Sure.

Quantum Electronics physicist, nice to meet you!

I built a system to measure the quantum state of a superconducting circuit very fast.

Here's the arXiv paper.

thanks. let me check

it doesn't open :/ but thats so cool

Why doesn't it open?

and then it goes blank without that blue loader.

Your browser's pdf reader must be screwed up.

Here's the main page.

no I'm pretty sure pdfs open on my browser :) and nothing is wrong with my internet connection

Could be because one of the figures is really really big (in terms of data).

Group website in case you're interested.

wow, you guys are building a quantum computer :O

and partnered with google

thats great

?

Well... I'm actually not technically at UCSB any more.

I am a "Googler" as they call us.

So technically, I'm partnered with UCSB :-)

which google location do you work at ? SV?

Santa Barbara.

New office ;-)

wow :) u must be loving it

Yeah.

It's sweet.

Good weather, amazing research, awesome co-workers.

What's not to like?

Science is cool.

Do science.

;)

nowadays even people who love physics are skeptical about studying that because there are less jobs.

Hahahahahahaha

That's ridiculous.

Ok, well, wait a minute.

Are you talking specifically about theorists?

no generally going into studying physics.

Who told you that?

It's hard to get a job in academia, but getting a job with a physics PhD is not hard at all.

everyone, nowadays the younger generation avoids subjects like physics, math, and bio....

oh yes u have to go till a phD

They're morons then.

i dont understand how do people support themselves until a phD to get a job.

You get paid to be a PhD student.

Anyone who pays for their own science PhD is doing it wrong.

Only humanities programs make you pay.

(and even most of them give you a TA job or something)

oh i see...

I came out of my PhD with more money than when I went in, by a lot.

You have like zero expenses as a PhD student.

Just rent and food.

Maybe a car.

oh :)

so where did you make the money for the food and rent?

Food is turbo cheap if you can cook for yourself.

i see :) which university did you do your phD at?

Like I said, if you work in a science lab your adviser gives you money.

I did my PhD at University of California Santa Barbara (UCSB).

oh i see , thats a really good university btw....

Yeah, but any research university will pay you as a PhD student.

Don't do a PhD if you don't get financial support.

(At least, that's my opinion. Everyone's situation is different, of course)

oh i see :) thanks for telling me, i learnt something new today

Yeah, I hear that. I didn't know any of this as an undergraduate.

And nobody every really explains it to you.

Actually, yeah, that's kind of dumb. You should ask me and other people here lots of questions.

When you get to the university, frequently for the first year you work as a TA to get money for rent and food and whatever else.

i see. yes i will :)

Then often during the first summer you work in a lab (only talking about experimentalists now).

ohh :) thank you a lot for the info . really appreciate.

oh

i seee

That's when you get on the prof's grant.

also did you do masters or honours?

ah

This is a big deal because then you don't have to spend time teaching classes.

TA is a lot of work.

hmmmmm

What do you mean "masters or honours"?

Oh, wait a minute, are you in Europe?

like in for example australia or europe, after you do your bachelor of whatever. you either do masters or honours, and then you do your phD

I'm from australia

Ah, yes, the systems there are a bit different from the US system.

We don't really have the masters/honours thing.

Thats so unbelievably great :O

In the US, after your undergraduate degree you apply directly to the PhD program.

coz over here, you do 4 years undergrad, then 2 years master/ or 1 year honours....then u have to go on a phD

Right.

We don't really do that.

Both systems have a benefit.

thats so great! thats why i guess people over here avoid , since they have to stayy for 6 years and support themselves maybe

no i guess ur system is really good.

Ours is harder in some sense because you don't get the masters as an adjustment period.

but it takes less time, and thats very efficient and good

Six years is normal for an experimentalist to finish the PhD here.

No, not less time.

Guys

no i mean 6 years to even apply for a phD

@Jimmy360 Yes?

yes?

Have you ever noticed that

physics SE has fewer elections that most SE's

@TheArtist Dude, let's be specific. Here in the US you spend 4 years as an undergraduate, then around 6 years in PhD program.

Ten total years.

What about your system?

@Jimmy360 No. Does that matter?

Well, I just think that it's weird.

I like our mods though, and I dont want to change them or anything like that.

@Jimmy360 Meh.

I know almost nothing about SE politics.

@DanielSank oh , phD over here is like 3-4 years

@TheArtist See? Same total time.

so same total....but you dont fund your masters / honours .

yes yes :) so its just one advantage over there

The first ~3 years of PhD here are sort of like a masters, and you can get a masters degree at that time if you want, but for the most part it's really all just one PhD.

@TheArtist Oh, that part is not so good.

We have an Aussie in our group, actually.

Austin Fowler.

@DanielSank yep

Cool guy.

I've learned some Aussie slang.

oh haha :)

yes the slang ;)

It was surprising to me that "rooting" means something impolite in Australia, while here it means "cheering".

@DanielSank its like swearing :)

Yeah.

I just found that a funny overlap.

We'd say were are "rooting for a team" in the World Cup... Austin flinched when he heard that.

hahah

lol

just imagine someone coming unto you and saying in a serious tone " we are f**king for a team" in the world cup

:D

Yeah.

@DanielSank what is the first thing you tell when you meet someone in US ? like "hey whats up" ?

That's a common one, yes.

What about down under?

over here , almost everyone says " Hey, How's it going"

Yeah, that's common here too.

"What's up" and "How's it going" are probably the two most common.

Unless it's a good friend.

Then you can say "Hey, Jerkface".

oh i see :)

i thought its only over here that its mostly used

and how do you guys reply to "Hows it going" ?

@ChrisWhite Learn anything interesting yet?

@ACuriousMind Suppose we have an infinite sequence of sets of the form $A_0\supset A_1\supset \cdots\supset A_b\supset\cdots$. What can we say about the intersection $\cap_{b=0}^\infty A_b$? Apparently it's only guaranteed to be nonempty if the whole thing is a subset of a compact set $S$.

@0celo7 I don't think you can generically say something about the intersection (unless the space is compact, as you say, or, even stronger, Noetherian, for example).

@ChrisWhite Do you need an adapter? (I googled Chinese outlets.)

@ACuriousMind Intuitively, the intersection should just be the smallest set!

@ChrisWhite $\lim_{b\rightarrow\infty}A_b$

This proof in HE relies only on the fact that the intersection is nonempty.

I don't see how $A_0\subset S$, $S$ is compact, guarantees this.

@0celo7 Compactness is equivalent to "Every collection of subsets with the finite intersection property has non-empty intersection".

A chain of subsets trivially fulfills the finite intersection property, so the intersection over the whole chain is non-empty if the chain is inside a compact set.

Oh, you want the proof that compactness is equivalent to that?

I see how I might have missed the point^^

Yes pls :3

Hm...wait, are the $A_i$ closed?

@ChrisWhite When I lived in Germany, we had an incredible number of transformers and adapters.

@ACuriousMind Yes.

It's the set of limit points of something irrelevant here.

@0celo7 Good. Suppose the intersection is empty. Then, take the open cover $\{X - A_0, X - A_1, X - A_2,...\}$. By compactness, there is a finite subcover $\{X - A_{i_1},\dots,X-A_{i_n}\}$, which means $\bigcap_{j = 1}^n A_{i_j} = \emptyset$, which is a contradicition to the finite intersection property. Hence, the intersection of the collection is non-empty.

I see I can leave the other direction be :P

All I needed was something to search for.

Ah yes, when they don't drop the phrase "finite intersection property", I imagine it is difficult to find something on that

My initial searches landed me a bunch of angsty MSE posts from analysis students.

@ChrisWhite Not to split hairs, but adapters rarely have any electronics in them (sometimes they are fused), so it's more about the kind of equipment you use: Laptops and mobile phones have transformers in the chargers, and those are usually built to handle the electric supply of any country. If you buy electronics from China which plug directly into whatever the voltage is you get from the wall (say power tools, kitchen appliances the like), those might blow up back home.

Did you take the maglev train? If you do, make sure you take it during the fast hours (from what I remember, most of the day it runs a bit slower than the max speed).

@ACuriousMind GDI, why does a finite number of them have empty intersection? Are we allowed to choose the finite subcover so that their intersection is empty?

Or do we write $S=\cap_{i=1}^n (S-A_i)$ and deduce it from there?

@0celo7 If $\{X - A_i\}$ is a cover, then you must, for every point $x\in X$, find one $i$ such that $x \in X - A_i$, which implies $x \notin A_i$.

Hence there is no point that lies in every $A_i$, and so the intersection is empty.

I.e. by choosing the finite subcover you automatically get that the intersection is empty, no special choice required.

Ahhh

Drawing a picture helped.

@ACuriousMind In the second part of the proof, they say "If no finite subcollection sovers $S$". Doesn't there have to be a finite subcollection that covers $S$ due to compactness?

@0celo7 Don't we want to prove that $S$ is compact in the second part?

Oh

@ACuriousMind So...we assume that there is no finite subcover, and this leads to a contradiction. I don't see what part of the contradiction implies the theorem.

I'm also not sure why $\cap_\alpha(S-O_\alpha)\ne\emptyset$.

@0celo7 You assume there is no finite subcover, and you get a contradiction. Thus, there must be a finite subcover, and so $S$ is compact.

@0celo7 Because collections with finite non-empty intersections have non-empty intersection by assumption.

@ACuriousMind But how does the finite intersection property come into play?

@ACuriousMind Hmm

@0celo7 You use it to show that the original cover wasn't a cover to begin with (since you derive the non-empty infinite intersection from the non-empty finite intersections).

Their proof could be written clearer

Let $U_i$ be an open cover of $S$. Assume $U_i$ has no finite subcover. Then, the collection $\{S - U_i\}$ has non-empty finite intersections. By assumption, $\bigcap_i (S - U_i)$ is then also non-empty, implying $\bigcup_i U_i \neq S$, which contradicts $U_i$ being a cover. Thus, $U_i$ must have a finite subcover, and $S$ is compact.

This should be clearer, hopefully.

@ACuriousMind Yes, thank you.

Saying "by assumption" next to assumptions is nice.

Great. Now HE is constructing a new topology using intersections of chronological sets.

Too much topology.

Other things to do, cya!

@0celo7: I have just proven the formula $\chi_{S^2 V}(g) = 0.5(\chi_{V}(g)^2 + \chi(V)(g^2))$. Does that remind you of something?

Something something string theory something something $\mathrm{SO}(n)$?

Maybe?

Oh, $\chi$ is the character!

Did you figure out the trace formula!?

@0celo7 Yes, my group theory homework has been to prove the above and $\chi_{\Lambda^2 V}(g) = 0.5(\chi_V(g)^2 - \chi_V(g^2))$.

So now we need only use that the adjoint of $\mathrm{SU}(2)$ is the...symmetric part of the product of the fundamental with itself, I think, and we have what we wanted to prove, I think.

Could also be the antisymmetric part, but anyway, the trace formula itself does, surprisingly, not really rely on the properties of $\mathrm{SO}(N)$ or $\mathrm{SU}(N)$.

Care to post your derivation?

@0celo7 Well, it's just pushing a few indices...see if you can decipher the second half of page 5 of this.

Y u no latex problem sets

@0celo7 Coz not allowed.

I'm not sure what $S^2V$ or that other thing is.

@0celo7 $S^2 V$ and $\Lambda^2 V$ are the (anti-)symmetric tensors in $V\otimes V$.

Gotcha

Where $V$ carries the representation $\rho$.

In the case of our trace formula, $V$ would've been the fundamental.

Lol, Americans can't into cursive

You mean you can't read my handwriting?^^

What is the double g thing

...what double $g$? There's a bunch of $\rho(g)$

My $\rho$ is a bit...curvy.

Wat

That's a g my man

$g_\otimes (g)$

What's the otimes for?

I denote the rep induced on $V\otimes V$ as $\rho_\otimes$.

But it's a g...

:P

Lol...no, it's not, can't you see that the gs look different?

Ahhh

That's a really high res scan

Did you write this digitally?

Not a scan, I write that on a tablet.

@ACuriousMind Why S^2\Lambda in the trace?

On mobile btw

@0celo7 Because I'm tracing the matrix I got by applying $\rho_\otimes$ to the basis of $S^2\Lambda$ - that is the definition of the trace on $S^2\Lambda$.

What I thought you were doing S^2V

Oh, damn

That's a typo

Everything after $S^2$ should be a $V$.

That was surprisingly simple.

@0celo7 My thoughts, precisely.

You should write that as an answer on that question

@ACuriousMind Did you learn something in that class to help you with that or did it just come to you?

@0celo7 Well, the task was just "Prove that this formula holds", so, for lack of any other approach, I just wrote down the naive basis for the symmetric part and tried to trace over that. And, to my own surprise, it worked.

@0celo7 I think I will. This is so much simpler than what Lubos does.

@ACuriousMind Simpler? Yours actually makes sense!

@0celo7 Thanks ;) Here's the answer, inform me of typos/errors.

Parenthesis in the first equation using sigma

And more parenthesis in the sun in the last set of equations

+1

The proof of Prop 6.4.9 is two and a half pages :/

I guess one a a half of you take away the fun pictures

Is there a way for us to make it more obvious that posts here should contain one question?

Multiple questions don't bother me as long as they are related.

I was discussing the weak energy condition of GR with my supervisor a while ago (energy densities must be measured non-negative by any observer). I remember him telling me that pointwise violations of the WEC aren't a concern as it is not "physical" to measure an energy density at a point. What is meant by "not physical"? Is it that a density is an average over a volume of spacetime, thus a pointwise value bares no real meaning or is it that a pointwise measurement is not practically achievable?

Could someone take a look at my question:

?

@Rammus I think it is the latter - you can't measure things at a "point", every experiment will measure things within an interval. We measure energies, not densities, and the energy is the integral of the density, but a point is a zero measure set, so the integral of the density over a point, or a set of discrete points, is always zero - it carries no physical meaning at all.

@ACuriousMind Thank you for the explanation, very much appreciated.

@DanielSank thx for sharing! wondering, did the google acquisition affect salaries at all over there? also there are rumors martinis wants to get into adiabatic QM computing somehow but it seems like the lab doesnt have any papers on that...(so far)?

It's been too long y'all

@Rammus I think something along those lines, yeah. But then you'd have to show something like $\forall \epsilon >0 $ the density averaged over the $\epsilon$-cube is larger than zero, to make it be okay

@Danu Yeah, I was looking at violations of the strong energy condition for the real scalar field so I was specifically looking for negative values when averaged over world tubes/lower bounds for this. Similar to what Hawking and Ellis did in The large scale structure of spacetime pp. 95-96

@ACuriousMind Does "neighborhood" always imply open?

@0celo7 Depends on who's writing, but more often than not, yes.

And here I am wasting time not reading.

@Icosahedron I have a shitload of homework to do, but I want to finish one last proof.

I don't know what homework is.

@ACuriousMind Ok, then check out Prop 6.5.1 on p. 202. They say "neighborhood" in the second sentence, but it clearly seems wrong. For instance, take $q$ to be in $\bar{\mathscr{S}}$. Then an open nbd of $q$ will intersect the set.

@0celo7 In most textbooks, yeah

(I hate mathematicians being lazy with stuff like that!!)

Whoops. It says $\mathscr{S}$ is closed on the previous page.

Reading.

Also, I used to get so confused about people not properly using the signs for subsets and proper subsets!!!

Say I have a point in a set. Furthermore, suppose I may always find an open nbd of said point. Does that imply that the whole set is open?

No

Unless it's an arbitrary point

(and the open nbd is understood to be contained in the set)

For every point in the set.

@0celo7 You mean you can find an open neighbourhood inside the set

@ACuriousMind Yes.

Then yes, since the set is then the union of the open sets around the points, and the union of open sets is open.

For every, and the nbh's contained then yeah

@ACuriousMind Got it.

Y U IGNORE :'(

@Danu Got it.

@Danu Slow internet. ;)

That doesn't make sense.

Your face doesn't make sense

@Icosahedron You doesn't make sense!

:D

Dammit.

@0celo7 Rudeboi!

@Danu HE never does this or use a separate equality for definitions. The latter leaves me wondering if there's something obvious I missed or they're defining things with desirable properties.

@0celo7 Almost nobody does the subset thing (even I've stopped doing it by now lol)

@Danu Too lazy to type or too lazy to figure out when each is appropriate?

@0celo7 The former

The latter is usual trivial :P

(which is why people get lazy about it in the first place)

My initial confusion about it was mostly because I was no good at math (still am not)

Time for bed now :(

Bye guys

Tomorrow hanging out with that one girl again :D

Night.

How is Qmechanic the first one in the chat avatar thingie yet totally faded out?

@0celo7 : Beats me.

@Qmechanic Robot confirmed.

What did you study? Physics? @DanielSank

@MaryStar In undergraduate I majored in physics.

I was one course short of a math minor.

and I took a lot of language courses (Spanish, modern Greek, and Russian).

That's a funny story.

@MaryStar The day I graduated grad school, Google announced that they were starting a collaboration with the lab in which I did my grad school work.

I got a job doing basically the same thing I did as a grad student :-)

My official title is something like "Quantum Electronics Engineer".

Whatever. It just means that I work on superconducting qubits.

Great!! That sounds interesting!! :-) @DanielSank

@MaryStar It is.

I really like my job a lot.

It's one of my favorite things to do

That is good!! @DanielSank

@MaryStar How about you?

I am a graduate student of "Mathematics in Computer Science"... @DanielSank

@MaryStar Cool!

:-)