Anonymous
9:00 PM
@DanielSank That's an approximation....

Anonymous
Okay

Anonymous
If the density function has jumps I doubt that would be much useful

What?
I'm defining the process by saying that there's a probability per time of an event.
The whole point of the post was to show that a process defined that way has Poisson statistics.

hmm. I'm pretty sure I'm making some obvious mistake, but these C-G coeffs are throwing me for a loop
I"m trying to do an example I set for myself as review so I could help my students

@Semiclassical is that the point at which you realize you're in for a rough ride?
or does that come later?

9:05 PM
lol

C-G are confusing if you let them take over.
Gotta keep them in line.

If I take a pair of spin-1 particles with $m_1=m_2=0$, then the C-G table gives me $|1,0\rangle_1 |1,0\rangle_2 = \sqrt{\frac23}|2,0\rangle_{12} -\sqrt{\frac13}|0,0\rangle_{12}$
which seems sensible enough.

btw here is an example of SE being awesome
note the authors of the last paper my answer cites

@EmilioPisanty I don't get it.

9:07 PM
@DanielSank then compare to the author of the accepted answer

the comments are also worth noting...

Author of marked as answer question is... (dun dun dun...)

Ah.

(continuing) the two particle case I just gave seems simple enough. But now I wanted to take a third spin-1 particle, with $m_3=0$ as well

Anonymous
@DanielSank My point is that, in a continuous probability distribution if the probability of an event happening in unit time (say $1s$) is $\lambda(1)$, then the probability of the event happening in any arbitrary $0.1s$ within that $1s$ may not be $\lambda(0.1)$. So if you are defining $\lambda$ for unit time, it's not totally correct to conclude that the probability of that event happening in $\Delta t$ time is $\lambda \Delta t$.

9:10 PM
@Semiclassical I'm reasonably sure that C-G coefficients do not directly answer that question.

so I figured I could just use the table again to get $|2,0\rangle_{12}|1,0\rangle_3$ and $|0,0\rangle_{12}|1,0\rangle_3$ and in terms of $|s,0\rangle_{123}$

@Semiclassical I am pretty sure that doesn't work.

You may well be right, because if I crunch that from the C-G table I seemingly get
$$|1,0\rangle_1|1,0\rangle_2|1,0\rangle_3=\sqrt{\frac25}|2,0\rangle_{123}-\left(‌​\sqrt{\frac{4}{15}}+\sqrt{\frac13 }\right)|0,0\rangle_{123}$$

Anonymous
So, I'd rather talk in terms of probability density function rather than probability per unit time, which is quite an ill-defined term.

which isn't normalized to one.
however, 2/5+4/15+1/3 = 1.

9:14 PM
@Blue IMHO, that is a willful misunderstanding of what $\lambda$ means.

is the answer just "that's not what C-G is for"?

I already said that $\lambda$ is the probability per time of an event, as long as $\lambda \Delta t \ll 1$.
@Semiclassical I'm not sure. There's probably a way to do the 3-body case with CG coefficients, but I think it's not 100% trivial. Search the main site. There are posts on this.

kk
though there is says you can j1,j2,j3 by first coupling j1,j2 to J12 and then coupling J12,j3 to the total J
which sounds a lot like what I was trying to do :/

Okay so if I have a wavefunction $\Psi(x,0)=Ae^{-\lambda x}$ and I want to normalize it I think I should do $|A|^2\int_0^{\infty} e^{-2\lambda x} dx$ but the answers says that it's $2|A|^2$ before the integral sign. Why is this? (A is a positive real constant...)

probably they mean $\Psi(x,0)=Ae^{-\lambda |x|}$ for all real $x$.

9:19 PM
14

In A History of Western Philosophy, Russell argues: I conclude that the Aristotelian doctrines with which we have been concerned in this chapter are wholly false, with the exception of the formal theory of the syllogism, which is unimportant. Any person in the present day who wishes to ...

'Where Aristotle says "Some B1 are B2" this is clearly to be read as an intersection of types, what in modern categorical logic is called a fiber product. '
Of course Aristotle was all about fiber products
Used to get the urge to read his physics until I realized it's written in language like this

@Semiclassical That... that would make sense.

@Blue I think you're confused about the meaning of "per unit time". This does not mean anything about time=1.

'One way to express the critique of Aristotle’s physics in his own terms is to say that, while he was right that every action or change involves act and potency, he thought this meant there were agent-patient relations in the inanimate when in fact there are only interactions. '

The phrase "per unit x" is just a formal phrase that means the same thing as "per time".
It's a bad phrase. I shouldn't have used it.

Anonymous
It's way better to just say $\lambda$ is the average of the probability density function. Avoids a lot of confusion

Anonymous
9:23 PM
Also you don't have to hand-wave by saying things like $\lambda \Delta t \ll 1$

@Blue That's not hand-waving.
@Blue Ok fine, but how do you know the form of the distribution?

The usual way to obtain the distribution is to take a limit of the binomial distributions (which are more physically meaningful)

The whole point of the post is to show how you go from a physical principle of something having a probability per time of ocurring to the Poisson distribution.

@DanielSank think i found the error upon looking at this note: web.pa.msu.edu/people/pratts/phy851/lectures/lecture26/…

I must say it's not entirely clear to me what $\lambda$ being probability per unit time means. It's the expectation of the probability density function to me (but I am open to enlightenment if someone wants to tell me what that means)

Anonymous
9:28 PM
@BalarkaSen Exactly. I've been trying to say that one should define $\lambda$ as the "expectation of the probability density function"

@Blue WHAT density function?

No I mean if he's saying it surely it means something I don't understand

key point: "In order to couple the three angular momenta $j_1$ and $j_2$ were coupled to $J_{12}$ before $j_3$ and $J_{12}$ were coupled to $J$. Thus $J_{12}$ survives as a quantum number, which is necessary as the original state had six labels which requires six labels for the final state." (emphasis added)

@BalarkaSen Consider a process where some event may happen. In a small amount of time $\Delta t$, the probability that the thing happens is $\Delta t \lambda$, as long as $\Delta t$ is small enough such that $\Delta t \lambda \ll 1$.

Anonymous
@DanielSank We don't need to know the exact form of the density function.

9:29 PM
That's what I mean by "probability per time".

Anonymous
@BalarkaSen We're talking w.r.t Poisson distribution

@Blue Look, you can do what you're saying, but it answers a slightly different question than the one I asked.
My post explicitly asks "If I have a process where in each little bit of time there's a probability $\Delta t \lambda$ that a thing happens, please show that the probability of $n$ things happening in time $T$ is a Poisson distribution with mean $\lambda T$."

@Blue I am well aware

That is the question.
I am not asking a different question.

@DanielSank Hm.

9:30 PM
If you think the question is not well posed, then we can discuss that.

Anonymous
@DanielSank If you want to do it that way, then follow the Wiki definition: " The average number of events in an interval"

Anonymous
In probability theory and statistics, the Poisson distribution (French pronunciation [pwaˈsɔ̃]; in English often rendered ), named after French mathematician Siméon Denis Poisson, is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant rate and independently of the time since the last event. The Poisson distribution can also be used for the number of events in other specified intervals such as distance, area or volume. For instance, an individual keeping track of...

Anonymous
The word "average" is very important here

those aren't the same thing.

@Blue You're not understanding Daniel Sank's premise at all

9:32 PM
probability distribution of getting n events in small amount of time

Anonymous
@BalarkaSen Okay?

He's giving you a physical situation, giving you a $\lambda$ that comes from the situation, and telling you: it's the same $\lambda$ as in the Poisson dist.

@Blue you do realize that a real physical process always has a lower time limit below which the statistics aren't Poissonian, right?

You're begging the question by repeatedly telling him that he should define his $\lambda$ as the expectation of the Poisson dist.

@BalarkaSen Actually, be careful. Usually $\lambda$ is the mean of the Poisson distribution. That is not how I used it in my question.
@BalarkaSen Correct.

9:34 PM
@DanielSank That's literally what I said, I think

@BalarkaSen I think not :-\
> ...and telling you: it's the same $\lambda$ as in the Poisson dist.

You're obtaining that your distibution is a Poisson distrubution with mean $\lambda$ you're defining otherwise in the question, right?

@BalarkaSen No.
10

Suppose we have a probability per unit time $\lambda$ that something (e.g. nuclear decay, random walk takes a step, etc.) happens. It is known that given a time interval of length $T$, the probability that $n$ events happen is given by the Poisson distribution $$P(n) = \frac{e^{-\lambda T} (\lamb... for one, they'd have different units. \lambda as an expectation value would be a pure number. but then \lambda\Delta t can't be a probability Please read the post. 9:36 PM Oh, mean \lambda \Delta t. Yes As Lang used to say, "Your notation sucks" ;) But I understand now Thanks @BalarkaSen If there's a more standard symbol, I'll use it (and edit all the answers). Serg Lang was a nut. I met him once as a freshman, before classes even started, and he yelled at me about wave/particle duality. 9:38 PM Truly lolol @DanielSank that's incredible, what an honor @bolbteppa :-) I suppose the issue is that you can obtain \lambda in two ways Look for the purposes of this discussion, \lambda has dimensions of 1/time. No objections. \lambda is not the mean of the Poisson distribution. One is empirical: Let the process run for a long time, count the total number of events, and from that determine the event frequency 9:40 PM There's a story online I'm mis-remembering but apparently in a class he once said something and half the class tried to calculate it, and he said 'those that tried to calculate it will become mathematicians' or something 1/ \lambda is the mean of the exponential distribution of times before the next event, however! @DanielSank Perhaps a good idea to point the unfortunate coincidence of notation in the question as a NB? Anonymous @BalarkaSen In the physical situation he gave he says \lambda is the "probability per unit time". I'm just saying that that is not a well defined term. So you should go with average number of events in a time interval as wiki says. Or have a look at this post @BalarkaSen Why? Isn't \mu usually used for mean? Another one liner of his is that analysis is just number theory at \infty 9:41 PM @Blue Here's how I'd define it. Take that same long time run that I said, and break it into a bunch of small time intervals. In each of those, count the number of events per small time interval @DanielSank I think most people are used to \lambda being the parameter of the Poisson distribution (hence the mean). @BalarkaSen Oye. There aren't enough letters... That defines a probability distribution on the number of events per small time interval. Anonymous @Semiclassical *and take their average! Anonymous That's it 9:42 PM @Blue It's not the average, as DS points out. \lambda \Delta t is the mean of the pdf, not \lambda my example is bad, actually On the interval [t, t + \Delta t] @Blue Yes, you can do it that way. I'm not convinced it's necessarily better pedagogically, but I do see that it might be more robust a description. Maybe. I have to think about it when I'm not also drawing mechanical parts :-| 9:44 PM lol Anonymous @BalarkaSen Now we are back to square 1. In the physical situation we should start with defining lambda as the average number of events in an interval. I suspect it should boil down to: Instead of waiting a long time, you can wait a very short time and look for the probability that at least one event happens Oh I mean, sure, why not. Anonymous And then show it is equivalent to the \lambda in the Poisson distribution. and require that this probability be proportional to the time interval for sufficiently small times. 9:45 PM @Blue Nooo. \mu = \lambda \Delta t in DS's Poisson distribution. Do you realize that? ^ How is everyone not understanding this? In the standard Poisson distribution, \mu = \lambda. I think I'm understanding it, lol @DanielSank there are two derivations of the Poisson distribution in Landau's stat phys, sec 114, one from the Bernoulli distribution (which he also derives) and another from the Gibb's distribution, if that's any help Anonymous @BalarkaSen Ah, right. So we show it is equivalent to \mu 9:46 PM Using gas particles in a box to motivate it I'm really confused as to why \Psi(x,0)=e^{-\lambda |x|} having |x| instead of x means that 1=2|A|^2\int_0^{\infty}e^{-2\lambda x} dx instead of having 1=|A|^2... @bolbteppa ~sigh~ I made a post specifically asking to relate the Poisson distribution to a particular physical starting point. Now everyone wants to offer alternative approaches. Which is great! But please understand what I actually asked first. @CooperCape the domain is all real x Right... am I missing something here? I usually forget the small things and it ruins me. 9:47 PM @CooperCape It's because if f is even, then$$ 2 \int_0^\infty f(x) dx = \int_{-\infty}^\infty f(x) dx \, .$$\int_{-\infty}^\infty e^{-2\lambda |x|}\,dx=? Anonymous @BalarkaSen Hmm, I was talking about the standard Poisson distribution initially. But okay. Ohhhh right... thanks both of you :) That actually makes sense now... @Blue wtf do you mean by "standard"? Anonymous So, I guess we're done 9:48 PM The choice of variables does not make something "standard" or "nonstandard". Variables are just symbols for heaven's sake. If you think a conceptual object's standardness is tied to the notation used for that object, you're in for a world of sadness. lol Anonymous I could replace variables with elephant emojis. That would just add to the confusion. The way I'd put it is that the advantage of DS's way of approaching is that one derives the result from a single assumption: For very short times, the probability of having at least one event is proportional to time. @Blue Yes, but it would not change the meaning. Anonymous Better to stick to widely used variable notations. 9:49 PM @Blue Yes, usually. But do you realize that we just wasted half an hour because you refused to read how I was using the symbol? "Standard notation" is a fiction, albeit a very convenient one @DanielSank I read your post, you can use any derivation to derive it the way you are asking about, seems like the 'time' aspect is throwing you off in your modelling - but since probability per unit time times time is probability, so the time is irrelevant in the derivation until the very end @bolbteppa I do not understand. "unit time times time" ow lolol Anonymous 9:51 PM @DanielSank Well, now you can edit the post so that it becomes less confusing for later visitors. :) @Blue So what symbol would you like me to use? and how did you miss the gigantic equation$$\frac{e^{-\lambda T}(\lambda T)^n}{n!} \, ?
It's pretty obvious that $\lambda$ has dimensions 1/time there.

and that's definitely proportional to $T$ when $n=1$ and $T$ is sufficiently small.
(though to follow my statement above I should really talk about the probability of at least one event which is $1-e^{-\lambda T}\approx \lambda T$.)

Anonymous
@DanielSank I'll suggest an edit later, when I get time

@Blue You had 30 minutes to argue but no time to suggest a single symbol?

In this picture, at the end replace $\overline{N}$ 'average number of particles in the volume $V$ with 'average number of particles you find in a unit time $\lambda$ times time $T$ taken to measure', $\overline{N} = \lambda T$:

9:55 PM
:-P

@bolbteppa Yeah yeah Floris already wrote that up.

So what's the problem then

@bolbteppa I have no idea, other than @Blue complaining that "probability per time" isn't a well defined idea.

There's no problem, you're just not saying anything new. It's already in the answers.

9:57 PM
this conversation just feels like people talking past one another
2

It's still a cool derivation worth repeating :p

10:26 PM
Conversations are often like that

10:51 PM
@Cows Neat, that was my yearbook quote in high school

@EmilioPisanty thanks for the edit

Yo yo
dudes

11:10 PM
Hello

Is there any physics people that are working on "Instant matter transport"?

Seriously?

yeah like sending material from one place to another in an instant
just an idea

No, no one is working on that because it would violate causality, i.e. contradict relativity, and hence believed to be impossible

11:13 PM
Well theres quantum teleportation - it transports state.
But no matter is exchanged.

Also, quantum teleportation still needs a classical channel and proceeds therefore at most at the speed of light

But I have a question then about Bose-Einstein Condensate, doesn't matter behave like a wave?

Its unlikely to be instantenously for the reasons already point out.
both fermions and bosons have both particle and wave like character.

@EnlightenedFunky Quantum objects have features of both classical particles and classical waves, but they really behave like neither.

The Bose-Einstein condensate occurs because bosons don't obey the Pauli exclusion prinicple. So they can all be in the same state.
True, thats why some wit coined the word - wavicle - but it never really caught on.
At least I never heard any one use it.

11:17 PM
Ok but you can't increase the speed of the bose-einstein condensate cause you would increase its temperature right?

That looks like a reasonable assumption.

Well, you can certainly accelerate the entire condensate to whatever speed you'd like.

the other thing about condensates is that they bring quantum effects into the macroscopic arena.

but would it still exhibit wave properties as when at rest

I thought he was talking about accelerating the wavicles in the condensate?

11:19 PM
yeah what Mozibur said

@curiousone: are condensates why we get super-fluidity?
Actually hoo-haa with extra dimensions in String Theory, why has no-one pointed out that the usual formalism in quantum mechanics has infinite dimensions? I suppose they're not space-time dimensions...

because if so couldn't we make an instrument that reads that specific wave pattern interfence and turns it to energy wouldn't violate E=mc^2 and create an EM wave, and send it somewhere else and then bring it back to matter from said energy

Actually, with all the hoo-haa with ...

Apr 21 '16 at 20:42, by ACuriousMind
Are you perhaps confusing me with CuriousOne? :P

I'm not sure whos confusing who with who...

11:22 PM
wait
did you just call someone out in chat who doesn't use chat?

But someones being confusing ... curiouser and curiouser.

@MoziburUllah Yes, I think all superfluids are either Bose-Einstein condensates or fermionic condensates like the Cooper pairs of superconductivity, but I'm not entirely sure

acm is curiousone confirmed

I think I typed @curiousone instead of @acuriousmind
@ooolb:

@EnlightenedFunky I don't know what that sentence means, sorry.

11:23 PM
oh ok

there goes ACM again

I think I must be tired (yawn).

@MoziburUllah Yeah, you did. But CuriousOne is not active anymore, so it's not a grave confusion ;P

@EnlightenedFunky what are you on right now, my dude

you killed the competition :o

11:24 PM
i want some

Ok. ;).
Oh no!

@ooolb Why? I am on nothing... just thinking about something learned

@0celo7 That's one possible interpretation ;)

@0celo7 you don't have to kill them to get rid of them.
@EnlightenedFunky duuuuuuuuude
im gonna go learn stuff right now
bye

Couldn't an instrument be made that reads that specific wave pattern interference and turns it to energy wouldn't violate E=mc^2 and create an EM wave, and send it somewhere else and then bring it back to matter from said energy @ACuriousMind better?

11:29 PM
@ooolb do you want a more attainable exercise?

@EnlightenedFunky I don't know what you mean by "reading" a "specific wave pattern interference" and "turning it into energy". What wave pattern? You can't measure quantum wavefunctions, at least not like you do with actual classical waves. And how would you "turn it into energy"?

@0celo7 nah go big or go home m7
im working on it
@EnlightenedFunky is this ironic

                                      @ACuriousMind should I play Civ V?
whoa wtf

wat
what is that sorcery

11:32 PM
@0celo7 Sure, it's nice

I ᴛᴏᴏ, ᴀᴍ ᴄᴏᴏʟ.
ᴅᴏɴ'ᴛ ғᴏʀɢᴇᴛ ᴛᴏ ᴅʀᴏᴘ ᴀ ʟɪᴋᴇ.
ᴀɴᴅ sᴜʙsᴄʀɪʙᴇ ɪғ ʏᴏᴜ ʟɪᴋᴇ ᴡʜᴀᴛ ʏᴏᴜ sᴇᴇ.
ᗩᑕ⋒ᖇᓰᗢᘮᔕᙢᓰﬡᖱ
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that's a but much

hmm
was just getting started

@ACuriousMind want to play a game with us

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it broke

11:43 PM
@0celo7 Not today, I need to finish my Shadowrun prep for tomorrow

@ACuriousMind notice me :'(
i did that all for you

@ooolb But...why?

lol
bye

@ACuriousMind Are there any American adults here right now

I have no way of knowing that...why?

11:46 PM
United Airlines is being a hecker
they're not beating me senseless, but close to it

@ACuriousMind reassurance

@ACuriousMind how does having a crazy internet stalker feel?

@0celo7 Well, that sucks
@0celo7 I don't know

@ACuriousMind well you have one!!!
@ACuriousMind Should I make schnitzel or just sautee the pork

Depends. The better the quality, the less inclined I am to make Schnitzel out of it ;P

11:52 PM
it's good quality, but I can't get the right flavor on pork without a sauce

@0celo7 who? there can only be one. i'll eliminate the competition.

jesus dude
you need to fix your priorities

^I second that

what priorities?
:p
serious question does any of our lives have meaning anymore?
after ACM went to the IT side
i'm still going to learn physics though just to invent an interuniverse travelling machine and travel to the alternate universe where ACM is doing physics.