The h Bar

Stan Shunpike

00:02

What is a momentum operator?

user54412

@StanShunpike $g(\vec{e}_{(i)}, \vec{e}_{(j)}) = g_{kl} (\mathrm{d}x^k \otimes \mathrm{d}x^l)(\vec{e}_{(i)}, \vec{e}_{(j)}) = g_{kl} \mathrm{d}x^k(\vec{e}_{(i)}) \mathrm{d}x^l(\vec{e}_{(j)}) = g_{kl} \delta^k_i \delta^l_j = g_{ij}$

Stan Shunpike

Lol hang on I'm getting my laptop lol

Chajax

user54412

i've got to go, but digest that equation :)

user54412

also, am I the only one left in the room who doesn't use chatjax?

Stan Shunpike

Doubt it, but it is useful

user54412

00:05

also also, I was inconsistent with that notation

user54412

I usually place indices inside parentheses to indicate which vector/tensorial/form object I'm using, and indices without parentheses mean the component of the object

user54412

I did that for the vectors, but maybe I should have for the forms too

user54412

meh

0celo7

00:32

@ChrisWhite Eh, I don't like that notation. It should be clear from context if the index is a component index or a coordinate label.

Parentheses look ugly.

Stan Shunpike

@0celo7 Can every wave function be written as $\Psi = A e^{i(kx-\omega t)}%?

0celo7

@StanShunpike Naw.

Stan Shunpike

Didn't think so.

0celo7

That would make for a very boring world.

Stan Shunpike

But I don't get what the operators in the SE do. They make sense to me if the wave function is of that form. But the simple statement "operators map from one Hilbert space vector to another" doesn't.

@0celo7

Like how do I know which vector it maps to?

It could be any one.

0celo7

00:40

@StanShunpike Uh, it maps to the one you get when you apply the operator...

Stan Shunpike

How do we know which one that is? It just sounds like a vague thing.

@0celo7 like usually when I use an operator, I am given a matrix for it. So I know how to operate with it. I don't know what do do with for instant the Hamiltonian operator for the Coulomb potential

Like are they matrices?

0celo7

Infinite dimensional matrices, although the mathematical physicists might call me out on that.

alarge

@StanShunpike Maybe start think about simpler problems first. Like take the Laplacian operator in some simple geometry and find its eigenvectors. I suppose you could then discretize the operator to a matrix and numerically find its eigenvectors, too, just to see how they compare.

Stan Shunpike

@alarge What do you mean?

Like isn't the 3D wave function just the Laplacian?

0celo7

@StanShunpike The Laplacian is an operator, not a function.

Stan Shunpike

00:55

Sorry, I bungled my words

For the 3D wave function, isn't the momentum operator the Laplacian.

alarge

@StanShunpike Laplacian is the operator, $\nabla^2$ or however you want to write it. So you can trivially approximate that by a matrix, for example (with finite differences, say; but do remember to include the BCs).

Just forget QM for a bit, and just concentrate on computing the eigenvectors (eigenfunctions) of the Laplacian as a start (also numerically, if you think that is something that might interest you).

Stan Shunpike

Is the collection of all the eigenvalues for an operator called the spectrum?

dmckee

@Sean I had a staff scientist acting as a mentor when I first started full time research who used to introduce grad students as "the future doctor <name>". A lot of our high rep users are future doctors.

Mark Mitchison

@StanShunpike Thinking about the Coulomb potential or the Laplacian as a matrix might be a bit over the top. Does it make sense to you if I say that differentiation maps the function $f_1(x) = x^2/2$ to the new function $f_2(x) = df_1/dx = x$? Then you should be able to accept that $d/dx$ is an operator that maps from one function to another.

The only difference from what you wrote about operators mapping from one vector to another is that here we are talking about functions, not vectors.

So the only conceptual leap now is that functions $\approx$ infinite-dimensional vectors

alarge

@MarkMitchison When we're talking linear operators, I think there's some intuition in noting that they are in some sense infinite matrices (so some things, like the spectrum as pointed out by Stan, generalize nicely). But yes, there are some intricacies with infinite-dimensional spaces, so in that sense the matrix analogy is a bit "over the top". But again, a trivial numerical implementation (of say, solving the Poisson equation to take a simple example) might use matrices.

Mark Mitchison

01:11

@alarge Sorry, I didn't mean to tread on your toes there, and I agree that the idea of infinite-dimensional matrices is enormously useful. I just mean that one does not need to appeal to the idea of matrix representations in order to understand why people call $-\frac{\hbar^2\nabla^2}{2m}$ an operator.

It is just a thing that takes in a function and spits out another one.

That seemed to be the core of @StanShunpike's confusion, although perhaps I'm wrong

Stan Shunpike

01:29

@MarkMitchison I like your example. It works well. But how do I know what specific operation the operators in the SE perform on the wave function.

Mark Mitchison

@StanShunpike In exactly the same way. The Coulomb potential $1/r$ takes a wavefunction (vector) $\psi(r)$ and maps it onto a new wavefunction (vector) $\psi^\prime(r) = \psi(r)/r$.

You literally just take the function and divide it by $r$.

Stan Shunpike

Is $\psi'(r)$ a derivative or just denoting function after being operated on?

Mark Mitchison

Sorry, it's not a derivative

I just mean $\psi_{new} = \psi_{old}/r$

Stan Shunpike

What is the time derivative of the wave function? Like when I have taken partial derivatives they are usually of the form $f(x,y) = xy$ or something like that. So it's obvious how to take a partial derivative. But I thought the probability amplitude yielded a complex number not another function.

@MarkMitchison ^

Mark Mitchison

@StanShunpike I'm not sure I understand the question. A probability amplitude is a complex number.

If you want to find the time derivative you just differentiate with respect to time.

Stan Shunpike

01:43

That's it. But isn't that always zero?

Mark Mitchison

Are you asking about the physical interpretation of $\partial\psi/\partial t$?

No.

Stan Shunpike

Why?

Oh, because there is no time element?

Mark Mitchison

Well, if the wave function depends on time then the derivative is not zero

In almost all cases this is true

user105491

If there's no variable in time $t$, then the partial derivative is $0$. However, there are many interesting wavefunctions whose time derivative isn't zero.

Stan Shunpike

Okay, so how do we know the time dependency of $\psi$?

Mark Mitchison

01:44

That's what the Schrodinger equation is for!!

user105491

Yup. Look at the Hamiltonian operator. (I think.)

Mark Mitchison

@StanShunpike You have to solve it to find the time dependence.

Finding the time dependence of the wave function is the quantum mechanical problem. If you can solve that then you can calculate anything you want.

Stan Shunpike

So for particle in a box, what is the wave function? If you don't know off hand, that's okay I'll try to figure it out. I'm just curious because I don't think I fully understand the SE purpose. So the SE defines a PDE whose solutions are the wave function for the particle and allow us to find the probability amplitude at different times?

So really, it isn't nearly as complicated as I'm thinking it is.

I probably just haven't seen enough examples and therefore have never really seen what a solution looks like.

user105491

You might want to take the 8.05x course online at edX offered by MIT, @StanShunpike.

user105491

I'm taking it as well, and it's challenging enough to be fun.

Stan Shunpike

01:57

Great suggestion. It looks hard enough to be interesting. Thanks!

user105491

Sure!

Stan Shunpike

Are you a chat regular?

user105491

Me?

user105491

I come to the homotopy theory chat room often.

user105491

I'm an aspiring mathematician.

Stan Shunpike

01:59

Nice, so is my cousin. My uncle is a mathematician at Drexler.

user105491

Nice!

user105491

I used to do some physics some years ago but then got bored of manipulating equations. I found that I loved topology and algebra more, and got into it. So now I'm just trying to study some physics, recall a few things.

Mark Mitchison

@StanShunpike The wave function at some time depends on the initial conditions. Just like most problems in physics!

But the usual trick is to first find the energy eigenfunctions.

Kyle Kanos

@SanathDevalapurkar Then you probably weren't really doing physics

Mark Mitchison

Then any wave function can be represented as a superposition of these, each oscillating with a phase.

user105491

02:01

Yes, @Kyle, I probably wasn't! :-P

Mark Mitchison

(This is basically equivalent to solving the differential equation in the Laplace domain rather than the time domain.)

Stan Shunpike

@MarkMitchison What is an example of "initial conditions"?

Mark Mitchison

02:34

@StanShunpike For example, suppose I have a particle in 1D perfectly localised at the position $x_0$ at time $t = 0$. Then we have that $\psi(x,0) = \delta(x-x_0)$. Using the Schroedinger equation, you can then solve for the wavefunction $\psi(x,t)$ at future times. From this you can calculate the position-space probablity distribution or any other observable quantity you choose.

Generally speaking, you would expect the wave function to initially spread out. If there is a confining potential, this will tend to inhibit this spreading. For example if you have a square well you would expect reflection at the boundaries and so on. All of this information is captured by the Hamiltonian and its eigenfunctions.

Stan Shunpike

@MarkMitchison is the SE like the EFEs in that it doesn't just work for any old function...you have to specify the wave function in question and show that this equation solves for it.

And that may be different for different cases then.

0celo7

@StanShunpike Of course it doesn't work for every function...it's a differential equation to be solved for the function.

It seems I left a pot out in the garage when it was colder and it was OK to do so.

Now I have to get rid of a month's old half-full pot of pasta sauce.

Stan Shunpike

02:50

@0celo7 Well, smarty pants, I'm still learning! :p But yeah, now that you mention it, that was a very obvious statement. I just thought there was a general wave function it worked for. I didn't realize that you had to find specific wave functions for different cases. No wonder I have been confused!

@0celo7 Why would you leave a pot in the garage? Is your garage near your kitchen?

0celo7

@StanShunpike Well with $V=0$ it always works with plane waves.

@StanShunpike When it's colder the garage is like a really big fridge.

But it's been 70 degrees for like a week, so I'm scared what I'll find.

Stan Shunpike

For a week? Dude, lol. Who know what might be growing there....

@0

0celo7

Oh man, chance of 1-3 inches of snow

Pray for a snow day

::does the snow dance::

Stan Shunpike

We almost never got snowdays when I was in K-12

I went to the same school and we always had school.

Didn't matter how much it snowed

0celo7

State?

Stan Shunpike

02:56

IL

0celo7

In VA we melt in the snow

We're the northernmost southerners

Stan Shunpike

So you how much do you get on average? A few inches?

0celo7

Although some MD people consider themselves southern.

A few inches.

Kyle Kanos

@0celo7 They're below the canonical M-D line

0celo7

The problem is that the chilluns don't know how to drive.

@KyleKanos Some texts don't abide by that rule.

Mine did, but we learned both conventions for the AP test.

They will never ask if MD is a southern state because not everyone agrees.

Stan Shunpike

02:59

Are all the possible solutions to the SE known?

Kyle Kanos

It probably depends on the where in the state they're from

Like the more southern they are, the more likely they say they're Southerners and vice versa

0celo7

The county doesn't want to be responsible for damages to student vehicles. The first snow we had this year was 1 inch...300 accidents in the tri-county area.

~70 by 8 am in our county alone.

I got to school 2 hours late...I live 15 minutes away.

Kyle Kanos

That's awful

I was born & raised in NJ, I'm perfectly capable of driving in the snow

The idiots in SC cry for fear when we get 0.5" of snow

We literally closed campus this year with that much snow

Stan Shunpike

Both of those comments are crazy lol

70 accidents by 8am...wtf

closing a school for .5 inch of snow...

0celo7

Yo that pot is fucking toxic.

Most vile thing I've smelled this year.

user54412

03:03

Snow closings are great. I get to drive with no traffic into work, and I get to park wherever I want on campus.

Stan Shunpike

lol yeah there are definitely perks to them

user54412

(Yes, for grad students snow day = more work day)

NeuroFuzzy

Southern california...

sometimes the sky gets a bit grey?

:P

that can make driving sooo hard

Kyle Kanos

Well to be fair, I also went about 2 or 3 years without snow down here

NeuroFuzzy

And the moment I take a road trip I'm going to get into an accident haha

road trip in winter*

0celo7

03:05

@NeuroFuzzy My sister lives in Studio City and she sends out mass texts to everyone in the family when the temperature drops below 50.

"It's a Christmas miracle"

Kyle Kanos

My wife was talking to her mom (who still is in NJ) this afternoon and said it was a "bit chilly here." I commented loud enough (so the MIL could hear) that she meant 60 degrees F.

0celo7

That was horrifying.

There were gray tomato chunks in there.

NeuroFuzzy

Huh?

0celo7

@NeuroFuzzy Really old pot of pasta sauce had to be disposed of.

user54412

@0celo7 Oh so you've been to NJ I see?

Mark Mitchison

03:11

@StanShunpike Definitely not. There are lots of potentials for which the solution is not known. When you add more particles into the mix, it becomes even more complicated. Note that the Schroedinger equation is the most fundamental equation of Nature that we know how to write down. If you knew all the solutions you'd know a hell of a lot more than could reasonably be expected.

Kyle Kanos

@0celo7 Ok, that totally clarifies the Yo that pot is fucking toxic statement from earlier

0celo7

@KyleKanos Lol.

Stan Shunpike

@MarkMitchison Why does adding more particles create problems?

Mark Mitchison

We only know how to solve problems with two particles.

In general

No one knows how to solve the many-body problem

You want to know why physically?

In quantum mechanics it is especially complex because of the way quantum correlations work. The correlations between quantum mechanical particles are so strong that one requires an exponential amount of resources to even write down a general state.

So even using computers to solve by brute force fails once you have more than, say, 20 spin-half particles on a lattice.

Stan Shunpike

I have heard we only have exact solutions to the hydrogen atom. Is this why?

Mark Mitchison

03:17

@StanShunpike There are quite a few exact solutions known.

Kyle Kanos

Ooo: V=0

Mark Mitchison

The hydrogen atom is one of the most interesting, since it describes a real physical system (the Hydrogen atom) to a very good approximation.

0celo7

List of quantum-mechanical systems with analytical solutions

Much insight in quantum mechanics can be gained from understanding the solutions to the time-dependent non-relativistic Schrödinger equation in an appropriate configuration space. In vector Cartesian coordinates , the equation takes the form in which is the wavefunction of the system, H is the Hamiltonian operator, and T and V are the operators for the kinetic energy and potential energy, respectively. (Common forms of these operators appear in the square brackets.) The quantity t is the time. Stationary states of this equation are found by solving the eigenvalue-eigenfunction (time-independent...

Mark Mitchison

The harmonic oscillator is another good 'un

Stan Shunpike

I must be mixing up my words lol. So those are "exact" solutions. So there's nothing special about the hydrogen atom in terms of what we know about it relative to other atoms?

Mark Mitchison

03:19

No, I would say that hydrogen is pretty special.

Kyle Kanos

Hydrogen is the best element

Mark Mitchison

You can calculate most of its measurable properties to good accuracy using analytical techniques

0celo7

@KyleKanos You're not biased at all...

Mark Mitchison

No other atom has that property.

Although helium obviously is close.

0celo7

It is the simplest one.

Kyle Kanos

03:20

@0celo7 I'm a carbon-based life form, so I should be biased towards that

Stan Shunpike

There ya go! that's what I was trying to comment on. So that's what makes hydrogen special.

Mark Mitchison

And also Rydberg atoms.

Kyle Kanos

But I am (supposedly) an astrophysicist, so I should care about H a bit more than He and the metals

Mark Mitchison

Yeah hydrogen is the simplest element involving only two particles, and it is therefore special.

If you neglect the quantum fluctuations of the electromagnetic field, then the spectrum of hydrogen can be calculated exactly, unlike any other atom.

Ah, sorry. You were asking why we can only solve one atom exactly.

Yes, this is exactly why hydrogen is special in that sense.

i very much doubt that Wiki article gives a complete list of known solutions, by the way.

Those are probably just some of the most famous.

Stan Shunpike

Yeah, but it's till a lot. There were 20-30

0celo7

03:24

@MarkMitchison I'd imagine one could find a more complete catalogue on arXiv.

Mark Mitchison

If you wanted to parse a lot of data, sure.

I can already think of several important ones that are not on there, mostly many-body problems.

user54412

@StanShunpike I'll just remind you that "exact solution" and "closed form" and "analytic" are pretty poorly defined

Mark Mitchison

But it doesn't matter too much, the important point is that we know a few solutions but most interesting problems cannot be solved exactly.

And yeah, what ^Chris White said.

Stan Shunpike

@ChrisWhite Yeah, they seem to be. And I don't really accept or use a word until I have a specific definition.

user54412

It's @Waffle'sCrazyPeanut !

Waffle's Crazy Peanut

03:26

Hey @Chris! :D

How's your day going?

user54412

Welcome to Mathematical Physics Chat. What flavor do you like your category theory?

6

Waffle's Crazy Peanut

waffles didn't wake up in the morning to see math! Gah...

@KyleKanos: It seems like you're welcoming people to "Physics" itself, instead of "Physics.SE"

Welcome to Physics! Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. — Kyle Kanos 8 hours ago

user54412

@Waffle'sCrazyPeanut My code is free of known bugs... for the moment. So that's good.

Mark Mitchison

@ChrisWhite Ha!

DanielSank

@ChrisWhite Strawberry.

Kyle Kanos

03:30

@Waffle'sCrazyPeanut That's the AutoReviewer thingy

Waffle's Crazy Peanut

@ChrisWhite What does your code do?

@KyleKanos I see. But, what if the new guy is a physicist himself? :P

Kyle Kanos

@Waffle'sCrazyPeanut So I presume he'll be intelligent enough to know that it is the site that I am referring to

Waffle's Crazy Peanut

That's a lot of expectation there...

0celo7

Ugh, school will be on time tomorrow...

Waffle's Crazy Peanut

Seeing both profiles, is @ChrisWhite & @KyleKanos writing code on (more or less) the same specialization?

0celo7

03:34

The air needs to drop ~15 degrees.

user54412

@Waffle'sCrazyPeanut Evolves relativistic plasma in arbitrary coordinates.

Kyle Kanos

@Waffle'sCrazyPeanut Astrophysics is really the only commonality

user54412

yeah, we're a few orders of magnitude apart in size, timescales, and density I think

user54412

but hey, what's a factor of 10000 between astrophysicists?

Kyle Kanos

40 pc diameter, 1500 years, -1.0 <= log(n) <= 4

Waffle's Crazy Peanut

03:36

Heh... :D

user54412

less than 10^9 km, a few light crossing times, and um.. yeah, there's probably a physical interpretation of code units for density somewhere

Waffle's Crazy Peanut

Only when we reach the third year, we realize that there are literally no placements for aeronautics in our state (well, most of the state colleges at least), which enforces people to either go for higher studies, or pursue a totally different career like IT ... (sigh)

user54412

@Waffle'sCrazyPeanut do people leave the state looking for work?

Waffle's Crazy Peanut

03:51

@ChrisWhite Yearly statistics are that some of them leave the state looking for work and higher education while some others attend IT placements, get <some job> (no matter where) and settle down and the remaining few are either unemployed or <do something> inside the state...

user54412

random aside: I think the aviation people need a bit more physics:

user54412

9

Q: Were there any successful flying machines that did not rely on Bernoulli's Principle?

One of Leonardo da Vinci's flying machines appears to be a giant auger. This design would not have exploited Bernoulli's Principle, but rather used the blade's rotation to push the air below it down to keep the craft airborne. Image credit: Wikimedia Were there any flying machines built like...

aerodynamics

Stan Shunpike

Specialization of labour

Waffle's Crazy Peanut

To be honest, most of the mechanical-related engineering studies are boring. And, aeronautics is too boring, because one has to assume a lot. By lot, I mean assumptions are too greater than conclusions...

user54412

@Waffle I've always wondered: what do all the IT people do in India?

Waffle's Crazy Peanut

03:56

@ChrisWhite As far as I've heard, it's just Excel-based work - not much serious code, unless its somewhat reputed company which selectively picks the students...

Stan Shunpike

@Waffle'sCrazyPeanut where in India are you from?

Waffle's Crazy Peanut

@ChrisWhite You'll be a nice candidate for that... :)

@StanShunpike Chennai

Stan Shunpike

Btw your SE name is definitely way cooler than mine and I thought mine was pretty awesome to begin with :P

Waffle's Crazy Peanut

waffles is feeling proud

Stan Shunpike

Oh, I know a few Indian words

Not that this means much but I do like how Hindi is pronounced despite the fact there are like 1000 dialects

So theres like zero chance of me finding someone who speaks the tourist version lol

But its still awesome. Bahut danyevad is one I remember

Anyways, do you study physics there at uni?

Waffle's Crazy Peanut

04:05

@StanShunpike Hindi? I only know how to read & write Hindi. You may probably wanna ping @ManishEarth and he might give you a couple of lessons

@StanShunpike I study aeronautics (Manish fits your description in every way - He's the one studying physics) :P

@ChrisWhite: Speaking of aeronautics, it's nothing like physics! It's way too boring!

Well, as my BE lifetime nears the end, I feel more & more uncertain about my career.

user54412

@Waffle'sCrazyPeanut I like to think any subject can be boring with the wrong exposure, or awesome with the right teacher.

user54412

@Waffle'sCrazyPeanut There's been a lot of that in this chat room lately.

Waffle's Crazy Peanut

@ChrisWhite Oh yeah, that's probably right...

user54412

I have no idea where I'll be or what I'll be doing in 18 months, except that I won't still be here.

user54412

(or rather, I won't be in New Jersey. I'll probably be in this bar, like always)

Waffle's Crazy Peanut

04:08

@ChrisWhite What's that number "18"?

@ChrisWhite LOL! :D

user54412

@Waffle'sCrazyPeanut a rough estimate for graduation (or getting kicked out of the program)

user54412

we have a time limit, and mine is approaching

Waffle's Crazy Peanut

Hah!

I feel really bad about something (I only thought of that lately). In my 3 years of college life, I've been brutally forced to memorize stuff (like about a few hundred formulas so far). But, this time it nears my breaking point...

We have a subject called "Flight stability" and there are only 5 chapters. But, there are about 500 intertwined assumptions (which one has to memorize obviously)

... and only about 30 conclusions

user54412

sounds like all my friends in med school

user54412

I ask them why they have to memorize so much information rather than look it up when needed

user54412

04:14

and I've never gotten a good answer

user54412

@Waffle at least you're having fun in college? forging solid friendships? experiencing new things?

Waffle's Crazy Peanut

@ChrisWhite I've asked that directly to most of the professors - but they act like they were offended...

@ChrisWhite Sometimes, yeah. But, I'm pretty much isolated most of the time... Well, I dunno. Not much of them likes to speak with me (I guess), except a few.

@ChrisWhite On the brighter side, I get new experiences in physics & coding everyday... Oh, and I forgot to tell you - I got into QM lately (which consumes at least a few hours everyday), and I got into Mozilla (thanks to Manish), which consumes the remaining few hours. So, I don't really have much time to spend with real people :P

user54412

@Waffle'sCrazyPeanut It's not about all the many people you don't interact with for one reason or another, but the few you do :)

user54412

@Waffle'sCrazyPeanut Nice

Waffle's Crazy Peanut

@ChrisWhite Yeah, I agree...

Anyways, I always like that special coincidence. The ugly environment in college, my urge towards knowing physics, and the time when I got into Physics.SE (and of course, you guys) - it was these events that made me the ME now...

@ChrisWhite: I'm curious whether most of the equations in physics are linear (because I heard Feynman mentioning that sometime in the past)

user54412

04:33

@Waffle'sCrazyPeanut I feel like the really fundamental ones tend to be. The more emergent ones usually not. But it's a hard thing to count.

Waffle's Crazy Peanut

@ChrisWhite I asked that because I tend to bully some of my classmates saying something like, "we have linear equations most of the time unlike you unlucky insects!" :P

hwlau

it really depends on how you define linear

many differential eqns describe interesting effective physical phenomenon are nonlinear.

Waffle's Crazy Peanut

yeah...

Is it just me, or most of the time someone asks something about X, physicists tend to respond like, "it really depends on how you define X" :D

(No offense, I've asked it myself in various discussions, and people have started shouting!) :D

hwlau

because we need to define a problem precisely

and there are always some exception

Waffle's Crazy Peanut

heh, yeah...

hwlau

04:44

so th e phrase

I guess mathematian do the same

Waffle's Crazy Peanut

Random stuff - Am I right with my thought that physics enthusiasts appreciate music more than normal people, in the sense that they listen very well?

Because, I remembered that when I saw this...

hwlau

I have no idea

Waffle's Crazy Peanut

Listening to music while working helps create a more positive mood, and can help you to become more productive.

Tweeted by GoogleFacts on March 20, 2015 at 2:32 AM

Now, I'm curious whether both are true...

hwlau

I believe people in other jobs do the same

particular if the office space is small and narrow

just to filter out the noise...

Waffle's Crazy Peanut

No, I concentrate only on physics enthusiasts here...

hwlau

04:53

oops

no idea

Sound is a branch of physics, so they should have positive correlation

Waffle's Crazy Peanut

I seem to contradict the above tweet because, whenever I start hearing music (oh & yeah, I listen only to music), I try to listen to all the instruments used in that music, and that disturbs my productivity in work (in the sense, it takes time to finish it off than what would've happened otherwise)...

hwlau

You show some kinds of addiction :)

If your purpose is to concentrate then you should turn it off / lowing the volume at a later time

and pick some softer musics

Waffle's Crazy Peanut

@hwlau Then, I tend to get addicted to it in a few days... I'd probably need new music everyday :P

hwlau

I think hearing classical musics is better

not some musics that you want to keep finding new one

Waffle's Crazy Peanut

ah...

Stan Shunpike

05:21

I write awesome classical music :D

hwlau

05:32

@StanShunpike Sounds great. Write it yourself?

Maybe you can share :)

Waffle's Crazy Peanut

@StanShunpike Gravity was nice... :)

Stan Shunpike

06:06

Nice! Glad you liked it! Check out Sono Tuo Per Sempre. That one is my fav

@Waffle'sCrazyPeanut

@hwlau

soundcloud.com/gotlepton/sono-tuo-per-sempre

Physics Meta

0

Q: Is it possible to find the most voted answers?

I found a button to find the most voted questions, is there a way to find the most voted answers? Thanks

feature-request

Stan Shunpike

06:25

@hwlau @Waffle'sCrazyPeanut these are good too soundcloud.com/gotlepton/marche-turque-plus-exube-rance and soundcloud.com/gotlepton/oce-an-noir :D

Anybody gets a chance I got a question

0

Q: Can someone clarify what should and should not be an operator in my verification of the 1D solution to the SE for a free particle?

I just worked out the 1D free particle solution to the Schrödinger equation. My wave function was \begin{equation} \psi(x,t) = Ae^{i(px-Et)/\hbar} \end{equation} So I plugged this into both sides of the time-dependent Schrödinger equation and started to verify. I did LHS and RHS separately...

quantum-mechanics operators

Stan Shunpike

08:12

@JohnRennie re this question/answer, physics.stackexchange.com/a/171008/66165 you say orbital mechanics is very complex. Why? What makes them so complicated? I don't know much about it, but I am studying GR and curious why you think orbital mechanics is tricky.

Stan Shunpike

08:30

Also, my compliments on a nice answer.

: hat tip :: haha

Physics Meta

0

Q: Is this about Physics, or just theoretical armchair physics?

Because when it comes to real practical advice on specialist experimental techniques I may as well be talking to myself. It seems to be singularly useless.

discussion

John Rennie

09:06

@StanShunpike Hi Stan.

It just seems to me that you take the simplest oribtal calculation and it rapidly descends into a maze of Bessel functions or whatever.

GR is obviously far more complicated, but we tend to just give up on analytic solutions for all but the most symmetrical problems.

In both cases the equations of motion are simple, but the solutions aren't. But because we usually can't get solutions for GR it's more striking much pain regular orbital mechanics involves.

user929304

11:26

@JohnRennie Hi, have you had a chance to read the answers here?

tpg2114

12:10

Thanks to today's XKCD, I have a whole bunch of wikipedia articles open... The Dyatlov Pass one reminds me of the Mountains of Madness discussion we had awhile back:

Dyatlov Pass incident

The Dyatlov Pass incident was an event that resulted in the death of nine skiers in the northern Ural Mountains on the night of February 2, 1959. The incident happened on the eastern side of Kholat Syakhl, whose name in Mansi means "Dead Mountain." Since then, the mountain pass where the incident occurred is called Dyatlov Pass after the group's leader, Igor Dyatlov. Investigators determined that the skiers had torn their tents from the inside out in order to effect an escape from apparent threat. They fled the campsite, some of them barefoot, under heavy snowfall. Although the bodies showed no...

ACuriousMind

13:39

@ChrisWhite Do you miss the days when it was just you and Kyle discussing astro code? ;)

infinitesimal simplicio

13:55

Hi pal :-)

user929304

@ACuriousMind hi there )

@ACuriousMind got time for one small question?

ACuriousMind

@user929304 Hey :) Thanks for the bounty on my question, btw

And yeah, ask away

user929304

@ACuriousMind is the fact that both wavefunctions in QM and phase space probability densities in classical mechanics both evolve unitarily in time simply due to the fact that in both cases we have 1st order time derivatives? (Schrodinger eq, and Hamilton's equation of motion). Or is there some bigger fundamental reason behind?

ACuriousMind

@user929304 Well, intuitively it is because probability should be conserved in time - if the evolution weren't unitary, you would see the total probability to find the particle somewhere stray away from 1

But, mathematically, it is probably indeed the structure of the PDEs govering the evolution being first order in time, but I'm not so sure of that, my knowledge about differential equations is very limited

user929304

@ACuriousMind oh I see, so one should always associate conserved observables with unitary operators?

ACuriousMind

14:02

@user929304 Conservation laws are by Noether's theorem equivalent to symmetries, and symmetries are, by a theorem of Wigner, at least in QM to be implemented as (anti-)unitary operators, so yes.

user929304

@ACuriousMind and is all this supposed to hold (unitary evolution i.e.) only when all is in a stationary equilibrium state? meaning if there are interactions with external system, time evolution isn-t unitary anymore?

ACuriousMind

@user929304 If there is an interaction with an external system, the evolution of the total system is still unitary, but if you are losing energy/particles from the subsystem, then the restricted evolution operator of that subsystem can't be unitary, because that would, in turn, imply that no particles/states are lost from that subsystem (since probability would be conserved within it), hence it would not be "interacting" in the usual sense.

user929304

@ACuriousMind aha! nice! makes a lot of sense.

Danu

@ChrisWhite But there's only one guy who knows a bit of category theory :P

user929304

@ACuriousMind can I ask one last thing?

ACuriousMind

14:11

@user929304 Sure

@Danu So the preferred flavour is confusing?

Danu

@ACuriousMind Tastes... I'm not sure...

infinitesimal simplicio

Did you ever get that "lens space" question figured out? @Danu

Danu

@infinitesimalsimplicio Well you gave me the correct reference but I have no idea why that drawing thing makes any sense

ACuriousMind

The simplicio there took me by surprise

user929304

@ACuriousMind can one still define unitary evolutions in a grand canonical ensemble? if it is known to allow for E and N exchange...

ACuriousMind

14:15

@user929304 My statistical mechanics is weak, but I think the ensembles are supposed to be in equilibrium, i.e. time evolution is trivial on them.

user929304

@ACuriousMind ok thanks, food for thought, I ll figure it out. also thanks again for your time, very kind.

ACuriousMind

@user929304 No problem. I like your questions ;)

Danu

@Waffle'sCrazyPeanut For me certainly

@hwlau Depends on the person :) I listen to reasonably loud, upbeat music quite a lot while working

Jimnosperm

14:31

Update: I found out why I was rejected. I'm not yet published in a refereed journal. How ironic that I just submitted a paper to PhysRev this week

hwlau

@Danu Probably you need a strong one to do the modulation with your brain wave :)

alarge

@Jimnosperm That's sad. Such pointless metrics are routine in academia, though. Were you aware of the requirement beforehand, or did they rank the applicants based on this and it so happened that they had so many applicants that the ones who had published filled the entire quota?

infinitesimal simplicio

ironic :(

Jimnosperm

@alarge I was unaware of the requirement but I did mention in my application that I'd be submitting a paper around this time. But I think they just ranked applicants and gave preference to publishings over all other criteria

My master's supervisor was just as shocked to hear the news as I was. So at least it's not just my ego that thought I should have gotten accepted

hwlau

What are you applying for ?

Jimnosperm

14:44

PhD in theoretical cosmology

alarge

Which PhysRev did you submit to? A first author PRL publication should guarantee acceptance if they're that into (empty) metrics. That is, if you decide to apply again.

Jimnosperm

PRST

but I have another coming up that I intend to submit to PRD

infinitesimal simplicio

"Publish or perish" rears its ugly head again >8(

Jiminion

Your Supervisor should have known that was required?

hwlau

Are you sure your referee write very good recommendation letter for you

Jimnosperm

14:48

@hwlau That I can never know for sure. But I know my referees all very well and I can only imagine one not giving me a glowing review. But he said it was glowing, so that's what I believe

@Jiminion No, it's not usually required and most people don't publish during their master's degree. Maybe at the end, but not usually during

That said, I applied to the astronomy and astrophysics department and astronomers are usually published by the end of undergrad

Jiminion

My non-entry into a Phd program was depressing.....

hwlau

If the referees don't know you well, then it can reflect in your reference letter. and it is very important for your application

Jimnosperm

@hwlau My referees were my supervisor, the co-author of the paper I just wrote, who I worked closely with, and a professor I worked for for 5 years. They know me well

Jiminion

@Jimnosperm They should have told you what was up. It shouldn't have been a surprise. Unless they were shorter on resources than expected.

alarge

So now that you know the reason, do you think you'll apply again? (I would think that a PRD publication at least is solid enough to fulfil any publication requirements. PRST I've never read so I don't know; the IF is low, though, so it is not going to impress a panel that looks at pure metrics)

Jimnosperm

14:56

Yeah, I'm going to apply again. My potential supervisor over there was on a different review committee. Had I submitted to that one, I'd have been accepted. So I'm going to try to resubmit to the one he sits on next time

alarge

Good. All's well that ends well.

No need to "sell your soul" at this point then. You may want to reconsider once you're done with your PhD, though.

Jimnosperm

Oh, I'm still going to sell my soul to pay the bills for now

infinitesimal simplicio

You can always pick up a used soul on the Internet later ;)

alarge

Fair enough. I suppose the plan might be different though, if you're in a hurry to get employed, right?

Jimnosperm

Maybe. I'm really only in a hurry to pay rent

Jiminion

15:19

@Jimnosperm If you can commute to Windsor, you can dance at Danny's... :)

Jimnosperm

@Jiminion What?

Jiminion

I was joking. Danny's is an adult establishment catering to bachelorette parties....

Jimnosperm

Ah, a strip club

bolbteppa

15:39

@ACuriousMind does on ope like $J(x)J(y) = \dfrac{1/2}{(x-y)^2} + (JJ)(y) + ...$ ring a bell to you?

Danu

I feel like I'm going crazy: Why is it true that $(\mathbb{Z}_p^*)^2=\{a^2|a\in \mathbb{Z}_p^*\}$ is a subgroup of index 2

My book says it's 'clear' from the fact that $\mathbb{Z}_p^*$ is cyclic (which will be proven later---no need to worry about it for now)

ACuriousMind

15:59

@bolbteppa Not really

But I don't know many OPEs^^

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