Reputation
Next privilege 20,000 Rep.
Access 'trusted user' tools
Badges
2 25 47
Newest
 Nice Answer
Impact
~213k people reached

  • 0 posts edited
  • 0 helpful flags
  • 3,622 votes cast
1d
revised Particle on $S^1$ and $U(1)$-principal bundle
added 214 characters in body
1d
revised Particle on $S^1$ and $U(1)$-principal bundle
added 147 characters in body
1d
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.
1d
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.
1d
revised Particle on $S^1$ and $U(1)$-principal bundle
added 6 characters in body
1d
answered Particle on $S^1$ and $U(1)$-principal bundle
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
12
answered Practical Calculation of Geometric Phase
Apr
6
revised Instantaneous energy eigenstates for forced harmonic oscillator
deleted 5 characters in body
Apr
6
answered Instantaneous energy eigenstates for forced harmonic oscillator
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.
Apr
3
answered Spin state after boost
Mar
26
awarded  Nice Answer
Mar
23
awarded  Good Answer
Mar
13
answered Simplest Live Demonstration of Adiabatic Transport
Mar
12
comment How does a particle's spin z component changed under lorentz group
No, in general it will be another rotation not ($R$)