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Dec
16
answered Collection of histories vs. collection of momentary configurations
Dec
15
revised Braiding statistics of anyons from a Non-Abelian Chern-Simon theory
edited body
Dec
15
answered Braiding statistics of anyons from a Non-Abelian Chern-Simon theory
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
12
answered Separability axiom really necessary?
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
answered Vector potential $A$ on a 2-sphere $S^2$ of radius $R$ with some points removed
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
revised Magnetic Dipole: How to plug into Maxwell's equations?
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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.
Dec
8
answered Magnetic Dipole: How to plug into Maxwell's equations?
Dec
6
awarded  Revival
Dec
5
answered Fermi Walker vs. Fermi transport
Dec
5
answered Applications of analytic continuation to physics
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
revised Proof that eigenvalues of Fermionic creation/annihilation operators are Grassman numbers
added 981 characters in body