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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
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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
28
revised Proof that eigenvalues of Fermionic creation/annihilation operators are Grassman numbers
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Nov
27
answered Proof that eigenvalues of Fermionic creation/annihilation operators are Grassman numbers
Nov
21
revised Fermion boundary conditions at finite temperature
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Nov
20
answered Do we have a quantum field theory of monopoles?
Nov
20
revised Can we “trivialize” the equivalence between canonical quantization of fields and second quantization of particles?
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Nov
20
answered Why does adjoint representation matter in some field theories?
Nov
20
answered Can we “trivialize” the equivalence between canonical quantization of fields and second quantization of particles?