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It's because $$ \psi^\dagger \psi= \frac 12 (\psi^\dagger \psi-\psi \psi^\dagger)+const.\\ = \frac 12 (\psi^\dagger,\psi)\left[\matrix{1&0\cr0&-1}\right] \left[\matrix{\psi\cr \psi^\dagger}\right]+const\\ =\frac 12 \Psi^\dagger \tau_3 \Psi+const. $$


2

The magnetic field of a superconducting wire is generated by the electrons in the wire. When winding into a coil, the effect is self-reinforcing, the magnetic field of the coil increases. The magnetic field of the coil does not destroy itself by winding the coil. On the contrary, the magnetic field stabilizes itself. What can destroy the magnetic field of a ...


2

On very general grounds, the idea behind topological order is to classify a system by means of a topological, and as such non-local, invariant. The latter describes the bulk of your phase without needing to consider the system’s boundaries. However, this rather abstract invariant becomes of physical meaning when it comes to interfaces between topologically ...


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Suprisingly this is still a controversial topic. There are a number of different approaches that give different answers to the motion. A logically based theory is here Alan T. Dorsey Vortex motion and the Hall effect in type-II superconductors: A time-dependent Ginzburg-Landau theory approach, Phys. Rev. B 46, 8376 (1992) but this does model not agree with ...


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Let me try to provide a sketch of derivation. Consider the normal metal, so interaction with external field is $$H=\int d^3r\psi^{\dagger}\epsilon(p-eA)\psi,$$ where $\epsilon$ is dispersion law. Performing variation with respect to $A$, one find the expression for current, $$j=e\psi^{\dagger}\nabla_p\epsilon(p)\psi-e^2\psi^{\dagger}\nabla_p[(A\nabla_p)\...


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I can't plug in numbers right now but I doubt it would levitate. As you increase the superconductor volume, you increase its mass and therefore the weight. You should match this with the force from Meissner effect which probably scales with the surface area (since it arises from dissipationles surface currents). So for linear size $L$ gravity will increase ...


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