# Tag Info

0

A zero-field-cooled/field-cooled split in the magnetic susceptibility vs. temperature doesn't have to be superparamagnetism. In the case of superconductors, if we apply a field to the material and cool past T$_c$, some flux can be trapped inside, but if we cool first and then apply field, that flux will be shielded away, resulting in greater diamagnetism.

0

What a superconductor in a magnetic field does is create (by Lenz's Rule) a current that creates a magnetic field which in turn pushes the external field outside of the volume of the superconductor. Since such a material has no resistance, you have no heat losses that would decrease the internal current. To answer your question, a superconductor does produce ...

0

In addition to what Meng Cheng said, remark that the mean-field Bogoliubov-Gennes Hamiltonian, say $$H=\sum_{k}\left(c_{k}^{\dagger}H_{0}c_{k}+\Delta_{k}c_{k}c_{-k}+\Delta_{k}^{\ast}c_{-k}^{\dagger}c_{k}^{\dagger}\right)$$ can be diagonalised by the Bogoliubov transformation you mentioned : ...

3

Basically you are missing a term. Think of Lagrangian mechanics. If $q(t)$ is your generalized coordinate, and you make the change $q(t) \to q(t) + \delta q(t)$, then remember that the change in the action is $\delta S = \int_{t_i}^{t_f} \left(\partial_q L - \partial_t \partial_\dot{q} L \right) \delta q$. Thus you can basically think of the change in the ...

0

The localization of Majorana zero modes has a well-defined meaning: consider a Kitaev chain with two ends. Because of the zero modes, there are two nearly degenerate ground states, let us call them $|0\rangle$ and $|1\rangle$, which have opposite fermion parities. They are localized as "single-particle wavefunctions" in the following sense: if one computes ...

1

A quick answer: superconductivity is still an active field of research, but it has already been found in a lot of different materials. For instance, it has been found in the low temperature sector (and sometimes in the low-temperature and high-pressure sector) of materials which are usually classified at higher temperature as normal metal or insulator, from ...

0

It belongs to the symmetry class of no symmetry. i.e. the only symmetry is the fermion-number-parity conservation $Z_2^f$, which is always the symmetry of fermionic systems. See my paper http://arxiv.org/abs/1111.6341 for a discussion on the full-symmetry group $G_f$ for fermion systems.

1

Both ways are equally valid and yield identical answers. If you wanted to convince yourself, get the single complex $\psi$ equation, write $\psi = |\psi|e^{i\phi}$, and you'll find that you've reproduced the $|\psi|, \phi$ equations of motion. Varying w.r.t. $\psi$ or the real components are the same because of the product rule applied to $\psi$: ...

Top 50 recent answers are included