The h Bar

7:00 PM

@EmilioPisanty My current reading list is operator algebras, nonlinear regularity theory, abstract harmonic analysis. I don't exactly do non-dry math :P

Emilio Pisanty

@0celoñe7 exactly

life's too short for numerical analysis

0celoñe7

Wonder what's up with that

should be nice and round

dmckee

@SirCumference Short answer: yes.

Slightly longer answer: it depends a bit on your field on interested running from 'hard as hell' up to 'well nigh impossible'.

Sir Cumference

;-;

dmckee

Physical sciences tend toward the happier half of that range, but astrophysics tends toward the harder half of the physical sciences.

Especially if you are interested in pure theory.

Sir Cumference

7:03 PM

...i am...

0celoñe7

@dmckee That's assuming he can even get into a good PhD program

dmckee

Basically we train 7-25 PhDs for every tenure track slot (including the ones like I have).

Sir Cumference

@0celoñe7 So far I'm at a uni with great research opportunities and I'm managing in terms of grades. I'm a little confident in terms of grad schools

dmckee

@0celoñe7 You can bootstrap a solid but unremarkable PhD program to a good postdoc to a R1 if you are "all that and a bag of chips", but it is a harder track than the one faced by students ho get into a good PhD program in the first place.

Anonymous

Are there any jobs outside academia for cosmologists or astrophysicists ?

dmckee

7:06 PM

@Blue There are lots of jobs to which you can take the skills you developed in your PhD, but damn few industry job listing that say they are looking for an astrophysicist.

Sir Cumference

@Blue Depends on whether your asking about jobs using astronomy or jobs in astronomy

Emilio Pisanty

@dmckee that makes it sounds like those latter ones even exist

dmckee

I'm actually looking at a move to industry right now.

Anonymous

@SirCumference Yup, that. I guess it's a very niche area which industries don't need at all :/

dmckee

Things I'm looking at include medical radiation physics, remote instrumentation, big data, cyber security in SCADA systems in particular and so on.

Sir Cumference

7:07 PM

It's the difference between working for something like Google Earth, and researching why things in the universe work the way they do

Emilio Pisanty

@dmckee ~~how did you find that opportunity?~~

dmckee

As well as pure programming jobs, lab manager and lab saftey positions.

@EmilioPisanty I've seen one.

Just one, mind you.

Emilio Pisanty

@dmckee seriously?

0celoñe7

oh lord

dmckee

@EmilioPisanty At a private R&D firm that works with NASA. And they mean a interplantary plasma physicist.

Sir Cumference

7:08 PM

@dmckee Well, I'm wondering how difficult it is to become an associate prof

As in, is it common to be fired?

0celoñe7

@SirCumference Why are you doing this right now?

Considering switching to a better field?

Premed?

Emilio Pisanty

@dmckee to plasma-proof spacecraft or some such?

dmckee

That depends a lot on where you get on the tenure track in the first place. Places like mine it is basically enough to do the teaching job reliable and on time.

Sir Cumference

@0celoñe7 I'm in college, this is a very good time to reflect on where I'm going and whether I should pursue it

dmckee

Top tier places are a minefield more stressful than grad school.

@EmilioPisanty I think it was about comms in the van Allen belts around the giants.

Sir Cumference

7:10 PM

@dmckee Dear lord, and I thought grad schools were hard as hell

0celoñe7

@dmckee Allegedly some of the senior profs here sleep in their offices and always work weekends

They're writing grants for 30 people groups

and...stuff

Emilio Pisanty

@dmckee ah, yes, that sounds like a tough nut to crack

0celoñe7

Removed, why?

Sir Cumference

@dmckee Do profs often get housing at the uni?

0celoñe7

@EmilioPisanty cf. above. Does that look like a step size error?

dmckee

7:11 PM

@SirCumference Not really a thing in the US. A few of the oldest east coast school have some.

Sir Cumference

As in, do they save money on paying for a home?

dmckee

It is more common for the uni to subsidize housing loans in very high cost markets.

Emilio Pisanty

@0celoñe7 no idea what I'm meant to be looking at

Sir Cumference

Huh. I'm wondering because profs at places like Columbia still manage to afford living in Manhattan

0celoñe7

@EmilioPisanty The graph. Smooth solution is the analytical one.

dmckee

7:12 PM

@SirCumference I know a woman who had one of those jobs. First the salaries are pretty high and second the uni helps some.

Emilio Pisanty

@0celoñe7 and it does look like you're asking it to take that hard shoulder a bit faster than you oughter

dmckee

She doesn't have that job any more because she was poached by MIT. Not sure how she affords a place in Cambridge.

0celoñe7

Wonder what kind of math program MIT has. I've never heard of someone in my field from MIT.

Sir Cumference

That's gotta be very lucrative for somewhere like manhattan. Though it is probably equally competitive

The Dark Side

@dmckee Does "not tenured" mean "adjunct"?

dmckee

7:15 PM

@SirCumference Well, Janet and her Husband (who works as NMSU in Las Cruces) have a big house near Fermilab and they each keep a little place near their work.

0celoñe7

3 houses?

dmckee

@TheDarkSide No. "Assistant" AKA on the tenure track, but not yet achieved tenure.

@0celoñe7 Two incomes, no children.

0celoñe7

@dmckee Children must be terrible.

Sir Cumference

Hmm...I'm an undergrad at an R1 uni, yet we only have a handful of cosmologists. This is pretty terrifying

0celoñe7

What American says uni?

Anonymous

7:17 PM

@0celoñe7 One senior from my locality took up a course called "Mathematics and Computing" at MIT. Probably he is in the 3rd or 4th year now.

Sir Cumference

@0celoñe7 I'm on a phone and typing is a pain :P

0celoñe7

@Blue I don't mean the undergrad program.

dmckee

@SirCumference KSU has three particle physicists and one cosmologist.

0celoñe7

I mean the overall research program.

The Dark Side

@dmckee Is that time bound, or some other criterion?

Anonymous

7:17 PM

@0celoñe7 Oh, then I don't know

dmckee

Alabama had four particle types and zero cosmologists. But three astronomers.

0celoñe7

UTK has a billion particle physicists, and basically no one else.

dmckee

@TheDarkSide Tenure review is usually in the fifth or sixth year. Then you are either up or out. That's the scary part.

Emilio Pisanty

@dmckee dinc!

0celoñe7

@EmilioPisanty ?

Sir Cumference

7:19 PM

@dmckee I've heard. Having to repeat up to 6 years somewhere else and wait again for tenure sounds horrifying

Emilio Pisanty

@0celoñe7 dinc

0celoñe7

@EmilioPisanty aha

dmckee

"Dual income, no childred" is a demographic description that advertisers use.

Sir Cumference

It sounds like it could take forever to get tenure

dmckee

Janet and Vassili fit it exactly.

He's the kind of wine snob who will spend ten minutes consulting with a sommelier come back and suggest that you order steak or lobster.

Emilio Pisanty

7:20 PM

@dmckee is Vassili short for vacilón?

dmckee

She's about as jet-set as any physicist you've ever met.

@EmilioPisanty It's a Greek name. Reasonably common, I think.

Emilio Pisanty

Orquesta Aragon de Cuba - Que rico bacilon

►

so no relation, then?

dmckee

Doubt it.

Sir Cumference

@dmckee All things aside, is it worth it to be in a field you love?

Emilio Pisanty

OK, I'll take my tongue out of my cheek

dmckee

7:22 PM

@SirCumference I enjoy it. But the finances don't work out very well at the level I've ended up. Thus my desire to move to industry.

0celoñe7

@EmilioPisanty How am I supposed to know what a reasonable step size is?

Emilio Pisanty

oh, while this is still there, CC @DanielSank on the video above

@0celoñe7 you're not

0celoñe7

:(

Emilio Pisanty

find the observable you care about, and keep taking smaller step sizes until it doesn't change appreciably

0celoñe7

I guess that's as good of advice as any

Emilio Pisanty

7:26 PM

and then you say that your numerics are converged

(in the computational-physics sense)

just keep an eye out for rounding error

if you make your step size too small, then the error will creep back up

or, alternatively, go to the dry theorems and ask for formal error estimates =|

0celoñe7

@EmilioPisanty I feel like if I started learning another branch of analysis I'd go insane

user228700

@Blue @Abcd: I completely agree.

0celoñe7

I would need specific literature

Emilio Pisanty

@0celo7 amazon.com/Numerical-Recipes-3rd-Scientific-Computing/dp/…

user228700

@vzn: Yep, I'm taking CSE! I hope to pursue cognitive neuroscience after graduating. Also, hello! :-)

0celoñe7

7:29 PM

@EmilioPisanty Wait, what?

ιllιlılMostafaıllııllıl

get rekt

►

0celoñe7

@EmilioPisanty 1200 pages!!!!!!

@ιllιlılMostafaıllııllıl You watch h3?

Emilio Pisanty

@0celoñe7 it's the reference. You won't need all of those pages at once.

0celoñe7

@BernardoMeurer

Emilio Pisanty

seriously, if you want one book, that's the one

@0celoñe7 remember, you're not doing exact arithmetic with real numbers

Numerical differentiation

In numerical analysis, numerical differentiation describes algorithms for estimating the derivative of a mathematical function or function subroutine using values of the function and perhaps other knowledge about the function. == Finite difference formulas == The simplest method is to use finite difference approximations. A simple two-point estimation is to compute the slope of a nearby secant line through the points (x,f(x)) and (x+h,f(x+h)). Choosing a small number h, h represents a small change in x, and it can be either positive or negative. The slope of this line is ...

0celoñe7

7:32 PM

@EmilioPisanty Yeah, but I don't see how rounding matters more for really small steps.

Emilio Pisanty

that's the classic example

0celoñe7

Seems like it would be the other way around!

Emilio Pisanty

well, it depends on what you're doing

for numerical differentiation it's easy to see

0celoñe7

@EmilioPisanty This graph is confusing me.

Emilio Pisanty

you're doing $$f'(x) \approx \frac{f(x+\Delta x)-f(x)}{\Delta x}$$

0celoñe7

7:33 PM

Yes.

Emilio Pisanty

now, if $\Delta x$ is too close to the round-off error, then $(x+\Delta x)-x$ might be different to $\Delta x$

as far as the machine goes

ιllιlılMostafaıllııllıl

@0celoñe7 no. I don't know who is h3. just found this one

0celoñe7

@ιllιlılMostafaıllııllıl That is h3

The best stuff out there

dmckee

@0celoñe7 SHort version is the rounding error is fixed by your FP format, but small steps mean you accumulate many more of them.

0celoñe7

@dmckee Hmm, ok.

@EmilioPisanty I'm using a step size of 1e-3 right now.

Seems to be working much better.

Lines up exactly with the exact solution.

Emilio Pisanty

7:37 PM

@0celoñe7 so, machine error being at ~1e-15?

that'll depend on your system/language/configuration

this is all on double precision?

0celoñe7

hiding monkey emoji

I don't know

🙈

Emilio Pisanty

@0celoñe7 yeah, that's a hidingmonkeyemoji kinda thing to not know

what language is this on?

0celoñe7

@EmilioPisanty Matlab

Emilio Pisanty

es.mathworks.com/help/matlab/matlab_prog/…

> MATLAB® represents floating-point numbers in either double-precision or single-precision format. The default is double precision, but you can make any number single precision with a simple conversion function.

Double-precision floating-point format

Double-precision floating-point format is a computer number format that occupies 8 bytes (64 bits) in computer memory and represents a wide, dynamic range of values by using a floating point. Double-precision floating-point format usually refers to binary64, as specified by the IEEE 754 standard, not to the 64-bit decimal format decimal64. In older computers, different floating-point formats of 8 bytes were used, e.g., GW-BASIC's double-precision data type was the 64-bit MBF floating-point format. == IEEE 754 double-precision binary floating-point format: binary64 == Double-precision bina...

so typically 1e-15 of machine round-off error

0celoñe7

@EmilioPisanty So I should go to 1e-14 assuming my CPU doesn't explode?

Emilio Pisanty

7:45 PM

@0celoñe7 that'll make your computer explode

and also, refer back to that graph above

0celoñe7

Ah, yes.

So 1e-12?

Emilio Pisanty

that $h=10^{-16}$ on the left is the machine error

0celoñe7

-11?

Emilio Pisanty

that depends on how your problem scales

what exactly are you doing?

and how does computing time scale with step size?

0celoñe7

Uh, Matlab's ode45 function

@EmilioPisanty I don't know.

Emilio Pisanty

7:48 PM

and also, note (i) that the graph above is specific to numerical differentiation, and (ii) that the message is that the optimal step size depends on the problem and on the input

@0celoñe7 find out ;-)

one of the key things to keep in mind

Secret

Why Computers are Bad at Algebra | Infinite Series

►

actually, more like introduction to floating point arithmetic

dmckee

8:05 PM

@EmilioPisanty It might help if the vertical axis of the figure was labeled.

The horizontal axis is labeled $h$, which if I recall correctly is the step size used for computing the difference, but ... it's not clear to be what the vertical scale is.

0celoñe7

@EmilioPisanty how is this

Emilio Pisanty

@dmckee not my figure; source

@0celoñe7 typesetting-wise? generally OK

I'm not at all a fan of underlines, but to each their own

0celoñe7

@EmilioPisanty the underlines are a bit strange

Emilio Pisanty

@0celoñe7 I'll do your typesetting if you'll write the email I'm struggling with

0celoñe7

@EmilioPisanty I'm scared.

heather

8:12 PM

what are some good resources for learning about phase matching in a bbo crystal in the context of spdc?

Bernardo Meurer

@heather What on earth are you doing?

heather

working on an experiment - basically, to entangle photons, change the phase of one, and show that both are still coherent.

Emilio Pisanty

@heather ah, good question

0celoñe7

@BernardoMeurer fortran errors!

smooth bruckner broke again

heather

@EmilioPisanty the one thing I've found that's at all useful is this

but i don't understand it very well.

0celoñe7

8:20 PM

@BernardoMeurer I am reinstalling python

Emilio Pisanty

@heather as I understand it the standard reference for nonlinear optics is Boyd

but it's probably too technical and it might not be useful enough for you to warrant the investment of time

rp-photonics.com/phase_matching.html

^ good starting point

also similar articles at the same site

and their bibliographies

0celoñe7

@EmilioPisanty Dang, Matlab is limited to 1 core

This is going to take FOREVER

heather

@EmilioPisanty okay, thank you.

0celoñe7

8:36 PM

@EmilioPisanty Still going

This might have been a bad idea

ιllιlılMostafaıllııllıl

@0celoñe7 mathworks.com/help/gads/parallel-computing.html

0celoñe7

8:57 PM

@ACuriousMind Help, I have a quantum mechanics question

dmckee

@0celoñe7 The answer is in the state $(|\text{yes} \rangle - |\text{no} \rangle)/\sqrt{2}$.

0celoñe7

@dmckee It's not a yes/no question

ιllιlılMostafaıllııllıl

@0celoñe7 you can ask

dmckee

@0celoñe7 Spoilsport.

0celoñe7

@ιllιlılMostafaıllııllıl If the magnetic field $B$ has finite energy over $\Bbb R^3$ and satisfies the Maxwell equations distributionally, why can we choose the vector potential in Coulomb gauge to be in $L^6(\Bbb R^3)$?

ιllιlılMostafaıllııllıl

9:15 PM

*Maxwell's
The question is wrong

@0celoñe7 ok; what does it mean for the vector potential to be in $L^6$?

If you tell me this, I'll be able to answer your question

0celoñe7

@ιllιlılMostafaıllııllıl measurable and $\int_{\Bbb R^3}|A|^6\, dx<\infty$

@ιllιlılMostafaıllııllıl so u gonna answer?

ιllιlılMostafaıllııllıl

@0celoñe7 I understand the question now but can't (immediately) see the answer.

It's a complicated inequality involving some integrals of vector fields

also, I don't understand how is this a quantum mechanics question

0celoñe7

10:41 PM

@ιllιlılMostafaıllııllıl maybe it's not

@BernardoMeurer Michele is visiting while I'm home

heather

11:10 PM

hello @Mithrandir24601

Mithrandir24601

@heather Rytsas! PhD topic...

is going to be... ::Searches for exact title::

Simulating Models on the Boundary of Quantum Physics

[note: this isn't yet set in stone, and the final thesis title in 3 years will most likely be different]

Anonymous

@Mithrandir24601 Is that what you're working on?

heather

so - sorry, but what exactly does that mean?

Mithrandir24601

@Blue It's most likely what I'm going to be working, although I've got a backup choice of randomised benchmarking/t-designs

heather

what sorts of models?

Mithrandir24601

11:14 PM

So, all throughout life, you're told that Hamiltonians have to be hermitian

heather

hamiltonians -> equation of system, i know that. what does it mean for something that's not a matrix to be hermitian?

Mithrandir24601

you can write a Hamiltonian as a matrix in a given basis - it simply means that $H = H^{\dagger}$

So, why is this?

heather

well...reversibility, right?

you have to be able to "undo" an action

and see the previous state of the system before the action.

Mithrandir24601

OK - why?

heather

well, that was sort of the big thing of classical mechanics.

determinism, being able to figure out the state of the system from any point.

Mithrandir24601

11:19 PM

yeah, but quantum $\neq$ classical

heather

right...

hmm.

maybe because if you are unsure of the previous state, something bad happens with the equations?

but then, why is measurement okay?

well, okay, let me try something.

Mithrandir24601

There are actually a couple of answers to my first question - in this case, the reply to your answer is that the 'fundamental' principle isn't time (T) symmetry

0celoñe7

@heather Physicist abuse of terminology.

heather

@0celoñe7 heh, physicists seem to do that a lot from the point of view of mathematicians

Mithrandir24601

@0celoñe7 ::cough:: <_< >_> <_<

0celoñe7

11:21 PM

Hermitian does refer to matrices exclusively, but physicists will confuse it for "symmetric" in the sense of operators.

It means that $\langle y,Ax\rangle = \overline{\langle x,Ay\rangle}$.

heather

what does that bar thing on the top mean?

0celoñe7

Complex conjugation.

heather

and do the brackets have any special significance?

0celoñe7

@heather They are the inner product of the (pre-)Hilbert space.

heather

inner = dot product, right?

0celoñe7

11:24 PM

@heather A dot product is a special case of an inner product.

For a finite dimensional space, all inner products are dot products in some basis.

@Mithrandir24601 Do you need a lozenge?

heather

@0celoñe7 okay, thank you. =)

Mithrandir24601

@0celoñe7 I wouldn't mind a mint to be honest :P

heather

@Mithrandir24601 well, what is the fundamental principle?

Mithrandir24601

@heather In this case, I'm referring to CPT symmetry - charge, parity, time

0celoñe7

@heather Nature is clearly not invariant under C, P, or T. Somehow it is invariant under CPT.

Mithrandir24601

11:27 PM

so a system doesn't have to be T-symmetric, but CPT symmetric

0celoñe7

Very strange result.

heather

that sounds really whacky. how can something be CPT without being T?

0celoñe7

@heather The C, P, and T operators do not act "mutually exclusively" on everything.

Some correct things that were changed by others.

It's the combination that leaves the universe invariant.

Roughly, at least :P

heather

huh. so sometimes they kind of cancel each other out (crudely speaking)?

0celoñe7

Yes.

Mithrandir24601

11:28 PM

@heather It was apparently first discovered in some particle physics experiment

heather

wow, so is the measurement operator one of those things - CPT but not T?

okay, then a different question:

how is it that so many things are T and CPT?

under what circumstances do operators 'cancel'?

Mithrandir24601

(In slightly more mathematical terms, it's actually to do with the eigenvalues of the Hamiltonian, which would have been a response to a different possible answer to my first question)

Not necessarily the projection operator

(I don't really know if it is or not, I don't know if applying the projection operator is the same as making a measurement. It's a bit confusing :P )

heather

this sounds like something i need to read about!

Mithrandir24601

But anyway, typically, most people only consider time-symmetric Hamiltonians (with real, positive eigenvalues)

heather

so, i'm messing around with the matrix you gave, and i have a question: if you multiply that matrix by another matrix, does that represent measuring in a different basis?

@Mithrandir24601 so are you researching the ones that aren't time-symmetric?

Mithrandir24601

11:33 PM

@heather Yes!

heather

did you write about it for any previous papers? or are there any good books/papers you'd recommend to read about it?

(sorry i'm asking so many questions)

Mithrandir24601

The reason that I said it was related to the projection matrix was because you mentioned that it has to be unitary. Saying that a gate is unitary is the same as saying that the Hamiltonian is hermitian i.e. is T symmetric

So, while this is true that an entire circuit (excluding measurement stuff) is unitary, there are ways of looking at the circuit to get that part of the circuit isn't unitary (this is a terrible explanation :/ )

@heather arxiv.org/abs/quant-ph/0501052

heather

well, okay, one moment. checking if a gate is unitary is a little different - $U^\dagger$, that finds the hermitian, but then you multiply it by U and check if the result is equal to I. How are they the same?

Mithrandir24601

@heather Ah, so if you've got a Hamiltonian $H$, how do you time-evolve it?

0celoñe7

You don't evolve the Hamiltonian...

heather

11:37 PM

^

shouldn't the hamiltonian take in inputs? it describes the system at a given time

(right?)

Mithrandir24601

@0celoñe7 OK - fair enough - bad phrasing, sorry. If you've got a system with Hamiltonian $H$, how do you time-evolve the system? :P

(in my defence, it's late and I should really be in bed)