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2d
comment Particle on $S^1$ and $U(1)$-principal bundle
@ACuriousMind cont. In addition, the flux defines a distinct quanization of the classical problem of a particle on a ring, because these quantizations are classified accroding to $Map(\pi_1(S^1), U(1))$, i.e., the character group of the fundamental group, please see Doebner and Tolar arxiv.org/abs/quant-ph/0601176. Physically, these different quantizations correspond to different fluxes. The wave functions can be chosen as true functions on the circle only for the trivial $U(1)$ bundle, with vanishing flux.
2d
comment Particle on $S^1$ and $U(1)$-principal bundle
@ACuriousMind, thank you for your remark, indeed what I wrote about the flux being a Chern class was wrong and I corrected that in my answer (The Chern class must be integer and here we have $0<A\le 1$). However, the flux charcterizes flat $U(1)$ bundles over $S^1$ (sometimes called secondary characteristic class). The sections of the associated flat bundle must satisfy twisted boundary conditions, otherwise, they will not solve the Schrodinger equation with nonvanishing flux.
May
1
comment Large and small gauge transformations?
@Friedrich There is a single notion of large gauge transformations. Their description as configurations which do not tend to unity at infinity is a special case applicable to one point compactification of a flat space. On a general space there is no special point as infinity and you have to resort to the general definition.
Apr
15
comment Why do we assume local conformal transformations are symmetries in 2D CFT
This means that the symmetry group in contrast to the symmetry algebra will not be unitarily representable on the Hilbert space, and this is not the case that we need, since as mentioned above, we need to gauge this symmetry out.
Apr
15
comment Why do we assume local conformal transformations are symmetries in 2D CFT
in quantum field theory (in contrast to quantum mechanics), spontaneously broken generators drive the vacuum outside the Hilbert space of the theory since we can prove that minimum energy vectors of the type $e^{i\alpha G}|vac\rangle$ where $G$ is a spontaneously broken generator are orthogonal to the vacuum for all $\alpha \ne 0$ to the vacuum, thus cannot be represented by a unitary operator (the infinitesimal action $ G|vac\rangle$, however, exists as it is the Nambu-Goldstone mode.)
Apr
15
comment Why do we assume local conformal transformations are symmetries in 2D CFT
@ungerade, in string theory, the Virasoro algebra must be a true symmetry of the Hamiltonian and the vacuum. This is the only way one can gauge it out and obtain a reparametrization invariant theory at the quantum level. Thus, the string theory can be composed of various sectors (matter and ghosts) each having a non-vanishing central charge but the total central charge must be zero. The way to achieve this is to take the string theory vacuum as the tensor product of the various constituent vacuua. Each sector by itself needs not be reparametrization invariant....
Apr
15
comment Why do we assume local conformal transformations are symmetries in 2D CFT
@ungerade you are correct
Apr
4
comment Spin state after boost
@TylerHG I don't think that the rotation will end up along the x-axis, since it depends on both $p$ and the boost speed $v$. The expression you wrote for $\Lambda p$ is correct. So, I don't see how one can avoid performing the full computation by multiplying three 4-matrices. It is not so terrible because they are quite sparse.
Mar
12
comment How does a particle's spin z component changed under lorentz group
No, in general it will be another rotation not ($R$)
Mar
10
comment How does a particle's spin z component changed under lorentz group
@ Charlie The general formula of the representation matrices $D_j$ (in terms of the Euler's angles) is given for example in: en.wikipedia.org/wiki/Wigner_D-matrix. So, you would need to identify the Euler angles of the rotation matrix $R$. Luckily, you need to work in the special case of spin 1, in this case the representation matrix is the rotation matrix itself: $D_j(R)=R$.
Mar
8
comment Berry phase in 1D materials
@FraSchelle - Sorry for the late response. For a connection betwee the Zak phase and the existence of edge states, please see the following article arxiv.org/abs/1109.4608 by:Deiplace, Ulmo, Montambaux. For a connection bween the Zak phase and the $Z_2$ invariant, please see the following article arxiv.org/abs/1402.2434v1 by Grusdt, Abanin, Demler
Mar
2
comment Why does a Heisenberg magnet break the O(3) symmetry in stead of SU(2)?
@hongchaniyi Indeed, the point symmetry group of a nematic state is $D_{\infty h}$ , a subgroup of $O(3)$.
Feb
23
comment Why should the modes of the linearized metric perturbation be “wavefunctions” of gravitons (in the Randall-Sundrum model)?
@Danu For example a non-flat manifold can have more than one spin structure allowing different kinds of spinors to live on it. In this case there will exist a spinor bundle which is not a Cartesian product.
Feb
22
comment Why should the modes of the linearized metric perturbation be “wavefunctions” of gravitons (in the Randall-Sundrum model)?
... This phenomenon becomes even more interesting when the original product is only locally Cartesian in the form of a fibre bundle. In this case the Hilbert space tensor product structure is lost.
Feb
22
comment Why should the modes of the linearized metric perturbation be “wavefunctions” of gravitons (in the Randall-Sundrum model)?
Yes, this is basically, what I meant. Quantization of a Cartesian product of two spaces results a tensor product of the corresponding Hilbert spaces. What I said is that we sometimes use the intermediate stage where one of the spaces has been quantized and the other one is still classical, i.e., we use an outer product of a classical space with a Hilbert space. This is the case when we talk about a classical Fermi-Gas , for examle, where the space coordinates are classical, but the internal coordinates are quantized (in the form of the spin finite dimensional Hilbert space).
Dec
15
comment How do higher-order optical chiralities look like?
Please, see the following recent review by T.G. Philibin where the conserved quantities generalizing the zilch are given in equations (21-22). These quantities correspond to the Noether symmetries given in equation (19). arxiv.org/abs/1303.0687
Nov
26
comment Energy in dynamical variational principle
This method is called instanton calculus. It is really not trivial to implement and requires special expertise, but it led to tremendously important discoveries in physics. The article given at the end of this note is my favorite reference doing instanton calculus on geometrical manifolds of the type obtained in the variational approach. However, it treats a much simpler problem (a single spin) than the Bose-Hubbard model. I'll try to write for you a more comprehensive answer (with more references) in the next few weeks if I can. arxiv.org/abs/cond-mat/0111139v1
Nov
26
comment Energy in dynamical variational principle
The main reason for the existence of this correction is that, while classically, all the states on the classical solution path in the trial wave function manifold are degenerate in energy, quantum mechanically there should be a unique ground state, which is obtained from quantum splitting of this degeneracy. The computation of the correction to the ground energy, indeed involves imaginary time replacement. …
Nov
26
comment Energy in dynamical variational principle
If the trial wave function manifold is close enough to the ground state, then the conserved energy value on the classical solution is a good approximation of the ground state energy. (Quantum) corrections to this first approximation can be obtained from quantizing the effective theory defined by the Lagrangian $\mathcal{L}$. Since the manifold of trial wave functions is in general non-Euclidean, one needs more general quantization techniques than canonical quantization to perform this task (e.g. geometric quantization). …
Nov
24
comment Energy in dynamical variational principle
Certainly, $E(f_n, f_n^*)$ is the Hamiltonian of the mechanical system defined by the (time independent) Lagrangian $\mathcal{L}(f_n, f_n^*, \dot{f}_n, \dot{f}_n^*)$. Thus it conserved. Please see my answer physics.stackexchange.com/questions/197297/…, where you can think of the parameter vector $R$ as the set of coefficients $f_n, f_n*$. Technically, the conservation stems from the antisymmetry of the Berry curvature.