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Hi, I am a string theorist and a publicist.


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
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
comment Polchinski Equation (7.2.4)
Absolutely! Do you understand my - perhaps too concise - answer? It's been years since I was calculating it, so I slightly blindly parroted myself. ;-)
May
12
comment $su(1,1) \cong su(2)$?
It's actually possible, @kaiser. Could someone please settle this question? For example by an explicit form of the matrices in terms of Pauli matrices etc. It isn't quite needed to explain why the groups are different but it would be nice to fix errors in "redundant" formulae, too.
May
11
comment $su(1,1) \cong su(2)$?
@childofsaturn - no, SU and SO are never the same. $SU(m,n)$ is pseudounitary, i.e. complex matrices with $\det M=1$ obeying $MGM^\dagger=M$ where $G$ is diagonal with $m$ times $+1$ and $n$ times $-1$. $SU(1,1)$ ends up isomorphic to $SL(2,R)$ or also $SO(2,1)$.
May
11
comment Quantum Entanglement - What's the big deal?
The only Bell whom you may find in the history of quantum computing - the page above - is the Bell in "Bell Labs", and please be aware that his name was Alexander Graham Bell, not John Bell or what was the name of your "hero".
May
11
comment Quantum Entanglement - What's the big deal?
Terry, good to hear you. Bell's contribution to quantum computing was zero - i.e. vastly smaller than the string theorists' contributions. Quantum computing's links with string theory is actually a hot subject. Quantum computers are based on the regular quantum mechanics known from 1925, especially Pauli's insights about spin, and it started as an applied physics or engineering discipline in 1970 when quantum codes began to be constructed. Feel free to search through the timeline of quantum computing en.wikipedia.org/wiki/Timeline_of_quantum_computing
May
10
comment How many fundamental fields / constraints are in Maxwell's Equations?
What do you exactly want to know about them? "What about" isn't a well-defined question. I said that the auxiliary components are needed for the Lorentz-invariant dynamic of the spin-one field. They are needed for the most general situations. If you pick some less clever, special, e.g. static, situations only, you may become unable to prove what I proved (or sketched a proof) above. But that doesn't mean that my proof has a problem.
May
5
comment Basic question about superspace, Grassmann numbers and world sheet supersymmetry
In that case, $\theta$ and $\bar\theta$ must be treated together and be equivalent to $\theta_A$ with different components $A$. It's still true that the derivative of $\theta$ bilinear is $\theta$ linear.
May
4
comment Do light and sound waves have mass
I assure you that phonons are exactly as real as photons, they are conceptually the very same thing, obey the same quantization rules, form a Hilbert space isomorphic to a Fock space, and their behavior is described by quantum field theories. Sound is a vibration of the underlying material, and in the same way, light is a vibration of the electromagnetic field. They only differ by the substance that vibrates and the elementary degrees of freedom. The case of light only differs by the substance's being simpler or more fundamental/elementary -fields that exist even in the vacuum.
May
2
comment Basic question about superspace, Grassmann numbers and world sheet supersymmetry
Dear @leastaction, thanks for reading. Concerning $\bar\theta\theta$, I believe that there is a mistake in your formula. It seems like a chiral superfield that only depends on $\bar \theta$, look at the $\psi$ term, so that should be written as an argument on the left hand side and the last $B$ term should actually be $\bar\theta\bar\theta$, and that's meant to represent $\epsilon_{AB}\bar \theta^A\bar\theta^B$, perhaps with a factor of $1/2$ or $i/2$ or whatever is their convention. But these chiral fields should only contain $\theta$ or only $\bar\theta$.
May
2
comment entanglement status of late hawking radiation in AMPS thought experiment
Dear @ChrisHanney, the whole radiation you can see at +infinity results by unitary transformation from the initial state, so it's the same information "transformed". I wouldn't call it "entanglement". One isn't entangled with his own future. The early and late Hawking radiation also come after each other but for them, their entanglement makes sense because different directions generically make an early Hawking particle and a late Hawking particle spacelike-separated. So it makes sense to talk about them as existing at the same moment, and they are entangled.
May
1
comment What does it mean for a particle to have spin of 2?
Dear @Al.Ka - to really understand it, you have to comprehend the basics of quantum mechanics. Observables (quantities) like the angular momentum are given by operators... In particular, the angular momentum generates rotations around an axis,and rotation by 4.pi has to be trivial - 2.pi is allowed to change the sign. That's why the spin components are always multiples of hbar/2 or 1/2 in the usual conventions. The spin - without a component - is the maximum value of a component in the multiplet. For massless particles, we must measure th angular momentum componen along the direction of motion