The h Bar

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!

Anonymous

@JohnRennie I wrote it on a "dare" from Duffield (it has been deleted now) ;)

Anonymous

But thanks. It took a few days :). I luckily also found Scott Aaronson's very nice lecture series (notes) in the process

Avnish Kabaj

@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.

Anonymous

Thanks :P

Anonymous

7:38 AM

@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)

Avnish Kabaj

Ooooooh

Still

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

Q: Can a suspended user award a bounty which they started before getting suspended?

@EmilioPisanty Aha!

1

My 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?

support bounty account-suspension

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.

A: Suspended user cannot award bounty

9

We 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? :-)

Check Your Schadenfreude

►

No, more importantly

watch your bracket matching

@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 😊

Anonymous

11:01 AM

I prefer Kaomoji over emoji ;)

Anonymous

(＠＾◡＾)

@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

Anonymous

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

11:04 AM

@BalarkaSen actually I do like the tension produced by a (possibly) unmatched bracket.

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?

11:07 AM

@EmilioPisanty no, it can't possibly be a superposition because my posts are invariably incoherent

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

11:12 AM

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!

the horror

do these dad comedians have no natural predator?

Just arteriosclerosis :-)

Anonymous

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

Cup Fever

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.

Anonymous

@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

2:56 PM

@Slereah isn't everything bounded here?

@glS $p$ isn't bounded

@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

Anonymous

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

Anonymous

In case anyone wants to help us ^

Anonymous

3:09 PM

@EmilioPisanty Thanks, that's an excellent example :)

@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

Anonymous

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

Anonymous

(which is obvious I guess, but still stating it)

3:17 PM

You're implicitly assuming that A is normal

hmmmm

i buy that

Anonymous

But since real matrices can have complex eigenvalues

Anonymous

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

Anonymous

@EmilioPisanty Ah, right

3:22 PM

@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.

Anonymous

Diagonalizability makes the condition stronger, interesting

Anonymous

@glS You mean diagonalizable over $\Bbb R$?

@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

@Blue you are right, we have to distinguish between real matrices and matrices which have real eigenvalues

Anonymous

3:34 PM

@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 $\{t: e^{iAt} \text{ is unitary}\}$

Anonymous

Umm, we haven't found the accumulation of those points either

Anonymous

Gotta think about it

A: Under what conditions is the exponential map on a Lie algebra injective?

there might be an answer to this question here

16

There 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

Anonymous

We better learn Lie algebra soon :P

Anonymous

3:39 PM

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

Anonymous

@BalarkaSen Good point!

I was badly mistaken up there

That’s skew-symmetric

Anonymous

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

Q: When is the exponential map between matrices injective?

I asked this on math.SE:

0

Consider 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 $2\pi i\mathbb Z$. ...

linear-algebra matrices exponential-function matrix-exponential