps you might find this answer interesting as well : http://physics.stackexchange.com/a/144026/16689 – FraSchelle Nov 28 '16 at 19:02
  • This answer is likely what you are looking for

    http://physics.stackexchange.com/questions/243534/invariant-polynomials-of-the-landau-theory-of-phase-transitions-crystal-symmetr

     – John M Nov 28 '16 at 19:21
  • @JohnM I don't get how the lattice structure can determine the free energy expansion for superconductors since the expansion is always the same. – Diracology Nov 28 '16 at 19:30
  • @FraSchelle I read the answer you mentioned just before posting my question. If I interpret the order parameter as the energy gap I don't see why it must appear as even powers in the free energy. I also read in Landau-Lifschitz's book (statistical mechanics) some argument about the invariance of the free energy under phase changes. But any power of $|\psi|$ is invariant as well. – Diracology Nov 28 '16 at 19:35
  • If it is an energy gap, it must be of even power, because it must be symmetric with respect to the exchange of the sign of energy. In any case you must have products of $\Psi$ and $\Psi^{\ast}$ in order to produce something real. To get the real and imaginary part of $\Psi$ would mean the observable are phase dependent, a clear contradiction with the principle of gauge symmetry of quantum mechanics. The Josephson energy for instance depends on the phase difference only, not the phase. In a bulk superconductor nothing should depend on the phase. – FraSchelle Nov 29 '16 at 10:42
  • @FraSchelle Can we associate $|\psi|^2$ with $n_s$, the density of superconducting electrons? I think $n_s$ would be a nice order parameter, $n_s=0$ above the critical temperature, $n_s\neq 0$ below the critical temperature and it is a smooth function of $T$. If the relation is valid then we could write the free energy as a Taylor expansion of $n_s$ which means an expansion in terms of even powers of $|\psi|$. – Diracology Nov 29 '16 at 12:36
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    @Diracology Clearly you can associate $n_{s}\propto \Psi\cdot\Psi^{\ast}$ without trouble. The Landau expansion will contains only even powers of $\Psi$ then, even for odd powers of $n_{s}$. Any real quantity (like $n_{s}$) made of a complex one (like $\Psi$) is necessarily squared in one way or an other. Clearly you can think of $|\Psi |^{2}$ as being the density of Cooper pairs if you wish. The phase is important though, and so a superconductor can not be described by a real quantity. It must be described by a complex order parameter. Observables are always real quantities. – FraSchelle Nov 29 '16 at 18:13

2 Answers2

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The other commenters have given various reasonable explanations for this. Here is another one:

Since we approximate the free energy as something like a functional version of Taylor expansion in $\psi$, we want the expansion to be differentiable close to the transition (i.e. where the order parameter $\psi$ goes to 0). Note that odd powers of $|\psi|$ are not complex-differentiable functions of $\psi$.

Here's a reference that makes an argument of this nature.

EDIT: Another reference that makes the same argument is Tinkham's book on superconductivity (Chapter 4)

Arkya
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I am not sure that you want more formal explanation but let me sketch. To develop GL expansion by order parameter $\Delta$, one should start from original 4-fermion theory with contact attraction (BCS theory). Then one should perform decoupling in Cooper channel which means Hubbard-Stratonovich transformation, $$\Delta\bar{\psi}_{\uparrow}\bar{\psi}_{\downarrow}+\bar{\Delta}\psi_{\uparrow}\psi_{\downarrow},$$ and $\Delta$ plays role of Cooper pairs field. After decoupling, we can safely integrate out fermion fields by introducing Nambu spinor, $$\Psi=(\psi_{\uparrow}\,\,\bar{\psi}_{\downarrow})^T$$ and its conjugated $\bar{\Psi}$. Integratin over fermion fields gives effective action of theory which is nothing more than GL expansion. Effective action has form $$\mathrm{tr}\ln(1+\mathcal{G}_0^{-1}\Delta),$$ where $$\mathcal{G}_0^{-1}=\begin{pmatrix}-\partial_{\tau}+\partial^2/(2m)+\mu & 0\\ 0 & -\partial_{\tau}-\partial^2/(2m)-\mu\end{pmatrix}.$$ Expanding log term and taking trace, you can easily verify that all odd terms have zero contribution.

To see that effective action coincides with GL expansion, it is worth mentioning that we consider only static configurations, so $\Delta$ does not depend on time and effective action will be $$S_{\text{GL}}=\int d^3x\left[A|\Delta|^2+B|\partial\Delta|^2+C|\Delta|^4\right].$$ Coefficients can be calculated explicitly, gradient term appears if you consider static but spatil inhomogeneous order parameter field.

For me, it is also important that statement about vanishing of odd terms of $\Delta$ in GL expansion does not come from the fact that $n_s\propto |\Delta|^2$. I assume that you derive $n_s$ and I am not sure that quantity $|\Delta|^2$ can be measured.

Artem Alexandrov
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