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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
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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 ($R$)
Mar
11
awarded  group-theory
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
10
revised How does a particle's spin z component changed under lorentz group
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Mar
9
revised How does a particle's spin z component changed under lorentz group
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Mar
9
answered How does a particle's spin z component changed under lorentz group
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
8
answered $SO(4,2)$ symmetry of the hydrogen atom