In the article Fast Factoring Integers by SVP Algorithms the author claims that he discovered classical algorithm for factoring integers in polynomial time. The Quantum Report mentioned here that it has similar performance to Shor algorithm which is often considered to ignite interest in quantum ...
@NikeDattani cc @fqq here's a several days old HNQ from crypto.SE that's largely critical of the claims in that paper
for the record, by our established policies I would consider equivalent questions on physics.SE as off-topic since they are more requests for review and not questions about established science
yes, I saw several people discussing it on twitter as well (not the best platform but I don't know enough about the topic to follow the details anyway), it was linked on Scott Aaronson's blog
not surprising, given that RSA does not seem to have been destroyed
@ACuriousMind the actual question I think is OK, it's just a weird starting point to ask "is Shor's algorithm optimal?"
@eryceriousmatherfunker pick up a book and "cogitate"
> work through as many exercises as possible
> Knowledge cannot be acquired all at once; it must be gained step by step. Don’t try to spell ‘catastrophe’ if you cannot spell ‘cat’! Start with the basics before you move on to the difficult stuff.
I have an irreducible representation of a finite group. How can I find the basis functions of this representation? It seems like guesswork is the most common approach, but I am not very good at guessing
Is it reasonable to say that when you have some particle travelling through space (in say scalar qft), we represent this in Feynman diagrams as just a propagator (which is just a line), and the fact that the higher order corrections can be rewritten in such a way that we just get a propagator with a correction to the rest-mass tells us that there is some "self interaction" happening which increases the energy of the particle by increasing its rest mass?
Or rather, modulo complaints about what Feynman diagrams actually represent, is there anything obviously objectionable about what I've written above?
The "correction to the rest-mass" being the sum of all 1PI diagrams
What are the reasons why we usually treat Quantum Field Theory in momentum space instead of position space? Are the computations (e.g. of Feynman diagrams) generally easier and are there other advantages of this formulation?
@PM2Ring Ah no I wasn't using position in that sense, I meant it more hand-wavily. That the fact that the non-interacting propagator has corrections in the interacting theory that simply algebracially shift its rest-mass can be interpreted as (ignoring the usual complaints that "Feynman diagrams don't represent real processes") "the particle interacts with itself"
You seem to know how the 1PI contributions shift the pole of the propagator and hence the mass. What does trying to express this technical fact with vague statements like "the particle interacts with itself" add?
note that the dressed mass is a renormalization parameter, so it's not as if the mass is a prediction of QFT, it's an input
you can't say "I start with a free particle and now I add some interactions and here's how heavy it gets" because the mass "added" by the interactions will be "infinite" in most theories!
I guess you don't really gain much by intentionally muddying it with over-physical interpretation, I just wondered if what I wrote above is wrong in any very obvious way except for objections about what exactly we think particles do
oh yeah I hadn't thought about the divergence
I'm just getting onto renormalisation, I haven't really got far into it yet
@ACuriousMind I am even more confused now .... suppose I have an apply in front of me, and I give it a displacement vector [2, 1]. Then I move it around, and now it is [6, 10], and since a =3, b = 10; but it is no longer a (displacement) vector? What am I missing?
I think I am okay with the math, but some concept of mine is wrong, and I am not sure what is it
probably [a 0, 0 b] is not an operation of "moving it around"
@Shing The thing is that moving something around is not the same as multiplying the individual components with scalar $a$ and $b$! When you have something at position $\vec v$ and you displace it by $\vec w$, then the new position is $\vec v + \vec w$. Note how this is adding another vector, not multiplying by scalars.
What are the reasons why we usually treat Quantum Field Theory in momentum space instead of position space? Are the computations (e.g. of Feynman diagrams) generally easier and are there other advantages of this formulation?
