The h Bar

00:03

@Obliv yes

SnoopyKid

02:19

Hello, between a 96 horsepower 2200 lbs car and a 96 horsepower 800 lbs motorcycle, the 96 horsepower 2200 lbs car will generate more powerful impact force than the 800 lbs motorcycle with the same horsepower if both are hitting a wall. Is that correct?

Obliv

02:37

Not necessarily. Depends on the speed of each vehicle, material & what's actually hitting the wall

and how much is hitting the wall, etc. But in general, F = ma.. so the 2200lb car has greater mass and therefore more force generally

user 70432

> Horsepower refers to the power an engine produces. It's calculated through the power needed to move 550 pounds one foot in one second or by the power needs to move 33,000 pounds one foot in one minute.

Horsepower

Horsepower (hp) is a unit of measurement of power, or the rate at which work is done, usually in reference to the output of engines or motors. There are many different standards and types of horsepower. Two common definitions used today are the imperial horsepower, which is about 745.7 watts, and the metric horsepower, which is approximately 735.5 watts. The term was adopted in the late 18th century by Scottish engineer James Watt to compare the output of steam engines with the power of draft horses. It was later expanded to include the output power of other types of piston engines, as well as...

user 70432

03:58

Pope Francis blesses special WBC belt on the line in Tyson Fury vs Oleksandr Usyk undisputed clash

Sanjana

06:23

I am having a question I had before too: when exactly can we say that a particular type of generating function does not exist given a canonical transformation? I thought it happens when the variables in the argument of the generating function of that particular type are independent, but I found some cases in which they are not and yet a generating function of that type exists

Here $Q$ and $q$ are not independent, but still generating function of type 1 exists. I even found the correct option to be $(c)$. But, then what is the condition for non-existence of type 1 generating function, for example?

ACuriousMind

10:15

@Sanjana you're essentially asking the same question as physics.stackexchange.com/q/699266/50583, see also physics.stackexchange.com/a/329443/50583 - I don't think there are a lot of explicit conditions here one can write down beyond Qmechanic's condition that if the resulting embedding can't be a graph of the independent variables of the transformations, it doesn't have a generating function of that type

Sanjana

10:36

@ACuriousMind Hmm... yeah it seems that's the only way

Btw, I came to know that the transformation $(q,p) \to (a, a^\dagger)$ for harmonic oscillator is singular, i.e. there are no config space coords corresponding to $a, a^\dagger$... Does this manifest in some way in the QM world?

ACuriousMind

I don't know what you mean by "singular" - that transformation just doesn't make any sense in ordinary mechanics since phase space isn't a complex manifold and so going $a^\pm = x\pm \mathrm{i}p$ doesn't actually mean anything

Sanjana

11:05

@ACuriousMind By singular I meant that you can't invert to get a Lagrangian because the Hessian is singular.

ACuriousMind

invert what to get what Lagrangian?

again, this isn't even really a well-defined map on phase space because you generically have no complex structure on phase space

Sanjana

@ACuriousMind Can you act for a while that you don't know this please :)

ACuriousMind

no, because the conversation makes no sense to me this way :P

Sanjana

@ACuriousMind Okay fine, then let's settle this: I am complexifying the phase space so as to include canonical transformations of that kind

@ACuriousMind And if I define $a= \frac{x+i p}{\sqrt{2}}$, then $\frac{\partial^2 H}{ \partial a^2}=0$ which is why I can't get a Lagrangian for the Hamiltonian $H=\frac{1}{2}(p^2+q^2)$

and btw I was avoiding the complexification issue, because I think that does not actually interfere with the question I am asking.

Ryder Rude

@Sanjana hi. also see this en.wikipedia.org/wiki/Segal%E2%80%93Bargmann_space

Sanjana

11:15

and oh...btw this transformation is discussed in Jose-Saletan (which is a physics book and ofcourse they didn't discuss the complexification issue, so I refrained from taking the shelter of saying "Hey! The book says it...":p)

@RyderRude Yes, I know 'bout this

ACuriousMind

@Sanjana I'm still not quite sure what's going on here - in order to get a Lagrangian from a Hamiltonian you must perform a Legendre transform on the "momentum" part of the Hamiltonian. The $a,a^\dagger$ coordinates are not such position/momentum polarization coordinates (since their Poisson bracket is not the $\{q,p\} = 1$ of Darboux coordinates), so trying to do a Legendre transform here would not be expected to yield anything good anyway

I mean, you're right that it doesn't work, but...I wouldn't have tried that in the first place :P

Sanjana

@ACuriousMind Why is this so natural to you?

And why do you say $\{a, a^\dagger\}=1$ is not that of $\{q,p\}=1$ of Darboux coords?

Ryder Rude

@user85795 really interesting

ACuriousMind

Quantum mechanically I'd say this just manifests in $a,a^\dagger$ not being self-adjoint - the position and momentum operators are self-adjoint with continuous spectrum, $a,a^\dagger$ are not, they're just two different kinds of things

@Sanjana careful: $[a,a^\dagger] = 1$ is the quantum commutator - the Poisson bracket is $\{a,a^\dagger\} = \mathrm{i}$

Ryder Rude

but note that this is still "almost" a canonical transform despite the $i$

Hamilton's equations pretty much hold in such co ordinates, with slight modification

Sanjana

11:21

@ACuriousMind Wait a sec...I missed an $i$ in the definition. Does this change anything? I will just give a snapshot from the book

Ryder Rude

we have $\frac{da}{dt}=\frac{1}{i} \frac{\partial H}{\partial a^{\dagger}}$

Sanjana

@ACuriousMind How is this related to the singularity of the Legendre transformation :o

ACuriousMind

@Sanjana it isn't

I think I misunderstood the angle of your question

Ryder Rude

but $a$ and $a^{\dagger}$ cannot really b thought of as independent, I think becuz the complex conjugation is a constraint

so we r not using the whole complexified phase space, but a constrained one

Sanjana

@ACuriousMind As you can see from the screenshot, they said that it is impossible to obtain the Lagrangian formalism for the generalized coordinate $Q$. So I was just wondering what does it mean for QM...

ACuriousMind

11:26

@Sanjana I think nothing?

The only place you need the Lagrangian formalism in QM is if you do the path integral, but the usual path integral is in terms of ordinary position and momentum where the Lagrangian does exist, so why would it matter?

Sanjana

Idk... If I get a Hamiltonian system which in certain canonical coordinates admit no Lagrangian description, does it mean nothing special when this system is canonically quantized?

I mean, will there be no effect in such a quantum mechanical system?

ACuriousMind

but no one "canonically quantizes" the QHO in the $a,\mathrm{i}a^\dagger$ coordinates!

we quantize real (not complex!) phase space in the $x,p$ coordinates

(in geometric quantization this would again be related to having to use coordinates with appropriate polarizations)

Sanjana

@ACuriousMind arghhh... Ok, but I already knew that. I guess, there isn't nothing new here...

Btw @ACuriousMind, I realised that the way you use the word constraints and the way QMechanic uses the word is different. To you, constraints are those equations which hold as offshell identities. To QMechanic, constraints are those equations which don't have time in it. E.g. Faraday's law is a constraint to you but not to QMechanic. Is this just a matter of semantics/personal taste or is something deep going on here?

(Or maybe I misunderstood/misread something...which is always an option :p)

ACuriousMind

@Sanjana I'm not really sure what you mean - the non-trivial Hamiltonian constraint in EM is Gauß' law, not Faraday's law - Faraday's law is just part of $\mathrm{d}F = 0$, which in turn is a consequence of the definition $F = \mathrm{d}A$, not a Hamiltonian constraint

also I'm pretty sure Qmechanic and I should use the word the same way in the Hamiltonian context, you just need to be careful that there are also instances outside of the Hamiltonian formalism where one might talk about constraints, such as the (an)holonomic constraints in other parts of mechanics

Sanjana

11:42

@ACuriousMind You didn't use the word as a "Hamiltonian" constraint. It was in a conversation with another user. Look at your reply to this message. I am sure you weren't meaning "Hamiltonian" constraint

This was the Qmechanic answer where he calls only the Gauss laws constraints (Now even you seem to say the same)

ACuriousMind

12:14

@Sanjana yeah, outside of the Hamiltonian context I don't think I use that word particularly consistently

Obliv

14:37

@ACuriousMind do you think learning module theory would make understanding QM easier?

since modules over a field F and vector spaces over F are the same

ACuriousMind

I'm not sure what "module theory" is - but if the question is whether for this particular case learning more generally about rings and modules than about fields and vector spaces, the answer is no

what you need for QM is the theory of infinite-dimensional Hilbert spaces and operators on them, i.e. functional analysis, not modules

Obliv

hmm Idk anything about functional analysis :\ I wonder if there are introductory texts

Mr. Feynman

First you need some measure theory

And now fall down the rabbit hole like I did

ACuriousMind

if you're asking if you should learn this before first learning QM - probably not

elementary QM is perfectly fine with linear algebra + some Hamiltonian mechanics under your belt

Mr. Feynman

Oh, sure. Even if you decide to learn about it, you should do like the red diamond user above suggests

ACuriousMind

14:45

(plenty of courses/texts try without classical Hamiltonian mechanics, but this tends to confuse students about what's really quantum about quantum mechanics :P)

Obliv

I'm not sure if I should be grateful or not for physics texts being terse about the mathematical definitions and usage. I thought if I go out of my way to understand the math then the physics would follow more naturally but idk

I guess not. More familiar maybe but not easier to understand

this computational quantum mechanics book says that analytic solutions are pretty useless to study and that in this day and age numerical techniques are more important.

Makes it seem like undergraduate QM is pointless or something

ACuriousMind

only if you think the hydrogen atom is pointless :P

Obliv

Yeah I guess it depends on what field you are in and what you're doing

ACuriousMind

also while courses vary wildly it's not infeasible to do a bit of perturbation theory at the end of a QM intro

also you can't call the basic underpinning of a field of study "pointless" just because modern research has exhausted what you can do with just the basics :P

Obliv

very true

There was a company at a job fair developing fusion energy via stellarator and I was like man that seems so cool but one would need a lot more than undergraduate QM to work at such a place

bolbteppa

15:00

@Obliv Absolutely not, you will only get more confused, just read a good book on it

Mr. Feynman

@ACuriousMind I really can't imagine myself using a Hamiltonian operator without knowing what a Hamiltonian is

Really, that should be a crime

user 85795

a crime against what, exactly

Obliv

15:28

@bolbteppa isn't LL really dated?

bolbteppa

QM is really dated, the most important stuff was worked out a hundred years ago

Obliv

Sure but I'd probably rather read a book written in 2010 on a subject 100 years ago than a book written 100 years ago :P

also I don't speak russian so the appeal of reading something old lessens when it's been translated

user 85795

Have you read Shakespeare :P

(I completely agree about the translation part)

Obliv

15:49

@user85795 Yes I do enjoy old works of literature and poetry. When I'm reading math/physics though I prefer modern texts

Obliv

16:11

what the heck does unit norm and $\delta_{ij}$ mean

like norm is 1?

@SillyGoose maybe ballentine wasn't the play D:

hmm seems like notation used in inner product spaces

ohhh ok nvm i get it

Mr. Feynman

16:44

@Obliv yes, unit norm means that the norm is one. In this case we are talking about the norm induced by an inner product $v\mapsto||v||=\sqrt{(v,v)}$

$\delta_{ij}$ is a Kronecker delta

DIRAC1930

@Obliv I would throw that book in the bin lol. The whole of many-body theory and qft literally relies on the simple harmonic oscillator

Silly Goose

17:04

@Obliv u could try other tests like townsend, griffiths, or sakurai

@DIRAC1930 balletine still covers the harmonic oscillator

DIRAC1930

L&L 3 is a complete masterpiece but not an ideal place to learn qm for the first time IMHO

There are probably some fantastic lectures by Allan Adams online that I would recommend

Silly Goose

Idk L&L 3’s treatment of angular momentum seems a little meh

Well ideally one would read ballentine and another qm text. Since ballentine is more foundations focussed and perhaps something like Sakurai will give more breadth of application

but if one understands ballentine presumably one would know enough to understand any other works of QM

“Breadth” as in more discussion on perturbation theory and ifs application to scattering i guess…

Silly Goose

17:34

L&L 3’s initial discussion of what a measurement is also seems to be a bit outdated

bolbteppa

I think I'm going to have an aneurysm

user 85795

Why

is this discussion giving you a headache? @bolbteppa

Silly Goose

Is there a method of quantization that takes observables as the starting point? Roughly, construct quantum operators whose expectation values are the classical observable values

ACuriousMind

@SillyGoose what does "classical observable value" mean?

most quantum states don't describe a state with a meaningful classical limit, so there is no sense in which the expectation values of observables in those states could be "classically observable values"

Silly Goose

I guess match up $O(x) = \langle x \lvert \hat{O} \lvert x \rangke$. Where this procedure is limited to classical observables dependent only on x or p

ACuriousMind

17:48

@SillyGoose yeah, that's the wrong thing to do

you'll never get a classical limit out of $\lvert x\rangle$

wait

if you limit this to observables only dependent on $x$ or $p$ this is trivially true for all operators that are functions of the operator $\hat{x}$ as soon as you say $\langle x\vert \hat{x}\vert x\rangle = x$

Silly Goose

Oh

ACuriousMind

that's just how functional calculus works - if you have eigenstates $\hat{x}\lvert x\rangle = x\lvert x\rangle$, then $O(\hat{x})\lvert x\rangle = O(x)\lvert x\rangle$

in fact, this is more or less the definition of $O(\hat{x})$

Silly Goose

I see

ACuriousMind

nothing to do with quantization or classical limits

bolbteppa

18:06

> It is as good an idea to read the masters now as it was in Abel’s time. The best mathematicians know this and do it all the time. Unfortunately, students of mathematics normally spend their early years using textbooks (which may be, but usually aren’t, written by masters) and taking lecture courses which are self-contained and make little or no reference to the primary literature of the subject.

> The students are left to discover on their own the wisdom of Abel’s advice. In this they are being cheated.

Slereah

Reading 19th century textbooks is a bit of a chore

Mr. Feynman

But the masters don't use $\LaTeX$

Silly Goose

So if we have an observable O(x), then classically this observable is independent of momentum p in the sense that a state can be specified by (x,p). Quantumly, the observable would be sensitive in general to different levels or momentum being present in the state. Can this be thought of as an example of a classical sort of symmetry not lifting to a quantum one (in this case necessitated by CCR)

bolbteppa

Apart from Dirac's book, you have L&L written by the people who discovered most of this QM stuff, there are other QM books by those guys which have good stuff but nothing like those two

Slereah

The Masters typically wrote the books when people barely understood the topic

and usually using methods which may not be the best

Silly Goose

18:13

Improving a students scientific literacy seems a good idea but I’m not really sure how one is supposed to read physics papers without knowing basic physics. For other subjects (other than perhaps math or philosophy) enough of the basics can be understood in one semester in normal language. I don’t think this is the case for physics.

ACuriousMind

@SillyGoose no

where's the symmetry supposed to be?

Silly Goose

Perhaps a student could read a paper from the 1800s on classical mechanics. But i’d be surprised if a student can read a quantum optics paper or quantum computing paper as their quantum education

ACuriousMind

or rather, the classical observable is invariant under translations in momentum, and so is the quantum observable, since $x$ generates translations in $p$ and $[O(x),x] = 0$

Sir Cumference

18:26

anyone here happen to be good with sheet music (specifically piano)?

ACuriousMind

what does "good" mean here?

I can read sheet music, playing it is another matter :P

Sir Cumference

well, if anyone could tell me which of these looks easier to sight read:

with the difference being on the last beat (this is in 4/4)

ACuriousMind

that's above my pay grade, sorry :P

Sir Cumference

all good lol

Silly Goose

What does the bracket on the top one mean?

Sir Cumference

18:31

a triplet (i.e. splitting a note into three smaller notes)

Silly Goose

the bottom one looks easier to sight read because it lacks the bracket :P but that’s because i don’t know what the bracket symbol means XD

Oh is the bracket supposed to make clear that those are the notes that should be grouped together

Sir Cumference

kind of

the inclusion of the bracket does affect how long they are held for, but yeah that's kind of the gist

user 85795

18:49

You could post it on the musicology se

Sir Cumference

yeah i just did now, tho the chat seems kind of dead in comparison

and whenever you post a question on a main SE site there's a fear of getting downvoted for it being a dumb question

well maybe that's just my anxiety speaking lol

user 85795

Nah, how else are we gonna learn :-)

Sir Cumference

true enough

user 85795

Do you play the piano regularly :^)

Sir Cumference

well i started getting into it when I was in college, but only recently have I started to get more deeply into it

wish I knew earlier how interesting music theory could be. changes the way you think about songs

kind of in the same vein that physics changes the way you see the natural world

bolbteppa

19:03

@Slereah When you start reading stuff that writes $xx$ for $x^2$ you know you've gone too far back

Slereah

@bolbteppa as our forefathers did

user 85795

Indeed the music of the spheres sounds differently to those who appreciate mathematics.

bolbteppa

I remember massively massively failing on

Disquisitiones Arithmeticae

Disquisitiones Arithmeticae (Latin for Arithmetical Investigations) is a textbook on number theory written in Latin by Carl Friedrich Gauss in 1798, when Gauss was 21, and published in 1801, when he was 24. It had a revolutionary impact on number theory by making the field truly rigorous and systematic and paved the path for modern number theory. In this book, Gauss brought together and reconciled results in number theory obtained by such eminent mathematicians as Fermat, Euler, Lagrange, and Legendre, while adding profound and original results of his own. == Scope == The Disquisitiones covers...

More Anonymous

19:54

Why are people still studying string theory

?

bolbteppa

20:05

It's one of the most important/promising research programs in all of physics

Slereah

useful for guitars

3

Sir Cumference

it also just seems mathematically cool

granted I haven't gone deep into it

More Anonymous

Hmm ... I thought I explained it had problems with its vaccum no?

Ages ago

chat.stackexchange.com/transcript/149450/2023/11/2

@bolbteppa Do respond to this^

bolbteppa

I don't know what that's talking about

More Anonymous

So we can define an acceleration operator no?

Using the Hesienberg picture

Actually never mind

Its painful to recall these shambles

Silly Goose

20:27

What is the big deal about bose einstein condensates?

It seems implied that they’re a more “macroscopic” quantum effect, which is an interesting fact but are there other reasons?

ACuriousMind

@SillyGoose superconductivity involves Bose-Einstein condensation and superconductivity is both pretty cool and technologically pretty important

also, just abstractly, BECs being a "new state of matter" is interesting in its own right, since it does not fall into the traditional gas/fluid/solid distinction

Vincent Thacker

Hi @ACuriousMind Do you have any idea what happened to my flag on this answer? physics.stackexchange.com/a/814252/174766

It has gone unanswered for 4 days

The LQ review passed but it didn't get deleted because someone upvoted it

ACuriousMind

20:42

@VincentThacker we usually leave "not an answer" flags pending for the LQ review to deal with it; unfortunately the mod UI isn't very good at showing us these kinds of flags where the review has already completed

I've handled it now

More Anonymous

@ACuriousMind are u familiar with Nagarjuana's philosophy?

ACuriousMind

nope

More Anonymous

Im trying to contrast his way of thinking about the self with Descartes

@ACuriousMind Any idea where there would be a lecture series about this kinda stuff?

Vincent Thacker

@ACuriousMind Thanks, I see. What happened was that the initial LQ correctly completed with 6 recommend deletion votes, but when the 6th vote arrived it had gotten an upvote so the deletion didn't occur. I reflagged it as "Not an answer" afterwards but no new review task was created

So I guess it went under the radar

ACuriousMind

@VincentThacker yes, what happens in this case is that a "disputed LQ review" flag is raised automatically - this is the mechanism that's supposed to alert us mods to it, but if you've looked at the answer before, it's still greyed out in the mod UI as "already looked at"

so I've been scrolling past it for the last few days :P

Vincent Thacker

20:49

Oh I see

Sanjana

22:01

Given a Hamiltonian, when does a Lagrangian description exist? The usual answer is given by the Hessian condition, but is there any way to know that in a coordinate invdependent way: I ask because... Can it happen that writing the Hamiltonian in some other coordinates, the Legendre transformation becomes non-singular, so as to yield a Lagrangian description?

ACuriousMind

@Sanjana There is no truly coordinate-independent notion of "Hamiltonian" or "Lagrangian", because the notion of switching between them requires you to split a phase space $P$ as $P = T^\ast Q$ for the configuration manifold $Q$ (the "space of generalized positions") in order to perform the Legendre transform to the Lagrangian on the space $TQ$

that's the deal with the "polarizations" I kept talking about earlier: You need to choose a n-dimensional subspace $Q$ of "positions" in your 2n-dimensional phase space for these notation to make sense

Sanjana

@ACuriousMind Why do you say "split" a phase space? I didn't come across that terminology yet...

ACuriousMind

my second message explains this - I mean polarizations

Sanjana

oh alright. I dunno about polarizations in this context. I will try to learn a bit about it before talking

ACuriousMind

it's not an accident that we call the n-dimensional submanifolds on which the symplectic form vanishes "Lagrangian submanifolds" and that a polarization is a foliation by such manifolds

the reason is precisely that that is what you need to have in order to define the Legendre transform to the Lagrangian formalism

surprisingly it's hard to find sources that state this explicitly, but at least this MO comment agrees with me

Mr. Feynman

22:34

After choosing the conformal gauge ($g=\eta$), we can still perform gauge transformations that scale the metric and compensate them with Weyl transformations. This is more easily understood if we write in lightlike coordinates $$ds^2=-d\tau^2+d\sigma^2=-d\sigma^+d\sigma^-.$$ Then, separate one variable transformations $\tilde{\sigma}^\pm(\sigma^\pm)$ rescale the metric as we wish. I was wondering whether we can be sure that such factor is non negative, as a conformal factor should be

ACuriousMind

infinitesimally (as physicists often implicity argue) yes

since you'd have to continuously pass through factor 0 to get to negative factors, but the transformation with factor 0 won't be proper transformations

Mr. Feynman

@ACuriousMind and is that enough to access the lightcone gauge?

I mean, up to the conformal gauge we could use "finite" transformations. It is not obvious to me why for the residual symmetry only infinitesimal ones would be considered

ACuriousMind

I can't tell you off the top of my head either

Dannyu NDos

22:52

My dad permitted me to re-enter the univ, into the department of applied physics, for next year.

He's going to support 7 thousand USD of the registration fee, but I'm paying the rest.

And he insists that I must get another Master's degree, or even a Doctor's degree.

And heck, I'm gonna take neo-material chemistry as the double major.

*Sigh* Money is everything, it seems.

Dannyu NDos

23:43

Regarding that, my greatest fear is whether the experiments are gonna involve hydrofluoric acid.

After all, the department of applied physics is about semiconductors.

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