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Hi, I am a string theorist and a publicist.


Apr
13
comment Autocorrelation Functions <---> Pair Correlation Functions
This seems to be a misunderstanding. Autocorrelation function is derived from the correlation between $y(t)$ and $y(y+\Delta t)$ and this autocorrelation is a function of the lag $\Delta t$. It is a pair correlation function between two things, namely the value of a function and the value of a function which is obtained from the first one by a shift in $t$. If you ask about "pair correlations", what are the two things in the pair? Obviously, if something in the pair is something completely different than the original function or its shifts, the pair correlation will have nothing to do with..
Apr
13
comment Why is Mosely screening in heavy atom K-shell 1 unit?
I am confident that your explanation is at least on the right track. At any rate, even the Wikipedia article does say that $(Z-1)$ arises because of differences in the electron-electron interactions between the initial and final states and holes could be very useful to quantify this difference. It must be possible to justify the number 1 in some way.
Apr
13
comment What is a non linear $\sigma$ model?
Oops, you're right, +1.
Apr
12
answered Is there a general physics simulator for learning purposes?
Apr
12
revised What is a non linear $\sigma$ model?
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Apr
12
answered What is a non linear $\sigma$ model?
Apr
12
comment Where does the wave function of the universe live? Please describe its home
Dear @Jim, I agree with David. In QFT, you may try to determine the spacetime dimensions by isolating one-particle states and finding that the Hilbert space of one-particle states has a simple basis diffeomorphic to $R^{d-1} $, the spatial momentum, or in an equivalent way. But the Hilbert space is much greater than the regular space. Every basis vector of the Hilbert space in a basis corresponds to one mutually exclusive state in which the whole physical system may be. There are usually infinitely many.
Apr
12
comment How to combine the error of two independent measurements of the same quantity?
Excellent answer, +1.
Apr
12
revised How to derive addition of velocities without the Lorentz transformation?
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Apr
12
answered How to derive addition of velocities without the Lorentz transformation?
Apr
12
comment What are the limitations of the superspace formalism?
There is also an interesting twistor-like transform for the 10D super Yang-Mills by a guy called Witten, sciencedirect.com/science/article/pii/0550321386900908
Apr
12
comment What are the limitations of the superspace formalism?
Dear Dilaton, good question. Ron and Simon have already answered to some extent and I will only offer a different extent, extending Ron's comment in particular. If you want to make N=1 SUSY in 4D i.e. 4 real supercharges manifest, you need 4 superspace fermionic coordinates. With 16 supercharges, you would probably need at least 16 fermionic coordinates in the superspace but then the fields would have $2^{16}=256$ components which is pretty high give that you only need 16 on-shell components only. Most of the component fields would have to be auxiliary, linked to deritives of others, etc. Hard
Apr
12
comment How to derive addition of velocities without the Lorentz transformation?
Very true, Ron, +1.
Apr
11
revised How do we measure $i[\hat\phi(x),\hat\phi(y)]$ in QFT?
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Apr
11
comment How do we measure $i[\hat\phi(x),\hat\phi(y)]$ in QFT?
The commutator of $X(Monday)$ and $Y(Tuesday)$ is nonzero - you don't really need quantum field theory to discuss these things, any QM theory with time is enough - but it doesn't correspond to any operation done with simple fields located on Monday and Tuesday slices. Moreover, if you talk about operators $X(t)$ for all $t$, they're not really independent from one another. By the Heisenberg dynamical equations, operators at one moment are functionals of operators at any other moment. There are many ways to measure them by procedures at various times one may choose.
Apr
11
comment How do we measure $i[\hat\phi(x),\hat\phi(y)]$ in QFT?
"Except that time-ordering makes the algebra of operators a commutative algebra (and not a C⋆-algebra unless we also introduce anti-time-ordering)." - A commutator with time-ordering inside is really meaningless; I am not defending anything of the sort. I am just saying that if there is a process that does something with $X(Monday)$ and with $Y(Tuesday)$, then $X$ will be manipulated with one day ahead of $Y$, and if the manipulations are expressed as measurements or operations done with a ket vector, $X$ will be inevitably on the right side from $Y$.
Apr
11
comment How do we measure $i[\hat\phi(x),\hat\phi(y)]$ in QFT?
If you had a QFT without an explicit Lagrangian, like e.g. 2-dimensional model's minimal CFTs, then one couldn't evaluate the commutators by any methods based on the classical limit. But there would still be quantum operator formulae that determine the commutators of operators in terms of some other operators from the full basis of possible local fields. Those would be encoded in the OPEs.
Apr
11
comment How do we measure $i[\hat\phi(x),\hat\phi(y)]$ in QFT?
Apologies if I am wrong but you may be misunderstanding how the commutators are determined in the process of quantization. For Lagrangians that depend at most at first derivatives, one writes some coordinates - spanning a configuration space - and their velocity. They fully specify the initial conditions. The canonical momenta are $\partial {\mathcal L}/ \partial (\partial_t x_i)$. The $[x,x]$ and $[p,p]$ commutators vanish while $[x_i,p_j]=i\hbar \delta_{ij}$. These rules are true even for interacting QFTs.
Apr
11
comment How do we measure $i[\hat\phi(x),\hat\phi(y)]$ in QFT?
Dear Peter, I think that I have answered your question about thrice already. The commutators of two operators are always some functionals of the "basic operators" such as the fields and their velocities (or canonical momenta); there is nothing "third" they may depend on. And in fact, the equal-time commutators between any operators in SM-like QFTs are completely identical to the commutators you get in the free theories where you just drop the interaction Hamiltonian. There is nothing nontrivial waiting in this evaluation.
Apr
11
comment How do we measure $i[\hat\phi(x),\hat\phi(y)]$ in QFT?
Dear Peter, concerning your "i take my... to say otherwise", it is also incorrect. I've added an explanation of it to the answer, too. The equal-time commutators in QFTs such as the Standard Model are completely unaffected by the interactions. The interaction terms add zero to the equal-time commutators because the renormalizable interactions in the Standard Model are non-derivative terms, which therefore don't change the formulae for the canonical momenta at all. So you're wrong that the interactions change anything.