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Hi, I am a string theorist and a publicist.


Dec
11
comment Convergence of quantum effective action to finite loop order
Dear @Squark, as I said in the previous sentence, it's not possible. If you compute any particular amplitude (Green's function, scattering amplitude, or a term in the effective action), a large number of vertices may only be achieved by adding a high number of loops. You can't add vertices without adding loops, unless you are adding applies and oranges which you shouldn't. As some other people mentioned, you could also compute inclusive cross sections with may include a very high number of soft photons (without loops), but you said it's not what you meant, either.
Dec
10
comment Higgs Field - Is its discovery truly “around the corner”?
Dear @Matt, nice mini-answer, +1. One could also say that if it is "perfectly compatible with the SM", the mass is lighter than one would expect for a "typical" non-SUSY Standard Model (by 15 orders of magnitude, in fact, to add a funny twist to it). Incidentally, there will be evidence and "candidates" but the press release won't contain the word "hint". Do you want to make a bet? ;-)
Dec
10
comment Virasoro constraints in quantization of the Polyakov action
A $Q$-cohomology is the class of all vectors $|\psi\rangle$ that satisfy $Q|\psi\rangle=0$ and that are identified by the equivalence $|\psi\rangle \sim |\psi\rangle + Q|\phi\rangle$ for arbitrary vectors $|\phi\rangle$. In the cohomology, you may find some vectors with no "positive" excitations by the $b,c$-ghosts and a certain proper ghost number. If you take these states and ignore all the $b,c$ ghosts in them, you get states of the old covariant quantization. Also, I wanted to stress that the fact that the Hilbert space is interpreted in a stringy way plays no role: it's still a QFT.
Dec
10
comment Virasoro constraints in quantization of the Polyakov action
Dear @Squark, when we're happy to know the BRST formalism, it's a good idea to start with it (there is no central term in it, so no obstruction to impose the annihilation by all generators!), and view the old covariant quantization as a particular convention to choose the representatives from the BRST cohomologies. Because the ghosts-$bc$-free Virasoro generators commute to the central extension, one would need to talk about a complicated representation theory of these "extension" groups and it's not necessarily an insightful approach.
Dec
10
comment Convergence of quantum effective action to finite loop order
OK, then sorry, @Squark, but in that case, I have no idea what you could be possibly asking. You are talking about a convergence of a "sum". The only sum that is at risk of being divergent (and it, indeed, is divergent) is the loop order expansion. There is no "infinite sum" at a finite order. At a finite order, there is always a finite number of diagrams/terms and the sum of a finite number of finite terms is always convergent. It seems that you think that some methods to "complicate" the diagrams (ext. legs?) don't add any $e$; but all of them do.
Dec
9
answered Convergence of quantum effective action to finite loop order
Dec
9
answered Virasoro constraints in quantization of the Polyakov action
Dec
8
awarded  Enlightened
Dec
8
awarded  Nice Answer
Dec
6
awarded  Nice Answer
Dec
5
awarded  Popular Question
Dec
5
reviewed Approve suggested edit on Free Body Diagram of Fluid Statics Problem
Dec
5
reviewed Approve suggested edit on How does a change in temperature affect relative humidity
Dec
5
reviewed Approve suggested edit on Can I use an antenna as a light source?
Dec
4
comment Heisenberg picture saves locality in QM?
Heisenberg picture is physically equivalent to Schrödinger's picture and in relativistic theories, namely quantum field theories and string theory, locality exactly holds in both and can be easily proved (and is a direct consequence of the Lorentz invariance). It's more manifest in the Heisenberg picture, indeed, because the equations are the same as the classical field equations which preserve locality. All comments about "nonlocality" are due to some people's fundamental misunderstanding of quantum mechanics. Locality was proved long before Deutsch learned physics. QFTs are called local QFTs
Nov
27
comment Hidden observers in Double Slit experiments - Do they matter?
Dear Colin K, I am writing the answer not only for the OP but for everyone who asks the same question which may be asked, believe me, even if one has been exposed to double-slit experiment. Fraggle, nothing "causes" probabilities to shrink. Probabilities by definition always describe "shrunk" outcomes. For example, if you throw a dice, the probability distribution is spread over the numbers 1,2,3,4,5,6. It is completely spread. But it doesn't prevent "3" from being the result. In fact, it is guaranteed that one sharp number will be the result if you throw dice.
Nov
26
revised Hidden observers in Double Slit experiments - Do they matter?
added 1110 characters in body
Nov
26
answered Hidden observers in Double Slit experiments - Do they matter?
Nov
26
comment Why $\displaystyle i\hbar\frac{\partial}{\partial t}$ can not be considered as the Hamiltonian operator?
Also, it's true that it would be "useless" to define the operator to be the same because then you would have no nontrivial dynamical equations that could predict the future. That's what you are probably saying. However, even if it would be useless, one may still ask whether it would be legitimate. I think the answer is No, see e.g. Qmechanic's answer. $[H,t]=0$ is different from the nonzero $\partial_t,t=1$ so the operators can't be the same. ($\partial_t$ isn't really an operator acting on the states like $\psi(x,y,z)$ but "extended" states $\psi(x,y,z,t)$ etc.: another "distortion".)
Nov
26
comment Why $\displaystyle i\hbar\frac{\partial}{\partial t}$ can not be considered as the Hamiltonian operator?
Right, your point understood, Hans. When one says "Hamiltonian is by definition XY", there are different angles what the "definition" may mean. Of course, one may constructively define the Hamiltonian for particular systems, like $p^2/2m+V(x)$, in which case it's this expression by definition. More generally, we want to define it as whatever is needed for Schrödinger's or Heisenberg's equations to hold. The latter approach is more general. Still, when we say "the equation holds", it is not the same thing as saying "the operators are the same" because $\psi$ can't be canceled.