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12:06 AM
I'm reading Greene's Elegant Universe and he presents tunneling as being related to the energy-time uncertainty. Is that really the case? I mean they're certainly both tied to the Schrodinger equation, but I've never heard tunneling being presented in that way
 
0
Q: Why can I only accept my own answer after 24 hours?

zyyWhy do I have to wait 24 hours to accept my answer to my own question? What is the site trying to avoid?

 
Interestingly, these two answers seem to be different
https://physics.stackexchange.com/a/395721/24839
https://physics.stackexchange.com/a/54220/24839
 
12:21 AM
On a similar note, are there any experiments that can make a position measurement during tunneling rather than making a measurement on the opposite side of a barrier?
 
 
5 hours later…
5:02 AM
0
Q: Can I ask an answerer to Fix a broken link in their post?

DMacIn this post: How can I put a permanent current into a superconducting loop? There is only one link at the bottom of the only answer. Can I ask the answerer to fix the link? How?

 
 
1 hour later…
6:20 AM
@EmilioPisanty You had not shared that.
 
 
1 hour later…
7:25 AM
@Blue I finally got around to reading your answer to Duffield's question in QC. That is an awesome answer. It must have taken you ages to write!
 
@JohnRennie I wrote it on a "dare" from Duffield (it has been deleted now) ;)
But thanks. It took a few days :). I luckily also found Scott Aaronson's very nice lecture series (notes) in the process
 
@Blue you're one of the highest rep users over there
Nice
 
@Blue It seems obvious that Duffield wasn't interested in having his question answered. He just wanted a platform to shout about how stupid QC researchers are. Just like his posts here are a platform to shout about how stupid GR researchers are. Oh well :-)
But I have to say your answer is a masterpiece and elegantly puts Duffield in his place.
 
Thanks :P
@AvnishKabaj The rep corresponds to the fact that I spend a lot of time there rather than a measure of what I know ;) (I just started learning QC for 3-4 months whereas there are people with 15+ years of experience on the site e.g. Daft Wullie, John Watrous, etc)
 
7:46 AM
Ooooooh
Still
 
 
1 hour later…
9:10 AM
@JohnRennie oh, f*s
> But you’re forgetting that I’m into fundamental physics in a big way
no.
He's into grandstanding about fundamental physics in a big way, in a big way
@DanielSank thoughts?
@DanielSank see also this related piece
 
@EmilioPisanty he's banned on QC now, as well as here :-)
 
9:37 AM
@JohnRennie huh
does he still get to award that bounty?
how does that work?
 
Good question, though I think he's only banned for three days more.
But he won't award the bounty to the best answer (Blue's) because he's ideologically opposed to the truth.
 
@JohnRennie bounty runs out tomorrow
 
@EmilioPisanty don't know then.
 
@JohnRennie true dat
@JohnRennie ::fetches popcorn::
this'll be interesting to watch, then
 
@EmilioPisanty I've been in the same position. There are questions where I have the most upvoted answer by far but Duffield has awarded a bounty to a different answer because he wouldn't admit I'm right to save his life.
 
9:40 AM
@JohnRennie oh, for sure
no, I just want to watch the system handle the rules conflicting with each other
maybe the bounty will auto-award itself at 50% to the highest-voted answer?
that'd be hilarious
 
yeah it does that if the person doesn't award it to someone else
 
@EmilioPisanty Aha!
1
Q: Can a suspended user award a bounty which they started before getting suspended?

fresherMy friend's account is suspended for a month in one of the Stack Exchange sites. He asked a bounty question and someone answered it. However, before he could award a bounty to the answer, his account got suspended. Is there any option to award the bounty for that answer?

So Blue will get half the bounty. That's ... excellent! :-)
Duffield gets to award half the bounty to answer that he hates :-)
That's karma for you.
2
 
9
A: Suspended user cannot award bounty

wafflesWe have said this before, I will say it again. We do not intend to put any extra effort supporting outlier users that have abused the system. Consider being unable to assign a bounty part of your "punishment".

heh
 
9:56 AM
@EmilioPisanty Couldn't happen to a nicer chap.
(Schadenfreude? Me? :-)
 
@EmilioPisanty :-)
The young things don't have this problem since they use emojis not emoticons. You just have to wait for the current generation of old time computer nerds to die off.
 
10:50 AM
@JohnRennie well, we just have to ask SE for an emoji keyboard on chat, don't we?
I reckon that with a convenient emoji keyboard and a nice-looking font* we can get even some of the old farts to switch over to the modern stuff
* yeah, I know, fat chance, right?
 
I find myself massively unimpressed by the current obsession with emojis. Why the unicode code page has to be cluttered up with thousands of basically pointless images completely escapes me.
 
@JohnRennie ah, lighten up, you grump 😊
$\Huge 😊$
learn to love the emoji
 
Bah bumhug
 
@JohnRennie you sure that's in the right order?
 
:-) or even 😊
 
11:01 AM
I prefer Kaomoji over emoji ;)
(@^◡^)
 
@Blue I do find them very clever. The hours of time people must have spent poring over code pages to put them together!
 
john is just jelly because emojies solve his bracket-smile dilemma
 
Lol
 
he wants to live in an eternal angst fueled by the choice of putting a bracket after the smile or putting the bracket as the smile, both being equally morally reprehensible acts
2
 
@BalarkaSen actually I do like the tension produced by a (possibly) unmatched bracket.
 
11:05 AM
John Raskolnikov
 
I've always found it enormous fun to play with language - possibly that's an English trait.
 
everytime he puts a smile it is both a crime and a punishment to the readers
 
I guess I find emojis unsatisfying precisely because they are generally used to reduce ambiguity not exploit it.
Indulge in it even.
 
@JohnRennie it's a Schrödinger bracket, both matched and unmatched at the same time
@JohnRennie are they, really, though?
 
@EmilioPisanty no, it can't possibly be a superposition because my posts are invariably incoherent
 
11:08 AM
or is it that you don't yet speak emoji well enough to use it joyously ambiguous ways?
 
@EmilioPisanty OK I'll concede that point. It may well be that I'm criticising out of ignorance.
 
$\Huge 🙃$
 
Actually it's not superpositions that are destroyed by decoherence is it? It's entanglement. Superpositions are destroyed, or not, by the measurement process.
I keep getting this wrong.
 
@JohnRennie I shiver at the thought that almost all of the retired unmarried Englishmen of age over 50 turns into uncles who are basically dad comedians throwing dad jokes every two seconds hoping to get some claps from their nephews and nieces
Such is the fate of Great Britain
 
You underestimate us. Dad jokes are told as a premediated attempt at torture, not because we expect anyone to laugh at them.
Niece: who do you think is going to laugh at your jokes?
John: Me!
 
11:14 AM
the horror
do these dad comedians have no natural predator?
 
Just arteriosclerosis :-)
 
 
2 hours later…
1:24 PM
@BalarkaSen Need some help here: Can exponential $e^{iA}$ of a non-Hermitian matrix $A$, be unitary?
 
I'll work with real matrices because I'm paranoid. $\exp$ commutes with transpose (just write down the series expansion). So $\exp(A)^T = \exp(A^T)$. If $\exp(A)$ is orthogonal, $\exp(A)^T = \exp(A)^{-1} = \exp(-A)$, so $\exp(-A) = \exp(A^T)$. That would mean $A^T = -A$.
Similar arguments should push through complexly
(This means exponential of a real matrix is orthogonal iff it's skew-symmetric)
 
@BalarkaSen $\exp(-A)=\exp(A^T)$ implies $A^T=-A$? maybe for real matrices
not for complex ones though
 
Ah yeah
See why I was paranoid?
 
$\exp(2\pi i \hat\Pi) = \exp(0) = \mathbb I$ for any $\hat \Pi = \hat \Pi^2$, I should think
 
does that still fuck up Hermitian-ness? Naw
 
1:33 PM
you can probably still get $\exp(A^\dagger) = \exp(-A)$
but at the very least that allows for $A^\dagger = -A +2\pi i B$ where $A$ has a shared orthonormal basis with $B$ (but where $A$ might have complex eigenvalues) and $B$ has a spectrum in the integers
or $\exp(iA^\dagger) = \exp(iA)$ implying $A^\dagger = A +2\pi B$ with the same conditions
in that case $A$ is normal, at least, but you can see how things could get much uglier
hmmmm, wait, that probably doesn't work. If you conjugate that then you get $A=A^\dagger + 2\pi B$ or $A^\dagger = A-2\pi B$, which breaks things
 
modulo $2\pi$~
 
1:49 PM
$$\exp\left(i\begin{pmatrix} 2\pi & 1 \\ 0 & 0 \end{pmatrix}\right) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ according to Mathematica
@Blue @BalarkaSen
 
France looks nervous.
 
ditto with $$\exp\left(i\begin{pmatrix} 2\pi & a \\ 0 & 0 \end{pmatrix}\right) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ for arbitrary $a$, it seems
 
@EmilioPisanty Ah, so the $2\pi$-periodicity thing does mess that up
 
@BalarkaSen apparently?
I don't know why that would happen so I'm reluctant to believe MM at face value
ah, got it
the bottom row of zeroes means that when multiplied with itself the matrix $\begin{pmatrix} i\theta & a \\ 0 & 0 \end{pmatrix}$ just gets powers of $i\theta$ on the diagonal and feeding into the top-right element
so
$$\begin{pmatrix} i\theta & a \\ 0 & 0 \end{pmatrix}^n = \begin{pmatrix} (i\theta)^n & -ia(i\theta)^n/\theta \\ 0 & 0 \end{pmatrix}$$
and therefore $$\exp\begin{pmatrix} i\theta & a \\ 0 & 0 \end{pmatrix}^n = \begin{pmatrix} e^{i\theta} & a(e^{i\theta}-1)/i\theta \\ 0 & 1 \end{pmatrix}$$
so yeah, there you have it
(damn. extraneous ^n on that last equation.)
 
vzn
2:08 PM
@danielunderwood re greene, top popsci author, ever heard of the so-called black hole electron? ran into it recently & think it might align with fluid paradigm.
 
@BalarkaSen Umm, I was especially concerned about complex matrices :P I didn't read through what you wrote though. I'll be back in half an hour
 
Emilio gave a counterexample so that answers your question
I don't think about complex matrices much
3hard5me
 
vzn
> Currently, Greene studies string cosmology, especially the imprints of trans Planckian physics on the cosmic microwave background, and brane-gas cosmologies that could explain why the space around us has three large dimensions, expanding on the suggestion of a black hole electron, namely that the electron may be a black hole. en.wikipedia.org/wiki/Brian_Greene en.wikipedia.org/wiki/Black_hole_electron
 
what about $i I$
pretty simple complex matrix
or even $0$
It is a complex matrix
too
 
hi-yo
I had an argument in the QC room that $e^{i A t}$ unitary for all $t$ implied $A$ hermitian, but that's evidently a stronger condition than just $e^{i A}$ unitary
with Emilio's example such a case where $e^{i A}$ is unitary but not $e^{i A t}$
 
2:22 PM
@Semiclassical : Look up unbounded operators
me thinks
 
(more precisely, it's unitary for integer $t$ but not for non-integer $t$)
sounds legit
 
unbounded hermitian operators are usually nasty
Brian Hall probably has an example somewhere
 
glS
2:56 PM
@Slereah isn't everything bounded here?
 
@glS $p$ isn't bounded
 
glS
@Slereah yes but I mean the question was about matrices, which are essentially linear operators on finite-dimensional spaces
or at least, I believe @Semiclassical argument was mostly referring to matrices not general (possibly unbounded) operators
 
oh right
Hm
I don't remember the condition for $e^{iA}$ to be unitary
 
in The Classical Channel, 25 mins ago, by Blue
Now what we need to look for is 1) The exact set of conditions for which the matrix exponential $e^A$ of a complex matrix $A$, is unitary 2) The exact set of conditions for which the matrix exponential $e^{iA}$ of a real matrix $A$ is unitary
In case anyone wants to help us ^
@EmilioPisanty Thanks, that's an excellent example :)
 
glS
3:11 PM
@EmilioPisanty @BalarkaSen is it true even when restricting to real matrices though? How do you see it? I mean $e^A=e^B\Longrightarrow A=B$ for real $A$ and $B$
 
Should be reducible to the question of when $e^A=1$ is possible
in which case the answer is "not very often"
 
@Semiclassical no, that's not true
 
One thing to note is that the eigenvalues of $e^A$ are $e^{\lambda_i}$ if the eigenvalues of $A$ are $\lambda_i$. Also, they share all eigenvectors
(which is obvious I guess, but still stating it)
 
You're implicitly assuming that A is normal
 
hmmmm
i buy that
 
3:20 PM
But since real matrices can have complex eigenvalues
And $e^a = e^b$ does not imply $a=b$ if $a,b$ are complex
 
@Blue ... which implicitly assumes that $A$ has a basis of eigenvectors, again implicitly assuming that $A$ is normal
 
@EmilioPisanty Ah, right
 
glS
@Blue that for sure. The gist I believe is that the complex logarithm is multivalued, but if you restrict its output to be real it's not. But in that case is also only defined for positive reals though, which might complicate things. (also, dealing with matrices complicates things because this kind of argument only works when dealing with diagonalizable matrices)
so... if $A$ and $B$ are reals and diagonalizable, is it the case that $e^A=e^B$ implies $A=B$?
 
You're all right. It is not injective in general.
 
Diagonalizability makes the condition stronger, interesting
@glS You mean diagonalizable over $\Bbb R$?
 
3:31 PM
@Blue are you sure that the only thing you know is that $e^{A}$ is unitary? i.e. are you sure you don't also know e.g. that $e^{\lambda A}$ is unitary for $\lambda$ in some open interval around $\lambda=1$?
'cause that would change things significantly
not that the pointwise problem isn't interesting
 
glS
@Blue you are right, we have to distinguish between real matrices and matrices which have real eigenvalues
 
@EmilioPisanty Well, it's an imaginary problem anyway. In physics we mostly care about $e^{iAt}$. If $e^{iAt}$ is unitary for all $t$, then $A$ is Hermitian. But one problem here is that we do not know for which set of values of $t$, $e^{iAt}$ would be unitary
 
@Blue that's a not-great way to look at it
you just need to know the accumulation of points of $\{t: e^{iAt} \text{ is unitary}\}$
 
Umm, we haven't found the accumulation of those points either
Gotta think about it
 
glS
there might be an answer to this question here
16
A: Under what conditions is the exponential map on a Lie algebra injective?

YCorThere is a complete characterization, in a large part due to Dixmier and Saito (both independently in 1957): If $G$ is a real Lie group with Lie algebra $\mathfrak{g}$, then the following are equivalent: $\exp$ is injective $\exp$ is bijective $\exp$ is a real analytic diffeomorphism $G$ is so...

but it would require someone able to translate the Lie algebraic formalism for the rest of us
 
3:39 PM
We better learn Lie algebra soon :P
But that looks interesting
 
This is too hard. Look at exp(0, 2*pi; -2*pi, 0)
The power series computation runs the same way it runs for exp(2*pi*i)
because (0, -1; 1, 0) is essentially the real matrix that represents i as a complex linear operator on C
Lots of matrices A such that exp(A) = 0
 
@BalarkaSen Good point!
 
I was badly mistaken up there
 
That’s skew-symmetric
 
Ah
 
3:44 PM
I think the harder question was whether you could have real but non skew-symmetric A such that e^A=1
 
(exp(A) = I, I meant)
@Semiclassical This I don't know but feels like you can get a 3x3 example by adding that skew-symmetric dude in a 2x2 block
 
I don’t see that right away, but I wouldn’t be shocked
 
(0, 2pi, 0; -2pi, 0, 0; 0, 2, 0)
 
I can’t tell by inspection if that works, but cool if it does
 
Mr pari/gp says it works but someone should check
 
3:52 PM
I guess that raises a question of why it can be done for matrix size n>2 but not n=2
Some statement about GL_n(RR) ?
No, that’s not it
 
glS
I asked this on math.SE:
0
Q: When is the exponential map between matrices injective?

glSConsider a complex matrix $A$. When does $e^A=e^B$ imply $A=B$? Is there any general statement that can be made as to when this holds? It is clearly not true in general, a trivial example being when $A$ and $B$ are diagonal in the same basis but with eigenvalues differing by $2\pi i\mathbb Z$. ...

 
4:07 PM
ok, now that there's a bunch of smart people here, I need some help
I have this observable that depends on $\sigma$
 
I can't view images unfortunately
 
and I would like to choose the point on $\sigma$ that maximizes the contrast between the yellow and the blue curves while also maximizing the signal
and, unfortunately, the ratio yellow/blue goes up to infinity at $\sigma\to 0$
 
@EmilioPisanty it sounds as if there isn't an obvious maximum
As in - you can't simulatneously maximise both
 
@JohnRennie yeah, I kinda need a compromise between contrast and signal
 
So it's just a question of what tradeoff is acceptable.
 
4:16 PM
it feels like a reasonably common question
 
You need some function that gives you experimental goodness as a function of contrast and signal (presumably signal to noise). Then see how that varies with sigma. But only you know what that function might be.
 
@JohnRennie yeah, I know, it's a puzzler, this one
 
Do some simulations
 
basically, I want to distinguish those ridges with the highest contrast + signal as possible
 
i.e. some mock data analysis with different tradeoffs between contrast and signal.
 
4:21 PM
@JohnRennie well, this isn't data analysis as such
it's just functions I can churn
 
don't have much to do today hmmm
 
@enumaris You can always eat and then eat some more ;)
 
hmmm, not sure I can just leave to go eat all day lol
 
Order pizzas to your office
:P
 
@Blue that's pretty much what I do at the weekend :-)
 
4:25 PM
That's obviously what all healthy humans should do (to keep being healthy) ;)
 
I don't you'd stay healthy very long if you ate continuously ...
 
The google Starbucks WiFi wants my name + email + postal code (and agree to receive Starbucks promos)
Neeewp
 
I have a sacrificial e-mail account I use for that
 
Jio is too good (and cheap). We don't need to use the shop WiFi anymore (avoids all those annoyances) :P
 
Yeah, my response is just to not use my laptop for the time being
Not going to trade my email for an hour or so of WiFi
 
4:31 PM
Get a fake email id
 
do it
in palpatine's voice
 
@Blue that's ... pretty damn worth bookmarking! :-)
@Semiclassical try that web site Blue linked
 
do it
 
Lol, it's useful and works most of the time. In case it doesn't there are a few others too
 
80% of the time, it works every time?
the office is uber quiet today...
I feel like nobody came in lol
 
4:40 PM
@Semiclassical I take they don't take mailinator?
@Blue pretty good, but mailinator.com still beats it
you can choose the address before going to check
it'll also receive any emails sent to spamhereplease.com and a bunch of other domain aliases
 
I need a fake mobile number to bypass those pesky registration forms of game sites, except none of those fake mobile generators work
 
5555555?
 
@enumaris Well, most of those websites send a One Time Password (OTP) as SMS to your number. So while you could make up any random number, you wouldn't get hold of the OTP
I too would like to have a fake "mobile number generator", like Secret :P
@EmilioPisanty Looks good!
 
glS
@EmilioPisanty @BalarkaSen
 
5:09 PM
hmmm
 
mmmh
 
this sql querry is taking too long -.-
 
It's telling you to take a break, just like when you have to compile something
 
time to go read some more L+L I suppose
started up 150 epochs of training on my deep LSTM...time to just chill lol
one computer running LSTM another computer running a SQL query...not much else I can do atm
 
5:43 PM
oooo my order has been estimated to arrive early...nice
 
What's in it?
 
some magic cards and like a binder to keep my nice cards in :D
 
cards?
 
yeah, magic the gathering
 
lol, you still collect those
 
5:46 PM
why not
 
Magic: The Gathering is a both a trading card and digital collectible card game created by Richard Garfield. Released in 1993 by Wizards of the Coast, Magic was the first trading card game created and it continues to thrive, with approximately twenty million players as of 2015.Magic can be played by two or more players in various formats, which fall into two categories: constructed and limited. Limited formats involve players building a deck spontaneously out of a pool of random cards with a minimum deck size of 40 cards. In constructed, players created decks from cards they own, usually 60 cards...
 
gotta have hobbies
 
@enumaris just pleasantly amused ;)
Not many collect cards anymore
 
I just restarted lol
my interests change pretty quickly tho XD
 
I wanted to play d&d when I was 12ish
But sadly my friends weren't interested
 
5:51 PM
hmmm, I've never tried d&d
seems complicated heh
 
I used to collect duel master cards
When I was 9
Had a huge pile
But parents dumped them somewhere when we shifted our house :P
 
awww
I still have some cards from when I was little
but definitely I lost a lot of cards as well
I find that rereading physics after some time I feel like I have a better understanding...mmmhm...
conceptual understanding anyways
maybe no better at doing problems lol
 
6:22 PM
@Blue That is a fantastic game.
@enumaris What did you order?
 
magic cards and equipment :D
 
@enumaris Yes yes, what exactly?
 
like which cards or?
 
Yes.
 
I got some cards to round out my deck, mostly stuff like counter spells...and then I got like a binder to put my cards in, and a deck box, and some sleeves
 
6:24 PM
What format?
 
Standard
 
ah
ok
Wanna see something neat?
 
sure
do you play mtg as well?
 
@enumaris Casually.
 
same here heh, tho I'm prolly more of a collector than a player tbh
I got a few Approach of the Second Sun as my win condition...see if I can make a deck use that as my win condition lol
a few seal aways, a few cast downs, some syncopates...black,white,blue deck for funsies
all control XD
 
6:28 PM
@enumaris You may appreciate this.
 
hold on
google drive is banned here lol, gotta open with phone
 
Why would someone ban Google Drive?
 
it's all banned
what are these cards...o.O
 
I designed them.
:-D
 
ahhh
so yellow is your new color?
desert mana?
 
6:30 PM
yep
 
when did you design these? Unfortunately Amonkhet already has some deserts as land so you might have to do something about that heh
oh I think my door dash is here
brb
 
I designed them years ago.
 
6:48 PM
@Sid lololol 2-0
 
Stunning
 
Eh?
 
Belgium is playing excellently actually
 
@enumaris Amonkhet ripped off my ideas.
 
European football has truly ascended
 
6:49 PM
They deserve to win the first half
 
Yo @enumaris ping me if you get a chance to look at those cards.
 
Have you see those passes? Man they're amazing
 
7:24 PM
@DanielSank just finished my lunch, taking a look now :D
 
7:37 PM
@DanielSank first thing is I think the (1) for colorless mana might be better as a <diamond> colorless mana. Now there's cards that can only take colorless mana and for those you need to add a <diamond> mana.
@DanielSank All Is Sand seems way powerful...a lot of players utilize a lot of non-basic lands - negating all of that and turning them into basic deserts for only 4 mana is like...insta win
Bandit seems a bit weak...2 drop for a 1/1 and 2 more for vigilance...so like 4 drop for a 1/1 with vigilance...
@DanielSank where did you get the illustrations? They are quite good :D
Corrupt Courtesian...3 drop for a 1/2 with no abilities is too weak isn't it?
 
7:53 PM
Anyone knows what circuit.comment and circuit.h mean here?
Can't find it in the documentation
Tried looking into the qiskit.QuantumProgram class
 
What just happened
 
RIP Brazil
 
Good, imo. I found their game horrible.
 
last South American team out too?
 
7:57 PM
yup
 
only Europeans left lol
 
it's all European now bikhes
 
now it's more of a eurocup lol
but maybe Russia is Asian Confederation?
I can't recall
are they UEFA as well?
oh yeah they are part of UEFA
so yep...basically just a Eurocup now
 
The real man's football
 
Every 20 years the home team's won
1978 1998
2018?
That's only the past two times this but still
 
7:59 PM
Latin American football became football for (actually maybe not PC) quite a few years back
 
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