@BernardoMeurer Consider $B_\epsilon(a)$. Let $x\in B_\epsilon(a)$. Let $\eta=\epsilon-||x-a||$. Claim: $B_\eta(x)\subset B_\epsilon(a)$. To show this, we must show that if $y\in B_\eta(x),$ then $||y-a||<\epsilon$. But $$||y-a||\le ||y-x||+||x-a||< \epsilon -||x-a||+||x-a||=\epsilon.$$
@BernardoMeurer So now we have: A set $X$ is open if and only if $$X=\bigcup_{\alpha\in J}B_{r_\alpha}(x_\alpha),$$ i.e. if $X$ is a union of open balls.
For one thing, it's not productive. The people with the authority to override suspensions are the SE team. If one believes one has been unfairly suspended, one should take it up with them.
It means that moderators generally recognize that each others' decisions are made for valid reasons and will not override them out of respect for those reasons. If a decision is to be reversed, it should be done because the reasons for that decision are not actually valid, and nobody is better placed to make that determination than the mod who did it in the first place.
@0celo7 Counterquestion: Could you please use formulations that cannot be mistaken for crude sexual implications? Honestly, if you don't see that "I'll shove these balls down your throat" will likely get you suspended, there's no point in discussing this issue.
@0celo7 because you are chatting with other people. (To clarify: to whatever extent you are being punished for other people's dirty minds, it is because you're chatting with other people. I don't mean to comment on whether or not it's accurate to say that this qualifies as a punishment.)
@0celo7 I have no idea what that is supposed to mean. Are you seriously claiming "I'm allowed to make innuendo as long as 13-year-olds don't get it"? That's not how respectful conversation works.
my name is Yordan Yordanov and I am a bachelor in molecular biology but I would really like to ask a question about physics and to be more precise-about non-equilibrium thermodynamics. Please, excuse me if I got some terms wrong or messed up a thing or two. It is just that I don't feel like "I am...
@0celo7 There are no execution squads and this is not a court of law. Nevertheless, I think that that particular legal principle encodes a more general prinicple of human interaction that intent does not matter as much as the perceived meaning.
@ACuriousMind My friend @skullpetrol would beg to differ on the execution squad front. And I refuse to accept that legal principle when we do not have free discussion of suspensions, i.e. as long as there are secret courts.
If I'm finding $\langle \psi | H | \psi \rangle$, where $H$ depends only on $r$ (the radius), what bounds do I use for the integrals? For example, I'm doing this for the hydrogen atom, but I don't know how to decide the bounds for the radius. Is it $r=[0,R]$ or $r=[0,\infty]$?
Thousands of people chatting on this server, and somehow you manage to step on the equivalent of a rake hidden in the grass more often than just about anyone else?
@loltospoon You integrate over all of space, always, the only reason to do otherwise is if you can leave out parts where the wavefunction is known to be zero. (Or if you have a cyclic dimension or some such nonsense :-P)
@0celo7 be that as it may, you should probably consider that most moderators would just suspend you at the drop of a hat at this point, purely for efficiency.
@Shog9 I don't understand. We were having a pleasant conversation then you said most people would execute me at the drop of a hat? What kind of thing is that to say to a person?
@0celo7 Think of it this way... You're some random moderator in chat here, and you see a flag pop up. "Hmm, there's a flag, I wonder if this person just made an innocent mistake and... Oh, it's you again."
If I want to use the variational principle to estimate the ground state energy of the hydrogen atom, what hamiltonian do I use? Is it $H = -\frac{\hbar ^2}{2m}\frac{d^2u}{dr^2}+(\frac{-e^2}{4\pi \epsilon _0}\frac{1}{r}+\frac{\hbar ^2}{2m}\frac{l(l+1)}{r^2})u$
Anyway, I have potatoes boiling over so I gotta run, but... Yeah, you're both pretty well ranked on the top-100 troublemaker list for chat, so I wouldn't be too cavalier about it all. Everyone makes mistakes, but most folks don't make the same ones repeatedly.
@loltospoon Yeah, whatever the actual Hamiltonian is. I suppose you could also do it with an approximate Hamiltonian, but that's going beyond the variational principle as it's usually taught.
@0celo7 Counterquestion: Could you please use formulations that cannot be mistaken for crude sexual implications? Honestly, if you don't see that "I'll shove these balls down your throat" will likely get you suspended, there's no point in discussing this issue.
Why is it that in my book they define the variational principle as $E_{gs}=\langle \psi | H | \psi \rangle$, but on this website it defines it as $E=\frac{\langle \psi | H | \psi \rangle}{\langle \psi | \psi \rangle}$
Ok guys, I'm really not convinced with what I wrote above. In this other example, the person uses different values of $T$ and $V$ in using the variational principle to estimate the GSE of the hydrogen atom.
He/she doesn't use my obnoxiously large and complex Hamiltonian
