The h Bar

DanielSank

12:00 AM

@JaimeGallego The answer is simple: physics.

Why? Because if you pick physics you get to do engineering and math anyway.

Jaime Gallego

@DanielSank Yes, the "first principles" physics approach to engineering is quite attractive.

DanielSank

@JaimeGallego Well also note that once you get to real life physics, at least as an experimentalist, you do plenty engineering anyway.

For the last couple weeks I'm working on a circuit board that processes RF signals.

It's all circuit design.

Plenty engineering.

The cool thing is that those RF signals carry the information about the quantum state of a superconducting qubit. So there you go: physics lets you do quantum mechanics and engineering at the same time!

'sup @rob?

rob

@DanielSank Howdy

Emilio Pisanty

> Cosmic Bell Test: Measurement Settings from Milky Way Stars. Johannes Handsteiner et al., Phys. Rev. Lett. 118, 060401 (2017)

> Bell’s theorem states that some predictions of quantum mechanics cannot be reproduced by a local-realist theory. That conflict is expressed by Bell’s inequality, which is usually derived under the assumption that there are no statistical correlations between the choices of measurement settings and anything else that can causally affect the measurement outcomes.

> In previous experiments, this “freedom of choice” was addressed by ensuring that selection of measurement settings via conventional “quantum random number generators” was spacelike separated from the entangled particle creation. This, however, left open the possibility that an unknown cause affected both the setting choices and measurement outcomes as recently as mere microseconds before each experimental trial.

DanielSank

Oh goodness, this again.

Emilio Pisanty

12:14 AM

> Here we report on a new experimental test of Bell’s inequality that, for the first time, uses distant astronomical sources as “cosmic setting generators.” In our tests with polarization-entangled photons, measurement settings were chosen using real-time observations of Milky Way stars while simultaneously ensuring locality.

> Assuming fair sampling for all detected photons, and that each stellar photon’s color was set at emission, we observe statistically significant $\gtrsim7.31\sigma$ and $\gtrsim11.93\sigma$ violations of Bell’s inequality with estimated $p$ values of $\lesssim 1.8\times10^{-13}$ and $\lesssim4.0\times10^{-33}$, respectively, thereby pushing back by ∼600 years the most recent time by which any local-realist influences could have engineered the observed Bell violation.

↑ now that is one cheeky abstract =P

@DanielSank yeah, I imagined you'd have that kinda reaction to this one

I found the ~600 years thing a nice touch though

DanielSank

@EmilioPisanty I may just not appreciate the importance.

Emilio Pisanty

@DanielSank no, I kinda side with you tbh

DanielSank

I think Bell violation experiments are important! I just don't necessarily find it interesting to think up loopholes and then debunk them.

I dunno, maybe I'm missing something.

Emilio Pisanty

I mean, it's cool and all, but to a large extent this is Zeilinger bragging about the fancy stuff his group can do

but on the other hand, superdeterminism is way weirder than QM, I think

DanielSank

It just so happens that Marissa Giustina is in my office right now.

2

Emilio Pisanty

12:19 AM

I think this sort of work just tries to make that sword sharper

DanielSank

She was Zeilinger's student. She did their recent loophole-free Bell violation.

I will ask her to educate me about these matters.

Emilio Pisanty

@DanielSank any wisdom you glean, pass it on ;-)

DanielSank

nature.com/nature/journal/v497/n7448/full/nature12012.html

I think that's her big paper.

Oh wait no that's not it.

Emilio Pisanty

heck, ask her if she wands to do an AMA here =P

DanielSank

journals.aps.org/prl/abstract/10.1103/PhysRevLett.115.250401

Emilio Pisanty

12:21 AM

you probably mean journals.aps.org/prl/abstract/10.1103/PhysRevLett.115.250401

DanielSank

I think that's it.

Yeah.

@EmilioPisanty I will.

loltospoon

Why are eigenvalue equations so counter-intuitive? At least for me, a beginner. $ab=cb$, but $a \neq c$. This absolutely blows my mind!

DanielSank

@loltospoon Huh?

I promise you they're not counterintuitive ;-)

Emilio Pisanty

@loltospoon It's just saying that two completely different geometrical transformations can still agree on certain directions.

loltospoon

$a$ is the operator, $b$ is the eigenfunction, $c$ is the eigenvalue.

Emilio Pisanty

12:27 AM

@loltospoon yes. they're wildly different types of object, and your notation obscures those distinctions.

loltospoon

When I was told to think of $a$ as the operator, it began to make a little more sense.

DanielSank

Oh you mean like $\hat{T}|v\rangle = t|v \rangle$?

Emilio Pisanty

you can rephrase it as $f(x)=g(x)$ if you want

loltospoon

@DanielSank yep. In the QM context.

DanielSank

@loltospoon You don't "think of $a$ as an operator"... $a$ is an operator!

@loltospoon There is absolutely nothing quantum about that equation.

Emilio Pisanty

12:28 AM

$f:\mathbb C^n\to\mathbb C^n$ is an arbitrary linear function, $g(x)=\lambda x$.

DanielSank

Bra-ket notation is for vectors and linear transformations. You don't need to be talking about quantum mechanics to talk about vectors and linear transformations.

loltospoon

But let's look at $5x$ where $5$ is the operator acting on some variable $x$. How can you tell me that the result is $cx$ where $c \neq5$???

Emilio Pisanty

@loltospoon ...so?

DanielSank

@loltospoon woah woah woah

Emilio Pisanty

matrices and numbers are different

DanielSank

12:29 AM

Think of this differently.

Define a function $\text{TIMES_FIVE}$ which is defined by the equation $\text{TIMES_FIVE}(x)=5x$.

Now it's obvious that $\text{TIMES_FIVE}$ and 5 are not the same thing.

loltospoon

@DanielSank but then do I still think of $TIMES_ FIVE(x)$ as an operator or no?

Can functions be operators?

DanielSank

Functions are operators, yes. They're not necessarily linear operators though.

Use \text{...}.

loltospoon

So the for eigenvalue equation using your function it would be $\text{TIMES_FIVE(x)}x = 5x$ ?

DanielSank

@loltospoon Why are there two x's on the left?

Danu

@DanielSank Functions are operators? Wu-what?

DanielSank

12:38 AM

@Danu I probably should have said operators are functions.

Because I realize now that I have no idea what mathematicians think "operator" means.

Danu

Functions on what? :P

DanielSank

I'm not having this discussion.

Danu

For me, a function means a map $X\to \Bbb R$ or $X\to \Bbb C$

Operator typically means "thing that maps from one space into itself"

loltospoon

Yea I'm only a beginner physicist, so deeper mathematical definitions will go over my head anyway

DanielSank

For me, a function means a map $\text{WHATEVER} \rightarrow \text{WHATEVER ELSE}$.

@Danu o

Danu

12:39 AM

@DanielSank Ah. That's what I call a "map".

loltospoon

^ that's the physicists way

Danu

I also used to do physics

loltospoon

to @DanielSank 's comment

Danu

But okay

skill patrol

Nice work in there Danu :-)

Danu

12:40 AM

Glad you approve, skill :P

skill patrol

:P

Danu

@DanielSank Slightly more precisely, an operator is typically an endomorphism too. So that's a map $X\to X$ that preserves the relevant structure on $X$

DanielSank

Whatever. The point is that you can always think of an operator as a function/map/I don't care what word you use.

@Danu Ok.

Danu

For instance, an operator on a vector space is typically taken to be a linear map $V\to V$.

DanielSank

Nobody ever said than in my however many years in physics education ;)

Danu

12:41 AM

@DanielSank That's right :)

I know; I like establishing good terminology for mathematical concepts ;)

I think I had a discussion with ACM about what the fuck "functional" is supposed to mean, once upon a time

Now that's a lame word :P

DanielSank

It's a great word!

Functor and functional are awesome.

Danu

I think my conclusion was that it's just a function, but on a space of functions.

Functor is great.

DanielSank

Finite Simple Group (of Order Two)

►

Danu

Functional, not so much :P

obligatory, haha

WITHOUT LOSS OF GENERALITY

That line is a little... sexually aggressive :P

DanielSank

It is?

Danu

12:44 AM

Not really

But when taken out of context it could be misconstrued.

"without loss of generality, I will assume you feel the same way"

rob

@DanielSank Now, that's an AMA that I'd be interested in.

Danu

^

Actual physicist AMA is the best AMA

DanielSank

@rob Ok, I've asked her if she's interested. We'll see. She gets a lot of requests for speaking etc.

...and a lot of crank email, so I'm given to understand.

Pichi Wuana

Is there a reason of why world lines are inverse to classic mechanics when talking about graphs? I learned in high school that the x-axis is time and the y-axis is position.

rob

@DanielSank I can't even begin to imagine.

Danu

12:49 AM

God, it must be terrible, publishing on QM.

One more advantage of doing really mathy stuff... No one will bother you :p

Bell's inequality papers probably maximize your crank attraction.

DanielSank

@Danu I dunno, travelling for free sounds fun.

Danu

Right, the official speaking invitations are nice, I guess.

skill patrol

@PichiWuana that's a convenient way to display what is going on.

rob

@PichiWuana I'm familiar with both conventions in different contexts.

Pichi Wuana

In other words, there is not a real reason?

skill patrol

12:52 AM

Yup.

Context is king!

Pichi Wuana

Oh

It's a bit confusing because now lower slope is higher speed

skill patrol

You'll get used to it :-)

Clear thinking is the important thing.

Pichi Wuana

I hope. I'm learning world lines / theory of relativity alone through Coursera, while at the same time my works are the inverse way

skill patrol

Make things easier for yourself and follow their convention.

Danu

It might be a good exercise to transform the different types of pictures into each other a few times.

Pichi Wuana

12:59 AM

Which convention are you talking about?

skill patrol

The one in Coursera.

Pichi Wuana

@Danu Good idea

@skillpatrol Ooh yeah

rob

1:17 AM

@PichiWuana I wonder if this tradition dates back to early thought experiments in relativity about simultaneous things happening in different parts of trains?

You would draw a train on your paper, labeled "now"

another train above or below it, labeled "later" or "earlier" depending on what you're doing

then when you start measuring coordinates for events, you find that you've got a vertical time axis just because it was perpendicular to your horizontal train.

Pichi Wuana

Never heard about that

skullpetrol

Yup, that's how Einstein originally did it.

rob

@PichiWuana books.google.com/…

Pichi Wuana

Wow

rob

@PichiWuana A great book to read, if you're teaching yourself relativity.

Pichi Wuana

1:22 AM

I thought that I would read it after the course so everything is clear when I read it

Jaime Gallego

Gotta go, buh bye

rob

@PichiWuana It's an explaining book. The author was something of an authority on the original development of relativity.

skullpetrol

It was originally written in German and loses a lot in the English translation.

rob

@skullpetrol I'm prepared to believe that ... but I got a lot out of it as an undergrad.

Pichi Wuana

Well I saw a lot of math when I saw it in iBooks

Seemed to be things that would be hard to understand

skullpetrol

1:27 AM

He presents special relativity and general relativity in such a concise manner in that book.

rob

@skullpetrol Are you a German speaker?

Pichi Wuana

Oh

skullpetrol

Nope @rob

rob

@skullpetrol Just curious whether your opinion of Einstein-as-translated was based on personal experience. I think languages are neat, but I don't plan on getting to German for a while yet.

skullpetrol

1:40 AM

Yes, on personal experience.

rob

@skullpetrol You thought the original in German was better than the English translation, but you don't know German? That's confusing to me.

loltospoon

Could someone help me understand what this phrase means in this context?

imgur.com/a/u8lus

it's from Griffiths, Introduction to Quantum Mechanics, 2nd ed., pg. 307

skullpetrol

@rob I meant the English sentences are missing something throughout.

loltospoon

He writes the the final value for the Hamiltonian is in units of $-E_1$, but I don't get what that means

skullpetrol

@rob and I attribute that to translation.

Sorry for the confusion.

rob

1:51 AM

@skullpetrol I suppose that's possible. I often find translated text to be more spare, in a sense, than text written originally in English, but a lot of the time the effect is charming. I don't remember the language in that book being especially remarkable either way.

But I think I read it in 1998, and perhaps again in 2006, so it's been a while.