The h Bar

David Z

00:15

@BernardoMeurer cc @JaimeGallego it looks fine...

Was there an old one? I actually don't remember it

Jaime Gallego

10

A: Community Promotion Ads - 2016

David Z

Ah, gotcha, from last year's ad thread.

I do like that your new one is cleaner than the old one.

Though there does seem to be quite a bit of whitespace between the different pieces of the image

Jaime Gallego

I need to add a border though

vzn

@JaimeGallego re nighthawks, was wondering, why an "incredible" painting?

Jaime Gallego

00:31

@vzn I couldn't explain. It's probably the colors, the simplicity, the atmosphere.

And painted in 1942.

It was very innovative for its age.

vzn

@BernardoMeurer looks great thx! looking fwd to it... ps didnt know youre a P vs NP fan, have an entire section on it... also recommend fortnows recent book :) vzn1.wordpress.com/category/p-vs-np ... your toaster cpu reminds me of a hackerspace prj, an IoT mousetrap that sends a msg when it catches something :)

@JaimeGallego so have you studied some art history? was enjoying some art myself yesterday at fair...

user228700

05:03

Hi, everyone :-)

John Rennie

Morning

user74961284659036576321890754

Hello.

4 hours later...

pZombie

Water isn't compressible, right? So if I was to put an open barrel of water at room temperature inside a tube with a little less pressure than the atmosphere has, would the water remain at the same height inside the barrel?

John Rennie

05:20

Water is compressible, just not very compressible like a gas.

pZombie

So the water would rise slightly?

John Rennie

We measure the compressibility of a material by its bulk modulus. A totally incompressible material would have a bulk modulus of infinity.

Water has a high bulk modulus but not infinite.

pZombie

My next question would have been, if it takes energy to get a barrel of water in and out of a pipe which has slightly less pressure

John Rennie

@pZombie The water would rise slightly, but by an amount so small the rise would be hard to detect.

pZombie

yes, but even if it rises slightly, it would seem to mean that I would have to put energy into getting it in and out of the pipe with slightly less pressure

John Rennie

05:24

Yes, but in practice the difference would be undetectably small

pZombie

well, I was thinking about putting some water at room temperature at about 25°C, inside a pipe with low pressure where water boils at about 20°C.

Connect the steam at the top to some other curled up pipes which are led through the water itself, and open valves which push several different sized magnetic balls through those pipes.

the energy for pushing those magnetic balls around some copper coils and generating energy, would have to be subtracted from the water. Hence the water would cool down

When it cools back down to initial levels, I was thinking to pull out the water and replace it with new water at 25°C

Since if this would work, I assume it would already have been done. So I am trying to figure out where my thinking went wrong

basically a pipe with low pressure that has a valve of shutter at the bottom. You place a barrel below it, then open the shutter, connecting the barrel to the "low pressure chamber"... shut the valve again once the water inside the barrel has cooled down, and pull the barrel out... repeat

Since this probably does not work as I imagine it, there has to be some error in how imagine the dynamics of this system considering the energy would be.

Slereah

07:45

"The signature is taken to be +---"

>:|

Balarka Sen

angry reacts only

Slereah

Hey @BalarkaSen

Balarka Sen

yo

Slereah

I'm trying to learn about string theory, but then they start throwing around the orthosymplectic group

So I have to learn more SUSY

pZombie

did string theory predict the outcome of any experiments more accurately than current other theories?

Slereah

07:50

Not as far as I know

Balarka Sen

sounds like crazy shit

pZombie

are there any predictions string theory makes which diverge from predictions of current theories and which could eventually be tested in experiments?

Slereah

I am learning about it, so I can't tell

@ACuriousMind might know

Though of course, there's the whole compactification thing

So we would probably need to narrow down the compactification before we can really make good predictions

Slereah

08:12

Is there a supersymmetric principal bundle?

Some kind of superframe bundle

Danu

08:43

Probably

@Slereah Hurr durr transport coefficients from AdS/CFT haha

Slereah

I'm sure that's a hilarious joke to someone

Danu

there is some way of calculating some weird statistical physics quantity

using AdS/CFT

and it gives some outcome which is better than the standard model

Slereah

neat

Danu

but it's hardly a direct prediction of string theory

The Dark Side

Hello $\hbar$!

Does anyone around know how to insert footnotes in answers?

Unless I'm missing something, the formatting page doesn't mention that !+?

John Rennie

08:56

There is no proper footnote facility - you have to cheat

The Dark Side

How sir?

John Rennie

I generally manually insert the footnote number using ${}^n$, then at the bottom of the answer insert a horizontal line, and below that the footnotes.

The Dark Side

Hahaha...

Slereah

or just put it in parenthesis you walnut

The Dark Side

:: Sigh ::

Slereah

08:59

I guess the super principal bundle would have the superspace bundle as an associated bundle :p

The Dark Side

^ I assume that's a response to Danu, and not to this walnut ...

John Rennie

@Slereah Walnut? Is that the insult du jour

The Dark Side

OK Thanks guys.

Slereah

09:15

What is $\mathcal N$ in SUSY anyway

Is it the grading of the algebra or something

Bernardo Meurer

Morning folks

Mithrandir24601

@BernardoMeurer Morning :)

John Rennie

Morning :-)

The Dark Side

Morning for you :P

:)

Bernardo Meurer

What are y'all up to in this rainy morning?

Slereah

09:25

Looking at supergravity

Mithrandir24601

@BernardoMeurer Currently making porridge. May or may not get round to doing some revision for my (last ever!) exam on Wednesday. Also may or may not get round to doing all the other stuff (essay, project, a bit of writing, another bit of writing etc.) I have to do, as it's a Bank Holiday :)

The Dark Side

@BernardoMeurer Gazing at Physics Meta in amazement at a person who ran Linux on a toaster ...

I hope by a "toaster" you don't mean a laptop which generates a lot of heat :P

Mithrandir24601

@TheDarkSide He didn't really run Linux on a toaster - he just plugged a raspberry pi in, which I consider cheating :P

Bernardo Meurer

@TheDarkSide Glad you liked it, lol

@Mithrandir24601 To be entirely fair it wasn't even a raspberry pi, it was just some ARM SoC-based board like an RPi

The Dark Side

@Mithrandir24601 Hehehe.

Bernardo Meurer

09:29

And the nice thing was that the toaster was controlled by the board!

I had my toasting managed by a cron job!

Mithrandir24601

@BernardoMeurer Hmm... That's not so badly cheating then, I suppose... :P

Bernardo Meurer

@Mithrandir24601 I dream about doing my last exam ever every night

Actually that's not true, I dream that I solved P vs. NP more

Mithrandir24601

@BernardoMeurer There will come a time in your life when it actually happens. And it's pretty exciting after 9 years of some form of exams :D

Bernardo Meurer

I bloody hate exams

I hate them so much I broke my professor's exam format last semester

Mithrandir24601

So do I :/ I'm terrible at them. Just cannot do them in time. At all

@BernardoMeurer What did you do?..

Bernardo Meurer

09:35

Our grade was divided in a lab part, which was some really freaking hard VHDL projects and an exams part

I scored a 19/20 on the lab

and a 10/20 on the exam

so the professor was SURE that I was somehow cheating

and they all went nuts

but it was just me being bad at exams :P

Mithrandir24601

09:47

Yeah, in my (undergrad) finals, the examiner made a remark that my computational physics exercise mark was high, but most of my exam marks were mediocre, with a couple of acceptable marks... I then did an MSc and my marks exploded upwards :P

Slereah

10:04

I am looking at naked things

"A spacetime is nakedly singular if there is a point $p \in M$ and a future-incomplete timelike or null geodesic $\gamma$ such that $\gamma \subset I^-(p)$"

Ger Wyn

Hello. Do anyone here know the best video channel to learn Chemistry for grade XI and XII?

SBM

Chemistry SE has a resources section maybe @GerWyn

Bernardo Meurer

10:19

Wohoo GCC 7 is out!

SBM

5

Q: Resources for learning Chemistry

Based on various other Stack exchange site (Mandarin Chinese, Russian and German), we adapt this project here for chemistry, since it's a great idea to have all kinds of resources in one place. This is a specifically created Community Wiki which gathers resources for learning Chemistry. The list...

GCC is 30 years old

now

Though installing GCC 7 on cygwin is awful

Bernardo Meurer

Cygwin is black magic

Just use the Linux subsystem

Or just run Linux

pZombie

you mean the new linux subsystem they added in win10 when you enable developer mode?

SBM

I wish I had windows 10

Bernardo Meurer

@pZombie Yes.

pZombie

10:32

I am not sure if you could do anything with the WLS you can do with cygwin

Bernardo Meurer

@pZombie You can do literally everything you do in cygwin with the Linux Subsystem

It's actually pretty nice

The only bad thing is it runs on Windows

pZombie

then again, I don't really know exactly what cygwin or mingw does other than me having used them at times trying to compile some programs from source to work under windows

Bernardo Meurer

cygwin is a POSIX compatibility layer for the WIN32API, that's all I know

SBM

I'm not sure I could dual boot or run VMs

Bernardo Meurer

Why couldn't you dualboot?

pZombie

10:35

for simple stuff, why would you want to dual boot if you can just run virtualbox?

or even simpler stuff that requires no desktop , just use the linux subsystem

SBM

Dual boot'd slow the system down.
WLS is available only on Windows 10

Bernardo Meurer

@SBM Dual boot has no impact on your system speed

@pZombie Because you don't want to run Microsoft malware :)

SBM

oh okay will need to borrow a distro

Bernardo Meurer

borrow?

SBM

Distros are big downloads. I have a 2gb data limit every month

Slereah

10:54

A Riemannian metric tensor defined a Riemannian metric, but is the converse true?

Secret

I think the Cygnus A supermassive black hole binary will be a good place to check for gravitational waves for the next step

John Rennie

11:07

@Secret Did you mean Sagittarius A* or Cygnus X-1 ?

Oh, you mean the galaxy Cygnus A?

Why would the galaxy emit gravitational waves?

Slereah

Uh-oh

Paper I want is in Доклады Академии Наук СССР

Secret

11:21

@JohnRennie public.nrao.edu/news/vla-reveals-new-object

John Rennie

@Secret the newly discovered object, whatever it is, is 1500 light years away from the galaxy core. We're going to be waiting a long time for that to merge with the central black hole.

Slereah

Do we have any soviets on the chat

Rokhlin's theorem

In 4-dimensional topology, a branch of mathematics, Rokhlin's theorem states that if a smooth, compact 4-manifold M has a spin structure (or, equivalently, the second Stiefel–Whitney class w2(M) vanishes), then the signature of its intersection form, a quadratic form on the second cohomology group H2(M), is divisible by 16. The theorem is named for Vladimir Rokhlin, who proved it in 1952. == Examples == The intersection form on M Q M : H 2 ...

nice

Secret

@JohnRennie ah yes, i must have misread some numbers and get a small separation

Secret

12:04

arxiv.org/abs/1705.04620

But, suppose this experiment really get the result as it intended, what will become of relativity. The moment you can show that human minds can violate the violation of bell test, then causality becomes a really complicated business where you have to take account of all the minds scattered through the globe for each decision?

Sha Vuklia

12:25

Guys, I don't understand why they plotted $\phi(k)$ and therefore concluded that the spread in momentum is large? I know that $p=\hbar k$, but I don't see how $p$ and $\phi(k)$ are related. In the case of the plot of $\Psi(x,0)$, I can see that if we were to calculate $\sigma_x$, we would get a small value, because the spread is small. However, if we were to calculate $\sqrt{\langle\Delta p\rangle}$, I don't see how we can use $\phi(k)$ directly?

Oh, maybe I have an idea: a wave packet is a superposition of waves whose amplitude depend of $\phi$. Hence, if the spread in $\phi$ is small, we have a wave packet that consists of a small range of waves, whose momentum doesn't differ a lot. So the overal spread in momentum is much smaller.

rob

12:42

@ShaVuklia Isn't the momentum proportional to the wavevector, $p = \hbar k$?

Sha Vuklia

@rob yes it is

but that alone doesn't explain why we plot $\phi(k)$, does it?

rob

@ShaVuklia So the second plot shows that all $k$ are equally likely

Sha Vuklia

oh, so $\phi(k)$ or $\vert\phi^2(k)\vert$ is sort of a probability density?

rob

Exactly: $|\phi(k)|^2$ is a probability density in momentum space, just like $|\phi(x)|^2$ is a probability density in position space.

Sha Vuklia

ah I did not know, thanks @rob

rob

12:50

@ShaVuklia Glad to help

Secret

Suppose I give the hamitonian of a molecule to a physicist to solve (and suppose for the sake of argument, that the solution (numerical or not) is actually found), what physical quantities will he/she/it interested in in the solution of that hamiltonian?

Jaime Gallego

13:15

G' afternoon

Secret

14:31

SLOW MOTION: AGT 2017 Close-Up Magic Act Works With Cards and Coins

►

Bernardo Meurer

14:43

@vzn Remember to share the AMA post with the CS people

Slereah

"A supermanifold $X$ of dimension $p|q$ is a ringed space $(|X|,O_X)$"

Save me

Secret

reddit.com/r/Magic/comments/6dh9o0/…

Bernardo Meurer

15:11

@JohnRennie Man, this laptop with 16GB of ram is a beast

John Rennie

@BernardoMeurer Cool :-)

I've been using my new (old) Precision m6700 in anger for the first time now I'm at my Mum's and I'm very pleased with it as well.

It's the same CPU as your Latitude and also 16GB RAM. Proprietary malware though :-)

Bernardo Meurer

@JohnRennie The only problem with 17" laptops is they're a bitch to carry around

John Rennie

@BernardoMeurer Damn yes! This thing is built like a tank!

I had to buy a backpack to carry the thing in :-)

Bernardo Meurer

To be fair, my E6530 is also a damn Panzer. It's literally indestructible

Mithrandir24601

@BernardoMeurer It's worth it though :)

Sha Vuklia

15:17

Guys, if we have a symmetric potential about $x=0$, then why is $\langle x\rangle=\langle p\rangle=0$? I would think we should be looking at how $\psi$ looks instead of $V$?

Bernardo Meurer

@Mithrandir24601 Meh, my 15" screen is good enough. I do wish I had another monitor though

John Rennie

@ShaVuklia the symmetry means $\psi$ has to be either symmetric or antisymmetric. Either way the expectation value is going to come out zero.

Sha Vuklia

@John is it because the higher the potential, the less likely we will find our particle there?

Mithrandir24601

@BernardoMeurer I don't have space for another monitor :/ Nor would I be able to carry it anywhere very easily :P

John Rennie

@ShaVuklia Suppose $\psi$ is symmetric, then $x\psi$ is antisymmetric so $\psi^*\,x\,\psi$ is antisymmetric.

And when we integrate an antisymmetric function we get zero because the values either side of the origin cancel out.

Bernardo Meurer

15:21

@Mithrandir24601 Well, I mean mostly for home use. I don't spend enough time outside my house to need a second screen on-the-go

Sha Vuklia

I understand what happens after we've concluded that $\psi$ is symmetric @John but I don't see why it had to be symmetric in the first place. is it because the higher the potential, the less likely we can find the particle?

Mithrandir24601

@BernardoMeurer Fair enough - I don't use it much outside/bring it with me anywhere that much either, but then moving house every year means that I'd rather have the bigger laptop screen than a second monitor

John Rennie

@ShaVuklia If the potential is symmetric we're saying $x$ and $-x$ are physically indistinguishable.

And the symmetry of the wavefunction has to reflect this.

Sha Vuklia

oh right of course

that makes sense

thanks! @john

John Doe

@ShaVuklia You are using 'john' with reckless abandonment...hence you have summoned another John...

@Mithrandir24601 Yo how's it going?

John Rennie

15:25

@BernardoMeurer the Precision has the same 1080p resolution, but for my (aging!) eyes 1080p on a 15" screen is just a bit too small. The 17" screen is that bit easier to read.

Sha Vuklia

oh haha oops :P @JohnDoe sorry

Mithrandir24601

@JohnDoe Pretty much same as yesterday - a bit of revision for Wednesday and a bit of tidying/cleaning to do...

Bernardo Meurer

@JohnRennie I've been dying to see a 4K screen. ONE DAY

John Doe

@Mithrandir24601 For probability density function $\rho$ in QM are $|\alpha^{(i)} \rangle$ the eigenstates of $\rho := \sum_{i} w_i | \alpha^{(i)} \rangle \langle \alpha^{(i)}|$ which diagonalize it?

@ShaVuklia No prob :)

John Rennie

@BernardoMeurer The main reason I wanted the Precision is the keyboard is so nice. The gen 3 i7 Latitudes and Precisions use a proper keyboard not a chiclet keyboard, and it's much nicer for typing on.

@BernardoMeurer the ultra high res screens use a higher dpi so you don't get any more info on the screen, it just looks nicer. Personally I don't see that as a huge advantage.

Bernardo Meurer

15:28

@JohnRennie My favorite keyboard ever is the Thinkpad one, GOD that keyboard is nice

The old ones, not the new ones

Mithrandir24601

@JohnDoe They might be - you can write/define it so that they are the eigenstates, but not necessarily (I'm not sure if that's clear or not...)

Bernardo Meurer

@JohnRennie Sure, you don't get more real-estate on the screen BUT I can watch my movies better

John Rennie

Meh. Watching films on a laptop isn't that great anyway compared to a big TV with the sound going through your hifi.

John Doe

@Mithrandir24601 Okay thanks

Mithrandir24601

@BernardoMeurer :D :P

Bernardo Meurer

15:36

@JohnRennie I do not own a television :P

I do not own anything with a screen apart from my laptop and my phone

John Rennie

@BernardoMeurer I guess it would be silly to buy a TV if you're going to be leaving Portugal at the end of the year. The cost of shipping it would be ridiculous.

Bernardo Meurer

@JohnRennie Yep, I also do not have space in my room for a TV :P

vzn

16:04

1

Q: CS angle AMA "ask me anything" jun13 Bernardo Meurer

earlier this year Michael Wehar proposed a CS AMA "ask me anything" session that had strong support but which has not gone further/ advanced so far beyond conceptual stage. this is to announce not exactly that event but one related. on jun13th Bernardo Meurer will host the Physics AMA. he has str...

discussion chat

Fawad

@DanielSank you didn't add upcoming event of bernanado in this physics.meta.stackexchange.com/questions/7783/…

vzn

@BernardoMeurer just saw a demo clip of grand turismo ps4 4k at best buy, think it was. breathtaking! whats this about leaving portugal? only for the summer?

Secret

16:37

> Even if this experiment does not lead to a violation of quantum theory
it
shows how we can talk scientifically about mind-matter dualism. There
ex-
ists a class of scientifically testable theories invoking duality that are
open to
falsification.

Slereah

@ACuriousMind I need a superman

A man who knows about supersymmetry

Jaime Gallego

ACuriousMind

@Slereah What ails you?

Jaime Gallego

He needs to have God killed

ACuriousMind

At least it's a Nietzsche and not a Hitler reference. Our niveau is improving :P

Slereah

16:45

@ACuriousMind 1) does the $\mathcal N$ in SUSY refers to the dimension of the Grassmann numbers $\Bbb R_ m$ in $\Bbb R^n_c \times \Bbb R_a^m$

2) is the tangent plane of the supermanifold supposed to be some kind of spinor + vector thing?

3) what is the canonical map between spin manifolds and supermanifodls

Bernardo Meurer

@ACuriousMind You better have liked my AMA meta post thrice

Slereah

4) also what's the link between $OSp$ and the super Lorentz group

$OSp$ rotates the (super)frames which is weird because that's usually the role of the Lorentz group

ACuriousMind

@Slereah 1. Yes, it denotes their representation as spinors, and therefore also their dimensionality. Note that you can have things like $\mathcal{N} = (2,0)$ and $\mathcal{N}=(1,1)$ in dimensions where you have Weyl spinors, which would have the same number of supercharges but a different structure. 2. Yeah. 3. No idea what that is. 4. It appears...somehow, but I'm not really familiar with it.

Slereah

So I'm guessing that $\mathcal N = 0$ is just ordinary GR

Well ordinary field theory, even

What is the structure of the tangent space, exactly?

ACuriousMind

@Slereah Yes, $\mathcal{N}=0$ would be a strange way of saying "no superthings here" :P

Slereah

16:49

I'm guessing that it is as well $\Bbb R^n_c \times \Bbb R^{\mathcal N}_a$

But how do you extract spinors from that?

Also what changes does the value of $\mathcal N$ do here?

I know that there is some mysterious link between grassman numbers and spinors, but I'm never too sure what that is

I guess it relates to Clifford algebras

ACuriousMind

$\mathcal{N}=1$ is not the "dimension" of the Graßmann part.

$\mathcal{N}=1$ means "minimal amount of supersymmetry", which means your supercharges assemble into the lowest-dimensional possible spinor, often a Majorana spinor.

Slereah

Is there a link between the two?

ACuriousMind

Sure - but it depends on the dimension of the bosonic manifold.

Slereah

Is "the bosonic manifold" just a fancy term for "the spacetime"

ACuriousMind

Yes

Slereah

16:56

Makes sense I guess

Would that be 2D for $\mathcal N = 1$ in 4D?

Since majorana spinors are 2D vectors

ACuriousMind

yep

Slereah

Alright

How does the tangent space "split", exactly?

You've got a vector in $\Bbb R^n_c \times \Bbb R^2_a$

It represents both a vector and spinor, if I understand

Is it just that the first half is the vector component and the second half the spinor component?

ACuriousMind

I'm not sure I understand the question - the tangent space is spanned by derivations as susual

Slereah

But a SUPERVECTOR FIELD represents both a spinor and a vector field, right?

ACuriousMind

uhhhh

I think a "supervector field" is just a field $f^i(x,\theta)\partial_i$ where $\partial_i$ runs over both bosonic and fermionic indices.

And yeah, I guess the fermionic part still transforms as a spinor.

But generally, I'd avoid wading into the murky waters of supergeometry unless you really need to :P

Slereah

17:08

I guess I should find this fabled canonical correspondance between spin manifolds and supermanifolds

It might help

Also does that mean that for very high $\mathcal N$ the correspondance is not the same?

Like you might get vector fields associated with Rarita fields

ACuriousMind

I have no idea which correspondence you mean

Slereah

You know

$Q \Phi = \Psi$

or somesuch

ACuriousMind

I know not what that has to do with spin manifolds

Are you trying to ask what the supermultiplets are? :P

Slereah

Could be!

Also according to some paper, there is a correspondance between spin manifolds and supermanifolds, which I guess is maybe something like a mapping between supervector fields and vector fields + spinor fields, mb?

ACuriousMind

So, please state a bit more clearly what you want to know and maybe I can tell you

@Slereah Yeah, I've never heard of such a correspondence. Doesn't mean it isn't there, but it's at least not "well-known" in the SUGRA papers I'm reading

Slereah

17:21

I'm guessing that most people don't even use supermanifolds

I haven't seen a lot of them before looking for them specifically

ACuriousMind

You can get by without using them.

Slereah

found the theorem

"For any $p \in M_0$, there is a canonical isomorphism of $\mathbb Z_2$-graded vector spaces $\iota_p : T_p M_0 + S^*_p \xrightarrow{\sim} T_p M$"

With $M_0$ a spin manifold and $M$ a supermanifold

ACuriousMind

Ah, yeah, that's what I said above - the supervector fields are spanned by $\partial_i$ for both bosonic and fermionic coordinates, and the bosonic part is of course $T_p M_0$, and the fermionic part transforms pretty much by definition in the spinor representation.

Slereah

And $S$ the spinor bundle

ACuriousMind

Note that that is not a correspondence "between spin manifolds and supermanifolds". To me it just says that the tangent space of a supermanifold is the direct sum of the bosonic and fermionic pieces.

(Under "correspondence" I initially thought you were asking for some sort of bijection between spin manifolds and supermanifolds)

Slereah

17:34

Well I'm guessing no for that

You still need some "structure" for the supersymmetry

Some supergauge bundle, I dunno

I just wanted to read about string theory ;w;

But there isn't a lot of non SUSY string theory left

So i must learn the ways of SUSY

I'm looking in DeWitt's book for the exact meaning of $\mathcal N$ and in a mocking twist there is literally no N in the index

He's got a whole chapter on groups, though

let's give it a look

dmckee

17:56

@TheDarkSide I use html superscript mark up (<sup>1</sup>) to indicate the footnotes and set the notes off from the text with a horizontal rule (---). Numbering and ordering are manual.

Other people do different things.

I see that John R. uses MathJax to get the superscript, but I got started on Stack Overflow where there is no MathJax, so I got into the habit of plain html.

Note that there are several html tags that are accepted and processed that do not have markdown equivalents. You can also use strikeouts, for instance (but not in chat, sigh).

Slereah

18:19

@ACuriousMind can the supermanifold be non-trivial?

Like what are the restrictions on the topology of the supermanifold

user74961284659036576321890754

18:42

Sounds like a poker game with "superman."

;-)

Slereah

Basically everything about supersymmetry is usual physics with "super" in front

super vector spaces, super vector fields, supermanifolds, super lie algebras

ACuriousMind

@Slereah Look up Batchelor's theorem

If you were asking if the fermionic part can be "non-trivial", then no, not really, it's always sort of a vector bundle over the bosonic part.

The Dark Side

@dmckee Thanks (for method number 2). :)

Also, on an unrelated note:

(Sorry for the spam comment, but really ...) Haha hahaha ... — The Dark Side 10 hours ago

^ To whomsoever it may concern,

I'm deleting this comment now.

Apparently, 0celot got banned and I wasn't aware of the entire development ...

user74961284659036576321890754

18:57

yup

The Dark Side

So, inadvertently, this comment may appear crude/rude/insensitive to people.

Apologies.

Slereah

@ACuriousMind makes sense

would be weird otherwise

I guess that's partly why you don't really need supermanifolds

user74961284659036576321890754

@TheDarkSide accepted

The Dark Side

?

Wat?

user74961284659036576321890754

the apology is accepted

The Dark Side

19:01

Are you 0celot in disguise?

user74961284659036576321890754

no

The Dark Side

OK.

John Doe

@ACuriousMind With regard to density operators: If we express our states by say $$| \psi_m \rangle = \sum_{j}c_{jm}|j \rangle$$ where $\{ j \}_j$ are our basis states, if we have a completely mixed state then as I understand this implies that each $| \psi_m \rangle$ is equally likely, but am I correct in stating that this does not necessarily imply that each $|j \rangle$ is equally likely?

My reasoning is that we have two types of probability at work here, the quantum-mechanical probability coefficients $c_{jm}$ and the probability $w_m$ from the probabillity density function $\rho = \sum_{m}w_m |\psi_m \rangle \langle \psi_m|$, hence for a completely mixed state we have that $w_m$ is equal for all $m$ but this does not imply that $c_{jm}$ is equal for all $m$. What do you think?

Danu

19:29

o/

ACuriousMind

19:51

@JohnDoe No, you are not correct. A completely mixed state is simply $\frac{1}{N}\mathbf{1}_N$, where $N$ is the dimension of your space. A multiple of the identity is a multiple of the identity in every basis, so indeed saying that all $\psi_m$ are equally likely is equivalent to saying all $j$ are equally likely (if the $\psi_m$ form a basis).

John Doe

20:08

@ACuriousMind
Oh yeah I see. Would this be an example $\rho = (1/2)|+ \rangle \langle + | + (1/2)|- \rangle \langle - |$?

ACuriousMind

Yes, that is a completely mixed state.

John Doe

@ACuriousMind Okay thanks. One thing, would you agree that since the density operator is hermitian that it can be written as a linear combintation of eigenstates and eigenvalues $$\rho = \sum_j \lambda_j | \lambda_j \rangle \langle \lambda_j |$$ where $\lambda_j$ are the eigenvalues. Hence this again gives the probability $\lambda_j$ of finding the system $| \psi \rangle$ in state $|\lambda_j \rangle$?

ACuriousMind

@JohnDoe Yes. As I said, I would pretty much define a "density operator" to be of that form, with that specific interpretation of the $\lambda_j$.

What is your definition of "density operator" that this is not immediate?

Are you just defining it by a list of its properties, like unit trace, Hermiticity, non-negative eigenvalues?

John Doe

@ACuriousMind My definition is that $\rho := \sum_{i}w_{i} | \alpha^{(i)} \rangle \langle \alpha^{(i)}|$. But $w_i$ are not necessarily eigenvalues of $\rho$ or that $| \alpha^{(i)} \rangle$ are eigenstates of $\rho$.

ACuriousMind

@JohnDoe ...that they are follows immediately from that definition if the $\lvert \alpha^{(i)}\rangle$ are orthogonal to each other.

John Doe

20:22

@ACuriousMind Yeah I know but why would that be guarenteed. In fact in Sakurai it states that "need not be orthogonal". This is the definition I am using. From Sakurai.

@ACuriousMind He doesn't mention writing it in the eigenvalue eigenstate form at all.

@ACuriousMind He mentions the eigenvalues and eigenstates but the equation of the density operator in the form as you mentioned never appears.

ACuriousMind

Wait a moment, what are the $w_i$ in your definition, if not probabilities?

@JohnDoe Which form? I'm still a little bit confused what exactly your issue is.

John Doe

@ACuriousMind They are probabilities but not eigenvalues of $\rho$.

ACuriousMind

Yeah, sure, if your states are not orthogonal they won't be

John Doe

@ACuriousMind Yes I understand. I have no issue right at the moment :) Are you going offline now?

ACuriousMind

@JohnDoe I'll be around for a while

no_choice99

20:32

hey guys a quick question basically

John Doe

@ACuriousMind So you can start with a state $\rho = \sum w_i | \psi^{(i)} \rangle \langle \psi^{(i)}|$ and then rewrite $\rho$ in an eigenstate eigenvalue form $\rho = \sum \lambda_j | \lambda_j \rangle \langle \lambda_j |$...and then you know the probabilities of the system for the eigenstates $| \lambda_j \rangle$ as well as $| \psi^{(i)} \rangle$. That seems like you are getting something for nothing...

no_choice99

about the heat eq. it's linear right? its operators are linear. so a sum of 2 solutions should be a solution to the eq. right? Well I'm unable to show this holds for the function T(x,t) = A exp(-x/d1) *cos(omega1 *t -x/d1 -phi1) + B exp(-x/d2) *cos(omega2 *t -x/d2 -phi2) while each term do satisfy kappa d^2T/dt^2 = dT/dt

i even used Maxima CAS to check up my math and it tells me this sum doesn't satisfy the heat eq while each term does

so i really dont know what's going on

ACuriousMind

@JohnDoe Why would you get "something for nothing"? Both sets of probabilities encode the same statistical information - the same density matrix.

John Doe

@ACuriousMind I was discovering the magic of physics there for a second, and you are saying it's an illusion...But yeah I see what you are saying.

ACuriousMind

@no_choice99 Without you being more explicit about what the problem is, I'd wager you're making some sort of computational mistake. The heat equation is indeed homogeneous and linear when written as $D f = 0$ for $D = \partial_t - \kappa\partial_x^2$. The principle of superposition can't fail.

no_choice99

20:46

thanks @ACuriousMind that's reassuring. I really tried with both hands and Maxima

John Doe

@ACuriousMind Could you maybe give me a hint of how to show that $\langle \psi| \rho | \psi \rangle = Tr[| \psi \rangle \langle \psi| \rho]$...

no_choice99

each term indeed do satisfy the heat eq. but not their sum... i will retry but i have little hope

@ACuriousMind the problem doesnt matter much. i've basically seen that function in a book to describe the temperature of something, and i had a doubt on whether it would satisfy the heat eq. so i tried to show it by hand, failed. tried with maxima, failed again

it turns out it was because i couldn't factor /cancel some contants because omega2 and omega1 differ and other things like that

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