The h Bar

Emilio Pisanty

00:04

@dmckee that's the funniest bone name I've ever seen =P

but that sounds extremely painful

Dunno if shock-suppressed pain is the best descriptor for that one, though. It's just that I'm relatively used to being bashed about (many years of martial arts), and this one was quite similar to other instances of fall-induced shock to the thorax, including inducing some heavy coughing. However, persistent sharp pain after the fall is another story.

Bernardo Meurer

What on earth are y'all talking about

@dmckee Look how fucking awesome salmanarif.bitbucket.io/visual

Emilio Pisanty

00:28

hey SE, where's my badge?

I want my badge

The Dark Side

02:48

Epic @EmilioPisanty :

@Strawberry "The laser broke because of a power spike and half our mice died and our algae had an unexpected and unrelated bleaching event, so we couldn't obtain results to any level of statistical significance above that previously reported in the literature to either confirm or reject any hypothesis" is a mode of failure that is distinct from a null result and it's still not publishable. Academia still needs a kick up the bum to publish negative results, but that's not synonymous with failed experiments. — E.P. yesterday

Nailed it.

peterh

02:59

@EmilioPisanty How can you see 18 edits? The edit approval queue wasn't ever so long on this site.

@EmilioPisanty :-) :-) :-)

@EmilioPisanty Now I can see your badge on your user page. You got it 34 mins ago.

Secret

03:33

-1

Q: Black hole condition (suck and blow)

I think a black hole has a vortex inside that sucks in space like a tornado. However, I also think space is being blown as a bundle of straight lines as well into the tip of the tornado (blowing itself is not vortexed). The effect is that each string being blown in circulates independently into t...

black-holes

But then, there's this waterfall analogy in one black hole website (to be refugees)

SBM

04:44

Hello everyone

PhyMan

@SBM hey

SBM

Hey @PhyMan

Sir Cumference

05:20

@EmilioPisanty Did you get your badge?

Secret

05:37

Torsken kommer II - Torske kroner nå

►

SBM

06:50

What does that mean @Secret?

Secret

It's an advertisement by the swedish government on printing a new type of bank note. The music later went viral because of the catchy tune and (at least as seen in facebook) is shared everywhere

SBM

Oh

And the kanji from yesterday isn't the congee on the Wikipedia but more like dal.

I don't know what you call it in English, yoghurt masala dal?

Secret

@SBM where?

SBM

Some chat discussion from yesterday with @blue @Kaumudi.H and @PhyMan

Secret

07:05

Kanji (drink)

Kanji is a fermented drink made in India for the festival of Holi. Kanji is made with water, black carrots, beetroot, mustard seeds and heeng. it may be served with boondi sprinkled on top. During Holi, kanji is often added to urad dal dumplings or vada; the latter is then called kanji vada. == See also == List of Indian beverages == References... ==

It's literally a category in itself

SBM

Not that either

One word with hundreds of meanings yikes

Secret

yesterday, by ACuriousMind

Words can have more than one meaning. Live with it.

dmckee

@BernardoMeurer Sometimes the body's reaction to a traumatic injury suppresses the pain involved and allows that person to continue to function at relatively high efficiency for a time.

This has its upsides, of course, but also its downside as you may worsen the injury by continuing to stress that part of the body.

Both of my "go straight to the emergency room, do not pass go, do pay $200+" event have gone that way.

pZombie

08:27

If I take a syringe, suck some water in that is close to the boiling point at let's say 99°C. Then I shut the syringe at the front and pull a little on it, so some of the water inside the syringe starts boiling. Will at any point the boiling water inside the syringe start pushing the syringe further or do I have to keep pulling for more water to boil?

John Rennie

@pZombie When you pull on the syringe you reduce the pressure inside it.

The water boils until the pressure returns to ambient pressure, then it stops boiling.

pZombie

If the water wasn't close to boiling point and I would pull on the syringe, then, since water isn't compressible, I would end up with water + some vacuum inside the syringe, right?

@JohnRennie But does that answer my question? Do I have to keep pulling the syringe, or will there at any point the steam push the syringe?

John Rennie

@pZombie No, because the water would boil in the reduced pressure and fill the syringe with water vapour.

@pZombie The water would boil until the pressure rises back up to ambient pressure, then it would stop boiling. So the pressure inside the syringe will never rise above ambient pressure i.e. the boiling water would never push on the syringe plunger.

pZombie

I think I can see now what you mean. That makes sense.

pZombie

08:53

Now if we imagine a syringe inside a vacuum chamber. Only the tip sticking out and being sealed. We dive the tip into a container of water, and open the seal or valve to allow some water to be sucked into the syringe.

sorry hit enter too early

The syringe has also a weight attacked on top of the plunger, which gains potential energy if the water pushes against it. I assume that there will be a certain height the weight will reach until the pressure from the water will stop pushing against the plunger. Am I correct so far?

What I am trying to understand is the dynamics of this whole process. From some videos I watched, some of the water seems to be turning into ice, while another part into steam. Yet, we have stored some of the energy in potential energy in the weight that has been pushed up. That energy is missing from the water+steam system or so I believe. So the water+steam system is colder than before or so I believe.

So if we were to choose the exactly right pressure inside the vacuum chamber, it seems to me that the water would end up as just colder water if we waited long enough

PhyMan

@Secret we were not talking bout that

pZombie

09:14

Or let me put it differently. We have a syringe with some water inside. We heat the water up beyond 100°C but keep the plunger locked for some time until the water is at 120°C. Then we release the plunger. The syringe is very long and we ignore any friction. Eventually, when the syringe is extended far enough, the pressure inside should be equal to outside, and it would stop extending, correct?

But what if while the plunger was extending, we had a weight attached on top of it, hence the weight gaining potential energy. Where did this energy come from then?

I assume it comes from the water, hence, if we lock the weight at the highest position. The water inside the syringe should be colder than if we just let it extend without the weight on it. Is this correct or am I thinking it wrong?

user314159

Walter Lewin, MIT professor: "All of you have now lost your virginity... in Physics!"

Emilio Pisanty

10:01

@user314159 just to check, you know that Lewin is a complicated figure, right?

I'm not saying don't watch the lectures, just be aware that they have a darker side from some perspectives

Swapnil Das

i don't like his lectures

He tries to entertain the audience much more than teach physics.

Kenshin

that's cos physics is boringh

Secret

What on earth did I just read...?

Kenshin

@Secret, @SwapnilDas can I share a game Idea with you

Secret

ok

Kenshin

10:07

It's a gambling game and you have to pick one of these categories

<1% : 100x

<10% : 10x

<20%: 5x

<50% : 2x

Now there are 100 players

each player selects one of the categories

Now if they select the <10% category and 10% or less peopel chose that category, they get a pay out of 10x

Emilio Pisanty

@Kenshin that's... way more info than I thought was publicly available

Kenshin

Payouts begin from the 1% down, (i.e. only 1 payout can occur per round, only the first category that satisfies the condition gets a payout)

How should one play the game?

oh and if there is no winning category, all players get their money back (minus commission)

Emilio Pisanty

But yeah, it's something to be aware of if one wants to go down Lewin lane

Swapnil Das

@EmilioPisanty D'you like The Yale guy? Ramamurti Shankar?

Emilio Pisanty

@peterh you mean the review counter on top right? That thing is irreparably broken

@SwapnilDas never heard of him

pZombie

10:13

Kenshin do you know the answer?

Swapnil Das

Ooh ok.

Kenshin

@pZombie no is there an answer?

I invented the game

pZombie

that's a very interesting game

Kenshin

ty

pZombie

mind boggling

Secret

10:17

< 1% for 100 people means this category is only satisfied if there are exactly zero people choose it ?

Kenshin

<=

Mithrandir24601

@Kenshin I would think that the best bet would, on average, be the 1%, as it's the first option on the list, so the other categories actually have a lower overall probability of winning than it first seems thanks to your 'only 1 payout can occur per round' rule. Just my thoughts

Kenshin

@Mithrandir24601 but if 2 or more poeple followed that strategy, you would always lose

because it will only payout if <=1 % choose that category

My thoughts are, is that you should choose the 1% category maybe about 1% of the time or just over (potentially jsut over given Mith's point that only 1 payout occurs and we start by checking the 1% category)

Mithrandir24601

Ah, I get the game now - it's a game theory game, not a probability game :P

Kenshin

yea

pZombie

10:20

that's why it is so mind boggling. If you follow what is supposed to be the best strategy, you end up losing, because given than people would not play random but also follow the best strategy, it becomes a loss, so there is no best strategy

so the answer must be that the best strategy for this game is to play random

Secret

Expected value of the game:

$$\frac{1}{\sum_{i=1}^{100}{}^{100}C_i}({}^{100}C_1 (100)+\sum_{i=2}^{10} {}^{100}C_i (10)+\sum_{i=11}^{20} {}^{100}C_i (5)+\sum_{i=21}^{50} {}^{100}C_i (2))=$$

Bernardo Meurer

@ACuriousMind Where are you when I need you?

pZombie

but if playing random was the best strategy, and you assumed that everyone else was smart enough to figure that out, knowing that everyone would play random, you could actually then choose another category to max your profits

but again, that would be what everyone else would be thinking as well....

Kenshin

yes pZombie

Mithrandir24601

@pZombie Hence, it's game theory :P

Kenshin

10:29

for instance let's say we assume you should pick the first category 1% of the time

then if everyone does that

you can calculate that this category will actually be vacent 65% of the time

(1 - 0.99^100

)

so then a smart player will chose this category 33% of the time

but then so will everyone else lol

so then they change

etc.

pZombie

Love your game. You should post it as a question

on stack

Secret

I am not sure if this game can have a nash equilibrium, given that the strategy seemed to change on the fly

Mithrandir24601

@Kenshin There'll probably be an ideal probability with which to pick the 1% option that means, on average, one person wins every time?

Kenshin

@pZombie maybe I don't want to release this just yet

user314159

@EmilioPisanty indeed I do know, hence the bold-facing on the word "virginity." :-)

His conviction came as shock to me.

ACuriousMind

11:25

@BernardoMeurer What ails you?

Slereah

hello

Bernardo Meurer

@ACuriousMind What's the word for multimeter in german?

ACuriousMind

...*Multimeter*

@Slereah heyhey

Bernardo Meurer

.-.

What about voltimeter and amperimeter?

ACuriousMind

Same but without the i, i.e. Voltmeter and Amperemeter

Bernardo Meurer

11:26

Well, that was simpler than expected

Thanks :P

ACuriousMind

If you want to be very German, you might call the multimeter a Vielfachmessgerät, the voltmeter a Spannungsmessgerät and the amperemeter a Strommessgerät

Slereah

why would you want to be that German, though

Bernardo Meurer

@ACuriousMind Oh YES now that's what I was looking for

Danu

12:09

@ACuriousMind Do you want to help me understand a short computation in Milnor's book that I'm just blanking on currently?

ACuriousMind

@Danu Which one?

Danu

Morse theory; page 98, bottom. I can explain the context

ACuriousMind

@Danu Okay, I'm looking at it, what's the issue with it?

Danu

@ACuriousMind So I don't understand much about it---the first equality already weirds me out a little (what's up with that exchange of $u$ and $t$?---but I mostly just want to understand how to conclude that you get something nonzero in the end.

Especially since we only know $\partial_u v(u)=X\neq 0$, not its covariant derivative!

If you want some context let me know

Bernardo Meurer

@Danu Context plz

Danu

12:22

Not you guys.

(you should learn Riemannian geometry first; I can't give you a book's worth of background)

ACuriousMind

@Danu I'm not even sure what the $\frac{D}{\partial u} v(u)$ is supposed to be

I mean, $v$ is a path in $T_p M$, how is he taking a "covariant derivative" there?

Danu

Pull it back

ACuriousMind

Pull what along what?

Danu

So this $D/du$ stuff just means $\nabla^*_{\partial u}$ where $u$ is the parameter along the interval $I$

Along $V$?

I'm weirded out about this because in the first expression

Bernardo Meurer

Hey I understand what's going on!

Damn

I learned something

In uni

Amazed

Danu

12:34

we're dealing with a vector field along $\gamma_v$, i.e. the derivative there is $\nabla^*_{\partial_t}$ where $t$ is the parameter along $I$, the domain of definition of $W$.

@ACuriousMind Btw, let's denote this path by $\cal V$ from now on to avoid confusion with the tangent vector $v$

So yeah it is pretty weird to me---the last expression doesn't seem to be a covariant derivative of a vector field along $\gamma_v$ any more

ACuriousMind

@Danu What is $\nabla$ there?

Danu

LC connection on $M$

ACuriousMind

But $\mathcal{V}$ is not a path in $M$, nor is it defined along a path in M!

Danu

No, exactly

47 secs ago, by Danu

So yeah it is pretty weird to me---the last expression doesn't seem to be a covariant derivative of a vector field along $\gamma_v$ any more

But the first thing is that

ACuriousMind

It's just a path in $T_p M$, i.e. a collection of tangent vectors at a single point, I don't see what the covariant derivative is supposed to be.

But yeah, it starts with the strange appearance of $\frac{D}{\partial u}$ one step earlier

Danu

12:37

@ACuriousMind You can still get one from the LC I guess, but I don't think that's what one is supposed to do

ACuriousMind

Should be $\frac{D}{\partial t}\frac{\partial}{\partial u}$

Danu

@ACuriousMind What if we start by just assuming that's a typo

Yeah, that's what I'd hope too

My problem is that even then I don't really know what to do

ACuriousMind

@Danu I.e. $u$ and $t$ are switched by accident? Then the second equality makes even less sense :P

Danu

Yeah... just put in $t$ instead of $u$ in the 3rd expression?

ACuriousMind

But how would $\mathcal{V}$ depend on $t$?

Danu

12:39

yeah, nope :P

Bernardo Meurer

So the whole book is a typo is you guy's assumption?

Danu

But this definitely makes sense: $$\frac{D}{\partial t}\frac{\partial}{\partial u} \exp(t\mathcal{V}(u))\bigg|_{(0,0)}=\frac{D}{\partial t}(d_v \exp tX)\bigg|_{t=0}$$

that's chain rule

But then...

ACuriousMind

Well, you know that $\partial_u \mathcal{V}(0) = X$

Danu

We need to use that $d_v\exp$ is the identity at the origin

@ACuriousMind Edited that in

But I don't know how to take cov. der's without passing to coordinates

ACuriousMind

Wait...isn't $d_v \exp(X) = 0$ by assumption?

Your $d_v\exp$ is Milnor's $\exp_\ast$, right?

Danu

12:44

Yeah, this is garbage :P

Bernardo Meurer

Watching this is great :P

ACuriousMind

Well, I have no idea what is happening there

Danu

0/10 Milnor notation

Milnor's style is so hard for me to comprehend. I had the same problem with Characteristic Classes.

(though I guess that was mostly Stasheff's work?)

Bernardo Meurer

I hope it's just a Milnor setback in your studies

Danu

That doesn't really work, but okay

@ACuriousMind I'll ask around in the math chat room sometime later, I guess.

ACuriousMind

12:57

@Danu I have a feeling it should be easier to see that $\exp_\ast(tX)$ is not constant.

Especially since every source I can find that doesn't just cite Milnor for this theorem just declares it obvious this field is not constant :P

If the $W$ was constantly zero, then the "variation through geodesics" would not be a variation at all. That doesn't look as if it can happen.

Slereah

Apparently $\Bbb R^4$ with an exotic smooth structure has completely different properties in GR and it's absolutely awful

although ironically its tangent bundle is the same as regular $\Bbb R^4$

Danu

13:22

@ACuriousMind Yeah, it is kind of a dumb thing. But I have to think about it for a bit.

Physics Meta

1

Q: How I learned to stop worrying and start asking questions

As has been noted in the past, particularly in Question self-destruction: why don't experts ask more questions? and in Does reputation correlate with the question-to-answer ratio?, and also here, this site has the peculiarity that, in overwhelming proportion, our 'expert' users, by multiple measu...

discussion questions

Sha Vuklia

13:57

Guys, I’m trying to understand why we can take the time-independent wave function to be real. I can follow what they do mathematically, but I don’t understand the last remark on energy. Why can we choose which wave function we want if they have the same energy? Griffiths says that there is a different wave function for each allowed energy. But does that mean that the ones that are associated with the same energy, must be linearly dependent?

Emilio Pisanty

@ShaVuklia it depends on the situation

if your energy level is non-degenerate, then $\psi$ and $\psi^*$ will be linearly dependent, and you can then make $\psi$ real by multiplying it with an appropriate complex phase.

if your energy level is degenerate, then it gets interesting

particularly if $\psi$ and $\psi^*$ are linearly independent

Sha Vuklia

I have no idea what it means for the energy level to be (non-)degenerate.

but

Emilio Pisanty

this is the case e.g. for the $m=\pm 1$ states with $l=1$ in hydrogen

Sha Vuklia

this question is like the second question in the first paragraph about the time-independent Schrödinger equation

so I'm proposing to keep things as simple as possible?

Emilio Pisanty

@ShaVuklia a degenerate energy level supports multiple linearly-independent eigenstates with the same eigenenergy

a non-degenerate level supports only a single independent eigenstate

Sha Vuklia

14:06

yea right, well I haven't had eigenstates and eigenenergies and eigenfunctions and the like

I'm guessing then that

Emilio Pisanty

ok, fair enough

Sha Vuklia

I should assume for now that they simply are linearly dependent?

$\psi$ and $\psi^*$

Emilio Pisanty

@ShaVuklia it's a reasonable assumption (just keep in mind that there are important situations where that fails)

where your quote says

> Since they have the same value for $E$ as $\psi$...

it just says that both $\psi$ and $\psi^*$ are solutions of the TISE with the same energy

Sha Vuklia

yes I noticed that

Emilio Pisanty

and the two linear combinations are therefore also solutions of that same equation, so they also have the same energy

Sha Vuklia

14:08

so for now I just assume that if solutions have the same $E$, then they are linearly dependent?

Emilio Pisanty

@ShaVuklia no

sorry

ACuriousMind

@ShaVuklia I think the point here is not that $\psi$ and $\psi^\ast$ are necessarily linearly dependent, but that $\psi + \psi^\ast$ and $\mathrm{i}(\psi - \psi^\ast)$ are real and span the exact same subspace as $\psi$ and $\psi^\ast$ no matter their dependence.

Emilio Pisanty

you're looking at the case where they are independent

@ACuriousMind @ShaVuklia Yeah, that can also work

Sha Vuklia

@ACurious that was my first guess too

in fact, I also showed that an even potential can have even/odd solutions as their "basis"

in that case, I could write $\psi$ as a linear combination of an even an odd function

so I should be able to do that here too?

Emilio Pisanty

@ShaVuklia sort of, but not really

in one dimension, even functions and odd functions have different energies, pretty much always

a general wavefunction is always a linear combination of a linear and an odd function

but solutions of the TISE typically have definite parity

However, that structure normally breaks when you go to more general situations

Sha Vuklia

14:16

right, I guess. But what ACurious said solved my problem. I'll stick to the spanning argument! (sorry, I wish I could think along with you, but I've only read chapter 1 and 2 from Griffiths, which is not the formal part of the book)

Emilio Pisanty

@ShaVuklia yeah, take it at your own pace ;-)

Slereah

15:15

The big O notation doesn't make sense

Obviously we can't write $f(x) = \mathcal O(x)$

It should be $f(x) \in \mathcal O(x)$

user228700

Oh @JohnR: Hi! :-) Safe drive?

user228700

Ah, geez, you're gone again :-/

Emilio Pisanty

15:32

guys

the Earth is flat

Flat Earth: Spirit Level Flight Experiment

►

there's actually scientific evidence

I never knew

Slereah

15:56

simple evidence

Take a level, put it on the ground

u will notice it is flat

unless it's not

Sid

Of course, I mean how hard is it to understand that we are all actually living in a giant pancake?

Slereah

more importantly if he's on a plane, why not look out the window of the plane

That will be a slightly more informative experiment

Sha Vuklia

16:36

Guys, I'm suddenly confused by what the de Broglie formula actually says:
$$
p=\frac{h}{\lambda}.
$$
Griffiths is explicitly talking about the wavelength of $\Psi$ here, so how can we speak of "the" momentum of the particle, when the momentum is undefined? Why isn't the de Broglie formula about $\langle p\rangle$ or something like that?

@Slereah the notation makes sense, because it's simply convention.

Slereah

The de Broglie formula is for plane wavefunctions

Sha Vuklia

ugh i don't even know what those are:l

Slereah

Wavefunctions are $L^2$ functions, so they can be expressed as Fourier transforms

$$\int e^{ixp} \hat f(p) dp$$

In those circumstances, you can treat each $e^{ixp}$ individually

And for those "wavefunctions", you find that $p = h/\lambda$

This may not be apparent in the actual wavefunction, which is a mixture of many momentum states

SBM

What does the hat on $f(p)$ mean? And what is L^2?

Laplacian?

Sha Vuklia

the hat means it's an operator

SBM

16:41

oh

Sha Vuklia

and $L^2$ roughly means it's square integrable

Slereah

Actually not here

It means it's a Fourier transform

Sha Vuklia

oh really

SBM

interesting

Sha Vuklia

well tbh, I didn't follow anything you said

haven't come across that in griffiths

i've only read chapter 1,2

Slereah

16:41

Have you done wave mechanics first

Wave mechanics is pretty important to do before QM

Sha Vuklia

i know a bit about waves

enough to follow griffiths in any case

(I'm not doing formal QM yet)

ACuriousMind

@ShaVuklia 1. The deBroglie formula/hypothesis is a statement prior to full development of quantum mechanics, you should not expect it to make straightforward sense when attempting to interpret it in actual QM. 2. In terms of QM, deBroglie just "claims" that the wavefunction of a particle with definite momentum $p$ is $\psi_p(x) = \mathrm{e}^{\mathrm{i}\hbar p x}$, where you may convince yourself that this works out to $p = h/\lambda$.

SBM

Waves and QM looks intriguing

Sha Vuklia

@ACurious oh right, then. I'll try to verify that. Thanks

ACuriousMind

Now, no actual wavefunction has definite momentum, but due to Fourier transform, as @Slereah said, you may think of any wavefunction being the superposition of such plane waves - the integral is "just" a more general sum.

(Don't ask for mathematical rigor regarding the latter part :P)

Slereah

16:45

yeah I didn't want to bring up the rigged Hilbert space with a beginner :p

ACuriousMind

The game is rigged

Sha Vuklia

Is the superposition of the plane waves similar to the superposition of the stationary states?

because that's the only thing I've encountered so far

Slereah

Well they are both superpositions of eigenstates

ACuriousMind

"superposition" is just a silly physicist word for "linear combination" :P

Slereah

The first one is eigenstates of the momentum

The second one is eigenstates of energy

Sha Vuklia

16:46

@ACurious lol:P

SBM

eigenstates ...

Slereah

eigenstates are states with a definite value with respect to that operator

Sha Vuklia

I can't wait until I read chapter 3 from Griffiths (but first my exams). I want to talk too about eigenstates and such :P

SBM

oh I was thinking linear algebra

ACuriousMind

@ShaVuklia It's funny 'cause it's true

@SBM It is just linear algebra, except when it's not but we physicists pretend it is, anyway :P

SBM

16:49

Must be fun, right?

ACuriousMind

@SBM It is :)

SBM

could you suggest a good resource?

ACuriousMind

Bunch of recommendations here, I can't personally vouch for any because I haven't read any

Slereah

I don't know a whole lot of basic math books because during my degree we didn't have any books

It was just classes and xeroxed notes

MathematicalRain

Hey there. Does someone have any experience with the Creutz algorithm (also known as the Demon algorithm?)

John Doe

16:58

@Mithrandir24601 Hi

Slereah

Sorry this is a good christian channel

We do not deal with demon algorithms

SBM

Thank you @ACuriousMind

Mithrandir24601

@JohnDoe Hi

John Doe

@Mithrandir24601 How's it going?

Mithrandir24601

@JohnDoe Let's see... Just back from Kung Fu. Currently having some hot chocolate and I've got my last ever exam on Wednesday. So not too bad :)

SBM

17:04

Thank you @ACuriousMind

ACuriousMind

@SBM I heard you the first time, you're welcome ;P

SBM

oh sorry, poor internet

John Doe

@Mithrandir24601 Yeah that does sound good, especially the last exam part. Before I mention somethiing about what we discussed yesterday. Did you watch the boxing last night? Kell Brook vs Errol Spence?

Mithrandir24601

@JohnDoe Nope. I do kickboxing (as part of KF), but I don't really watch boxing :P

What did you want to ask/mention?

John Doe

@Mithrandir24601 Will type it now.

@Mithrandir24601 If you are busy don't worry about it.

Mithrandir24601

17:10

@JohnDoe I'm not busy... No guarantee I'll know the answer though

John Doe

@Mithrandir24601 Okay it's just an idea that you can comment on, not really a question.

Mithrandir24601

Fair enough

John Doe

@Mithrandir24601 I just want to find you think of the reasoning of the following. My understanding of a text that I amusing So the proabbility of function of knowing nothing about say using is that. For a say case of measuring the position of an object in space, when having no knowledge, the probability is given as a constant function $P(x) = c$. But for the case where $x > 0$ we consider the function as described yesterday $$P(x) = \frac{c}{x}.$$The reason for this is as follows:

@Mithrandir24601 We consider that certain probability functions should be invariant under transformations which belong to transitive groups. Hence for the case of the positive quantity we have that it is not invariant under the translation group hence we can't consider a constant probability function. Does this make any sense?

Mithrandir24601

@JohnDoe I'd say (but I'm no expert) that it depends on what you want to do with the probability distribution - under translation/addition, the above isn't a Haar measure over the positive reals, but it is under multiplication

John Doe

@Mithrandir24601 Yeah I understand. The book I'm reading is Quauntum Measurment Theory with Applications. The idea seems to be that you consider the situation, like an object in space (where the probability is with respect to position) or say a lake (where the probability is with respect to volume) and then you consider which transitive group you require it to be invariant under tranformations. So it the object case it should be invariant under translations...

@Mithrandir24601 For the lake we want it to be invariant under the group of scaling tranformations...and so on.

@Mithrandir24601 The purpose is just to motivate a probability function. I think the idea is much more complicated because they give references to other books which deal with it in detail.

Mithrandir24601

17:28

@JohnDoe So if you want it invariant under scaling, then $$\frac{1}{x.y.z}dx\, dy\, dz$$, should be reasonable (in 3D)...

If I haven't mentioned it already, a Haar measure is a measure on a group, so that's why it's different for multiplication and addition - is that what you're asking?

(although there is a way to make it invariant under both addition and multiplication)

John Doe

@Mithrandir24601 I don't know. Since for the one dimensional case you have as you stated yesterday that $\int_{c}^{d}p(bx)dx = \int_{c}^{d}P(x)dx$, so in the one dimensional case it is invariant in that sense I guess...

PhyMan

Is there any way of getting latex on The mobile website?

John Doe

@Mithrandir24601 No I'm not asking anything. I just wanted to see what you though of that idea of defining probability functions based on their invariance with respect to groupd.

Philip Cherian

Looks around for the nearest field theorist Anyone here with some experience solving loop integrals using Mathematica? I doubt my question is worth an actual post on the site, so I figured I'd ask here...

Mithrandir24601

@JohnDoe Oh, OK - it's a good idea. It's what I'm working on at the minute though, so of course I'm going to say that :P

John Doe

17:36

@Mithrandir24601 Yeah it sounds interesting, unfortunately it's not the point of this book so it's going to move on to other stuff. But I want to learn more about group theory, and their applications in QM.

Mithrandir24601

@JohnDoe Fair enough. So you want to know my thoughts on why it's useful?

John Doe

@Mithrandir24601 Yeah

@Mithrandir24601 I read further now they mention the Haar measure as an alternative to defining the probability in terms of coordinates...

Mithrandir24601

@JohnDoe Ah, OK... I don't know what it's use is in maths, but I'll go with what I'm currently using it for, keeping things as simple as possible (otherwise it might get messy): Say you've got a single qubit that you can do arbitrary operations (gates) on to get a single qubit circuit. This circuit is represented [not sure if this is the right word or not] by SU(2). So, if you want to integrate over the group, you use the Haar measure

(i.e. if you want to integrate over all possible circuits)

John Rennie

@Kaumudi.H Thanks, yes, I'm now safely in Somerset :-)

user228700

Wokay :-) Good to know.

John Doe

17:48

@Mithrandir24601 Oh okay. Yeah I want to read up on that since I'm doing a project on two level systems later this year. Does 'gates' refer to any operator acting on a quibit?

SBM

good night

Mithrandir24601

@JohnDoe It can also be used to integrate over a subset of that group

user228700

@SBM 'Night! :-)

Mithrandir24601

@JohnDoe Any unitary with determinant 1

John Doe

@Mithrandir24601 Kewl very interesting

Mithrandir24601

17:50

@SBM Night!

Philip Cherian

Open question: does anyone know of any mathematica scripts that compute Feynman parametrizations other than this one? I'm almost certain that they must exist, but I can't for the life of me find them anywhere on the internet...

Mithrandir24601

@JohnDoe so, yes that would be any ideal evolution of the state (things like decoherence probably messes this up a bit though, so you might have to use U(2) instead of SU(2) or something, depending on the processes involved, but I'd have to look at this further to be sure)

Sid

@blue results tomorrow?

All the best!

Anonymous

@Sid Oh yeah. Thanks. I don't care about it though :P. How was your CBSE result?

Mithrandir24601

::minor lightbulb moment:: Is the Haar measure also related to the whole intensive properties thing? :O

Sid

17:59

@blue English showed me the finger. 91%

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