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bio website motls.blogspot.com
location Czech Republic
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Hi, I am a string theorist and a publicist.


Jun
24
comment meaning of Higgs coupling is flavor conserving
No. Well, the Majorana neutrino masses are typically flavor-violating - they cause flavor-changing oscillations - but those mass terms are usually not considered "parts of the Standard Model" because they must arise from some beyond-the-Standard-Model dynamics as effective terms (they are not gauge-invariant etc. classically). One may talk about the CKM matrix, too - it also changes flavor. But mass terms aren't really "interaction terms", anyway, because they are quadratic.
Jun
21
comment Atomic Physics: stimulated emission
Right, it's a technicality to achieve a given state - and the method isn't necessarily unique. The laser starts to operate when this state is achieved and the laser actually emits.
Jun
21
comment What experiment would disprove string theory?
That's what Polchinski et al. are saying but they are wrong, as shown in numerous papers, like those by Raju and Papadodimas. But this is a way too extensive question that is off-topic in this general thread.
Jun
16
comment AdS/CFT-duality: How does the $U(1)$ decouple form the $U(N)$?
Up to global identifications, U(N) is the same thing as SU(N) x U(1), so it is not a simple group. Even for purely field theoretical reasons, the gauge multiplets in the SU(N) and U(1) factors are living completely independently. Moreover, all the matter in SU(N) is really in adjoint and all the adjoint is neutral under the U(1). So nothing couples to the U(1) gauge field, it is decoupled! It is a bit subtle question why the AdS bulk space is really equivalent just to the SU(N) and not the whole U(N). But if it were the whole U(N), the bulk physics would have to be made of 2 indep. parts too
Jun
14
comment Separability in quantum physics
If I understand you, then yes.It means that local realism makes the wrong prediction (inequality) - and is therefore falsified - whether or not you assume "separability". So separable local realist hidden variable theories are excluded; and inseparable local realist hidden variable theories are excluded, too. Quantum mechanics is really the unique theory that allows maximally determined states to be entangled i.e. non-separable while the fundamental laws are local.
Jun
14
comment Separability in quantum physics
I totally agree it is not sensational - and Bell's theorem is not sensational and irrelevant for physics (because it proves things given assumptions that are known to be wrong in physics). It is a heavily overrated theorem because what it excludes are things that were known to be excluded from the mid 1920s when QM was born. Still, it is a theorem that proves something and it is possible to correctly talk about the content of it, isn't it?
Jun
14
comment Separability in quantum physics
OP: Check the relationship between the locality and separability in the theorem e.g. here arxiv.org/abs/1302.7188 - the paper also sensibly explains why separability isn't really a required assumption of the theorem, so your idea cannot be a loophole.
Jun
14
comment Separability in quantum physics
Sorry, @CuriousOne, but you totally misrepresent the point of the theorem and related comments. The conclusion isn't that local hidden variables are not needed. The conclusion is that they are not possible.
Jun
10
comment What is the meaning of the dlog integrations in the on-shell/grassmannian representation of N=4 SYM scattering amplitudes?
Dear Giulio, an example of a contour is a real axis. Each complex variable may be integrated independently along real axis, for example. The variables effectively are decoupled to independent ones because they describe internal momenta. But there is never ever any 2-dimensional integral here, it would be neither holomorphic in the required variables nor finite or well-defined due to the ill-defined 2D integrals of the poles.
Jun
7
comment Why $c$ is $3$ x $10^8$ times faster than a $1$ $m/sec$ car?
Yes, it makes sense, and by equations governing the motion of the electron, we can show that the question is equivalent to the question why the fine-structure constant $e^2/4\pi\epsilon_0 \hbar c$ or what is the formula is equal to $1/137$. Today, we can't calculate this number from the first principles but we know how to compute it from more fundamental numbers, couplings of the electroweak theory at the high scale, and renormalization group running. A full theory - which really means string theory plus the information about the right vacuum - in principle allows one to calculate it exactly.
Jun
7
comment Why $c$ is $3$ x $10^8$ times faster than a $1$ $m/sec$ car?
Thanks and the answer to your question about editing is pretty much No, I don't understand it. Whether two questions have the "same" $c$ is a physically meaningless question because $c$ is dimensionful. Whether they have the same $c/v_{\rm car}$ depends on the identification of "the car" in both Universes. As I already described, there is no way to decide what is the right way and what is the wrong way. Adult physicists work with units with $c=1$ so $c$ is the same in all (relativistic) Universes because mathematical constants like 1 are constant across universes, independently of physics.
Jun
6
comment What is the essential difference between a resonance and a particle?
Dear @AcidJazz - to talk about a resonance, you have to decide about a particular decay/fusion channel. But the particle that is seen as a resonance in this decay channel may also decay in other channels, or enter other processes. You can't distinguish a/the unstable particle and a resonance as an object because they may be the same object; but you should still distinguish the words "unstable particle" and "resonance" because they express different ideas. It's like asking how to distinguish Barack Obama from the U.S. president. Well, it's the same person, now, but it doesn't have to always be.
Jun
4
comment Why is a 1mW laser dangerous?
Dear @no_choice99, a reflected beam - even laser beam - is basically indistinguishable from the original one, except that it's weakened by 1-30% and it's apparently coming from the mirror image of the source. A mirror just doesn't matter. The anisotropy of the LED source only increases the power by a factor of 2-5, the laser still wins by a rather large factor.
Jun
1
comment Tensor product of operators in QM
Yes, absolutely, @SebastianHenckel
May
22
comment Fermionic path integral on the disk - Recovering the vacuum state
Dear @childofsaturn, it's the integration over the variables that appear as factors that give you 1, like $\int d\theta_1 \,d\theta_2 \,\theta_1\theta_2=-1$, OK? If some thetas are missing or excessive, the integral is zero. ... Alternatively, you may try to decompose the integral to infinitely many fermionic 1D or 2D integrals whose result is one of the infinitely many factors. Note that $S$ is a sum over $n$, so $\exp(-S)$ is a product over $n$, and the boundary conditions may be written in terms of the Fourier modes, too. Try to compute one of these factors, e.g. for $\chi_5,\chi_{-5}$
May
22
comment Fermionic path integral on the disk - Recovering the vacuum state
There's no known "in principle" calculation that would yield the result in the form of the infinite product. But you may verify that it's the right result by integrating over all the $\chi_{-n}$ and $\bar \chi_{-n}$ to get one - the linear factors cancel against the integration measure.
May
22
comment Fermionic path integral on the disk - Recovering the vacuum state
Right. Your edits didn't fix either of the "small" errors you're not interested in but the form of the ground state wave function is OK. You see that the wave functional is simply independent of some (1/2) Fourier modes, and linear in others (1/2 of them). It is not just $exp(\chi_{-n}\chi_n)$ that would be similar to the bosonic case. In principle, this result follows from the boundary-condition calculations, too. But you simply derive it by checking the extra integrals over the boundary Grassmann variables. Some of them are zero, some of them are not.
May
22
comment Fermionic path integral on the disk - Recovering the vacuum state
Dear @childofsaturn, if you had the idea to do "the same" for the fermions by yourself, you shouldn't be surprised by these "more trivial" results. This will be discussed in sections like 4.4, 5.3, 6.3 etc. especially for the $bc$ ghost system, which is really the same as your $\psi$'s on a flat world sheet. The simplest partition sum is zero, and you need certain insertions to get a nonzero result. Those will be generated automatically by the string formulae, and they may also be seen to depend on the Grassmann boundary conditions if you choose to have them.
May
21
comment Fermionic path integral on the disk - Recovering the vacuum state
There are many small issues here. When it's a disk, the area is $t\cdot dt\cdot ds$ and not without the $t$, right? Also, if you work with a Euclidean-signature disk, the kinetic terms for the fermions should use the complex $\partial_t \pm i \partial_s$ and not the "light-cone-like" sum and difference, right? But when you correct all these things, it should be straightforward to compute all such Gaussian integrals, by completing squares etc. You will need to find the classical solution with the boundary conditions.
May
18
comment Traces in different representation
Hi, I did use the property to get (5.122) but I don't think that I could have used "nothing else". Maybe you can. Now I am reading the text, they clearly assume that you use the Taylor expansion for exp(iF). But it is hard to reconstruct exactly what solution they had in mind if they don't write it down. It's enough to have a solution.