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1

I am the first author of the Wikipedia page of Matsubara frequency and the maker of the summation table therein. The result of your summation is $$\frac{1}{2} \eta \left(-\frac{\left(\xi _1-\xi _2+\xi _3\right) n_{\eta }\left(\xi _2-\xi _3\right)}{\xi _3 \left(\xi _2^2-2 \left(\xi _3+\xi _4\right) \xi _2+\xi _3^2+\xi _4^2-\xi _5^2+2 \xi _3 \xi ...


1

Essentially, yes, gauge fields emerge when you fractionalize your elementary particles. Suppose you have some sort of a constraint on your system e.g., no double occupancy at each site. $$\sum_{\sigma} a^\dagger_{i,\sigma} a_{i,\sigma} \leq 1$$ You can now write the electron operator in terms of spin and charge degrees of freedom $$c^\dagger_{i,\sigma} = ...


0

I assume your ordering is such that $k<l<j<i$ and $S_m = n_1 + n_2 + \dots + n_{m-1}$ are phase factors calculated before applying any creation/annihilation operators. You do take into account, correctly, that applying an annihilation operator lowers the phase factor for the action of the next operator by $1$, but should also account for the fact ...


4

It's a good question and I don't think there is a simple way of seeing it from the action. I just checked Fradkin (section 7.7 and 7.8) and through an RG analysis he shows that any half-integer spin behaves the same as the spin-$\frac{1}{2}$ case. But for the latter he then refers to the exact Bethe ansatz solution to show it is gapless! However, I wanted ...


2

The ratio of emissive power to the absorbitity is constant when the substance is at thermal equilibrium with surrounding. Or The emissive power of a substance is equal to its absorbtivity under the same conditions.


2

I'll start by saying that correctly stating Kirchhoff's law is quite tricky. "Emissivity equals absorptivity" in a certain sense, but they may depend on wavelength, and angle of incidence (or emission), and polarization. In magneto-optic materials, you can have high absorptivity from one direction balancing high emissivity into a different direction!! (This ...


-1

As an experimentalist, I know that black body is dependent on materials, which is taken into account by the emissivity . I always took the cavity model of the black body formula as an idealized tool to fit observed experimental distributions, the only contact with reality being the oscillators envisaged. The formula fits , and the experimentally best fit ...


2

It depends a little on what you mean by "extreme" electric current, but the answer is probably no. The energy scales are wrong. Electric current in a metal is a sub-electron-volt process: a potential difference of much less than a volt can displace electrons all the way through a piece of metal. The weak interaction is a keV- or MeV-scale process. And ...


2

The criterion for the Fermi liquid theory to be justified is that the imaginary part of the self-energy must be vanishingly small around the Fermi surface (both in the energy scale and in the momentum deviation). $$-2\Im\Sigma_\text{el}(k,\omega)\to 0\text{ as }k\to k_F\text{ and }\omega\to 0.$$ The broadening of the quasi-particle peak in the spectral ...


0

"Because it does not have a crystal structure, it is hard to find physical similarities with a solid" simply proves that you're using the wrong definition of "solid". Relatively few solids have crystalline structure, so a good definition of solid simply can't depend on that. Let me see if I can offer a better definition. Rigid body (Idealization) The ...


0

The poor man's scaling is a perturbation method. The polynomial series that are produced by this method are not convergent (zero radius of convergence). This serie are called asymptotic serie. They give you good results until some small corrections ($\sim D_0 e^{-\frac{1}{\rho J}}$), when the series start to diverge. This is typical in Many-Body Physics and ...


2

Glass is a typical amorphous solid. Amorphous materials typically show no melting point but do have a Glass Transition Point ($T_g$). Below it, the material behaves like a solid, with a glass-like fracture surface when fractured. Typical amorphous materials include several types of elastomer (rubber) like natural rubber (NR), with a $T_g$ of around ...


2

Sputtering deposition is not normally preformed at ultra high vacuum pressures, thus the films tend to be polycrystalline while e-beam evaporated metal films could be done at much lower pressures resulting in a more uniform film, even single crystalline depending on other conditions like the substrate, lattice mismatch and so on. This is just one difference. ...


1

Noting the simple relation $$ \frac{\Delta E_{i,j}}{\Delta E_{i,j}-\hbar\omega-i\eta}-\frac{\Delta E_{i,j}-\hbar\omega-i\eta}{\Delta E_{i,j}-\hbar\omega-i\eta}=\frac{\hbar\omega+i\eta}{\Delta E_{i,j}-\hbar\omega-i\eta} $$ and by Sokhotski-Plemelj theorem $$ \lim_{\eta\rightarrow 0^+} \frac{i\eta}{\Delta E_{i,j}-\hbar\omega-i\eta}=0 $$ because, if the limit ...


1

I don't think we need Sokhotski-Plemelj for this. Think of $E_j - E_i$ as a fixed value $E$. Then the formula is re-written as $$\frac{\hbar \omega}{E - \hbar \omega - i \eta}\, .$$ Now let $x \equiv \hbar \omega$ and you get $$\frac{x}{E -x - i \eta} \, .$$ This integral is dominated by the part where $x \approx E$ so let's try shifting the variables $y ...


2

Chiral $p$-wave superconductor and He A phase can be considered being equivalent phases of matter, for the following reason: The fact that the superconductor is charged does not make too much difference in this regard, because it only affects the electromagnetic response (one has Meissner effect, the other does not), but electromagnetic field is an external ...


0

when the water evaporates it goes up into the sky and comes back down as rain, this is commonly known as the 'circle of nature' no, the water would heat up The chlorine is denser at the bottom of a swimming pool copper would condense, the chlorine and sodium would disolve SOURCE: Jake Burn, Milk Churn, University of South Bradford


4

It is definitely a Kronecker sum. Take the case where there are only two different states $+$ and $-$, then, for example, $$ \hat H =E_+ \hat a^\dagger_+ \hat a_++E_- \hat a^\dagger_- \hat a_- .$$ What does $\hat a_+$ means ? Well, if we label the states with the number of excitations in the states $+$ and $-$ by $|n_+,n_-\rangle$, then we understand $\hat ...


0

If it is still of interest, next to all the excellent suggestions above there is a book from 2013 which I found rather helpful as it makes some neat observations I could not find in other texts: Nonequilibrium Many-Body Theory of Quantum Systems - Stefanucci and Leeuwen


1

Confinement is usually discussed in the context of gauge theories. Here you do not have any specific information about the properties of the insulator, so it is not clear why you put "confined states" and "localized states" on the same footing. Insulating states must have a gap for charge carriers. The gap can come in two types: 1) in a clean system, or ...


0

The "only" thing you need to do is to establish a mapping. You have a basis function at $$ \vec{R} = a\vec{X} + b\vec{Y}$$ with index i. In other words, your basis is $\phi_{abi}$. Since Matlab only understands (well) vectors and matrices, you need to map this to a continuous index n. For example, a square with sides $N_a$, and $N_b$ and $N_i$ basis ...


2

The torque exerted by $\vec B$ is perpendicular to it, so the $z$ component of angular momentum is conserved.



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