# Tag Info

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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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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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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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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 ...

2

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 ...

2

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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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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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 ...

2

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 ...

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 ...

1

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 $T\to ... 1 You did not include the proper infinitesimal imaginary part for frequency in the first equation for the Green's function, which would have given the correct pole structures. So it is not clear whether this is time-ordered, retarded or advanced. The retarded Green's function is$G_R(\omega,p)=\dfrac{\omega+\xi}{(\omega+i\delta)^2-\xi^2-\Delta^2}$You can ... 1 It seems that you thought$\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$the symmetry ... 1 Ok, finally solved it in a very simple geometrical way. IF we take a square slanted lattice in the hexagonal lattice, like in the image, which is$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 ... 1 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, ... 1 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 ... 1 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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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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