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bio website motls.blogspot.com
location Czech Republic
age 41
visits member for 4 years, 6 months
seen 19 mins ago

Hi, I am a string theorist and a publicist.


Jun
6
answered What is the essential difference between a resonance and a particle?
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
4
awarded  Enlightened
Jun
4
awarded  Nice Answer
Jun
1
revised Tensor product of operators in QM
edited body
Jun
1
comment Tensor product of operators in QM
Yes, absolutely, @SebastianHenckel
May
31
awarded  Nice Answer
May
29
awarded  Enlightened
May
29
awarded  Nice Answer
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.
May
17
comment Hidden observers in Double Slit experiments - Do they matter?
Right. One should also explain why "too weak" won't destroy the interference pattern. A beginner may be tempted to think that any, arbitrarily weak influence will "measure", and that may really be the OP's case. Well, the interference pattern is only broken if the interaction evolves the two possibilities into (nearly) mutually orthogonal, and therefore mutually exclusive, states. A small which-slit-dependent modification of the state will only modify the states so that their inner products are still essentially the same, and they (mostly) interfere.
May
17
comment Traces in different representation
Dear @0celo7, your question is exactly what I was answering in my answer. If you want to calculate the trace over the symmetric, antisymmetric, or adjoint representations via "factorization only", you will fail because these representations do not factorize to tensor products. They are parts of the tensor products, as you wrote, so you must use at least some less trivial methods to disentangle the "parts" of the tensor products.
May
15
revised Polchinski Equation (7.2.4)
added 917 characters in body
May
15
comment Polchinski Equation (7.2.4)
Yes, of course, there's a way to determine this only good $\omega$.I am adding it to the answer.
May
14
awarded  Nice Answer