2
$\begingroup$

Blundell's book on Magnetism, talks about the generalized rigidities as a general consequence of spontaneously broken symmetries. In this context, it mentions that in a superconductor the phase of the macroscopic wavefunction, $\phi$, is uniform across a sample. Twisting this phase across the sample produces a supercurrent proportional to $\nabla\phi$. In the first place, I wonder, is there any way to create such a spatial variation of $\phi$ in a single superconducting sample? If so, how would you achieve that in a laboratory?

Please understand that I am not interested in two different superconducting samples joined by an insulator as in Josephson junction but in spatial variation of $\phi$ in a single continuous superconducting sample.

$\endgroup$
8
  • $\begingroup$ Ion implants, local laser heating, ...? $\endgroup$ – Jon Custer Jun 15 '19 at 15:30
  • $\begingroup$ Wouldn't simply running a current through a superconductor give you the varying phase you're looking for? $\endgroup$ – KF Gauss Jul 5 '19 at 3:08
  • $\begingroup$ @KFGauss So a SC doesn't conduct in its ground state? $\endgroup$ – mithusengupta123 Jul 5 '19 at 3:19
  • $\begingroup$ Sorry I don't understand what you're asking and how it relates to my comment. $\endgroup$ – KF Gauss Jul 5 '19 at 3:50
  • $\begingroup$ @KFGauss Supercurrent, in absence of a magnetic field, is proportional to the gradient of phase $\phi$ i.e. $\nabla\phi$. But $\phi=constant$ in the ground state. So according to your comment, the ground state cannot support a supercurrent. $\endgroup$ – mithusengupta123 Jul 5 '19 at 4:53
1
+50
$\begingroup$

Yes, there is a very simple way to introduce a gradient in the phase $\nabla \phi$, all you need is to have a current running in your superconductor (i.e. a supercurrent). If you have a constant current $\mathbf{J}$, and then you have a constant phase gradient.

$\endgroup$
1
$\begingroup$

This can be most easily understood in light of Landau-Ginzburg theory. In the ground state their is some order parameter, for ferromagnets this is the magnetization $\textbf{M}$, for superconductors (in Landau-Ginzburg theory) this is some wavefunction with some phase $\phi$. He states that in the ground state there will always be some term proportional to $(\nabla \phi)^2$ which represents the energy cost to have a non-uniform order, assuming that uniform order represents the lowest energy state (ground state).

In Landau-Ginzburg theory, you have divergence of the coherence length $\xi_{GL}$ as $T \to T_c$ in the absence of an external magnetic field (keep in mind that GL theory is only valid close to $T_c$). This coherence length describes the length scale over which the order parameter (and phase) changes. Furthermore, this coherence length strongly depends on the electron mean free path, i.e. it strongly depends on the purity. This degree of purity can even allow one to change a type-I superconductor to a type-II superconductor.

Now to answer your question, in reality, we always have spatial variation of the phase due to impurities. In the ideal case (conventional superconductors), we have no spatial variation of the phase below the critical temperature. So when we have spatial variation of the phase, it will induce a supercurrent, which will try to decrease phase inhomogeneity, because it wants phase coherence (uniformity), because this gives the lowest energy state.

Ps. I have been talking in the absence of any field. In the presence of an external magnetic field, the supercurrent will also have a second term, which gives rise to the Meissner effect. \begin{equation} \textbf{j}_s =\frac{q\hbar}{2mi}(\Psi^* \nabla \Psi - \Psi \nabla \Psi^* ) - \frac{q^2}{m}|\Psi |^2\textbf{A} \end{equation}

$\endgroup$
2
  • $\begingroup$ Last sentence of the first paragraph: First you're saying that the ground state will always have non-uniform order (i.e., non-uniform $\phi$) and then you're saying that the ground state will have uniform order (i.e., uniform $\phi$). $\endgroup$ – mithusengupta123 Jun 28 '19 at 20:16
  • $\begingroup$ @mithusengupta123 I don't get the point. Does the energy of a SC depend on the spatial variation of its phase? $\endgroup$ – SRS Jun 29 '19 at 9:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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