LuboŇ° Motl

Czech Republic

motls.blogspot.com

Age: 41

Hi, I am a string theorist and a publicist.

11h
comment In QFT do we always use normal-ordered Hamiltonian?
This question of yours pushed me to write the following essay about the "primarility" of quantum mechanics: motls.blogspot.com/2015/08/…
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awarded Enlightened
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awarded Nice Answer
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Jul
31
comment Spontaneous symmetry breaking and time-reversal symmetry
I would say Yes - but what is the cause and what is the consequence or a description is debatable... But generally Yes, states in different superselection sectors have "very different" macroscopic consequences for the environment, so the information about the superselection sectors is being copied as classical information as these states decohere - so decoherence is why the separation to the sectors becomes visible.
Jul
31
comment Upgrade to Windows 10
I surely do enjoy it! Ho1: Win 10 is called 10 but it's actually a 9, too. ;-) If Stephen Wolfram sends me an upgrade again, I won't refuse, however.
Jul
29
comment Upgrade to Windows 10
Thank you, I just upgraded from Win 7 to Win 10 and computed 2+3 on Mathematica 9. ;-)
Jul
29
awarded Popular Question
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29
awarded Yearling
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29
accepted Upgrade to Windows 10
Jul
29
comment Polchinski Equation (10.4.7)
Similarly, the OPE you are asking for is not "equal" to zero, but it is $\sim 0$ because the OPEs had no singular terms to start with, so one couldn't have gotten any singular terms by differentiation, either. But again, $b(\sigma_1) \partial_\sigma^7 b(\sigma_1) b(\sigma_2)c(\sigma_2)$ has no reason to be zero, has it?
Jul
29
comment Polchinski Equation (10.4.7)
You are probably confused by the difference between the $=$ sign in the OPEs, and the sim sign $\sim$. The latter indicates that the right hand side only lists the singular terms, i.e. is correct up to additive terms $+O(1)$.
Jul
29
comment Polchinski Equation (10.4.7)
Dear MaPo, the product of the two operators is surely not equal to zero. Just map it to the cylinder, if the plane distracts you. $b(\sigma_1) b(\sigma_2)$ is just a product of two anticommuting operators, right? Why should it be zero? It is $O(z)$ indeed because there's no singularity for $z\to 0$ and the value for $z=0$ is zero due to anticommutation. But for $z\neq 0$, it cannot be zero.
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29
awarded Nice Question
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29
comment Upgrade to Windows 10
Just to be sure, I think that even "partial" answers may be very helpful. Most likely, no one knows everything about possible issues with the upgrade of all possible versions of the software. But much of the partial knowledge may be generalized etc.
Jul
29
comment Upgrade to Windows 10
Mathematica 8 but I would love to know the answer to Mathematica 9 - and perhaps Mathematica 10 - as well. ;-)
Jul
29
asked Upgrade to Windows 10
Jul
28
comment Graviton saturation, alternative explanation of galaxy rotation curves?
Magnus, you would need to write some equations. This kind of "spirit" of yours is nice, and probably basically shared by some original minds, but what is missing are the equations - what actually makes the gravity bounded and how? Moreover, what you need is kind of the opposite. You need to make gravity stronger than expected by Newton when the acceleration is too low (MOND) etc. At this vague level, this fixed theory of yours is pretty much "the same thing" as MOND, and if you mean something else, you need to describe it more accurately than you did.
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revised Polchinski Equation (10.4.7)
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28
answered Polchinski Equation (10.4.7)
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