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bio website motls.blogspot.com
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seen 14 hours ago

Hi, I am a string theorist and a publicist.


Apr
15
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Apr
10
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Apr
10
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Apr
9
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Apr
4
comment Klein factors and Conformal Field Theory
Dear Costa, the right prefactor (normalization) is clearly needed for the right normalization of the commutator, but that's it. The essence of the operator is the same, however. It's like if you used $X=x(a/L)^n$ instead of $x$, then the commutator $[X,p]$ will be $(a/L)^n \cdot i\hbar$ instead of $i\hbar$. But the special point about the exponential operators is that only the (divergent) overall normalization is affected by the normal-ordering effects. At least I hope so.
Apr
4
answered Klein factors and Conformal Field Theory
Apr
1
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Apr
1
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Mar
29
comment If the earth left the solar system for interstellar space. How long would it take for atmosphere to freeze
You're right, thanks.
Mar
28
comment Selection Rules in electron spectroscopy
Dear Laurent, this is really the general situation. The orbital angular momentum does not commute with the full Hamiltonian, and neither does the spin, so they are not "good quantum numbers". It's still possible to diagonalize the Hamiltonian. In a calculation method, we may diagonalize the Hamiltonian without the L.S interaction, and that part commutes with L and S separately. If we can't find a useful "bulk" of the Hamiltonian that commutes with L,S separately, then L,S eigenstates are useless in any search for the H eigenstates, but H still has eigenstates and they may be found otherwise.
Mar
27
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Mar
26
answered What is probability current in quantum mechanics?
Mar
24
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Mar
24
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Mar
22
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Mar
22
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Mar
20
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Mar
20
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Mar
20
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Mar
19
comment What's the $\ell$ in the Bicep2 paper mean?
The local peak from the primordial gravitational waves is normally expected around $\ell=90$. The spherical harmonic $Y_{90,90}$, for example, has the $J_z$ aligned "maximally vertically, so it is spinning maximally vertically among the $Y_{90,m}$ harmonics, and on this one, you see that the angular dependence contains $\exp(90 i\phi)$ which contains 90 maxima around the circle. So the "wavelength" of the component spans 360/90=4 degrees on the sky. The resolution has to be a bit better to actually "see" the shape of these waves.