The h Bar

4:00 PM

yeah, I remember that that's a weird one.

@TheDarkSide It "occurs" in any physical process where $\psi(x)$ occurs, since the two are indistinguishable.

And since I didn't put any requirements on $\psi(x)$ at all, you can just pick any "physically occuring" wavefunction $\psi$, and construct this horrible $\phi$ that also "physically occurs" in that same situation.

The Dark Side

HOW ARE YOU MATHEMATICIANS ABLE TO SLEEP AT NIGHT :P

ACuriousMind

@Semiclassical Yes. Yes they are.

Semiclassical

I think the distinction to be made: It's not that $\phi$ can be excluded, but that there's no reason to include it.

0celoñe7

@TheDarkSide We stop caring about physics.

The Dark Side

4:01 PM

@Semiclassical Exactly. Or maybe we could say, no "physical" reason :P

user685252

How about "SemiCondensed" @Semiclassical?

The Dark Side

@0celoñe7 Yes. That's some kind of bliss.

Semiclassical

@0celoñe7 Though it's not really surprising that that one is weird, since it's the boundary case of $V(x)=x^4+ax^2$. When $a>0$, there's a single minimum; when $a<0$, there's two symmetric minima.

@user685252 nah.

user685252

SemiQuantum?

ACuriousMind

@Semiclassical Yeah, that's what I meant above with asking when every state has at least one nice function representing it. But I want to argue against the idea here that the nice functions are somehow more "physical" than the horrible ones. Imo, all that we can say is that the nice functions are simply better for us to deal with.

Semiclassical

4:04 PM

pretty sure that's what semiclassical means already :)

user685252

:-)

Semiclassical

@ACuriousMind Ehhhh. Saying that the wavefunction is allowed to care about whether or not you're at an irrational seems pretty non-physical to me.

...except, then I remember the Hofstadter butterfly :/

The Dark Side

@ACuriousMind Wait a second. How doesn't this not mess up normalizability condition? (I'm clueless but I guess) we have an infinite number of rational numbers between $-\infty$ and $\infty$ ?

user685252

hi @Danu

Danu

hi

ACuriousMind

4:06 PM

@TheDarkSide The rationals have measure zero in the reals - the integral over any function that has some value at the rationals and is zero on the irrationals is zero.

Semiclassical

(one criterion for "is this potential going to give me a headache" is whether or not the minima are all quadratic. x^4 fails that, so it's not surprising that it's a pain)

Danu

@TheDarkSide Integration baby

The Dark Side

@Danu Hi :P

@ACuriousMind Let's put this in a language that I understand. The function isn't zero at the points other than the rational numbers...

Semiclassical

Here's a question, come to think of it. Obviously, one can write down examples where the confining potential isn't harmonic in the vicinity of the classical minimum.

The Dark Side

$\phi(x)$ is equal to $\psi(x)$

Semiclassical

4:09 PM

$\int_a^b \chi_\mathbb{Q}(x)\,dx=0$, I think is the point

ACuriousMind

@Semiclassical Heresy! Everything is a harmonic oscillator :P

Danu

Lebesgue integration sucks is the point

@Semiclassical Not that easy though...

The Dark Side

@Semiclassical Even when a and b tend to $\mp$ $\infty$?

user685252

@Semiclassical "Semi_ironic"?

Semiclassical

@TheDarkSide Yep.

That's not too far from semicynical :P

John Rennie

4:11 PM

Seminal?

The Dark Side

@Semiclassical Is that yep to me or to user685252?

Semiclassical

I'm not a measure theory guy, though, soooo

user685252

Semisarcastic

Semiclassical

Semitary :P

The Dark Side

Nevertheless, I don't know how we got stuck up into the behavior at infinities,

my original question was simpler.

59 mins ago, by The Dark Side

18 mins ago, by The Dark Side

The $V \rightarrow \infty$, $\psi \rightarrow 0$ argument looks fine, but is there any condition from the potential on the KE term of Schrodinger equation?

Sorry for the bad manners, but I'll have to rush for food. Anyone wanting to have something to say on that shall be most welcome.

Semiclassical

4:14 PM

@0celoñe7 With regards to Lebesgue integration, there's an old line (which you may have heard): "Does anyone believe that the difference between the Lebesgue and Riemann integrals can have physical significance, and that whether say, an airplane would or would not fly could depend on this difference? If such were claimed, I should not care to fly in that plane." - Richard Hamming

2

Emilio Pisanty

2 hours ago, by Emilio Pisanty

Thanks for your advice , Does Griffith has textbook on EM? — user46899 2 mins ago

sigh

Semiclassical

....

Emilio Pisanty

@user46899 Yes. This — Blue 2 hours ago

Semiclassical

google, people

Emilio Pisanty

Thanks a lot! Appreciated ! — user46899 2 hours ago

@Semiclassical exactly.

Semiclassical

4:16 PM

i mean, just doing 'griffiths em' is enough to find it

ffs

Anonymous

@Semiclassical He probably didn't know that em is same as electrodynamics...

ACuriousMind

"Teach a man to fish..." @Blue just gave him a fish :P

Emilio Pisanty

if a student is incapable of finding their way to a relevant textbook when all it takes is giyf'ing it, what are their chances overall?

Semiclassical

@Blue ...How?!?

DanielSank

Hi, everybody

Emilio Pisanty

4:17 PM

kinda reminds me of

yesterday, by John Rennie

Any child too dim to click Yes I'm over 13 probably isn't a great loss to science.

Anonymous

@Semiclassical Because I had that confusion a couple of years back :P

Semiclassical

thing is, they knew Griffiths and they had the acronym EM.

And searching "Griffiths EM" is enough to get that book.

ACuriousMind

@Blue You had that confusion at a point where you were able to ask a question about the derivation of Maxwell's equations?

Emilio Pisanty

@Semiclassical like?

Semiclassical

$V(x)=x^4$. That's the example @0celoñe7 quoted

Emilio Pisanty

4:19 PM

@Semiclassical ah, yes, well

Anonymous

@ACuriousMind Uh, probably no. I was just starting with basic electricity and magnetism course then in grade 11

Anonymous

Dunno...people are weird

ACuriousMind

@EmilioPisanty If the second dervative vanishes at that point, you don't get the quadratic term in the Taylor expansion.

Semiclassical

$V(x)=|x|$ is another example.

ACuriousMind

Well...or if it doesn't have a second derivative, I guess :P

Emilio Pisanty

4:19 PM

@Semiclassical I would argue that that's not a potential

Semiclassical

lol.

Pretty sure $V(x)=|x|$ is a textbook problem.

DanielSank

"problem"!="potential" ;-)

Emilio Pisanty

it's interesting enough, but the kink is not physical

DanielSank

@EmilioPisanty o/

Semiclassical

Oh, sure.

ACuriousMind

4:20 PM

@EmilioPisanty How do you feel about $-\delta(x)$ as a potential? :P

Semiclassical

That's where I was going, actually. Is there an example of a non-quadratic minimum which we'd consider genuinely physical?

Emilio Pisanty

@ACuriousMind as a model or as a physical thing?

Semiclassical

@ACuriousMind Eh, just replace that with a very narrow Gaussian :P

Emilio Pisanty

@Semiclassical cue catastrophe theory ;-)

Semiclassical

That actually is how I usually think of such things, though. There is some length scale for the potential, it's just much smaller than everything else in the problem.

DanielSank

4:23 PM

@Semiclassical Yes.

We can and have made this in our lab.

Emilio Pisanty

@DanielSank ::drumroll::

Semiclassical

What's the form of the potential?

DanielSank

You put a SQUID in parallel with an inductor and then if you put the right amount of magnetic flux in the SQUID you get a quartic potential.

0celoñe7

@EmilioPisanty why?

Semiclassical

Neat.

DanielSank

4:25 PM

The Hamiltonian for a capacitor/SQUID/inductor circuit, i.e. a flux qubit, is:

user685252

0

Q: My laptop fan has stopped working. Can we build any device to allow working of laptop without fan?

My laptop fan has stopped working, it gets hot quickly. Can we make any device using thermodynamics or any other concept, to make laptop work, without buying new fan? My laptop is the same model shown below, with fan below the power button.

thermodynamics

Emilio Pisanty

@0celoñe7 because a potential is a function $V$ such that the force is $F=-\nabla V$. If its derivative has anything worse than removable singularities, it's not a physically-realizable potential

0celoñe7

@EmilioPisanty In classical mechanics I might agree.

Although $\nabla V$ does exist for $V=|x|$, so...

Not sure what your point is.

DanielSank

$$H = \frac{\hat{Q}^2}{2C} - E_J \cos(2\pi \hat{\Phi}/\Phi_0) + \frac{1}{2L} \left( \hat{\Phi} - \Phi_\text{SQUID} \right)^2 $$

Semiclassical

@EmilioPisanty Ehhhh, that supposes that the force field has to always be meaningful in quantum mechanics

ooo, cosine potential

0celoñe7

4:26 PM

@Semiclassical This.

DanielSank

$$[\hat{\Phi}, \hat{Q}] = i\hbar$$

ACuriousMind

@DanielSank We know you like octopi, you don't have to put them in your formulae, too ;)

0celoñe7

@ACuriousMind Seriously?

Emilio Pisanty

@ACuriousMind what?

Semiclassical

Though there's potentially a loophole there: To get a quartic potential (zero second derivative), you need a specific relation between $E_J$ and $\Phi_{SQUID}$

ACuriousMind

4:27 PM

@EmilioPisanty ...it's a pun.

DanielSank

$L$ is the inductance of the inductor, $E_J$ is an energy that has to do with the Josephson junctions in the SQUID, $C$ is the capacitance of the capacitor, $\hat{Q}$ is the charge on the capacitor, and $\hat{\Phi}$ is the flux on the inductor.

0celoñe7

Since when are squids octopi?

Did I miss the memo

user685252

yes

DanielSank

@ACuriousMind squid != octopus, bro

Emilio Pisanty

@ACuriousMind ah, squid.

0celoñe7

4:28 PM

Oh

Semiclassical

So that requires an infinite precision for the applied flux.

Emilio Pisanty

yeah, squid are not octopi

ACuriousMind

I might have messed up the taxonomy there

user685252

same difference

DanielSank

@Semiclassical Well, sort of. The applied flux typically comes from a wire connected to a current source. It's a classical field.

Emilio Pisanty

4:29 PM

otherwise I would link to schneier.com/blog/archives/2017/07/friday_squid_bl_585.html & equivalents every week with CC @DanielSank

Semiclassical

Hmm.

DanielSank

@EmilioPisanty o_O

We couple the SQUID magnetic bias very weakly to the qubit.

Emilio Pisanty

@DanielSank schneier.com/cgi-bin/mt/…

John Rennie

@ACuriousMind they taste much the same to me ...

Semiclassical

I guess my point is that there's a distinction between being able to model the potential as quartic, and the potential truly being quartic

Emilio Pisanty

4:30 PM

@JohnRennie taste is a poor way to go about taxonomy

DanielSank

Weak coupling + strong classical field --> you can mostly treat the bias classically (i.e. I've never had to think about the fact that the bias is a coherent state, ever).

John Rennie

@EmilioPisanty No it isn't :-)

Semiclassical

presumably there's still some quadratic term, it's just so small as to be unobservable/irrelevant.

ACuriousMind

@JohnRennie I guess most animals are chickens to you, then?

4

DanielSank

@Semiclassical Pfft, well... this system is a circuit, so if you wanna break it down to basics you're going to have to model Avogadro's number worth of particles anyway :-)

Emilio Pisanty

4:30 PM

@JohnRennie so what's the taxonomical group for fried stuff?

Semiclassical

lol.

Emilio Pisanty

=P

DanielSank

@Semiclassical Well yeah in principle you're probably right.

Semiclassical

It's a bit of a slippery point, anyways.

John Rennie

@ACuriousMind I don't know where this everything tastes like chicken meme came from. That hasn't been my experience.

DanielSank

4:31 PM

::shrugs::

ACuriousMind

@EmilioPisanty These belong to the order "Kentucky"

Emilio Pisanty

@Semiclassical key word being irrelevant ;-)

Semiclassical

lolyes

Emilio Pisanty

@ACuriousMind dunno, I'd class them under Dallas

DanielSank

It is amusing that your name here is Semiclassical

ACuriousMind

4:31 PM

@EmilioPisanty (I was alluding to KFC. Why Dallas?)

Semiclassical

Even if one insists that the minimum is actually quadratic, it doesn't follow that that description of the problem is at all useful. If the term is that small, then probably one should just do the quartic case and not worry about the difference.

Emilio Pisanty

@JohnRennie will probably confirm

but it might have more variation within the UK than I thought

John Rennie

@EmilioPisanty never heard of it ...

Emilio Pisanty

@JohnRennie put google streetview anywhere in London and you'll find one soon enough

John Rennie

Nandos possibly ...

Emilio Pisanty

4:33 PM

@JohnRennie shhhhhh, there's americans in the room

DanielSank

@Semiclassical Certainly.

Duffing equation

The Duffing equation (or Duffing oscillator), named after Georg Duffing (1861–1944), is a non-linear second-order differential equation used to model certain damped and driven oscillators. The equation is given by x ¨ + δ x ˙ + α x + β x 3 = γ cos ...

John Rennie

@EmilioPisanty there are clearly subcurrents of which I am unaware :-)

DanielSank

^ General analysis of a forced oscillator with quartic term.

Semiclassical

Here's a question. As @0celoñe7 noted, $-\frac{d^2}{dx}^2+x^4$ is not a self-adjoint operator.

Emilio Pisanty

@JohnRennie buzzfeed.com/alanwhite/…

Semiclassical

4:35 PM

What happens if we deform that by including a term $\epsilon x^2$ with $\epsilon\neq 0$?

DanielSank

@Semiclassical I'm skeptical about that not being self-adjoint.

Emilio Pisanty

@Semiclassical isn't it? why not?

DanielSank

That looks rather self-adjoint.

ACuriousMind

It probably has domain issues

Philip Cherian

Why isn't it self-.. oh, everyone's asking the same question :P

Semiclassical

4:36 PM

I think the point has to do with the boundary conditions.

DanielSank

... which are?

user685252

Welcome @PhilipCherian

Semiclassical

0 at infinity. But I'm not the one who proposed that example originally, soooo :P

Philip Cherian

oh hallo. :) Also, where's the original thingie? Is it a post or was it on chat?

Emilio Pisanty

@Semiclassical wait, is that a $p^4$ there?

DanielSank

4:38 PM

@PhilipCherian eh?

user685252

in chat

Semiclassical

blah, first term should've been $-d^2/dx^2$

Philip Cherian

@DanielSank I was asking about the original example that Semiclassical mentioned.

Emilio Pisanty

38 mins ago, by 0celoñe7

@Semiclassical $d^2/dx^2-x^4$ is not self-adjoint.

Semiclassical

Yeah, that one.

0celoñe7

4:39 PM

@ACuriousMind Yes.

Semiclassical

Which corresponds to the Hamiltonian $H=p^2+x^4$.

0celoñe7

@DanielSank It is not essentially self-adjoint on its domain.

Emilio Pisanty

@0celoñe7 what does that mean?

Semiclassical

This is where my knowledge of the details is lacking.

0celoñe7

@EmilioPisanty It does not have a closed self-adjoint extension.

Semiclassical

4:40 PM

But I think the point is that the problem is not just that operator but also the boundary conditions we impose on it.

DanielSank

@0celoñe7 uh, so?

0celoñe7

@DanielSank It means one cannot apply spectral theory to it.

Emilio Pisanty

@0celoñe7 dx.doi.org/10.1119/1.17221 disagrees with you

DanielSank

@0celoñe7 That sounds wrong.

0celoñe7

@EmilioPisanty On which part?

Emilio Pisanty

4:41 PM

@0celoñe7 existence of a self-adjoint extension

dunno about closed

anyways, g2g

Semiclassical

4:58 PM

@0celoñe7 Yeah, I can't find anything in the literature that claims $-d^2/dx^2+x^4$ isn't self-adjoint.

Slereah

Riddle me this, Batman

Consider the prime decomposition of integers

Each factor of prime is encoded into a set of binary digits

Is it possible to have a full continuous set of integers using this greater than 2?

ie

If I put 1 bit for $2$, then I can generate $2^0 = 1$ and $2^1 = 2$

If I put 2 bits on $2$, I get $1, 2, 4, 8$

If I put 1 bit on $2$ and 1 bit on $3$, I get $1,2,3,4,6$

etc etc

I don't think it's possible

I think if I put 1 bit for the $n$ first primes, the product of all those numbers will always be greater than the $n+1$th prime

Except for 2

ACuriousMind

I don't understand your question

I do understand how you want to encode numbers - I just don't understand what you are asking about it

Slereah

Well, basically, can I represent numbers from $1$ to $n$ as a series of binary digits of their prime factors

In a way that it is a bijection

Every string is a single number and every number has a string

Is there a number $n>2$ where that is possible

ACuriousMind

How is one supposed to know whether any given string is just a large exponent of a single prime or a list of exponents of different primes?

Slereah

Well the algorithm I don't care about, I just want to know if there is such a number

ACuriousMind

5:12 PM

Then I still don't understand your question

Slereah

I guess I should write a little program to do it for the first few combinations

ACuriousMind

I can take any number and spit out the non-zero exponents $r_p$ in its prime factorization.

Of course there is a bijection between the set of all possible exponents and all numbers because prime factorization is unique.

Slereah

Yes, but the thing here is

It has to be represented by a fixed amount of binary digits

"between n and 2n always exists a prime"

Ah, I guess it is impossible

ACuriousMind

Then whatever you're doing depends on how you encode the exponents and the primes you belong to into these binary digits.

Slereah

The exponents are encoded the usual binary way

ACuriousMind

5:16 PM

But every encoding that crams the first $2^n$ numbers into $n$ bits is functionally indistinguishable from the standard base 2 encoding

I'm still not sure that whatever you're talking about is a well-defined problem

Slereah

It is a well defined problem, but apparently it won't work

ACuriousMind

If I have a list of non-zero exponents $r_p$, then you can't just store just these numbers - you must also store which primes $p$ these belong to.

Slereah

since there is a prime number between $n$ and $2n$, this means that if I take the largest prime $p$ with at least 1 bit of data for the exponent, then there will be a prime number before $2p$

So that the sequence isn't contiguous

ACuriousMind

Ohhhh, now I get what you were asking

Do I want to know why you were thinking about that? :P

Slereah

I was trying to think of a way to encode fractions that didn't waste any space

I guess that the "most efficient" way would actually be to make it a binary representation of the Farey sequence : en.wikipedia.org/wiki/Farey_sequence

But of course that would be terribly awkward to use

ACuriousMind

5:22 PM

@Slereah That does not really answer my question :D

Slereah

Does it not?

The easiest way to encode a fraction on a computer is just as a set of two integers

ACuriousMind

What sort of tiny memory are you working with that you have to care about how efficient your fractions are?

Slereah

But of course $(1,2)$ and $(2,4)$ are the same etc

You seem to be under the mistaken belief that this is for a practical application

I was just curious

I am also wondering about a general way to construct algebraic numbers now

or constructible, probably, since algebraic probably are shit to deal with

I think all constructible numbers can be represented as a composition of sums, products and square roots on integers, but I'm wondering if there's a canonical form that all obey

That can be represented simply

books.google.fr/…

YES

0celoñe7

@Semiclassical I don't know what I said, but I meant $-x^4$.

$+x^4$ is self-ajoint.

Well...essentially self-adjoint.

@EmilioPisanty They're not precise about the domain, so I don't have too much faith there. I'm taking the domain to be $C^\infty_c(\Bbb R)$.

0celoñe7

5:49 PM

@Slereah you seems excited

Slereah

Would u not be

0celoñe7

google won't let me look at that

I don't know what it is

Slereah

Representation of constructible numbers

0celoñe7

@Slereah What about it?

Slereah

I was looking for one

I am wondering if it's possible to represent floats as constructible numbers instead

to avoid any floating point rounding error

0celoñe7

6:00 PM

1

Q: What is the commutator of a Poisson bracket and the covariant derivative?

Consider a classical field $V^\mu$ on a curved background. We make a 3+1 split of coordinates into $t,x^i$, where $x^i$ are coordinates on spatial hypersurfaces $\Sigma$ and $t$ the parameter labeling them. Now consider a canonically conjugate $\pi_\mu$ such that the following Poisson bracket ho...

general-relativity field-theory commutator covariance poisson-brackets

How are delta functions appearing for classical things

Slereah

@0celoñe7 They pop up in field theories

0celoñe7

and who names the metric $d$

@Slereah That seems very wrong.

Slereah

does it

The EM field has delta functions

for its Green function

Hell the Poisson equation has it too

0celoñe7

@Slereah That's because point masses/charges are unphysical.

Slereah

the Green function is still there even without point masses

It's just a practical mathematical thing

0celoñe7

6:04 PM

@Semiclassical My dad is a fighter pilot/aeronautical engineer. I've tried to get him to understand measure theory :P

@Semiclassical But that reminds me of a question I've been meaning to ask on the main site

@ACuriousMind so what tags are off-limits now?

ACuriousMind

@0celoñe7 Hm?

0celoñe7

@ACuriousMind is the mathematical physics tag allowed or not?

ACuriousMind

@0celoñe7 Nothing with respect to that tag has changed

oh, wait

Yeah, use the version wiith the '+' at the end

0celoñe7

Now, how to write a post criticizing all physicists without being too condescending...

0

Q: Does mathematical sloppiness in quantum mechanics ever produce incorrect predictions?

Does mathematical sloppiness in standard quantum mechanics ever produce predictions that don't pan out? I'm not talking about things like the WKB approximation, but instead subtle functional analytic issues, such as assuming every Hamiltonian is self-adjoint, has an eigenbasis of bound states, et...

quantum-mechanics experimental-physics mathematical-physics+

@ACuriousMind @Slereah @EmilioPisanty @Semiclassical

Slereah

Can't think of a physical example, no

ACuriousMind

6:19 PM

I think that's a tricky question because I think in the examples where it would (like these states for which the uncertainty principle doesn't hold in its naive version due to domain issues) physicists typically add "just enough rigor" to get it right :P

0celoñe7

-1

Can you please ban whoever did that

Clearly vote manipulation

user685252

you're going down

0celoñe7

@ACuriousMind Such an answer would be acceptable if you can find a physical example (published!) where the uncertainty principle fails

Slereah

Maybe the particle in a box thing?

I don't know if the uncertainty was ever wrongly calculated

But it's a simple case where that's an issue

ACuriousMind

iirc it happens for "particle on a ring" states. Should be possible to find someone who's measured particles on a ring, but I'm not better versed in the experimental literature than you are

David Z

6:22 PM

@Slereah with infinite storage, sure, but in practice, there will always be some sort of discretization error

0celoñe7

@ACuriousMind Particle on a ring? Ring is compact, $L^2(S^1)$ is basically the best space...

Slereah

also I know that there are $x^4$ potentials used in quantum chemistry

@DavidZ Well no, they will just overflow :p

0celoñe7

@Qmechanic What does your comment mean?

Slereah

which is worse I suppose, but within those bounds, there's no loss

I think the issue with QM rigor is that people probably don't publish things when it goes wrong, because I think that when it does go wrong, it goes very wrong

You don't get factors of 2 usually you get divergences

ACuriousMind

@0celoñe7 See physics.stackexchange.com/q/233266/50583

Semiclassical

6:37 PM

I'm a little puzzled that the method of the last answer (treat the 1D ring as a limiting case of an annulus) doesn't resolve matters.

(in the question just linked, I mean)

Emilio Pisanty

@0celoñe7 how can an operator not be self-adjoint when a multiple of it is

?

Makes no sense

Semiclassical

$V(x)=-x^4$ is unbounded below @EmilioPisanty

So $x^4$ has a classical minimum but $-x^4$ doesn't. That's pretty bad.

Also, $-d^2/dx^2+x^4$ and $-d^2/dx^2-x^4$ aren't multiples of each other.

Emilio Pisanty

7:12 PM

@Semiclassical how's that relevant for self-adjointness, though?

Semiclassical

Well, suppose it's $-d^2/dx^2-x^2$ instead.

Emilio Pisanty

But yeah, I can see p2-x4 having trouble with adjointness, on classical grounds

You reach infinity too fast

Semiclassical

Right.

Physically I think you'd say that none of the states you'd write down would be stable.

plus, it's not even clear what the meaning of bound state would be in that case.

The weird case is $p^2-x^3+x$, since it's got a local minimum but no global minimum

TauNeutrino

7:34 PM

I thought of a question I think is pretty interesting. I was wondering why flags flutter in the wind as opposed to just staying straight in the direction in which the wind is blowing. I think I know why it is but I'd like to ask the Physics.SE community this question. Do you think I could phrase it to be on-topic here?

Dawood ibn Kareem

@Secret Wow! Is the hbar secretly some kind of weird vortex of people who have freakishly made the transition from pen-loser to pen-accumulator or something?

And the pedant in me is going to come out. People in this room keep using the non-word "octopi". I believe I've seen five people use this non-word in the last two weeks. There is no such word as "octopi". The plural of "octopus" is "octopuses". If you must give it a fancy foreign plural, "octopodes" is technically correct, but not in common usage. "Octopi" is just wrong. Always wrong.

Anonymous

@TauNeutrino The wind doesn't blow is a particular direction steadily. As for the fluttering that is due to unequal velocities/pressures on two sides of the flag (Bernoulli's principle). There might be some detail which I am missing. But yeah, your question looks fine. You can ask it on the main site.

Anonymous

@TauNeutrino Also have a look at this thread physicsforums.com/threads/why-does-a-flag-ripple-wave.277225

ACuriousMind

7:52 PM

@DawoodibnKareem We've had this discussion multiple times already :P

Dawood ibn Kareem

And people still haven't learnt?

OK, if this room is octopied, I can come back later.

2

Emilio Pisanty

@Semiclassical none of that is relevant for adjointness, though

Position and momentum are self-adjoint, and they're unbounded below and with no bound states

heather

@DawoodibnKareem lol

0celoñe7

@EmilioPisanty what?

I am claiming that $H=p^2-x^4$ is not essentially self-adjoint on $C_c^\infty(\Bbb R)$.

@EmilioPisanty To show this, one shows that the adjoint of $H$ is not symmetric.

Semiclassical

8:15 PM

I think the way that it sneaks in is that, to prove self-adjointness, one needs be able to do to integration by parts. If one can argue that the boundary term in there is zero, everything is fine; if not, though, then the self-adjointness doesn't go through.

That may be bollocks, though.

0celoñe7

9:04 PM

@Semiclassical Well, the issue is the formal definition of "adjoint". I can say more about it later.

For unbounded operators it's not nice.

Semiclassical

9:20 PM

I can buy that.

I imagine it has to do with self-adjointness not being equivalent to being Hermitian

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