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bio website motls.blogspot.com
location Czech Republic
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visits member for 3 years, 6 months
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Hi, I am a string theorist and a publicist.


Jul
11
reviewed Approve suggested edit on What is so good about diffractive optics?
Jul
11
reviewed Approve suggested edit on Why is gravitation force always attractive?
Jul
10
reviewed Approve suggested edit on Calculating time for a fully charged UPS
Jul
8
reviewed Approve suggested edit on Why glass is transparent?
Jul
8
reviewed Approve suggested edit on Madelung transformation
Jul
8
comment Geometric/Visual Interpretation of Virasoro Algebra
Sure, the central extension of an algebra is just a very subtle modification of the original algebra that doesn't change its physical meaning. For every central extension, one may obtain the original algebra simply by setting all the $c$-number generators to zero. This preserves the Jacobi identity etc. because the $c$-number generators commuted with anything else, anyway - well, that what it means that it was "central". ;-) In string theory, the Virasoro algebra is still the algebra of reparameterizations of the world sheet, even for $c\neq 0$.
Jul
8
comment Geometric/Visual Interpretation of Virasoro Algebra
Oh, I see. There is obviously no out-of-Hilbert-space visual representation of the central extension that would differ from the $c=0$ algebra. The reason is that the central extension has $c$-numbers in the commutators. ;-) Any $c$-numbers may only be represented as the transformation of phase in the Hilbert space, and a transformation of phase of a vector in the Hilbert space doesn't change the character of this state "physically" or "geometrically" - it's just about the normalization. So central extensions are just central extensions and they share the original visualizations with the $c=0$.
Jul
8
answered Geometric/Visual Interpretation of Virasoro Algebra
Jul
8
answered How is the complexification of spacetime justified?
Jul
7
reviewed Approve suggested edit on Why does the weak force distinguish left and right handedness?
Jul
7
awarded  Enlightened
Jul
7
awarded  Nice Answer
Jul
1
reviewed Approve suggested edit on How do I calculate the power consumed by a lightbulb?
Jun
30
reviewed Approve suggested edit on On a bicycle, why does my back tyre wear so much more quickly than the front?
Jun
28
answered Flux compactification
Jun
28
reviewed Reject suggested edit on Extended sound of thunder
Jun
28
reviewed Approve suggested edit on How is thermodynamic entropy defined? What is its relationship to information entropy?
Jun
19
comment Which derivative with respect to time is which in the Heisenberg picture of quantum mechanics?
Excellent, good news.
Jun
19
comment Which derivative with respect to time is which in the Heisenberg picture of quantum mechanics?
The latter two derivatives are equal to $v_{Heis} = p_{Heis}/m$ and $-\nabla \cdot V_{pot.\,energy} (\vec x_{Heis})$ in the simplest examples of quantum mechanics. These Heisenberg equations for the $d/dt$ derivatives mimic the classical equations of motion. That's of course no coincidence. All the time derivatives come from the $d/dt$ of the operators in the Heisenberg picture while $|\psi \rangle$ states are fixed. So the expectation values have to evolve, in the classical limit, just like the classical quantities: the operator equations have to agree in the classical limit. And they do.
Jun
19
comment Which derivative with respect to time is which in the Heisenberg picture of quantum mechanics?
Dear Qiaochu, even if $A_{Heis}$ meant nothing else, we still wrote its "definition" in terms of $A_{Schr}$, by the conjugation, so it is definitely not an undefined object. It is totally well-defined - if you decide that the conjugation is its definition - and all questions about it may be answered by pure thought. Yes, it is true that $\partial x_{Heis} / \partial t = 0$ and similarly for $p_{Heis}$. However, $dx_{Heis}/dt$ is not zero, and neither is $dp_{Heis}/dt$.