13,719 reputation
21740
bio website
location
age
visits member for 4 years, 6 months
seen 5 hours ago

Dec
19
comment Braiding statistics of anyons from a Non-Abelian Chern-Simon theory
@Ideat cont. Witten's approach is more "thermodynamical" and he prefers to see the traces of the non-Abelian statistics in the partition function. To be more precise,(and this fact was also written in not so much words in Witten's paper): A non-Abelian Wilson loop can be thought of as a particle moving on a group or a flag manifold stuck to the boundary in the limit where its mass vanishes. Then its dynamics restricts upon quantization to the lowest Landau level producing the correct Wilson loop insertion.
Dec
19
comment Braiding statistics of anyons from a Non-Abelian Chern-Simon theory
@Idear These two approaches are very similar (but not exactly equivalent). Actually, Witten in his Jones polynomial paper (on page 365) refers to this similarity and asserts that the Wilson loop can be "thought of" as the trajectory of a particle in 2+1 dimensions. Witten refers to a famous paper by Polyakov adopting the strategy that Lee and Oh used later. This is only one of the numerous issues that Witten only talked about in his Jones polynomial paper (even without giving a single formula), which proved to be very fruitful for subsequent research.
Dec
17
comment Collection of histories vs. collection of momentary configurations
@Nick Kidman 1) Yes, it is a quotient space often denoted as $ \mathcal{M}//G$ (with two lines) called the Marsden-Weinstein quotient or the symplectic quotient. $ \mathcal{M}$ is the unreduced phase space and $G$ is the gauge group.In a finite number of dimensions (where we can count), the dimension of the reduced space is $ \mathrm{dim] \mathcal{M} - 2\mathrm{dim] G$. The factor 2 comes from the fact that we need to remove one dimension per symmetry and another dimension for gauge fixing. 2) We can reduce only by the anomaly free subgroup of the gauge group.
Dec
15
comment Separability axiom really necessary?
@moppio89 cont. In quantum mechanics we are interested in computing probabilities. The Hilbert space is an auxiliary tool. The example that I had in mind is the case of bosonization where observable of the same physical system have representations on the Bose and Fermi Hilbert spaces.
Dec
15
comment Separability axiom really necessary?
@moppio89 Yes, I think that their main assumption that what is physically important can be deduced from a scattering process. In addition, the usual notion of vacuum restricts to a subspace of a nonseparable space. Also, may be the mathematical complications constituted a factor due to the need for more advanced measure theoretic analysis on the underlying space than in the case of a countable set.
Dec
10
comment Vector potential $A$ on a 2-sphere $S^2$ of radius $R$ with some points removed
@Hamurabi There is an error in the above analysis, the computation on the sphere is not correct. I'll post a correct answer very soon. I'll also answer your question in the comment, Sorry.
Dec
9
comment Magnetic Dipole: How to plug into Maxwell's equations?
You reached the (magnetic) continuity equation (the $4 \pi$ factor should not be there). The two solutions considered above automatically satisfy the (electric and magnetic) continuity equations. In the electric loop case, the sources satisfy $\mathbf{\nabla}.\mathbf{J}_e = 0$ and $\rho_e = 0$. In the magnetic charge case $\frac{\partial \rho_m}{\partial t} = 0$. (The charges are time independent) and $\mathbf{J}_m = 0$. I don't see an easy way to generate a magnetic dipole from a magnetic current.
Dec
9
comment Magnetic Dipole: How to plug into Maxwell's equations?
A magnetic current loop will give rise to an electric dipole and not a magnetic dipole solution, the same way that an electric current generates a magnetic dipole in the example above. These are the only solutions known to me to generate a magnetic dipole: 1) An electric current loop, 2)Two opposite magnetic charges.
Dec
9
comment Magnetic Dipole: How to plug into Maxwell's equations?
If magnetic charges would exist in Maxwell equations: $\mathbf{\nabla}.\mathbf{B} = 4 \pi \rho_m$ where $\rho_m$ is the magnetic charge density, then the magnetic dipole solution can be obtained exactly in the same way an electric dipole is obtained in the standard Maxwell theory; i.e., by taking two opposite infinitesimally spaced magnetic charges.
Dec
9
comment Magnetic Dipole: How to plug into Maxwell's equations?
@Hui Zhang Yes it is the gradient of a delta function. $m$ is the magnitude of the magnetic dipole moment vector. You can choose the radius of the loop $a$ tending to zero, the current $I$ tending to infinity, but the combination $I a^2$ finite, for example by taking $I =\frac{I_0}{\epsilon^2}$ and $ a = a_0 \epsilon$ and letting $\epsilon \rightarrow 0$. A similar trick is performed in the case of an electric dipole.
Nov
28
comment Proof that eigenvalues of Fermionic creation/annihilation operators are Grassman numbers
Cont. According to the first interpretation a coherent state can be considered as a function on $M \times M$ where $M$ is the supermanifold (whose coordinates are $\psi$ and $\zeta$ respectively) which depends only on $\psi$ and $\zeta ^{\dagger}$. One can then extend the inner product to this type of functions. Then one can identify the quantum states "super rays" differing by a multiple a Grassmann number in analogy with the bosonic case.
Nov
28
comment Proof that eigenvalues of Fermionic creation/annihilation operators are Grassman numbers
@user10001 Both interpretations can be made rigorous. In both cases one must be aware that a quantum state is not identified with a vector in a Hilbert space but rather with a ray (multiplication by a complex constant does not change the state). The second interpretation is the basis of the theory of "super-Hilbert" spaces which was studied by B. DeWitt and Rogers.
Nov
28
comment Proof that eigenvalues of Fermionic creation/annihilation operators are Grassman numbers
@user10001 I am answering your questions in a separate edit
Nov
28
comment Proof that eigenvalues of Fermionic creation/annihilation operators are Grassman numbers
@Echows I recommend to start from the following review arxiv.org/abs/math-ph/0202026v1 by Cartier, DeWitt-Morette, Ihl and Sämann
Nov
12
comment Boson calculus and Maximum Weight State
Yes, we use commutators to shift $b_2$. In fact all the commutators are trivial and the operators commute because the raising operators do not contain $b_1$ which is the only operator not commuting with $b_1^{\dagger}$
Oct
31
comment Is the third spin vector of a photon always suppressed?
@WetSavannaAnimal aka Rod Vance Electromagnetic waves in waveguides propagate in TE and TM modes where either the magnetic the electric field have components along the direction of propagation.
Oct
25
comment What are orbifolds and why are they useful and interesting for physics?
@Nick Kidman $\mathbb{C}^2/~$: $\theta$ ~ $\theta + \alpha$
Oct
18
comment Spin tensor and Lorentz group operator in bispinor case
If the problem is a sign problem, please notice that the $\Psi$s are Grassmann variables and they acquire minus signs when they are commuted.
Sep
13
comment Non-associative operators in Physics
This phenomenon was discovered by Roman Jackiw: inspirehep.net/record/204787?ln=en. It happens for a system of a particle moving in field of a magnetic monopole.
Aug
21
comment What is the algebraic property that corresponds to a topological term?
2) When you say that you want to go from quantum to classical, I assume that you mean that the quantum theory is defined by means of a path integral, but in general path integrals are ambiguous, and also it is hard to see from them the quantum linear structure that you want to dequantize, this is why I think that methods of geometric quantization are superior.