Tag Info

New answers tagged

0

This is actually exactly the same question I was asking myself a while ago, and it took me quite some time to figure it out. What I ended up doing was: diagonalise the BdG Hamiltonian in Mathematica solve the expressions for the eigenvalues for $\kappa$ neglect terms proportional to $\kappa^2$, $\kappa \Delta$ in the expressions for the eigenvectors, ...


2

Suppose we impose a current density $\newcommand{\j}{\mathbf{J}}\j$, then the resulting electric field $\newcommand{\e}{\mathbf{E}}\e$ is given by $\e = \rho \j$, where $\rho$ is the resistivity. In a perfect conductor, $\rho=0$. So in a perfect conductor with some fixed current $\j$, the electric field satisfies $\e = \rho \j = 0 \j = \mathbf{0}$. I don't ...


0

Can someone help provide me with an argument why the electric field must be zero in a perfect conductor? It's not clear exactly what you're looking for. In a sense, any argument attempting to prove that the electric field must be zero in a perfect conductor will beg the question. For example, here's an excerpt from "Electromagnetics for High-Speed ...


3

That's probably (one of) the biggest open question in condensed matter physics. High-T superconductivity is microscopically not well understood. Apparently, reduced dimensionality (CuO2 sheets in YBCO), interplay with magnetism (ground state tends to be antiferro magnetic), quantum fluctuations of spins, and other exotic concepts are parts of their ...


1

The electrical field $\mathbf{E}$ is an external field, which "drags" the conductor electrons through the conductor "lattice". The conductivity $\sigma$ describes the resistance of the "lattice". When the resistance is zero, a non zero current can exist in the conductor without necessity to support it with an external field. If $\mathbf{E}$ is non zero, the ...


1

A single photon mode is described by a creation operator. The problem is not complicated and really generic in bosonic problems. How to include the driving is just a matter of taste. Suppose your circuit is described by the degrees of freedom $x$ and $p$ (say position and momentum). A force $f$ would be added as a term $x\cdot f$ in the Hamiltonian for ...


1

Personally, I find it more intuitive to think in terms of the closely related quantity, the loss function, $-\text{Im}\frac{1}{\epsilon}$, rather than the optical conductivity. If one were to tune across a phase transition from a electronic liquid to an electronic solid, I suspect one would expect to see the softening of the free-carrier plasmon. Once the ...


1

Recall that the Fermi surface of the free electron gas, the total momentum is zero. How can electron gas be conductive? Well, when you apply the electric field, it will be non zero. The story is same for superconductivity.


1

When you write down the formulas A15-A18, apply this assumption to these definitions and put it into wave functions, which are first calculated by solving BdG equation. you can reach the Beenakker results.


1

To answer your question, one needs to understand a bit what is the Ginzburg-Landau (GL) formalism. Let us first recall the GL functional: $$F=\int dV\left[g\left|\left(\nabla-\dfrac{2\mathbf{i}e}{\hbar}A\right)\Psi\right|^{2}+a\left(T-T_{c}\right)\left|\Psi\right|^{2}+b\left|\Psi\right|^{4}+\dfrac{\left(\nabla\times A\right)^{2}}{2\mu}\right]$$ with ...



Top 50 recent answers are included