The h Bar

1:00 PM

$A\sin(wt-k^1x^1-k^2x^2-k^3x^3)$ 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

Ben Niehoff

@TheRaidersofLasVegas Not sure

1:03 PM

@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

1:06 PM

"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

Secret

and then, we have spherical harmonics...

Anonymous

1:28 PM

I think is should be of the form $z=(r^2-(x^2+y^2))\sin(?)$. I don't know what $?$ is.

Anonymous

geogebra.org/3d

Anonymous

1:43 PM

@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

1:49 PM

$\hbar$

@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.

Stone–von Neumann theorem

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 $\psi=A\sin(wt-kx)$

@Blue Try applying your operator to my function. What happens?

1:57 PM

"$p$ 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?

@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

2:00 PM

There are series where rearranging changes the value

For instance, $e^x = 1 + x + x^2/2! + x^3/3! + \cdots$

Anonymous

@rob Yes. I know that

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)

(It's a little much for me to type in chat, though)

Anonymous

2:02 PM

@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

Rearrange: $2i\sin x = e^{ix} - e^{-ix}$

@0celóñe7 dont worry, I am here to help

2:05 PM

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?)

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

2:08 PM

@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

oh no one can

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

2:10 PM

@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

@Blue What about waves whose amplitude isn't a displacement? I can give you some examples.

real part=vertical displacement; imaginary part=horizontal displacement?

@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

2:12 PM

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

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.

2:15 PM

how about an electromagnetic wave?

@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

Imagine you have a line of people standing in a row.

the trick is not to visualise anything

use symbols

The first one is facing to the north

2:17 PM

thats why maths are so powerful

Drawing pictures is always a mistake

Unless you are trying to define functions

The second one is facing 10° towards the east, the next one is facing 20° towards the east, and so on

Anonymous

@rob Okay?

Now imagine all these people are spinning at the same rate

There are two equivalent ways to describe what's happening

Anonymous

lol...fine. :-)

2:18 PM

one is "everybody's spinning"

@BalarkaSen this is triggering me, he writes vectorspace

But not metricspace

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?

This is a wave whose amplitude is orientation, "which way are you facing"

@Blue That's it

Anonymous

2:21 PM

@rob What do you mean by "moving along a line" ? He is not even moving...just rotating

@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

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

2:29 PM

@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

@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

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

2:34 PM

lol

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

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, $\boldsymbol E+i\boldsymbol B$

And even once you've simplified, you still have two wave polarizations whose amplitudes can be specified independently.

2:37 PM

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

$\LARGE🌚$

@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

2:42 PM

ACM is the new blue moon.

Secret

folks, just wait until tmr, he will be back from wacken

y u obsessed with blue, blue?

Another light-based example:

Circular polarization

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

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

2:45 PM

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

But that's just $e^{-i(kx-\omega t)}$

@Blue You're on the right track here

Anonymous

Yay ! :D

2:47 PM

@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

Someone tell me, why this is wrong: codechef.com/viewsolution/14812521

(Problem is to check if the number is prime)

@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

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

2:52 PM

BTW it makes no sense to consider a plane progressive 1D wave as $Ae^{i(wt-kx)}$ as some books do

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

@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

2:55 PM

So $e^{ikx}$ is a plane wave whose magnitude is the same everywhere.

@rob how is = 1 everywhere a probability density

@0celóñe7 blah blah limits and normalization

Anonymous

@rob What is $\psi^*\psi$ ?

Sid

@Blue Outside the loop? It's outside the for loop only. That's the only real "logic" part of the whole program.

2:56 PM

@Blue Wavefunction multiplied by its complex conjugate

:D

Anonymous

@Sid Just indent your lines properly. It's so difficult to read your code =P

^

Sid

Gah. Indentation is for kids. :P

and please, please, use spaces

Anonymous

2:58 PM

@rob How does that give probability density?

@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

Tabs versus Spaces

►

2

the -+++ vs +--- of computing

Anonymous

@rob Ok. So you are basically using $z \bar{z}=|z|^2$ ?

@Blue Bingo

Anonymous

Also why is $|\psi|$ not the probability function? Why $|\psi|^2$ ?

Anonymous

3:02 PM

@rob :)

@Blue We don't get to choose the reality we live in

Anonymous

@rob What?

@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 ?

@Blue The gold standard for deciding whether a thing actually happens

3:05 PM

@Blue my very first answer on this site :-)

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

$\hbar k e^{ikx}$ ?

@Blue Yes: your original function $e^{ikx}$, multiplied by a constant $\hbar k = p$.

Anonymous

$\frac{h}{\lambda}$ gives momentum of particle

Anonymous

Yup!

3:08 PM

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

So the operator gives you back your function again, multiplied by a constant

but, in essence, you need the Ehrenfest theorem.

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

It's an "eigenfunction" of the operator, and the "eigenvalue" is the momentum $\hbar k$

Anonymous

3:11 PM

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 $A\sin(wt-kx)$

@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

@Blue Yep

Anonymous

Phew....XD

Anonymous

3:14 PM

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

@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 :)

3:17 PM

you owe me a beer

@Blue Cool. Eat well.

Anonymous

@AccidentalFourierTransform I'll e-mail it to you. ;)

good

Heisenberg's principle is just a statement about growth of Fourier transforms

ba dum tss

no wait

what a stupid question

3:22 PM

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

3:32 PM

@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

@AccidentalFourierTransform That's just what my bartender says

@rob DELETE THIS

@AccidentalFourierTransform I hope that it's obvious I'm being silly.

yes, yes it is ;-)

3:34 PM

@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

3:37 PM

>:|

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

3:39 PM

I like my eyes

I cant even

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?

3:41 PM

@KyleKanos 10

Settings > personalization > colors should have some dark theme options

Original here

@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

3:45 PM

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

3:49 PM

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

3:51 PM

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

I see

@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.

3:53 PM

@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

3:58 PM

sure thing

5:39 PM

@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

Ack

5:44 PM

@0celóñe7 You need a cowplow