@S007 That it's "public" in the sense of being found on several websites doesn't tell you anything about its copyright, you need to find the original source and the copyright it's licensed under to be sure.
@Obliv yeah (except wth the additional property it is additive absorbing (but that's just's a convenient choice While the + cayley table is heavily constrained by the * cayleytable, a is not being constrainted by them). Since a is a two sided aobsorber, it has no inverses as expected. The point however is that the additive ientity 0 has a one sided multiplicative inverse, thus it is a legitmate division by zero
@heather It may well be that posting multiple ads to the community ads thread will make people think you're trying to hog the ad space and vote you down.
I would say that the girl should surely love that since they're the guy's girlfriend, but then again "My girlfriend always complains that I don't care about what she's doing, I don't like her post on Instagram / Wechat / Weibo... and I don't comment on her posts, bla bla bla." seems to contradict this
I guess there's no real reason why a nerdophile girl wouldn't want to be followed on social media
> So I build this. Although now she knows that I'm using this to like her post. But she's pretty happy about it (don't ask me why) and never complained again. So I'm very happy with this tool.
Wait, so you were talking about literal potatoes!? I mean, I've heard about a potato clock, but I'm pretty sure there's not enough in there for a computer...
I am, if I were a clueless help vampire (the kind which you surely don't want to attract to any healthy site), I'd be put off by all that text on the blackboard. That seems to say "work" which I would not like to do
@acuriousmind I still have your CFT notes sitting in my gmail in case I ever need to learn constructive fermionic field theory I'll look back at those with your permission 8^)
@heather Yes, the picture change is good. I'm not sold on the slogan, though, I'm not sure that I'm trying to "go" anywhere when I'm stuck on an exercises.
@0celo7 We developed the machinery necessary to construct 2D and 3D theories of fermions
Fermions are easier than bosons in some sense because the $\theta^2 = 0$ of Graßmann variables makes some things finite that would otherwise be infinite series
@MetaEd Huh? I'd be happy already if you can hand me a rigorous version of the non-unified Standard Model
Hmm, or maybe "Problem solving techniques in physics" but that's a tad impersonal. I also kind of want to make it funny like the first one. This is kind of maddening.
LANL had a neutron user facility like SNS, and like SNS it had mostly condensed-matter instruments but a couple of cold beamlines for nuclear physicists.
After SNS became operational the external funding for LANSCE shifted away from there ... the user program still exists but the operational schedule is spotty.
@0celo7 Yes. Integrated flux is about the same at HFIR and SNS, but to pulse the beam at HFIR you have to insert a chopper and throw away most of the neutrons.
I think the original proposal for my thesis experiment might have predated the cold guide hall at HFIR --- I don't know its whole history very well. I know that the beamline we considered using got snapped up by somebody else.
According to [this](https://arxiv.org/abs/quant-ph/9701019) paper, "If we wish to simulate the $n$-particle Schrodinger equation in d dimensions, one approach is to discretize space. If we discretize so that the particles move on a spatial lattice with $l$ lattice sites in each direction, the number of independent components of the $n$-particle wavefunction grows as $l^{dn}$. Even for $d = 3$, for reasonably large values of $l$ and $n$ this number becomes extremely large. If we wish to simulate the system on a classical computer, the number of independent components in the wavefunction is a…
@heather The "lattice" of the points where the particles are allowed to be. When you "discretize" space for a simulation, you only allow the particles to be in spots separated by a certain (often fixed) distance.
@heather In general, one hopes that in the limit $l\to\infty$ and simultaneously $\Delta x \to 0$ (where $\Delta x$ is the spacing of the allowed spots) one recovers the original continuum theory
but I don't quite understand the units here. What is meant by the "memory and computational requirements per time step"? Is this supposed to be thought of as bits/qubits, or what?
@heather You may think of it as bits, but it doesn't really matter - whatever it is you use to describe a single lattice point, you need $l^d$ times that to describe the whole lattice
@ACuriousMind, I guess the reason I'm curious is because I'm wondering if we are saying that for an arbitrary system where $n$ is much smaller than $l^d$ the number of bits it requires to simulate the system is $l^{dn}$ and the number of qubits is $l^d$, or basically a not insignificantly smaller number.
So that it is proven quantum computers are faster than classical computers in that aspect of simulation, I mean.
I've always been sceptical with such speed comparison claims, considering that the algorithms running on the two kinds of computers don't really compare. Right?
Not necessarily about the algorithms used. If the fastest algorithm on a classical computer takes some amount of time, and there is an algorithm that can be run on a quantum computer that is faster than that, well, quantum computers are faster than classical computers at something.
@heather Yes, but there's wall time and algorithmic complexity time.
If I have a gigahertz classical processor, and my quantum computer requires flipping some spins using long-ish pulses of megahertz microwaves, there has to be a lot of complexity speedup before the quantum computer has faster wall time.
What is the purpose of having a power supply in a photoelectric effect setup? Aren't you just concerned with the current induced by the cathode from the incoming photoelectrons (then you can derive the voltage)?
@Obliv I'm pretty sure you're talking about the power supply that's there to get the "stopping voltage" - you create an electric field against the motion of the electrons away from the plate to find out the maximal kinetic energy any emitted electron has (you compute it from the volate at which just no electron reaches your detector anymore)
@AndrasDeak They move away from the plate because they get the difference between the work needed to lift them out of the metal and the energy of the incident photon as kinetic energy. Their direction is somewhat random, but enough of them will move out of the plate
You may be confusing the setup with a cathode ray, where you need a voltage to accelerate the slow electrons emitted from a wire by the thermoelectric effect
