hallo

ohai

I am majorly confused about quantum again xD. So

so, 1) we only reasonably postulate the fact that any wave function can be represented by a linear combination of eigenkets if we are dealing with an eigenvalue problem such as H$\Psi$ = E$\Psi$?

2) These eigenkets exist arbitrarily (That is to say without representing them with respect to some basis). However, we can represent them in various bases. E.g. position space, momentum space, etc.?

3) To represent a ket as a vector, we project the vector onto each basis vector of interest? That is to say a basis vector in position space is an "eigenstate" of the operator, which here could be the hamiltonian if we're dealing with the TISE

@SillyGoose (Ignoring mathematical subtleties related to position/momentum) "ket" is just a silly name for "vector". Physicists have a tendency to say/write "ket" when they're talking about an abstract vector and "vector" or "wavefunction" when they're talking about the representation of a ket/vector in a specific basis, but the two things really are not different concepts

that every vector is a linear combination of eigenvectors of any self-adjoint operator is not a postulate, but a fact about linear algebra called the spectral theorem

your third point I don't really understand at all - "position space" is what you get when you express everything in terms of the basis of position eigenkets, i.e. to each $\lvert \psi\rangle$ you associate its components $\psi(x) = \langle x\vert \psi\rangle$ in the position basis.

oh oh oh so it's not that the TISE is an eigenvalue problem which allows us to write any vector in our hilbert space as a linear combination of the hamiltonian's eigenvectors

but that the hamiltonian operator is self-adjoint over the hilbert space?

also is there a resource which shows how to slowly generalize spectral theorem from finite to infinite case?

maybe ill check out hall to see if its in there

I wouldn't worry about the infinite-dimensional aspect too much

because when you do you pretty much need to do full functional analysis and worry about continuous spectra and what's actually going on with the position operator and so on, which is a distraction when you really just want to start doing QM

xD i need a hundred life times to learn the mathematics and then another 100 to learn the physics

well the original thing i was confused about was: consider the a particular state of definite energy for a particle in an infinite square well represented in position space. how do I represent this (Sin function) in momentum space?

Fourier transform it

the relation between position and momentum space is the Fourier transform because $\langle x\vert p\rangle = \mathrm{e}^{\mathrm{i}xp}$, so if the position representation of $\lvert \psi\rangle$ is $\psi(x) = \langle x\vert \psi\rangle$ then the momentum space representation is $\psi(p) = \langle p\vert \psi\rangle = \int \langle p\vert x\rangle \langle x\vert \psi\rangle\mathrm{d}x$ where we've used the completeness relation $1 = \int \lvert x\rangle \langle x\rvert \mathrm{d}x$

Hm, well I'm trying to transform an energy eigenstate, which I expect to also be a momentum eigenstate, and so expect to transform into a dirac delta distribution, but im not sure if that's right

but I end up with $\int{\Sin^2{px/\hbar}dx$

energy eigenstates are only momentum eigenstates when you have a free particle

In the infinite square well case do we not consider the particle to be free-particle like between the boundaries?

the infinite square well is a bit more subtle then what I wrote above :P

because in the infinite square well you don't have a continuum $\lvert p\rangle$, and the $\lvert x\rangle$ are restricted to being inside the well

Yeah I was very confused about energy being discrete

I'm essentially trying to prove that a definite energy state is a definite momentum state in the infinite square well, and I feel like I have a heuristic way of understanding it, but i really would like to prove it mathematically

and to learn more about the relationship between these discrete energy states and what that implies for the spectrum of momenta

@SillyGoose you can't prove what isn't true :P Just like in the unrestricted free particle case, not every energy eigenstate is a momentum eigenstate (consider the superposition of the momentum eigenstates with momentum $p$ and momentum $-p$)

oh mayn xD

as for the relation between position and momentum space in the infinite square well, there you need to think about Fourier series instead of the Fourier transform - functions on intervals $[a,b]$ are equivalent to periodic functions with period $b-a$, and those have discrete Fourier series, not continuous Fourier transforms, corresponding to the momentum being discrete and non continuous in the well

Well I was arguing that if you have a state of definite energy for a particle in an infinite square well, then you have 1) an energy value, which 2) can be converted into an angular frequency via de Broglie, which can then 3) be converted into a wave number, which 4) can be converted into a momentum

Ah okay okay

so my guess about the result being a dirac delta distribution is wrong xD

I mean...not so much wrong as not thought through to the end

what do you mean by that?

while for an unconfined particle both position and momentum space are square-integrable functions on $\mathbb{R}$, for the square well it's different - position space is square-integrable functions on the interval $[a,b]$ (the inside of the well) and momentum space is square integrable sequences indexed by $\mathbb{Z}$ - the coefficients of the Fourier series of the functions on $[a,b]$

so there isn't a "delta distribution" in momentum space because momentum space doesn't actually consist of functions on $\mathbb{R}$ here, it's just sequences of Fourier coefficients

so momentum and energy are discrete random variables in the language of probability now

and we compute a probability mass distribution

wait no

well XD

so what you get isn't a "distribution" but just an ordinary Kronecker delta - the momentum (not energy as we discussed above!) eigenstate with momentum $p_i$ is in momentum space given by Fourier coefficients $\delta_{ij}$

wait okay so i feel like it makes sense that you get a kronecker delta because

your state in position space is literally a single sin function of a single frequency

thus it is a single term in a fourier series indexed by n

yes, exactly

YES okay

and so this coefficient belongs to the nth state that we started iwth

or i mean the nth momentum? i guess

okay

So that confusion is squared away. But, I am still confused about your comment that not every definite energy state (for a particle in an infinite square well) is a definite momentum state. In particular, I'm first trying to figure out where my heuristic reasoning breaks down from above

i guess it assumes that such a state with the resulting momentum exists? when it doesn't have to?

it's not that mysterious: Your Hamiltonian is $H = p^2$, so two different momentum eigenstates with eigenvalues $p$ and $-p$ both have energy eigenvalues $p^2$

so their sum is an energy eigenstate with eigenvalue $p^2$, too, but it's not a momentum eigenstate

so the hamiltonian has a degenerate spectrum here

okay i see

wait so every energy state has two distinct momentum eigenstates? other than perhaps 0?

yes

Hm okay so when we were talking earlier about writing a state of definite energy in position space in momentum space we came to the conclusion that there was a single fourier coefficient. This corresponds to the energy state we started with

But I am thinking well shouldnt we be able to write the probability mass function which would be a kronecker delta function times the coefficient which gives the probability of getting a momentum value and since it is just a single kronecker delta, it will give a single momentum value?

i am trying to parallel what happens in position space with the state of definite energy providing the PDF as a function of position

Or are you saying that really our state of definite energy in momentum space should be written as a linear combination of momentum eigenstates somehow?

@PM2Ring Thanks :-) A student asked me about the proof in the mistaken belief I knew something about maths.

Mathematics books seem to have this habit of stating things in a way that seems strange but not saying why they've stated it in that way. It's one of the things that adjusted my trajectory away from maths and towards physics as a student.

Meanwhile physics texts will just state everything in a strange way and blather on endlessly why that's the only correct way to say it? :P

@SillyGoose Completely free of any particular basis, I'm saying that $\lvert p\rangle + \lvert -p\rangle$ is an energy eigenstate but not a momentum eigenstate. Purely mathematically, this is an eigenstate of $p^2$ but not of $p$. I didn't really understand the rest of what you wrote - I suspect you're still treating this as stuff one need to argue physically about but it's just linear algebra ;)

I understand why mathematicians state things so carefully. It's just that it makes a lots of books hard to understand for the beginner.

The great thing about physics is that physical intuition (usually) comes first, and we can worry about rigour later.

@SillyGoose In particular, the property of being a linear combination of other states is a property that's completely independent of "position space" or "momentum space" etc.

@JohnRennie depends on the beginner ;)

while the math may be "harder" in some sense, I often found it at least clearer what was going on precisely because the mathematicians state stuff carefully while the physicists make like 10 tacit assumptions in a single paragraph :P

The question was why do we introduce the natural numbers here. That is, why do we find the infimum of 1/n rather than of the positive reals.

My suspicion, and PM2Ring's, was that the naturals are in some sense "simpler" than the reals. But if so, would it have hurt the author to say so in the text?

Ah, that's not the "mathematicians are careful" problem, that's the "theorems fall from the sky" problem :P

From the excerpt alone, there is no reason at all why we consider this particular $S$. That doesn't make the statement less true, it's just not explained why anyone would care about this statement

typically such corollaries have one of two purposes (at least in textbooks): either they will be used in some proof later (so that the author can just say "using corollary 1.0.3.") or they are intended to show a particular proof strategy (and the content of the corollary as such is somewhat irrelevant)

The mathematicians I know seem a generally happy bunch so there exists a subset of homo sapiens not annoyed by this type of working. It's just not for me. Oh well :-)

I mean...mathematicians in general often think that facts about math have an intrinsic allure that needs no further justification - that's why they're mathematicians! :P

To be fair, the percentage of the population who find colloid science fascinating is small.

@JohnRennie No worries. :) I suspect we need to see more context of that proof. My guess is the author has been talking about properties of sequences of integers, and then gives that proof to demonstrate that we can use those properties to prove stuff.

Speaking of colloids...

It's surprisingly hard to make foams from non-polar solvents.

I have no idea if that would be practical or useful. But I must admit that I find the concept of rocket fuel with the consistency of shaving foam to be rather amusing. :)

Foams are only kinetically stable since the solvent-gas interface always has a higher free energy than the bulk liquid, so the stability of a foam depends on the kinetics of adsorption and desorption of the surfactant at the surface.

It seems that only water has the sort of kinetics that gives stable(ish) foams.

It's not totally non-polar, though. One of the components is a saturated aqueous solution of ammonium nitrate.

Ah, OK, if it's an aqueous solution you stand a chance. It still wouldn't be stable on a timescale of hours or days though.

Making shaving foam type foams with methane then igniting it is a standard pastime amongst colloid scientists. We are simple creatures and easily amused.

It just has to be stable for a short time: long enough for it to burn. The idea is to disperse oxidizer through molten paraffin wax.

As I'm sure you know, N2O is used in canned whipped cream. Its solubility in water is roughly similar to the solubility of CO2, but it's very soluble in oils.

Shaving foam is surprisingly complex. Typically you need two complementary surfactants to get an adsorbed surfactant layer that is very slow to desorb. It's the very slow kinetics that allows the formation of very small bubbles.

In whipped cream it's a network of flocculated casein particles that stabilises the foam.

I don't know offhand why they use nitrous oxide ...

Probably because it's the only cheap food grade propellant.

Years ago, I experimented with using N20 aerating a mixture of milk & cream (plus sugar & chocolate). You can get amazingly light foams with the right milk : cream ratio, but of course they collapse rather quickly.

That sounds like a thoroughly worthwhile use of lab time. I only wish I could have been there :-)

Nitrous oxide solutions have neutral pH. Plus it has a slightly sweet taste.

Curiously, N2O & CO2 have the same atomic mass, 44.

@JohnRennie Derek Lowe says "Never trust an organic chemist who can't cook".

A nice short interview with him: statnews.com/2016/03/05/derek-lowe-chemist-blogger

I don't read Derek's blog every week, but I do try to look at it semi-regularly. I discovered him via the XKCD forum, when someone linked Derek's classic post about chlorine trifluoride.

A friend of mine at Cambridge was studying coordination compounds made from a toxic heavy metal (I forget which one) and cyanide.

The cyanide was (probably) quite safe when it was bound to the metal...

@ACuriousMind It's a fair cop! :-)

@JohnRennie but society is to blame

@Slereah for what, exactly? That people do stuff for no reason or that people expect a reason? :P

@Slereah Sigh

Never

Also the problem of monty python quotes was already barely a thing when that comic was written, and it has basically disappeared by now :p

I think that was more of a 90's internet thing

I didn't think it was an internet thing, it was just some annoying thing nerdy friend groups tended to do :P

(ni)

btw, the German version of this is extended sessions of verbatim Loriot quotes

I believe the same era in France was marked by references to en.wikipedia.org/wiki/La_Cit%C3%A9_de_la_peur

"The city of fear" - ah, yes, classic title for a comedy

is Ehrenfest theorem the way to get equations of motion in QM?

@Relativisticcucumber no

Ehrenfest's theorem gives you time-evolution equations for expectation values. But the QM "equation of motion" really is the Schrödinger/Heisenberg equation, not whatever you get from Ehrenfest's theorem

okay i see thanks

woah this is weird

ACM and CM both present

who is CM

@ACuriousMind so is it accurate to say that in the infinite square well case that 1) energy and momentum and compatible because kinetic energy and momentum are compatible, so 2) energy and momentum operator share eigenstates, so 3) if we have an energy eigenket there COULD be two associated momentum eigenkets?

what does "compatible" mean?

do you mean that they commute?

also, I'm not making a hypothetical here that there "could" be two or more degenerate eigenstates. You can simply look at the eigenstates of momentum, pick one with momentum $p$ and one with momentum $-p$, add them and observe that they are eigenstates of (kinetic) energy

there isn't any "could" here, for $p\neq 0$ it's just a fact that there are two states

I am thinking of starting with an eigenstate of energy though and then applying a single momentum operator--i guess im not seeing how we know that this should return an eigenvalue

yes commute!

in general it won't, that's my whole point!

For any energy $E\neq 0$, the general eigenstate with that energy is $a\lvert p\rangle + b\lvert -p\rangle$ with $p = \sqrt{2mE}$ and $a,b\in\mathbb{C}$ freely chosen

this is only an eigenstate of momentum for the special cases $a=0$ or $b=0$

there was a user called curious_mind. a clout seeker. // can we say that momentum of massless particles comes from their spin?

no, why would we say that?

massless particles don't even need to have spin

well my professor said photons have momenutm but they dont have mass (not really weird to me bc relativity), but then he kind of made it sound like this relates to their spin

even tho massless particles dont need to have spin their mass can give rise to momentum so that's fine

or for some reason his statement implied that the spin can be potentially a source of their momentum that we say they have?

i cant recall the exact statement tho

Oh i see what you are saying i think now

also how does one know if they are an experimentalist or a theorist D:

if you don't know, don't worry about it :P

If you watch closely in a mirror, you'll see either a lagrangian or a collider in your eye

That is the answer

now that's just hep-ex propaganda!

there's lots of experiments that don't have anything to do with colliders

Of course, but I'm clearly biased :P

@ACuriousMind oh no but i am afraid of grad school with so little sense of direction -- well i thought i had direction but i am questioning

Today my Professor was discussing about density of states in crystals and he drew the graph in the case of cubium

@Relativisticcucumber I mean...just do what you like, why does it matter whether it's "theory" or "experiment"

Then he said "This is reminiscent of Sauron's tower, where the dark lord resides"

well what if i get into a program that only has one :( or like is only strong in one

I'm still laughing

we shouldn't fall into the trap of reifying categories - just because "experimentalist" and "theorist" are sometimes useful categories that doesn't mean they describe inherent properties of people

Also @Relativisticcucumber how many years of undergrad do you have left?

LOL

this is my last year DDDDDD:

The first version was kinda ambiguous :P

@Relativisticcucumber I think that individual universities' strengths are usually much more fine-granular than "experiment" and "theory"

@Feynman_00 lol

like, look for specific subfields you're interested in, not "theory"

i was really strongly going for hardcore theory but i have been obsessed with my advanced lab course -- generally i like field theory so that is what i have been planning for

but i am not sure if field theory even has an experimental component. i guess everything must have both so maybe i can investigate that

or like explore to see what that is like. i only know about theoretical field theory

what kind of "field theory"

You can be a "theorist" and like labs and you can be an "experimentalist" and like theoretical subjects

@Relativisticcucumber that's what I meant with not reifying the category: You don't need to exclusively like either theory or experiment, you're allowed to like both

@ACuriousMind i am very interested in field theory to model gravity, so i guess doing some work on the development of that is what i am currently leaning towards but i still have much to explore since my chances to really explore that in UG have been rather limited

"theorist" is a descriptor of someone who mainly does theory (it is operational, not essential, in pretentious terms), not necessarily of someone who only likes theory or is unable to do experiment

I mean, theorists and experimentalists do not belong to orthogonal spaces

@ACuriousMind ah wow

i guess i am interested in quantum gravity but i feel that is kind of like when I say that people react strangely :PPP

There are theorists who create models to adapt the data acquired in the labs

so hence field theory to describe gravity XD

@Relativisticcucumber every hopeful young student is interested in quantum gravity :P

LOL i know this is why i dont tell ppl bc they think im just an ignorant popsci person

not to discourage you - of course some of these must keep their interest to become quantum gravity researchers

and so i just hide my interests in the corner of my room to avoid shame

@ACuriousMind what kind of paradise do you live in? :(

@Feynman_00 I meant that in a "it would be kinda cool if I worked on that" way, not in the "man, I can't wait to read this 1000 page textbook on quantum gravity " way

a 1000 page textbook on quantum gravity sounds good to me

but I mean in that case you essentially consign yourself to being a theorist from the outset because QG experimental conditions are a bit hard to come by :P

@ACuriousMind Oh, I can't wait to be able to read a 1000 pages book on QG

you just have a choice between chalk-and-blackboard theorist and running-simulations-on-superclusters theorist

Carrying out extremely long computations is satisfying

And conceptual issues are even more fun

@Feynman_00 yes, I got that vibe from you :D

I consider that as a great compliment :)

I'm not judging

No I meant that

That's genuinely something I'm glad to read :P

So my understanding is that its fair to ask what is the distribution of tangent vectors (since they are involved in collisions). What about the distribution of $ v^\mu = T^\beta \nabla_\beta X^\mu$ where $T^\beta$ is the tangent vector and $X^\mu$ is the deviation vector?

Does this have any particular distribution?

@Slereah ^ (hope you don't mind me pinging you)

@Relativisticcucumber the word reify has come up today xD

is that a problem :P

nono just funny for some reason i assign some cool words to where i first learned them from

and reify for me comes from some paper about on considering the bad habit in physics of trying to interpret thoeries as real

something like that was the thesis

by mermin physicstoday.scitation.org/doi/10.1063/1.3141952

ah, I know that one

I'm largely aligned with him on that topic

i think i am also aligned

can you transform a function from an inertial reference frame to a non inertial reference frame with some transformation

like for a rotating frame you apply the rotation to the motion so that it becomes linear in the rotating frame

@ACuriousMind for one of my grad apps @SillyGoose was really encouraging me to use the word reify but it did not fit at all the sound of the sentence and it was amusing so now that word is a meme. coincidentally i had used like 2 other more sophisticated words in that sentence and nowhere else so it looked like i really got my vocab in gear for a second and it was just very awkward

i removed the sentence eventually

ah yes, you have to sound like you swallowed a thesaurus for the whole text or it's awkward, yes ;P