@Mr.Feynman ...... yes

@ACuriousMind yes i mean i already understand all of this logic regarding "quantum computers do stuff faster" and am vaguely familiar with the few existing algorithms that show this. what i dont get is why this extends to saying "quantum computers should be able to simulate any physical process. i just dont see the jump here. i mean even for computers (classical ones), (afaiu) we dont say they can simulate every classical physical process that exists.

thats just not what i thought computers do -- i thought they are computing machines. so i expect that they can perform tasks involving performing computations. but then to extend to what deutsch said seems to require saying that all physics processes can be written in terms of computable tasks. im just not sure why this would be true. especially if we are going for exactness.

i feel like it also implies that there is nothing beyond the quantum that we know, but that is a separate argument. its kind of like the argument that "if we connect enough items and make them talk to eachother like neurons, we would have a conscious brain" and idk thats not compelling to me XD

i guess the nuance is in "make them talk to eachother like neurons"

but even if i say that i believe the world is truly explainable by the quantum we know, i still have a problem w deutsch's thesis, so i guess the first question i raised is the one im most curious about

@SillyGoose ????

the definition of delusion: a false belief or judgment about external reality, held despite incontrovertible evidence to the contrary

@Slereah too many experiments to repeat for any one person. Science is a team effort.

Are the negative eigenvalues allowed values for the kinetic energy operator in 1D Qm?And how would one possibly measure these negative kinetic energy eigenvalues?

@Relativisticcucumber I think at least a debatable form of the principle amounts to the following: that not only every physical system can be described by mathematics (an assumption physics more or less always makes) but that at least to arbitrary accuracy the computable (by whatever kind of universal computing machine) subset of math suffices for this

@Arjun isn't the "kinetic energy operator" (positively) proportional to $\hat{\vec{p}}\cdot \hat{\vec{p}}$?

more precisely, what negative eigenvalues are you expecting $\frac{\hbar^2}{2m}\vec{\hat{p}} \cdot \vec{\hat{p}}$ to have?

is there a reasonable reason that using grand canonical ensemble to "find" the bose-einstein distribution works?

i.e., the procedure 1. suppose you have a grand canonical ensemble of one species of particle. 2. write the ensemble in the occupation basis. 3. take the expected value of the $k$th occupation number, this expression takes on the form of the bose-einstein distribution.

goodmorining

@SillyGoose Consider the kinetic energy eigenvalue equation ,$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=K\psi$,now $\forall$ real K,we have $\psi=exp(i\sqrt\frac{2mK}{\hbar^2}x)$ as an eigenfunction with the eigenvalue K,note this is true even for K<0. Hence the kinetic energy operator also has negative eigenvalues, my og question was if the eigenstates corresponding to these negative eigenvalues are allowed in physics and how one would possibly measure them.

is there a typo in the summation index? shouldn't it be summation over $n_k$?

($p_k$ is a function of $n_k$)

@ACuriousMind interesting. i agree w the first part, but im mulling over the latter

well i dont know enough about complex systems to have an idea of the accuracy of this claim i guess

@SillyGoose i feel like hes summing over the k on the p not the k on the n ? XD

@Relativisticcucumber but that would also still be wrong

@SillyGoose why? cant u call it $p_{n_k}$ and sum over $n_k$

p still should be indexed by $k$

it is also indexed by $n_k$ implicitly since the eqn before this one shows the definition of $p_k(n_k)$

this is what i think it should be

@ACuriousMind I've seen counterexamples

Although the initial conditions are probably not too realistic :p

@SillyGoose can u send a pic of the page

@Relativisticcucumber of the textbook?

yeah

i think its like he writes $p_k(n_k)$ and realizes these are the same $k$s

to be simple he just writes it as $k$

since $k$ fully characterizes that term as written

but what this represents conceptually is summing over $n_k$ so then we still have the remaining $k$ leftover (which is what i mean when i say hes only summing over one $k$, the $k$ on the $n_k$) so the $k$ on the $p$ is left

but i cant say for sure

in what ur saying are you also saying that the $k$ in the $n_k$ in the argument of $p_k(n_k)$ is getting summed over?

if so i agree

wait no

summing over $k$ for $n_k$ is not the same as summing over $n_k$ for fixed $k$

because summing over $k$ for $n_k$ is summing over occupation numbers for all $k$ energy levels. but what we wish to do is sum over possible occupation numbers for a fixed energy level, i.e., sum over the index $_{n_k}$

$\sum_k n_k = n_1 + n_2 +...$ is not the same as $\sum_{n_k} n_k = 0 + 1 + 2 + ...$

@SillyGoose yes i know

what does "seminal paper" mean? i keep seeing it around

i mean what i said originally that we write it as $p_{n_k}$ and sum over $n_k$ but i think he is just realizing that $p_k(n_k)$ uses the same $k$ so writing it simpler idk i mean i agree that what we r summing over is $n_k$

why is the energy gap hypothesis necessary? if it is not met for $E_1$, then we just have a degenerate ground state. Then, there could be an energy gap $E_2 - E_1 > 0$ in which case (2.23) still holds with the index starting at $k =2$

@lucabtz it means giving birth to lots of further work

seminal as in like a seed which gives rise to more and more, etc.

but just call $n_k$ $k$ and sum over only that k -- why cant we do this? @SillyGoose

@SillyGoose yeah makes sense

also makes sense it often describes Zamolodchikov's papers

@Relativisticcucumber oh that is a possibility. but i would personally confuse myself btwn energy level labels and occupation number labels for each energy label.

@SillyGoose "seed" ;)

@Relativisticcucumber Lol

@SillyGoose yesh

the root for seminal is probably the same root for another word indeed

@SillyGoose semen should just be seed in latin really

yeah indeed it does just checked

is latin commonly spoken/known in italy?

@SillyGoose 🤤

@SillyGoose it is studied if you go to a kind of high school called liceo

actually not even all liceos have latin in their programs

but i indeed had it myself

@SillyGoose latin isnt spoken, no? i mean nowadays?

like my hs had it for study but i thought nobody spoke it..?

@Relativisticcucumber only at college graduations

@lucabtz oh i see. i was just wondering because the aeneid is about the founding of the roman people who settle in what is today italy

and it is a latin epic

@SillyGoose yeah i studied the aeneid indeed

but not everyone really studies it

after all it is basically a propaganda book ordered by Augustus, the first roman emperor

indeedo

@Relativisticcucumber yeah it isnt spoken, it is just studied. But translation is difficult. a tipical assignment or test is translating a small excerpt of latin text from some latin author (often Cicero, Seneca, Livius, Ceasar, etc.). It is typically around 20 lines and you are oftne given more than an hour to do it

so most people will not be able to speak it if it takes more than an hour to translate twenty lines

@SillyGoose the mainly studied books here are I Promessi Sposi by A Manzoni and The Divine Commedy by Dante, not really the Aeneid

@lucabtz oh i have not heard of the first text. i have read the inferno and the aeneid (translated into english, not in latin :P)

@Arjun hello. the Hilbert space is of square integrable functions (or Born rule wudnt make sense), so $K<0$ is not allowed

note that $K>0$ is sort of not allowed either. plane waves r not square integrable

but these form a basis of square integrable functions, which r allowed @Arjun

@SillyGoose you are correct. It is a sum over $n_k$ here

@Relativisticcucumber that is just confusing notation. Note that after correctly summing over $n_k$ for fixed $k$, there is a next sum over $k$, so if you do not carefully so this, there will be too much confusion.

it shud say something like $p_k (n_k) n_k$ and a sum over $n_k$

but they just write $p_k$ instead of $p_k(n_k)$

@SillyGoose It is just nicer for exposition. You are correct, if the ground state is $r$-fold degenerate, then the excited states will just start their sum from $k=r$ instead of $k=1$, and the argument will continue as-is.

okay i see

@SillyGoose no, the only "person" I know who can actively speak Latin is ACM

Lol the added quotes

@ACuriousMind this time you can't fool me

@SillyGoose ;)

lol

@naturallyInconsistent That doesn't work so well. ;) If you suspect it's a dupe, look for a dupe target.

It's ok to post comments correcting misconceptions in the question, if that's likely to lead to the OP improving the question.

High rep members posting partial answers in comments sets a bad example. We have too many answers-in-comments on this site. Such comments bypass the normal voting machinery.

Some members are chronic offenders, and they use comments to post stuff that's not quite correct, or non-mainstream. We really don't want to encourage that behaviour.

BTW, the system doesn't usually delete closed dupe questions. IIRC, they need to be unanswered, and have a negative score. Dupe questions aren't intrinsically bad, unless they're really obvious dupes showing no prior research.

The point of dupe closure is to consolidate all the answers into one place, so they can be easily found, and easily compared. And so that they form a single pool that people can vote on. OTOH, that strategy isn't foolproof, if there are a bunch of old popular answers that are wrong. That causes issues on SO, but it's less likely to occur here on Physics.SE.

@MoreAnonymous he seems to be arguing for a completely relational universe

depending on what kinds of relations are allowed, this can mean a completely mathematical universe. category theory, for e.g., only cares about relations

i doubt theyre arguing for a category-theory type universe tho. they probably mean more general relations than in mathematics

but then the downside is that the philosophy becomes vague, as u cant know what exactly theyre talking about anymore, as is the case with most of philosophy

@RyderRude Yea I think so too

How does category theory make sense of units and dimensions? Like its one think to say a scalar field its another to say a scalar field of units kelvin

@RyderRude Then can't the k<0 solutions be used to form a basis for square integrable functions?Since apparantly they are eigenstates of the kinetic energy operator should they also not belong to the basis functions set?

@Arjun no, u wud need something like the Fourier inversion theorem, but using the $k<0$ solutions en.wikipedia.org/wiki/Fourier_inversion_theorem

there is the Laplace transform.. im not sure how it connects to this.

@Relativisticcucumber Oh, come on. I didn't...

but note that the k>=0 solutions already do form an eigenbasis because of the theorem

Why does this happen every time I reply to you @Relativisticcucumber D:

@Mr.Feynman wdym

note that the inverse of the Laplace transform is still an integral over complex exponentials, rather than real exponentials. it's just that there's also a decaying factor. so one wud hav to allow complex eigenvalues for this

@RyderRude Why can't the same theorem that guarantees that k>0 solutions form an eigenbasis used to show that k<0 solutions also work?

I don't understand why k<0 solutions are not legitimate enough to be considered as basis..afaik all eigenfunctions of an operator belong to the basis set right?

@Arjun ive never seen a rigorous proof of the theorem, but intuitively, any reasonable linear combination of real exponentials wud not be a square integrable function becuz real exponentials grow without bound

@Arjun the latter part of this statement as stated is certainly false

complex exponentials can cancel out their growth and stuff

consider a finite degenerate spectrum. you do not add the entire degenerate subspace to your basis. only the minimal amount to form a complete basis. however, every element of the degenerate subspace is an eigenstate of the operator.

@SillyGoose I agree with that..but in this case there are no degeneracies ..

i mean to correct the general statement you made. also, i believe the kinetic energy operator is degenerate, but that doesn't really change anything about the question you are asking. i just mean to point out the general fact i did

@Arjun note that "eigenfunction" is, by definition, something that lies in the hilbert space. so the momentum operator doesnt really have eigenfunctuons in the first place

but the complex exponential basis expansion works out because of the Fourier inversion theorem

it is not a "basis" in the strict sense..it is called a rigged basis @Arjun

@RyderRude So solutions corresponding to k<0 do not belong to the rigged basis set?

yes

intuitively, no reasonable linear combination of those functions would be square integrable

Hmm..that makes sense..hence one will never "measure" negative kinetic energy in QM right? @RyderRude

yes

@Relativisticcucumber this reply

laso note that the Fourier transform is unitary due to Parseval's theorem @Arjun

there is an interesting observation that states should really be normalizable and not just square-integrable. but there is some math result(s) which allow one to be sloppy with the language

the zero function is a trivial example of a square-integrable function that cannot be a state, yet is square-integrable

So to sum it up..we won't have pure complex exponentials since they are non normalizable ,we will only have square integrable combinations of them.And k<0 solutions NEVER produce a square integrable functions..so one never measures negative kinetic energy since exponentials are never present

but it is discounted because it is still not normalizable

oh

@SillyGoose yes

but note that the zero function is part of the hilbert space. normalizability is a restriction on physical states, but not the hilbert space

indeed the state space does not contain a $0$ ray

@SillyGoose @RyderRude when you guys say square integrable do you mean locally or over the whole line? And will either of them imply the other?

@Arjun also, the complex exponentials form a complete basis. so, any physical state u consider would have a Fourier transform spectrum with real momenta

over all of $\mathbb{R}$

integrating over $x \in (-\infty, \infty)$

If you allow $K < 0$, then your wave function is going to diverge as $x$ increases at some limit, so it's not going to admit a probability interpretation to allow such solutions, this is different to the non-normalizability which has a physical interpretation unlike the diverging solutions

your function could be $0$ everywhere except some "local" set $(a, b)$ where $a,b \in \mathbb{R}$. in this case you don't need to integrate outside of this "local" region since the contribution to the integral will just be $0$.

@Arjun so not only the real exponentials r ill behaved, they arent even required as complex exponentials span the entire space

square-integrable means that $\int_{-\infty}^\infty dx f(x) $ converges to some definite value.

@Mr.Feynman u mean i always nom or chomp?

@RyderRude If I understand it correctly..complex exponentials don't belong to the hilbert space..they belong to an extension called rigged hilbert space..but they somehow span all vectors in hilbert space?

If you have a free particle, it can be anywhere, including at infinity, there is no way you can describe a particle spending time at infinity with normalizable wave functions, you need these non-normalizable wave functions i.e. it is physically incorrect to do what rigor tells you to do and demand normalizable eigenfunctions

@Arjun but there r some subtleties with applying the Fourier inversion theorem on the whole L^2(R) space. see the "conditions on functions" section en.m.wikipedia.org/wiki/Fourier_inversion_theorem

@Arjun yes, but there r subtlties with them spanning all of $L^2(R)$ like in the link i gave. the theorem more nicely applues to Schwartz functions which is a subspace

but it applies nicely to any physical states u wud consider. so any physical state wud be spanned by complex exponentials

@Arjun if you are interested, there is a nice book called "Quantum mechanics" by Leslie E. Ballentine. the first chapter briefly introduces rigged Hilbert spaces (among other things) and the textbook is nice overall.

@RyderRude oh..so for all physical states that I'd encounter ..I can basically not worry about these subtleties right? That sounds greatly relieving lol

yes.. u can mostly ignore this stuff. this stuff comes under functional analysis if u ever want to study it @Arjun

@SillyGoose Thanks will check it out fosho

@RyderRude Oh..but I have to study linear algebra properly first lol

And then I'll study some functional analysis (for physicists)

okay :)

I don't know if I'd have the prerequisites to study FA after studying linear algebra say at the level of strang or axler

Not even clear that rigged spaces are actually correct for this simple example

it might help to know some real analysis and topology

dont trust @SillyGoose -- they r a math major

@Relativisticcucumber hey i was a physics major too

@Arjun to be clear, some trivial real exponentials combinations can sum to square integrable functions. i think the main crux is just that : 1. negative KE requires momentum measurements to give imaginary results 2. the fourier inversion theorem says the K>=0 form a basis, so no more vectors are required in the linear combination

@SillyGoose is ur main study no longer physics?

oh i am still studying physics @RyderRude

I'm still standing yeah yeah yeah

@Relativisticcucumber no, I mean those "....." before "yes" are kinda sus

@SillyGoose oh.

@SillyGoose but is ur main study math now?

@Arjun one of the assumptions of QM is that observables are Hermitian. so imaginary measurement outcomes for momentum must not be allowed

@RyderRude yes you are correct for k<0,we get imaginary p values..I should've checked this initially : ) and since momentum operator is hermitian and hence only has real eigenvalues..negative k values shouldn't be allowed..thanks for the help

@RyderRude Oh!Do you have any example at hand for $\sum Ae^kx+Be^-kx$ being square integrable ,because I can't think of any A's and B's for which the above sum is square integrable

@Arjun i only had A=B=0 in mind :P

@Arjun but i just thought that Taylor series can manage the growth of polynomilas using mixed signed coefficients and factorials, so maybe real exponential integrals can use similar tricks to achieve it....but im dont hav a concrete example

but most importantly we dont hav a general basis expansion theorem with real exponentials. Laplace transform can use exponential growth, but its inverse is not an integral over real exponentials

@PM2Ring thanks. that's much more clearer than what I had thought it was meant to be like. On myow end, it is more of a case of itchy hands: cannot help but type some throwaway answer-in-comment than not. Sigh.

@RyderRude Lmao

@Arjun Good university lectures and textbooks will cover some properties of what constitutes good solutions of the Schrödinger equation. In particular, the 2nd order derivative governs the curvature of the solutions for any $E$ and $V$ region, and the inability to make $\psi\to0$ at the infinite limits is scrutinised. Your questions here are all in that vein.

I should probably do my due diligence and check what Vixra says about quantum computing

vixra.org/pdf/1505.0027v1.pdf

I'm already hooked

It's about a quantum operating system and it talks about quantum gravity

I skimmed the Education section the other day, it didn't look ridiculous (unlike the physics areas), I was wondering whether the Chemistry section was full of nonsense or not I wasn't sure...

Well it's Vixra, you may have like enthusiasts posting about it

High school physics is probably within their grasp

there would have to be a lot of great research on vixra, but impossible to dig up

because of the crowd

Their own fault to send it to the garbage, surely

There is even the odd author on there who knows advanced math

You'll find the odd paper that seems competent but I don't think I ever found a paper I wanted to read on vixra

Usually people who are competent and want to publish something casually will do it on a blog, not on vixra

Last time I checked vixra someone "proved" gauss's divergence law for electrostatic fields was wrong XD

Knew it

lol

I have no words

there was this guy linked here who kept posting for 10 years about the kinetic energy formula being wrong

@bolbteppa It's either nonsense or an nlab article

The ones here, I mean I don't know if people are mistakenly using it for real or it really is just all nonsense

Comments: 8 Pages. Rejected by many top journals without review

lol

Some of them have multiple authors

@Slereah That might explain a lot

vixra.org/pdf/2401.0060v1.pdf

At least one of them there has a phd and is a lecturer etc and posted something that looks legitimate...

yeah there's a few legitimate scientists on vixra IIRC

usually as some protest against Arxiv

“crackpots” who were right

I mean in spirit it's not a bad idea, in practice you end up getting what this actually is...

The Stueckelberg one is interesting...

yeah I'm not gonna sift through this trash to find a cool paper :p

This must be discussed somewhere in the main site...but can't find it: given a canonical transformation, how do we find which type of generating functions can't produce such canonical transformations?

@Sanjana you'd have a far better experience and ease if you simply consulted Goldstein. The flow is the opposite: you check the 4 generating functions, and discover that each one fails to cover a certain type of canonical transformations.

The of course operator seems to be so called because if an inference is true, it is true after arbitrarily many evaluations

In the QI case that corresponds to the fact that coherent states are classical-ish and can be cloned, and therefore can be used multiple times in a quantum circuit

or approximate cloning, apparently

@naturallyInconsistent Ofcourse I have seen that. It's just that there's no systematic discussion of the issue. E.g. If I am given a canonical transformation, I wouldn't wanna waste time in choosing which generating functions to use. I think it depends on the invertibility of the transformations... In principle I could try to derive them on my own...but I thought... Why bother when u have pse :p

that's obviously the case. lol

So extensive parameters are those which add when combining systems like $U_{AB} = U_A + U_B$ or volume, etc

what are intensive parameters then? like an example would be nice

@Obliv Schroeder gives some examples

oh shit that's the one section I didn't bother reading :D

also the end of ch4

ty

np lol

Wait so entropy doubles but multiplicity doesn't right? Is it because entropy is $\log\Omega$

Idk

Oh yeah

$\Omega_{AB} = \Omega_A\Omega_B \implies \log\Omega_{AB}=\log\Omega_A+\log\Omega_B$

yeah, the last sentence talks about it

logs are just exponentials but I forget that sometimes

multiplicity is multiplicative rather than additive

so it's neither intensive nor extensive

entropy on the other hand is additive, so it's extensive

Reminds me I need to learn en.wikipedia.org/wiki/… at some pooint

@SirCumference Dude you would hate callen lmao.

It reads like anti-schroeder

I have to put on my thinking cap when reading anything in this book

sounds kind of similar to L&L

Yeah dude, it's got that dense graduate text feeling. Like dummit & foote for abstract algebra

which tbh is quite good, but i definitely wouldn't recommend it as a first introduction to thermo

Definitely not

the funniest book is probably Goodstein

Lol, that's where that quote is from

thought that was a question lol

the only stat mech book I really didn't like is Chandler

i think it was written by a chemist or something

Boltzmann got so much hate for his ideas I think is part of the reason why. idk about ehrenfest. I think they were speculated to have bipolar but that's just random guessing at some point

Oh so you took thermo more than once?

for undergrad stat mech i used schroeder, for graduate stat mech i used chandler

@Obliv ... Callen also gave examples...

yeah its in chapter 2.1

@Obliv No, that quote is very standard for stat therm textbooks and lecture courses

@Obliv There was a LOT of physics pushback against atomism. At the time, chemists had already long since accepted atomism as fact whereas physicists were still deluding themselves that classical continuum could have a chance.

Atomism was pretty controversial until fairly late

I don't understand what the alternative was.. like Leibniz believed in some inherent pointlike particles no? or was that newton

I think Leibniz believed everything was a fluid or something

I think it's time to bring back anti-atomism imo

Isn't that kind of like strings :D

Matter being infinitely divisible?

Yeah he called them something

monads

La Monadologie

plus in modrn days we think matter is fields

A continuous medium

We just switch idea every couple of century

@naturallyInconsistent Does callen go over blackbody radiation? Can't seem to find it

@Obliv Chapter 3.6 Electromagnetic Radiation. Chapter 16.8 Electromagnetic Radiation. Chapter 18.6 Non-Conserved Ideal Bose Fluids; Electromagnetic Radiation Revisited.

Hmm he uses h bar in 16.8

whoah he uses the riemann zeta function in the integral

@Obliv and your point is? I mean, by that point, it is more of whatever is nice and the prof likes, goes. There is presumably some simpler lecture module or textbook covering the simpler parts and all the different versions of Planck's Law and so on.

I was looking for an introduction to the issue raised by classical physics approach to blackbody radiation but it's ok the crux of the issue is u will get infinite energy by your choice of energy function for the partition function

if u dont set it up with multiples of h or whatever

I think because you get 1+e+e+e+...

if u take each antinode as kT energy

I guess u wouldnt have e^0 as a mode so just infinite e+e+...

Which I dont even know why we'd have kT as the assigned energy for an antinode but whatever I guess that's the amplitude of the classical E-field

I am not sure what it is you are trying to say. I have no idea what is the premise; I do not know what would 1+e+e+e+... correspond to, nor "each anti-node as kT energy", etc.

I was just saying that in the classical picture if we consider $Z = \sum\limits_s e^{-\beta E(s)}$ with the energy of each mode being $\frac{f}{2}Nk_BT$ you'd have $Z = 1 + e + e + ...$ (where 0 energy is the "ground" state)

which leads to $\overline{E} = -\frac{1}{Z}\frac{\partial Z}{\partial \beta} = \frac{\infty^2}{\infty}$ or something

in minor news, the astro team i'm on just won our uni's physics soccer tournament

so that feels pretty good

Oh cool, so it's like a soccer tourney for physics majors? :D

@Obliv that's not how that works.

@SirCumference oh nice yay winz

@naturallyInconsistent what does the classical picture look like, then? I don't think Callen covers it since it's wrong

@Obliv for the grad students, yep :)

@Obliv In old NR classical mechanics the KE is $\frac12mv^2$, which is then inserted into Maxwell-Boltzmann distribution, looking like $Z=\int e^{-\frac12\beta mv^2}\mathrm dv$, and then you get the $\frac12k_BT$ for this

for light?

@Obliv light is necessarily relativistic. It is easier to just take what Callen gives you.

@SirCumference oh, so you won the football tournament. Good job

@Mr.Feynman thanks :)

though, soccer is just short for "association football"

same way rugger is short for rugby football

tho people in the US don't usually use the latter as much as e.g. the UK would

@SirCumference I knew what I was doing :P

i figured lol

@Mr.Feynman come to think of it, when you say football in non-North American countries does it usually just refer to association football?

or do you gotta be specific

in the US it just refers to gridiron football since that's by far the more popular one here

@SirCumference I'm not sure about the usage outside of US (soccer) and UK (football) and where the rest of the world stands. For example, here we just use the italian word for it, which is "calcio"

And what in the US is football is called "American football" I guess

cosmology can b modelled using stat mech

@SirCumference congrats!

this means that stat mech can b used at many scales, while qm has an absolute scale

@RyderRude naw still doing physics mainly