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You are considering a system of fermions (SC hamiltonian) and a system of bosons (weakly interacting BEC). In order for the density to be finite, fermions must have a positive chemical potential $\mu>0$. On the other hand, the chemical potential of a system of bosons is less or equal to the energy of the lowest-energy state. In a non-interacting system ...

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How can a scattering process have bound states? We are familiar with bound states from our everyday experience. For example, two hydrogen atoms interact through the Coulomb force. This leads to the formation of a bound state, namely, the hydrogen molecule. The most simple model of this situation is the square-well potential. This potential has ...

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The reason that space group is not considered by Altland and Zirnbauer is that they were interested in the implication of the symmetries for transport properties of disordered electrons, and once you have disorder, spatial symmetries are no longer there. But you are right that the classification can be extended to include inversion symmetry and similarly for ...

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You miss first of all that Cooper pairs do not exist as some physical quantities. If you prefer, they are not particles as electrons. They are just correlations. The current is a collective response to a gradient of phase. You can generate such a gradient by a magnetic field, a voltage, a break of the condensate (like in Josephson system), with different ...

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In simplest terms, the presence of sub-gap zero energy localized modes (Majorana modes) makes a superconductor topological. A superconducting ground state is just a bunch of Cooper pairs and the BdG Hamiltonian describes excitations above the ground state. If the excitation spectrum has these localized modes then it is a topological superconductor otherwise ...

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In the context of ultracold Fermi gases, a BEC-BCS crossover means that by tuning the interaction strength (the s-wave scattering length), one goes from a BEC state to a BCS state without encountering a phase transition (thus the word "crossover"). It is also useful to know that the BEC state is a Bose-Einstein condensate of two-atom molecules, while the ...

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The below seems to be a candidate for the first use of the term 'Majorana fermion'. (I'm not sure if it satisfies your other criteria.) Salam, Abdus, and J. Strathdee. Super-symmetry and non-Abelian gauges. International Centre for Theoretical Physics, Trieste (Italy), 1974.

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There is a very direct relationship which answers your question, and I'll state it in the way I first learned about it (but you can derive a different connection by passing between dimensions): The 2-dimensional reduction of the Seiberg-Witten equations are the (abelian) vortex equations. The $SU(2)$-vortex equations on $\mathbb{R}^2$ are a ...

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The short, simple, and intuitive explanation is that in a superconductor state, electrons are paired (BCS case) because there is an effective interaction between them. To destroy such a pair and produce free electrons you need to invest a minimum of energy, which is this energy gap $\Delta$. This produces an excitation (2 free electrons), remember that SC ...

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A further investigation led me to the desired reference, which discusses this precise problem: Hard Superconductivity: Theory of the Motion of Abrikosov Flux Lines Work of Anderson and Kim, at Bell Labs, around 1964

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This is not a full answer to the question, but just to point out existing related studies. This kind of question was considered a lot in the context of Josephson junction, which is basically a superconducting ring but with a weak link (i.e. the junction), where intuitively vortices tunnel through the junction. The simplest model of such a system is just the ...

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The expression you wrote is strange. It looks close to the current-phase relation for a ballistic and short junction: $$j\left(\varphi\right)=2ev_{F}N_{0}\Delta\sin\dfrac{\varphi}{2}\tanh\left(\dfrac{\Delta}{2k_{B}T}\cos\dfrac{\varphi}{2}\right)$$ when $\varphi=\pi/2$, which is not the critical current at low temperatures, since $j_{c}=\max\left\{ ... 1 There is no fundamental difference between the signatures found in the two works (Kouwenhoven and Yazdani). Both are tunneling spectroscopy, which roughly measures whether there is a zero-energy single-particle state in the spectrum. Yazdani's setup allowed him to do measurement away from the edge, so that the localization of the edge state can be directly ... 3 There is no "Fourier optics" in superconductors in general, only in Josephson junction, and it's due to the Aharonov-Bohm effect. In bulk superconductor, the Aharonov-Bohm gives the so-called Little-Parks effect : the oscillation of the critical temperature with respect to the magnetic flux enclosed in a non-connected superconductor (= a superconductor with ... 3 After Fourier transform you get an equation$\displaystyle (E-\frac{k^2}{m})\alpha_{\vec{k}}=\sum_{\vec{k}'}V_{\vec{k}\vec{k}'}\alpha_{\vec{k}'}$Here$V_{\vec{k}\vec{k}'}\sim \int \mathrm{d} \vec{r} e^{i(\vec{k}-\vec{k}')r}V(\vec{r})$. Now we need to make some simplifying assumption about$V$. For$s\$-wave, usually we take ...

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