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2

You can derive the desired expression in the following way: \begin{align}\delta(R_{ab}R^{ab}) &=\delta R_{ab} R^{ab}+R_{ab}\delta R^{ab}\\ &=\delta R_{ab}R^{ab}+R_{ab}\delta R_{cd}g^{ca}g^{db}\\ &=\delta R_{ab}R^{ab}+R^{cd}\delta R_{cd}\\ &=2R^{ab}\delta R_{ab}\\ &=2R^{ab}\delta(R_{cadb}g^{cd})\\ ...

2

The value $R=\alpha^{\prime 1/2}$ is the self-dual radius under T-duality. One may indeed extract the massless spectrum – the spectrum of all fields much lighter than $\alpha^{\prime -1/2}$. Because the CFT has an $SU(2)\times SU(2)$ symmetry, as can be seen from the OPEs of the currents, the spacetime physics has this symmetry, too. Because one finds ...

3

While keeping the array page $9$ in ref1, already given, in mind, we add a new ref2, especially fig $1$ page $7$, paragraph $2.2.3$. $D = 6$, page $11$, table $5$ page $13$, and discussion page $12$ From fig $1$, page $7$, we see, that in $D=6$, the $N=2$ supersymmetry corresponds to a $(N_+, N_-) = (1,0)$ supersymmetry Looking at the discussion page $12$, ...

1

For static polarizability calculations it seems that both B3LYP and PBE functionals do a pretty good job; for benzene and napthalene I am getting numbers within a few percent of experimental values. What is much more important is the basis set. In particular, it's absolutely necessary to include diffuse functions. For benzene it's so extreme that ...

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Again, thanks to the $SU(2)$ PSG proposed by prof.Wen, I can answer my question now, $THT^{-1}$ is in fact $SU(2)$ gauge equivalent to $H$, and the statement "$H$ is also not SU(2) gauge equivalent to the time-reversal transformed Hamiltonian $THT^{-1}$" in my question is wrong. Let's rewrite the Hamiltonian as ...

2

I don't know the article you refer to, but I believe the Hamiltonian you discuss should get a $\pi$-phase shift after one turn around a (2D) lattice cell. So I guess it should read $H=F^{\dagger}\cdot H_{\pi}\cdot F$ with $$H_{\pi}=t\left(\begin{array}{cccc} 0 & e^{\mathbf{i}\pi/4} & 0 & e^{-\mathbf{i}\pi/4}\\ e^{-\mathbf{i}\pi/4} & 0 & ... 1 I would recommend to use a different functional, preferentially one having dispersion corrections and range separation. An experimentalist once asked me for polarizability data for organic molecular chains; I used \omegaB97XD and got results with an errors of 1% or less with respect to experiment. I didn't bothered to do B3LYP and PBE but I doubt they can ... 4 Topological degeneracy is only defined in the thermodynamic limit on a closed manifold. The ground state degeneracy of a finite-sized system or on an open manifold is not "topological", and can not be called topological degeneracy. Considering your examples. (1) The ground state degeneracy is ill-defined with open boundary condition. Because there might be ... 3 Answer: There is none. The issue at hand is that the Kaehler invariance is just that - an invariance, not a continuous symmetry of the fields. Most prominently the superpotential must transform as$$ \mathcal W \to \mathcal W e^{-h} $$A general superpotential that leads to consistent theories is$$ \mathcal W =\frac{1}{2} m_{ab} \phi^a \phi^b + \frac{1}{3} ...

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So the way these equations are solved is by looking what order we have at the equations, this means that when we put the ansatz inside the equations, we keep all the orders that are greater than the ones equating the equation. Then we collect all terms with the powers of $r$ (for $tt$ equation all the powers greater than $r^2$ are that of $r^3$, for $tr$, ...

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