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$i0$ is meant as $$\frac{1}{\epsilon - \omega_0 + i 0} = \lim_{x\to 0^+} \frac{1}{\epsilon - \omega_0 + i x}\, .$$ The idea here is that the Fourier transform of something containing a theta function is not well defined. This comes from the fact that the integral $$G(\epsilon) = -i \int_{-\infty}^{\infty} \theta(\tau) \text{e}^{-i \tau ... 3 In essence, you are taking the Fourier transform of the Heaviside theta function, \theta(t), to try and get$$ \tilde\theta(\omega)=\int_{-\infty}^\infty e^{i\omega t}\theta(t)\text dt=\int_{0}^\infty e^{i\omega t}\text dt. $$For nonzero \omega, this is perfectly fine and easily evaluates to {1}/{i\omega} once you discard the term at infinity (which ... 3 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 ... 2 This is why people start to abandon calling the zero modes in vortices "Majorana fermions" because they are NOT fermions. \gamma is a Majorana zero mode, which means it always has to pair up with another Majorana zero mode to form a 2-dimensional Hilbert space. Exchanging vortices generates a nontrivial unitary transformation on the degenerate space. 2 The Hamiltonian is time-reversal invariant: c_{k\uparrow}\rightarrow c_{-k,\downarrow}, c_{k\downarrow}\rightarrow -c_{-k,\uparrow}. You can check that explicitly. The ground state is also invariant, because Cooper pairs are all spin singlet. One of the significant implications of time-reversal symmetry for s-wave superconductors is the Anderson's ... 2 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 1 A short answer to questions 2 and 3: In Mermin-Wagner's paper the short-range condition is stated as \sum_{\bf R} {\bf R}^2 |J_{\bf R}|<+\infty. For interactions with (or more precisely majorized by a) power law decay |J_{\bf R}| \sim R^{-\alpha}, this requires \alpha > D+2, where D is the space dimensionality (i.e., \alpha >4 for D=2 ... 1 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 ... 1 The \mathrm{i}0 is short for$$\lim_{\epsilon\rightarrow 0^+}\mathrm{i}\epsilon$$and is used in the so-called \mathrm{i}\epsilon-prescription. It is necessary for some integrals so that the contour does not hit a pole and render the integral ill-defined. In this case, you must insert \lim_{x\rightarrow 0}\mathrm{e}^{-x\epsilon} into your Fourier ... 1 Exact statements The Hohenberg-Kohn theorems, which are the theoretical foundation of DFT, essentially say that the ground state properties of a many-electron system are only a function of the electron density. Any quantity you want to calculate can be re-expressed in terms of the electron density n(r), including the many-body ground state wavefunction, ... 1 The Green function you gave above is not the Green function for an insulator, which should generically has no pole if there is no Fermi surface. The Volovik argument is a bit circular, since you know from the beginning that p=p_{F} is the Fermi momentum, defining the Fermi surface. When p\approx p_{F} you can infer the form of the Green function as you ... 1 Band theory does not properly describe systems with strong electron-electron interactions. Specifically, the transition metal oxides you have mentioned are both examples of strongly correlated electron systems, which show all kinds of novel behavior which escapes description of band theory. In general strong interactions can induce phases such as Mott ... 1 Yes, the statement can be explicitly verified from the matrix representation of the spin operators acting on different spins. Acting on the spin-1/2 object, the spin operators read$$S^x=\left( \begin{array}{cc} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \\ \end{array} \right), S^y=\left( \begin{array}{cc} 0 & \frac{i}{2} \\ -\frac{i}{2} & 0 \\ ...

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The statement is not true. Counter example: quantum spin ice or U(1) spin liquid. In gapless spin liquid phase, the boson (spin excitations) are emergent U(1) photons in the deconfined phase, which are gapless but not condensed.

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Your statement itself is not quite right. What is not conserved is the chiral current, namely the current of fermions at one of the Weyl nodes. The physics can be understood essentially in one-dimensional version of the Weyl metal: consider a 1D electron gas. There are two Fermi points, and the low-energy theory is given by two "Weyl fermions" in 1D with ...

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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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You have to realize that STM is a surface technique. This means that you are probing only the surface, which can be dirty or have suffered from reconstruction, or oxidation. Moreover, it will give you rather limited statistics given the relatively small areas that you can probe. On the other hand, x-rays probe the sample deeper and with better statistics. ...

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