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Oct
9
comment Uniqueness of supersymmetric heterotic string theory
Hi @Yuji: Let me just copy Reimundo's reference. The paper is fully available here: pjm.berkeley.edu/pjm/2004/213-2/pjm-v213-n2-p05-s.pdf
Oct
9
comment The 2011 Nobel Prize in Chemistry
I think that the adjective "essentially discrete" doesn't mean just "discontinuous". It means a more constraining thing, maybe even a sum of a countable number of delta functions.
Oct
9
comment Cosmic bubble collision
The energy released is $V(phase_1)-V(phase_2)$ times $volume$ while the "rest mass" of the domain wall separating the two phases only grows as the $area$ which is much smaller for large radii, $r^2<r^3$, for large volumes. So this means that the $\gamma$ factor measuring the increase of the mass by near-luminal velocities is huge and the wall is quickly approaching the speed of light. See also motls.blogspot.com/2011/10/bubble-of-nothing-and-other.html
Oct
9
comment Cosmic bubble collision
Concerning the second point, the near-luminal expansion is obvious from the maths of the bubbles. A bubble is described by an instanton localized in the Euclidean spacetime, near $x^2+y^2+z^2+t_E^2\leq a^2$ or so. By Wick rotation, you may see that the transition is concentrated near the light cone $x^2+y^2+z^2-t^2\leq a^2$, so the point where the Universe switches is expanding nearly by the speed of light. Alternatively, the huge energy released from the difference of energy densities (in the bulk, difference between the two phases) is invested to accelerate the wall between the two phases.
Oct
9
comment Cosmic bubble collision
Dear @Siva, concerning the first point, one really has to be careful about the emotional meaning of words. If "something" grows exponentially, it doesn't mean that it will beat "anything" that grows linearly because we're comparing different quantities here. The shrinking of the space between the bubbles goes like $1-t$ where $t$ is time, that's the linear one because of the size growth, but it's corrected to $\exp(ct)-t$ because the distance inflates by cosmology. Still, $\exp(ct)-t$ crosses zero and goes negative for some values of $c,t$. The word "exponential" doesn't mean that it wins.
Oct
8
comment Sub and super multiplicativity of norms for understanding non-locality
Dear @Earl, I thought that my pathological counterexamples (their set of vectors with a norm smaller than 1) deeply fail to be convex, instead of being deeply convex! ;-)
Oct
7
answered Any use for $F_4$ in hep-th?
Oct
7
comment The 2011 Nobel Prize in Chemistry
With your definition, amorphous materials are not crystals (these "superstructures" make the diffraction diagram essentially continuous) but quasicrystals are crystals. Concerning the latter point, quasicrystals show that the assumption/proposition that "discrete diffraction diagrams imply periodic arrangement of atoms" was a widely believed myth.
Oct
6
comment Quasiparticles in Bohmian mechanics
The answer to your question, @Vladimir, is obviously No: every quantum system (whether it is a particle or quasiparticle) may always be found in a pure state. That's why any quantum system may exhibit 100% destructive interference etc., something that would indeed be impossible if an object were "forced" to be partly mixed, and this is one reason why any picture of the quantum phenomena that tries to be more classical than QM (including Bohmian mechanics) is inevitable inconsistent with the tests of interference and other empirically established features of the quantum world.
Oct
6
answered Calculating correlation functions of exponentials of fields
Oct
6
comment Quasiparticles in Bohmian mechanics
I think that your inherent suggestion that the Bohmian mechanics suffers from an inconsistency because of quasiparticles is right. QM shows that it doesn't matter whether one says that something is a particle or quasiparticle: the Hilbert spaces and dynamics and probabilistic predictions are isomorphic. In the Bohmian mechanics, it always matters because particles' degrees of freedom are "beable" while other e.g. quasiparticles' degrees of freedom are not. For the same reason, quantum computers can't simulate the reality in Bohmian mechanics. It just doesn't work.
Oct
6
revised The 2011 Nobel Prize in Chemistry
added 342 characters in body
Oct
6
answered The 2011 Nobel Prize in Chemistry
Oct
6
answered Cosmic bubble collision
Oct
6
comment Choice and identification of vacuums in AdS/CFT
On both sides, one may also consider the spectrum of other generators of the superconformal groups, and if there are ground states, they match as well. If you want to play with individual creation/annihilation operators in the bulk of the AdS, you won't find an easy dictionary in the CFT. The locality itself fails to be manifest on the CFT side even though it's manifest in the AdS bulk description. ... dS/CFT is a completely different issue and according to everything I can say, it's just wrong: we had long discussions with Andy Strominger about it and he would surely disagree.
Oct
6
comment Choice and identification of vacuums in AdS/CFT
Dear Disinterested, a random Bogoliubov transformation that renames combinations of creation and annihilation operators in the bulk as annihilation operators isn't a particularly natural operation in general and you shouldn't call the state annihilated by such new "annihilation" operators a vacuum. A vacuum is the lowest-energy eigenstate of an energy operator. In AdS/CFT, it's the global $H$ energy generating translations in time.
Oct
5
answered Uniqueness of supersymmetric heterotic string theory
Oct
5
comment Choice and identification of vacuums in AdS/CFT
The vacuum is an eigenstate of the energy operator corresponding to the lowest allowed eigenvalue. When there are two equivalent descriptions of a physical model, the vacua of both descriptions are obviously identified with each other because they satisfy the same condition I just described. If you're asking about something else than the trivialities I just said, then it could help if you were a little bit more comprehensible.
Oct
4
comment Sub and super multiplicativity of norms for understanding non-locality
Dear @Earl, I think that I have fixed the error in my argument and the conclusion is unchanged. Reduce the original norm to 1/1000 of it in a ray of vectors that are "nearly" tensor-factorizable. This doesn't spoil the super-multiplicativity because only strictly tensor products are constrained. However, the dual norm will be affected by this change, even the dual norm of factorizable vectors, and it will jump 1,000 times or so, spoiling sub-multiplicativity. Agreed? Some convexity or triangle inequality for the norm could be enough to ban variable norms of my type and revive your theorem.
Oct
3
comment Sub and super multiplicativity of norms for understanding non-locality
There may be a mistake in my argument, thanks for pointing it out. Will look at it again.