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12:30
@AccidentalFourierTransform Did Balarka clear up your doubt?
Anonymous
apparently this book does not exist in print books.google.com/books/about/…
Huh, the library has it
Anonymous
@0celóñe7 What?
@Blue some light calculus reading
12:45
Hi chat!
any resource which has some nice explanations of Einstein's Quote!
Einstein's quote?
Einstein's words or sayings!
@BAYMAX I am an expert on Einstein
what would you like to know
"I am truly a lone traveller and never belong to my country,my home,my friend or even my family with whole heart"
Like this quote
or saying
that doesn't sound like something he actually said
12:48
agreed
you do realize that the majority of Einstein "quotes" on the internet are completely made up
Anonymous
@0celóñe7 I won't be able to access the page without NetID and password
Anonymous
@BenNiehoff "You do realize that the majority of Einstein "quotes" on the internet are completely made up." - Einstein (1879-1955)
@Blue wot
did it take you to my school's login?
Anonymous
12:51
@0celóñe7 Yes
Anonymous
Now it works :D
actually if you really want to impress people link.springer.com/book/10.1007%2FBFb0068432
Anonymous
I won't be able to understand Calculus in Vector Spaces now I think. I'll have a look at the first one
@Blue "without norm" is the important word there
vector calculus = calculus on vector spaces with norm
Anonymous
12:56
Oh, okhay. Thanks. I'll have a look
abnormal vector spaces
hi everyone
Anonymous
Well. I have a doubt =P If electrons are considered as 3-D waves then how is the equation of the form $A\sin(wt-k^1x^1-k^2x^2-k^3x^3)$. Shouldn't it be a standing wave equation in form of a circle?
Anonymous
Anonymous
@AccidentalFourierTransform
13:00
$A\sin(wt-k^1x^1-k^2x^2-k^3x^3)$ is just an example of a 3D wave
there are more examples
the one in the picture is indeed not of the form $A\sin(wt-k^1x^1-k^2x^2-k^3x^3)$
Anonymous
What is the approximate equation of the one in the picture?
@Blue It's a sine wave as a graph over a circle
It's a meme picture
'It's not so easy to free oneself from the idea that coordinates must have an immediate metrical meaning.' -- Albert Einstein @BenNiehoff do you think he said that?
something like $\frac{1}{r}\sin(kr-\omega t)$, probably
@TheRaidersofLasVegas Not sure
13:03
@TheRaidersofLasVegas "The introduction of numbers as coordinates is an act of violence." - Hermann Weyl.
Numbers on a line are an act of violence?
Weyl was a hippie
How else does one "see" numbers?
Anonymous
@AccidentalFourierTransform What is $r$ ? If that is position vector then at r=0 it would blow up to infinity
LSD, for example
@Blue yes
13:06
"The next person to propose a new definition of connection should be summarily executed." - Mike Spivak.
Anonymous
@AccidentalFourierTransform How come is it infinite at r=0? It doesn't seem so from the image
Numbers pictured as points on a number line is useful.
@Blue you cannot see whats going in at $r=0$ in that image
you only see the plot at some fixed $r$, say, $r=1$
kinda irrelevant anyway bc all this stinks of de Broglie
the picture represents an old, irrelevant and inaccurate model
and then, we have spherical harmonics...
Anonymous
13:28
I think is should be of the form $z=(r^2-(x^2+y^2))\sin(?)$. I don't know what $?$ is.
Anonymous
Anonymous
13:43
@AccidentalFourierTransform How is $p_x=\frac{h}{2\pi i} \frac{\partial }{\partial x}$ ? hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
@Blue by postulate
And use hbar for Christ's sake
Anonymous
@0celóñe7 I don't remember the TeX command for that
Anonymous
Remind me
Anonymous
@0celóñe7 What is "postulate" ?
Anonymous
Oh got it
Anonymous
13:49
$\hbar$
rob
rob
@Blue Consider a plane wave $\psi = \exp{-ikx}$, with momentum $p = \hbar k = h/\lambda$. That's an eigenstate of your operator, whose eigenvalue is the momentum.
In mathematics and in theoretical physics, the Stone–von Neumann theorem is any one of a number of different formulations of the uniqueness of the canonical commutation relations between position and momentum operators. The name is for Marshall Stone and John von Neumann (1931). == Representation issues of the commutation relations == In quantum mechanics, physical observables are represented mathematically by linear operators on Hilbert spaces. For a single particle moving on the real line R, there are two important observables: position and momentum. In the quantum-mechanical description of...
@Blue =1, pls
Anonymous
@rob I don't understand eigenstates. Use simpler language please :P
Anonymous
Say we consider a plane wave $\psi=A\sin(wt-kx)$
rob
rob
@Blue Try applying your operator to my function. What happens?
13:57
"$p$ is a short-hand notation for $-i\partial$" is good enough of a justification for you?
cus thats all there is, to be honest
Anonymous
Okay, first of all...what does one mean by $Ae^{i(kx-wt)}$ is a plane wave? I know $A\sin(kx-wt)$ form. How do we get the former from the latter?
Anonymous
It's not even a superposition of sine and cosine....how does that imaginary term come in?
rob
rob
@Blue Oh, that's a great thing. Do you know Taylor series?
@rob you'd better explain why one can rearrange the terms
Anonymous
@rob Yeah. I know the taylor series for sines, cosines etc
14:00
There are series where rearranging changes the value
rob
rob
For instance, $e^x = 1 + x + x^2/2! + x^3/3! + \cdots$
Anonymous
@rob Yes. I know that
rob
rob
Now try the Taylor series for $e^{ix}$
All the even-powered terms are real, but their signs alternate
All the odd-powered terms are imaginary, and their signs alternate
You can compare series and you discover that $e^{ix} = \cos x + i \sin x$
Anonymous
Right. Then? (We just replace x with ix)
rob
rob
(It's a little much for me to type in chat, though)
Anonymous
14:02
@rob Yup, I know that part
Anonymous
Euler form
I can take it from here
first, forget everything rob said
On my phone, pretty hard to teach
@AccidentalFourierTransform can do it for me
rob
rob
Rearrange: $2i\sin x = e^{ix} - e^{-ix}$
@0celóñe7 dont worry, I am here to help
14:05
I think he knows the math behind going from sin(stuff) to e^i(stuff). Maybe he's asking what is the physical relevance in thinking of e^i(stuff) as a wave?
Anonymous
@rob Okay. So say $2i A\sin(kx-wt) = A(e^{i(kx-wt)}-e^{-i(kx-wt)})$
(I want to know the answer to that myself. Working in complex coordinates simplifies solutions to differential equations, but does that not obscure the physics?)
rob
rob
So, $e^{-i(kx-\omega t}$ is a superposition of a real plane wave and an imaginary plane wave.
Anonymous
@rob Yes. Now here's the problem. An imaginary plane wave makes no sense to me.
snip'd
rob
rob
14:08
@Blue Ah. Interesting.
@Blue does a real one make sense to you?
how?
its a formula/mathematica expression, with no (direct) physical meaning
Anonymous
@AccidentalFourierTransform I can visualize it. I don't know how to visualize the superposition of real plane wave and an imaginary plane wave
Anonymous
For example I can just plot $A\sin(wt-kx)$ on desmos and see how it moves
you can plot $(\sin x,\cos x)$ instead of $\sin x+i\cos x$
same thing
rob
rob
14:10
@Blue When you visualize a real plane wave ... how do you visualize its amplitude?
Are you imagining a wave on a string, where the amplitude is a displacement?
Anonymous
@rob Yes
rob
rob
@Blue What about waves whose amplitude isn't a displacement? I can give you some examples.
real part=vertical displacement; imaginary part=horizontal displacement?
rob
rob
@AccidentalFourierTransform That's one way to imagine it, but not the only one.
but if it works for blue...
after all, the question was $p=-i\partial$ :-P
rob
rob
14:12
QM suddenly makes a lot more sense if you can think a little more in the abstract.
it's kinda amusing how physicists think symbolically sometimes
my smiley has a funny wig
that's the job of weirdo mathematicians, i thought
rob
rob
Example you've seen before: sound waves. The amplitude is a pressure, rather than a displacement. Fair enough, @Blue?
Anonymous
@rob Ah, yes. But that just represents displacement of particles from mean position.
14:15
how about an electromagnetic wave?
rob
rob
@Blue Right, but it's mean displacement. Sound waves are messy, and you don't describe the motion of any individual particle the way you do a string
A displacement-free example: a "spin wave."
Anonymous
@rob Okay. But I still didn't understand how you'd visualize for example $A\cos(wt-kx) + iA\sin(wt-kx)$. I didn't understand what @AccidentalFourierTransform meant by horizontal displacement
rob
rob
Imagine you have a line of people standing in a row.
the trick is not to visualise anything
use symbols
rob
rob
The first one is facing to the north
14:17
thats why maths are so powerful
Drawing pictures is always a mistake
Unless you are trying to define functions
rob
rob
The second one is facing 10° towards the east, the next one is facing 20° towards the east, and so on
Anonymous
@rob Okay?
rob
rob
Now imagine all these people are spinning at the same rate
There are two equivalent ways to describe what's happening
Anonymous
lol...fine. :-)
rob
rob
14:18
one is "everybody's spinning"
@BalarkaSen this is triggering me, he writes vectorspace
But not metricspace
rob
rob
the other is "the person who's facing north is moving along the line"
he's missing the spacebar
Anonymous
Wait a second...spinning about which axis? Each person is rotating in his own position?
rob
rob
This is a wave whose amplitude is orientation, "which way are you facing"
@Blue That's it
Anonymous
14:21
@rob What do you mean by "moving along a line" ? He is not even moving...just rotating
rob
rob
@Blue Right, no one is moving, everyone is rotating.
But a legit question is, "where along the line is there someone who's facing North?"
And the answer to that question is moving along the line as everyone rotates in phase
Anonymous
Okay. Got it till here
rob
rob
You can ask a better question than "where is someone facing North": you can ask "what are the orientations of all the people along this line?"
The answer to that question is a sort of a wave, since everyone's orientation is similar to that of both neighbors but it changes continuously as you move down the line
And that wave travels as everyone turns
Anonymous
Ok. That's similar to a plane progressive wave
Anonymous
But I'm not sure if it would be correct to call that a wave
rob
rob
14:29
@Blue Exactly. But the "amplitude" isn't any kind of displacement
@Blue If its dynamics obey a wave equation, we can call it a wave.
Anonymous
Amplitude can be the angular displacement from any one direction...say North. Even that's a type of displacement :P
rob
rob
@Blue That's another way to think of it, sure.
So earlier, @AccidentalFourierTransform suggested electromagnetic waves as another example.
Anonymous
In EM waves the magnitude of field changes. Can be thought of as displacement of magnitude (with sign) from mean magnitude (i.e. 0).
Anonymous
rob
rob
Most drawings of EM waves lend themselves naturally to an analogy with a wave on a string, because you usually draw the electric field vector as a set of arrows originating from a line
Yes, exactly that sort of drawing
rob
rob
But an EM wave is an oscillation in two vector fields. There are six vector components to describe at every point in space, three for E, three for B
@ACuriousMind can you please stop enjoying yourself and come back? Ive got stuff Id like to discuss
rob
rob
It turns out that there are some simplifying relationships that let you specify fewer than six of these components. But the whole story has many coupled waves moving together.
lo and behold, $\boldsymbol E+i\boldsymbol B$
rob
rob
And even once you've simplified, you still have two wave polarizations whose amplitudes can be specified independently.
14:37
bam, a photon was born
Anonymous
@AccidentalFourierTransform I think that is a better way to think. Plot $A\cos(wt-kx)$ like the Electric field is shown in the diagram and $iA\sin(wt-kx)$ as the magnetic field is shown. Makes more sense
rob
rob
@Blue This is totally reasonable.
Other visualizations are also possible and might be more appropriate in other circumstances.
yeah where even is ACM???
i want to talk about gauge theory with him
ACM is the new chris white
he's the anti-chris cause no one likes him
Anonymous
14:42
ACM is the new blue moon.
folks, just wait until tmr, he will be back from wacken
y u obsessed with blue, blue?
rob
rob
Another light-based example:
In electrodynamics, circular polarization of an electromagnetic wave is a polarization state in which, at each point, the electric field of the wave has a constant magnitude but its direction rotates with time at a steady rate in a plane perpendicular to the direction of the wave. In electrodynamics the strength and direction of an electric field is defined by its electric field vector. In the case of a circularly polarized wave, as seen in the accompanying animation, the tip of the electric field vector, at a given point in space, describes a circle as time progresses. At any instant of time,...
You have vertically-polarized light and horizontally-polarized light with the same amplitude, but phase-shifted
Anonymous
rob
rob
When the vertical field is large and up, the horizontal field is zero
When the vertical field goes to zero, the horizontal field is maximized, say, to the left
When the horizontal field vanishes again, the vertical field is large and points down
rob
rob
You can encode the electric field for circular-polarized light as $\cos(kx-\omega t) + i \sin(kx-\omega t)$, where the real part gives you the vertical electric field component and the imaginary part gives you the horizontal electric field component
Anonymous
For example this is the Electric field (in the gif I posted). We could just superpose a Magnetic field perpendicularly. That would require 6 components to specify. We could then just represent it by $A\cos(wt-k^1x^1-k^2x^2-k^3x^3)+iA\sin(wt-k'^1x^1-k'^2x^2-k'^3x^3)$. Am I right? @rob
What are you on about
But quantum wave functions have nothing to do with EM fields
rob
rob
But that's just $e^{-i(kx-\omega t)}$
@Blue You're on the right track here
Anonymous
Yay ! :D
14:47
@BalarkaSen it's interesting when they think so symbolically that they don't stop to make sure the symbols make sense (really bad math in QM)
Sid
Sid
Someone tell me, why this is wrong: codechef.com/viewsolution/14812521
(Problem is to check if the number is prime)
rob
rob
@Blue Have another look at the electric field animation in the onebox for circular polarization above
@Blue I have a song for you: youtube.com/watch?v=78LMDe-78q4
rob
rob
Look at how the magnitude of the vector is the same everywhere, even though the orientation of the vector changes
The magnitude is $E = \sqrt{E_H^2 + E_V^2}$
Anonymous
@rob Yes. Yes! Gotcha!
Anonymous
14:52
BTW it makes no sense to consider a plane progressive 1D wave as $Ae^{i(wt-kx)}$ as some books do
rob
rob
Now if the horizontal component is the real part, and the vertical component is the imaginary part, the magnitude is $E = \sqrt{Re(E)^2 + Im(E)^2}$
Anonymous
For example a wave in a string
Anonymous
@rob Right
rob
rob
@Blue Now think of a Schrödinger wave $\psi = e^{-i(kx-\omega t)}$
You turn that into a probability density by taking $\psi^*\psi$
But the arithmetic is just the same as for the circularly polarized light
Anonymous
@Sid Obviously it is wrong. c could become 2 even when it is not prime in your code. You should check if c==2 outside the loop
rob
rob
14:55
So $e^{ikx}$ is a plane wave whose magnitude is the same everywhere.
@rob how is = 1 everywhere a probability density
rob
rob
@0celóñe7 blah blah limits and normalization
Anonymous
@rob What is $\psi^*\psi$ ?
Sid
Sid
@Blue Outside the loop? It's outside the for loop only. That's the only real "logic" part of the whole program.
rob
rob
14:56
@Blue Wavefunction multiplied by its complex conjugate
Anonymous
@Sid Just indent your lines properly. It's so difficult to read your code =P
Sid
Sid
Gah. Indentation is for kids. :P
and please, please, use spaces
Anonymous
14:58
@rob How does that give probability density?
rob
rob
@Blue That's a central hypothesis in QM ... it's the most common interpretation of the wavefunction.
That is, that the absolute magnitude of the wavefunction at some point in space is proportional to the probability of detecting your particle at that point
the -+++ vs +--- of computing
Anonymous
@rob Ok. So you are basically using $z \bar{z}=|z|^2$ ?
rob
rob
@Blue Bingo
Anonymous
Also why is $|\psi|$ not the probability function? Why $|\psi|^2$ ?
Anonymous
15:02
@rob :)
rob
rob
@Blue We don't get to choose the reality we live in
Anonymous
@rob What?
rob
rob
@Blue $\psi$ obeys a wave equation. $|\psi|^2$ gives you a probability density. Other options give wrong predictions.
Anonymous
@rob Okay. So that's an experimental observation ?
rob
rob
@Blue The gold standard for deciding whether a thing actually happens
rob
rob
Anyway, back to where we started: consider a plane wave $e^{ikx}$
Operate with $\frac\hbar i \frac{\partial}{\partial x}$.
What do you get?
Anonymous
$\hbar k e^{ikx}$ ?
rob
rob
@Blue Yes: your original function $e^{ikx}$, multiplied by a constant $\hbar k = p$.
Anonymous
$\frac{h}{\lambda}$ gives momentum of particle
Anonymous
Yup!
15:08
plausibility arguments aside, $p=-i\partial$ is just the definition of the symbol $p$. We call this the "momentum operator". The natural question -- how is this new operation related to the classical notion of momentum -- requires some maths to fully answer
rob
rob
So the operator gives you back your function again, multiplied by a constant
but, in essence, you need the Ehrenfest theorem.
rob
rob
That's special, but we want to sound smart, so we use the German "eigen" instead of "special"
Anonymous
@rob So performing $\frac\hbar i \frac{\partial}{\partial r}$ basically gives back momentum multiplied by the e^i{} stuff. Gotcha
rob
rob
It's an "eigenfunction" of the operator, and the "eigenvalue" is the momentum $\hbar k$
Anonymous
15:11
One thing which is still confusing me is "In quantum mechanics the solutions of the Schrödinger wave equation are by their very nature complex-valued and in the simplest instance take a form identical to the complex plane wave representation above. The imaginary component in that instance however has not been introduced for the purpose of mathematical expediency but is in fact an inherent part of the “wave”." Why? Why should the waves be complex valued?
Anonymous
Can't they be ordinary 1D waves moving in a specific direction like $A\sin(wt-kx)$
rob
rob
@Blue There are special, restricted problems where you can describe a system using a completely real wavefunction --- for instance, the particle in a box. But that turns out not to work in general.
Anonymous
@rob Okay. Is that again: experimental ? =P
rob
rob
@Blue Yep
Anonymous
Phew....XD
Anonymous
15:14
I wouldn't like to hear something like Heisenberg's Uncertainty Principle is also experimental. I'll lose my faith in QM then. lol
no, it's a theorem
rob
rob
@Blue It's a theoretical prediction with experimental support
Anonymous
@AccidentalFourierTransform That's good :D
Anonymous
@rob @AccidentalFourierTransform Thanks a ton for the help!
Anonymous
Going for dinner now :)
15:17
you owe me a beer
rob
rob
@Blue Cool. Eat well.
Anonymous
@AccidentalFourierTransform I'll e-mail it to you. ;)
Heisenberg's principle is just a statement about growth of Fourier transforms
ba dum tss
no wait
what a stupid question
15:22
your reputation is scarred forever for asking such a stupid qn
rip
i know i know
(ಥ_ಥ)
are these even alphabets in any language
like, m8...
What did he ask?
you'll never know
@AccidentalFourierTransform set tabs to spaces, problem solved
rob
rob
15:32
@0celóñe7 Abusing my moderator superpowers: he asked how many licks it takes to get to the center of a Tootsie Roll Pop
people who use tabs should be publicly executed
rob
rob
@AccidentalFourierTransform That's just what my bartender says
rob
rob
@AccidentalFourierTransform I hope that it's obvious I'm being silly.
yes, yes it is ;-)
15:34
@AccidentalFourierTransform I use the tab key, but it just inserts 4 spaces
that is twice as wrong
Saves 3 extra button presses :)
you dont deserve the time you saved
BTW no Silicon Valley fans here?
what a shame
I'm supposing Silicon Valley is more than a locale in Cali?
It's a shitty TV show
Oh, now that I think about it, that tabs v spaces link did have that shitty guy from the Verizon ads...he's an actual actor and not just a shitty awkward sales rep?
@BalarkaSen actually that version of HUP only works for wave functions that decay
@AccidentalFourierTransform Using 8 spaces for 1 tab is wasting monitor real estate with completely useless white space
(technically dark space since I use a dark theme)
> I use a dark theme
dude, what's wrong with you?
He's a quant
15:39
I like my eyes
I am a quant
He went to the dark side years ago
@KyleKanos last night I realized I need a dark theme on all of Windows
It's hard to play a dark game while my second monitor is the surface of the sun
How do I do that?
easy: don't play games at night
Which windows version? 10? 7? XP?
15:41
@KyleKanos 10
Settings > personalization > colors should have some dark theme options
@KyleKanos will check when I'm home, thanks
@0celóñe7 No Problemo
@0celóñe7 What I had in mind was $(\int_{\Bbb R} x^2 |f(x)|^2 dx)(\int_{\Bbb R} y^2 |\hat{f}(y)|^2 dy) \geq C$ for some constant $C$ (= $\|f\|/4\pi$ or something? I forget)
now, of course $f$ needs to have enough decay so that that integral makes sense
15:45
Alright, step 1 complete: new OS image loaded to flash drive. now comes the difficult step 2 & 3: backup everything and then install new OS
modulo that it's Cauchy-Schwarz really
in L^2
@KyleKanos what OS? Windows? OSX?
@JohnRennie TrueOS (BSD)
I'm one of "those" people that don't use common OSes
@BalarkaSen since when is y^2 integrable?
The problem with HUP is that the statement makes sense for many functions but the details require delicate function theory.
when did i say it was
15:49
You want to use Cauchy Schwarz
@KyleKanos ah I had heard a rumour that someone in the world was using BSD and now I know who it is.
Oh I didn't read that right, on mobile
@JohnRennie Haha, there are more than 1 of us
come on man
15:51
I bet you can all fit simultaneously into the same phone box though :-)
i don't know analysis and i am a hippie
but i don't do that much drugs
@JohnRennie If we can find one of those in the world....
@BalarkaSen really the delicate question is when those integrals converge
@KyleKanos weren't a whole lot of the old British phone boxes sold off around the world as collectors items?
There must be one near you.
15:53
@BalarkaSen but my retort about decay was in reference to the Fourier HUP on Wikipedia
Their proof uses integration by parts at infinity
I was referring specifically to that
Their proof uses integration by parts at infinity
bwikes
The internet here is shit
@BalarkaSen what?
a modified version of yikes
@ the fact that wiki uses integration by parts at infinity
everyone knows you can only use integration by parts at finity.
@BalarkaSen remind me about this later pls
Currently getting food materials
15:58
sure thing
 
2 hours later…
17:39
@BalarkaSen Ok, what exactly is your CS argument?
I'm busy with chemistry a little. Can we postpone this discussion?
@BalarkaSen Yes, I should be done with errands
My errands are mostly dull and uneventful
@BalarkaSen I live in DC so I've been in traffic for hours today
17:44
@0celóñe7 You need a cowplow
@BalarkaSen I would suggest Federer for general MT but his notation and definitions are nonstandard.
He defines step functions as functions having countable image, which seems to make things unnecessarily complicated because you have infinite sums and limits running around.

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