# Tag Info

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Bad question- the speed of light should be rephrased as "Speed of electromagnetic radiation" and then specify what wavelength of radiation we're talking about and which part of the neutron star is being discussed. Nobody has determined the EM properties of the inside of neutron stars so the question is not on target. Different parts of a neutron star have ...

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Is there any good definition of many body localization? Let's start with single-body (Anderson) localization. There, in a non-interacting system, a particle (e.g. an electron) becomes localized due to destructive interference with itself. This interference is induced by the presence of disorder. Turning interactions on brings us to the realm of ...

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All $\mathrm{SU}(2)_k$ with $k>2, k\neq 4$ are universal. For a proof see http://arxiv.org/abs/math/0103200.

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The string-net condensation is a general construction to obtain gauge fields and fermions. The chiral fermion problem refers to the fact that in the Standard Model (SM), the SU(2) gauge field only couples to the left-handed fermions but not the right-handed fermions. However in the (early version of) string-net condensation, the emergent gauge field will ...

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http://arxiv.org/pdf/1205.5792.pdf The first example in the paper is $C=3$ on a triangular lattice with two orbitals per site. It is essentially three-layers of Haldane's honeycomb lattice model, but stacked together in a clever way so the translation symmetry is restored. UPDATE: In fact two-band free fermion Hamiltonian on a square lattice can realize ...

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No, you can not write this transformation as $c_{i\uparrow}^\dagger|0\rangle=c_{i\downarrow}^\dagger|0\rangle$, because $c_{i\uparrow}^\dagger|0\rangle$ and $c_{i\downarrow}^\dagger|0\rangle$ are two orthogonal quantum states: they can not be equal. The transformation $c_{i\uparrow}^\dagger|0\rangle\to c_{i\uparrow}|0\rangle$ you start with is also wrong, ...

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That a Hamiltonian preserves a symmetry means $$[H, C] = 0 \Rightarrow CHC^\dagger = CHC^{-1} = H$$ For a unitary symmetry operator $C$ (or anti-unitary if it is time reversal). The Hamiltonian of a crystalline condensed matter system written in terms of the Bloch matrix is: $$H = \sum_{\vec k} \psi^\dagger(\vec k) H(\vec k) \psi(\vec k)$$ Where ...

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The trivial and non-trivial SPT states both are symmetric under on-site unitary symmetry transformations. The trivial and non-trivial SPT states can be mapped into each other by local unitary transformations (the $U$ in you question). Although such a $U$ is a local unitary transformation, 1. it is not on-site, 2. it is not the on-site unitary symmetry ...

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It seems that you thought $\Phi_0$ is a trivial SPT while $\Phi$ is nontrivial. This does not make sense without defining the symmetry transformation. The fact that $\Phi=U\Phi_0$ means both are product state ($U$ is a honest local unitary transformation). However, symmetry transformation is defined differently in the two states: for $\Phi$ the symmetry ...

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In Raman scattering, the molecule absorbs the photon into a virtual state, which doesn't actually exist. Unlike an excited state, the molecule can't stay in that state for longer than a time $\Delta t$ where $\Delta t \Delta E \leq \hbar/2$ - the Heisenberg uncertainty relation. A virtual state can have any energy level, though, and that's the reason for ...

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I think the difference between luminescence and Raman scattering lies in whether or not the mixed photon-molecule state maintains coherence with the exciting radiation. In Raman scattering, we imagine that coherence is maintained. In luminescence we imagine that coherence is disturbed by any one of a variety of interactions: for example, collisions with ...

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To kick things off: you're right. Molecular vibrational excitations are exactly the same as phonon modes. We don't use that language very much because the system is too small (so we can't have things like travelling waves which have momentum, and we need to work only with stationary waves) but the analogy is indeed exact. Now, as to what exactly is ...

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In this context, vibration usually refers to the relative motion of the nuclei. In a periodic solid (crystal), vibrational modes are called phonons. We can say, for example, that a normal vibrational mode of a branch $s$ with wavevector $\mathbf{k}$ is in its $n$th excited state, or equivalently, that there are $n_{\mathbf{k}s}$ phonons of branch $s$ with ...

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You did not include the proper infinitesimal imaginary part for frequency in the first equation for the Green's function, which would have given the correct pole structures. So it is not clear whether this is time-ordered, retarded or advanced. The retarded Green's function is $G_R(\omega,p)=\dfrac{\omega+\xi}{(\omega+i\delta)^2-\xi^2-\Delta^2}$ You can ...

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DFT is based on two important theorems: (1) Hohenberg & Kohn: the potential and the density are connected by a one-to-one map (2) Kohn & Sham: there is always a non-interacting reference system (map: V_xc: non-interacting <-> interacting problem) having the same density as the interacting one. In a nutshell: the potential and the density of the ...

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Consider a volume element $dV$ at a certain point $\vec{x}$. Let the strain tensor at it be given by $$e_{ij} = \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}$$ Let us diagonalize the strain tensor at this point and let its diagonal entries be $u^{(i)}$. Since the trace of tensor is invariant, ...

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Ok, finally solved it in a very simple geometrical way. IF we take a square slanted lattice in the hexagonal lattice, like in the image, which is $N$ particles along each side. Then the number of particles inside is $N^2$. The volume of that area is just $V=(Na)^2\cos(30)$, and so: $$\rho a^2=\frac{2}{\sqrt 3}$$. I'm sorry for asking, I was frustrating ...

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Your result only holds in the $U/t\gg 1$ limit. A brute-force method to obtain the result is by exact diagonalization in the Fock space basis, where all the operators are represented as matrices. Let us introduce the Fock states $|n_{1\uparrow}n_{1\downarrow}n_{2\uparrow}n_{2\downarrow}\rangle$ where $n_{i\sigma}=0,1$ denotes the fermion number on site $i$ ...

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It seems to me that you just cannot tell the difference between a Bose condensate and nothing in this case. What will change if you add some photons or phonons with zero energy to the system? No characteristics of the system will change. So it seems to me we have no criterion to decide if there is a Bose condensate in this case, and what's more important, it ...

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The zero energy states are localized at the boundary, and their wave functions decay exponentially in the bulk. So they are boundary modes (or edge states), which do not count as the bulk states, and do not contribute to the bulk gap closing. The only way to close the bulk gap in the SSH model is to tune $\delta t$ to zero. As long as $\delta t\neq 0$, the ...

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Duality transformation does not preserve topological order, and hence not preserving the topological entanglement entropy. The quantum Ising model has no topological entanglement entropy. See this related question for more discussions.

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A topological invariant in condensed matter systems is a number that doesn't change under a smooth deformation of the Hamiltonian. I like to think this as stretching a material that would change for instance the hopping constants. Now the way that one can take such a number may vary and I don't think there is a limitation in the definition neither that there ...

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A full introduction to the DMRG algorithm definitely does not fit here, and you can find many well-written introductory materials online. DMRG has been applied to simulate perturbed toric code model, which is the simplest example of string-net model, see http://arxiv.org/pdf/1205.4289.pdf. Generally speaking, DMRG for 2D spin models is indeed much more ...

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The Landau model for ferromagnetism has the following expression for the free energy density, as a function of temperature $T$ and magnetization $M$: $$F(T,M)=F_0(T)+\dfrac{a}{2}(T-T_C)M^2+\dfrac{b}{4}M^4+\dfrac{c}{6}M^6+\mathcal{O}(M^6)$$ First order phase transition occurs when the first derivative of $F$ (namely, the entropy) is discontinuous as $T\to ... 0 The answer is yes. Your Hamiltonian is $$H = t_{11}a^{\dagger}_1a_1 + t_{12}a^{\dagger}_1a_2 + t_{21}a^{\dagger}_2a_1 + t_{22}a^{\dagger}_2a_2,$$ and the retarded Green's function is defined as $$G^{R}(\nu t,\nu't')= -i \theta(t-t')\langle {a_{\nu}(t),a^{\dagger}_{\nu'}(t')}\rangle ,$$ for each$\nu,\nu' =1,2$. Notice that, for emphasis, I will use$G^{R}$... 1 I guess you are confused by the action on wave functions and operators. 1) The second approach to your first concern is "more correct." Notice that$t(k)$or$w(k)$is a$c$-number( or scalar, not an operator). When you write$w(k) \rightarrow w(-k)\$, I guess you treat it as a wave function (I am not sure, because if it is a wave function, it should ...

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This maybe the reference you want. http://www.sciencedirect.com/science/article/pii/0370157386901675# Fermion number fractionization in quantum field theory A.J. Niemi G.W. Semenoff

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In any case, for undoped graphene, the Fermi level of electrons and holes are symmetric so the Fermi Energy lies at the Dirac Point, so these two definitions would be equivalent.

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