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02:15
@User198 thats clearly just commutators
@imbAF why have you not just tried it? There is no difficulty because the artificial particle is massive.
02:28
@Arjun @SignorFeynman I think yall are reading it too uncharitably. In that offending paragraph, it says "In particular," before the parallel situation, which means that L&L intended for it to be a general statement. It is at most failing to state that this result is only true for the one spacetime point it is applied to. Yes, you have 6-2=4 DoF, but with Lorentz transformations, you have 3 DoF too; 4-3 = 1 DoF in the end.
By a Lorentz transformation, you can align the B field with an axis and its magnitude will be dictated by the $\vec E\cdot\vec B$ value. Then the E field will have a parallel component and a perpendicular component, and the original Lorentz transformation still had a degree of freedom (that is not used to fix the magnitude of B) to align this perpendicular component of E field to one other axis. So, out of E and B fields, you have zeroed three terms, leaving only 3 DoF to fit the 2 actual DoF
I think this is already quite a lot of "arbitrary". It is, however, not possible to be arbitrary with complete generality; you might want to check if latest edition of L&L removed this section because it is not a completely true statement.
 
3 hours later…
05:41
@naturallyInconsistent Yes I agree with this, we have one extra component left for E and H and this can't be arbitrary,since three of these will fix the Lorentz transformation, which in turn will fix the fourth component of E and H
@naturallyInconsistent I had troubles with this, he iirc has intended to say we could somehow make all four components arbitrary
 
3 hours later…
08:22
hi
08:52
morning
09:08
@Arjun yeah, it wont be completely arbitrary, but it is arbitrary enough up to magnitude. That's quite a lot!
 
1 hour later…
10:29
"So, you might think, as you possibly did, that QM has lost information vis-a-vis classical mechanics, but it is the other way round. The classical limit has higher, not lower, entropy than the original QM description. (You know that: think of the ℏ-dependent info suppressed in the classical limit. Think of quantum computing.) In a sense, the momentum information of the classical trajectory is embedded in the x-spread of the Wigner function of its quantum progenitor, which the formalism correlates to its p-profile in an eerie way, thus ensuring the uncertainty principle prevails."
"The classical limit has higher, not lower, entropy than the original QM description" aha
where is this quote from?
10:42
i think I've read this on PSE
14
A: Does quantum mechanics halve the dimension of phase space?

Cosmas ZachosThe most meaningful language to compare classical mechanics to quantum mechanics is the phase-space formulation of the latter, based on an invertible map between Hilbert space and phase-space: you want to compare apples with apples. In this formulation, a particle is represented not by a mere poi...

10:55
"This is all words, and won't make much sense before you do some simple illustrative calculations, for which it serves as a trail map. A free particle, a linear oscillator mode, a Morse potential state, etc... might illustrate the point."
thanks RR for the link
welcome
without precise definitions all of this is indeed "just words"
yes. one needs a definition of entropy
if one uses the usual entropy formulas on continuous probability distributions, one gets infinities
10:57
also, the Wigner distribution is not positive semi-definite
I guess one can make sense of this
but you want to make sure not to compare apples with oranges...
yes. Usually, one uses the Von Neumann entropy in QM. And it is just usual stat mech applied to QM states
the probabilities there are positive definite
@TobiasFünke yes
i think Cosmas's idea have some precise definitions behind them
In physics, the von Neumann entropy, named after John von Neumann, is a measure of the statistical uncertainty within a description of a quantum system. It extends the concept of Gibbs entropy from classical statistical mechanics to quantum statistical mechanics, and it is the quantum counterpart of the Shannon entropy from classical information theory. For a quantum-mechanical system described by a density matrix ρ, the von Neumann entropy is S = − tr ⁡ ( ρ ln ⁡ ρ ) ,...
it says that this is a generalisation of Shannon entropy, but it looks like the usual Shannon entropy formula
also, the density matrix is positive definite
11:31
@RyderRude well, it is a bit complicatedd
because convex decompositions of mixed states into pure states is non-unique
the von Neumann entropy in QM gets its operational meaning by the Schumacher compression (theorem)
but that's not my point about Wigner's distribution. I don't mean to say that Cosma is wrong or anything, I just mean that -- as usual -- words are ambiguous, and to make sense of such a statement one must define things in a clear manner.
12:11
@TobiasFünke oh
@TobiasFünke yes. Cosmas also says this
@TobiasFünke oh
12:52
First time I see a withdrawn paper on arxiv
13:08
I didn't even really realise it was possible
kudos Vincent Schlegel.
13:57
i also want to write a paper but im not sure about the topic
The usual order is that you have a topic and perhaps interesting results, and then publish a paper, no?
i dont have any good contribution to physics rn. i think it won't probably won't be a physics paper. and it is really hard to contribute to physcs
i will write some philosophy paper maybe. but first I have to make sure the ideas are even a new contribution. most of the time, my ideas have previously been thought of
like, i conceived of a compatibility condition between Bayesianism and Frequentism as a Bayesian who uses frequencies to form beliefs. but I later encountered this idea on philosophy SE
i will have to do extensive reading to make sure my ideas are even new
To write a research paper requires expertise in a specific subfield (usually). It is always hard, requires a lot of knowledge about the state of the art and what has been done already.
Do you want to write a research paper in a peer-reviewed journal?
what I like about philosophy is that there isnt much objectivity to the results. there are no "tests" of philosophy. anything goes
@TobiasFünke i think I want to write for arxiv
14:05
i mostly just want to do it for myself. like a paper diary that I can book back to
This sounds like you want to start a blog or so
i want it to also be serious contributions instead of something casual (or at least intended to be serious contributions)
but also, in theoretical physics, there are no "tests" either. Sabine says that basically anything goes these days
as I said, this is a hard undertaking. but you do you
like, u just have to put out some mathematics
but I have to read any field extensively before writing papers
i would want to add citations to the paper of previous peoples' works
May I ask, do you study physics?
14:08
not anymore...
i will have to read extensively
there was one post I made on PSE which I thought could be a contribution. but it hasn't gotten replies. it is far from rigorous
@User198 I don't think it's supposed to be taken literally (you can't take a some generic classical limit of a Wigner function and get a classical state). What Cosmas is saying is just that the classical theory - when you phrase it in terms of the limit of the Moyal bracket to the Poisson bracket, for instance - clearly "forgets information"(namely those terms of higher order in $\hbar$).
@User198 The Poisson algebra of functions is a Lie algebra, just like algebras of operators on a Hilbert space. I don't know what you mean by "other axioms remain the same". In particular, there is no Lie algebra isomorphism between the classical Poisson algebra and the quantum algebra of operators (this is Groenewold-van Hove, see chat.stackexchange.com/transcript/71?m=67081052#67081052)
14:40
@MoreAnonymous hi
15:10
do peer review journals take papers from unknown people seriously
@RyderRude consider something like a part time self-funded phd, drag yourself through the coals on how to do this stuff properly, consider the time investment you've already put into all this
@bolbteppa do u mean like going to university to get a phd?
@RyderRude Just save yourself time and commit to vixra if you're not going to do it properly
i somewhat just want to do it for myself. Like just get it out of my system
but maybe I could just write personal notes then. A lot of people do that
Well that's what vixra is for
15:12
lol
arxiv allows anyone to post, right?
maybe it would be better if I am also sharing the ideas instead of just making personal notes
@RyderRude hey
@MoreAnonymous hi
i read some physics papers from the 90s and they were speculations about quantum gravity. so not every paper is supposed to be a breakthrough
and they were from reputable physicists
like, a lot of the ideas are just for sharing instead of a breakthrough
i doubt those physicists themselves believe those speculations anymore. they later wrote revised speculations
Do you have some ideas on QG?
no
not anything precise
i sometimes feel if I just wanted to write a random theory, I could just do it. but even that is really hard
I had an investigation I intended to carry out ...Like when is it sufficient that QM theory satisfies the Equivalence principle
15:19
"i sometimes feel if I just wanted to write a random theory"???
maybe string theorists have a point when they say it is an achievement that they have any consistent theory at all
Is there any specific problem you want to solve or shed light on?
@TobiasFünke like, something that's at least mathematically consistent without worrying about experimenta
Pretty common in the science fan crowd
The desire to do physics without a goal
or to solve a fundamental problem that they know nothing about
@MoreAnonymous some people think QM violates the equivalence principle
i don't see it tbh
i am not deceiving myself
If you're going to do something, doing it properly is the obvious thing to do, unless that's the hard path, in which case one should ask what the whole point really is and what the most efficient way to do that is
@RyderRude This is not what scientists do. They don't write down "random theories". They don't go "I want to write a paper" and then do research on random stuff, they do research on what interests them and then at some point perhaps go "Hm, I could make a paper out of this". You don't even understand how the field you're allegedly trying to get into works.
A blog might achieve what you're hoping for
@bolbteppa for me, it is more about the joy of thinking rather than actually achieving the goal
15:24
(sure, some might do research less on what interests them and more on what they can get funded for, but that's still very different from the bizarre process you are describing)
this is why i at least want to be able to write a random theory
The reason people spend years doing a phd is because they want to do this properly and not go off the rails, you are basically guaranteed to fail without guidance from people who've done this before, if anybody could do it without guidance universities probably wouldn't exist
@ACuriousMind i think this is not how it works in practice. yes, people generally work in fields they are interested in. but many specific papers can be "throwaways", as in, attempts to survive the "publish or perish" attitude of academia
also, not every paper is a breakthrough. Plenty of papers are wild speculations. like throwing darts in the dark
Just use this handy list of motivations to write a paper
I mean as far as I've seen "gravity is locally indistinguishable from acceleration" the answer is no ... Like you can see the position commutes with acceleration (the double derivate of position in the Heisenberg picture) ..So you can talk about locality (/ position) and acceleration in the same sentence ... In fact you can do a QM version of the calculation here:

https://physics.stackexchange.com/a/121890/150174
15:27
Your goal makes no sense. Again, physicists don't "write down random theories". Are some papers speculative? Yes. Does that mean all speculation is publishable? No.
If you want to write a random theory you can just use snarxiv.org
much faster
i don't know about peer review journals. but there certainly are speculations papers on arxiv from reputable physicists
those papers intend to throw darts in the dark, like hoping to steer the field towards the right direction
@RyderRude I know a set of books that will challenge you to think for years about physics as it actually is, helping setting you up to be able to build on what we know (though even then you'd still need massive guidance...)
but most of those papers are missed darts in hindsight
@MoreAnonymous i will check it out
@MoreAnonymous i think that mass dependent effects are perfectly normal. I have a post about this
the post shows that thisnis nothing specific about qm
in fact, I could probably write a paper about this post of mine. i see this idea get confused a lot by professional physicists
See this post physics.stackexchange.com/a/771377/156987 @MoreAnonymous
i have read articles from professional physicists who don't make the argument I have made in this post
Well, then get to writing! You've talked orders of magnitudes more about wanting to write a paper than actually doing it. What's the point?
put your effort where your mouth is and actually do something
i will start with a paper about that post maybe. but the idea is really basic. it's just something basic that gets missed a lot
physicists make a fuss about mass dependent effects violating the equivalence principle
@bolbteppa did you just watch The Departed or something :P
Meow
@bolbteppa i am interested in these books.
15:37
Hello
Do you know the original movie, Infernal affairs?
The Course of Theoretical Physics is a ten-volume series of books covering theoretical physics that was initiated by Lev Landau and written in collaboration with his student Evgeny Lifshitz starting in the late 1930s. It is said that Landau composed much of the series in his head while in an NKVD prison in 1938–1939. However, almost all of the actual writing of the early volumes was done by Lifshitz, giving rise to the witticism, "not a word of Landau and not a thought of Lifshitz". The first eight volumes were finished in the 1950s, written in Russian and translated into English in the late 1950s...
It's obviously L&L. How can you have been around here for so long and not understand bolbteppa is going to say L&L :P
@TobiasFünke I haven't even seen The Departed!
@ACuriousMind then you should watch the original first anyway ;)
@ACuriousMind That is coincidental, that scene basically nails this discussion as you will see when you watch it! :p
@bolbteppa i know these books ofc. i do read physics books in general
or i used to read
i am not a pop sci person. i have read actual books
15:39
Hiya tobias!!!!!
I have a question
RR, with all due respect: what I tried to make clear before and what others have pointed out already: There is not much sense in "just writing a paper" about some "random theory". This is not how science is done. A blog seems what you want to do, i.e. sharing your thoughts on some topics
@TobiasFünke any other movie recommendations? :P (I'm sick and not going to do much else)
As I said: In the vast majority of cases you need a lot of expertise in a specific field to make actual and meaningful contributions
@ACuriousMind Oh. Get well soon. Yes, I like the director Martin McDonagh a lot!
Yeah that basically describes what you're trying to do, but you know few people are going to care about some idle speculations in a blog hence the goal of the legitimacy of a paper without any of the earned legitimacy
and then some south korean movies: memories of murder, oldboy... etc.
15:42
@TobiasFünke Ah, I've seen In Bruges
how did you like it? :)
@TobiasFünke Typical experience of trying to find a paper idea is discovering that someone already solved the problem 60 years ago
It was good (but I don't remember much of it, it's been a few years)
I see. It is actually one of my favorite movies. ^^
Movies are for some reason a kind of data I'm not very good at retaining :P
15:43
@TobiasFünke Maybe I should give it a chance given how many times I've seen the Scorcese one...
The answer is one the same page as you: "The answer, as it seems to me, is that they confuse weak equivalence principle with mass-dependent effects. See, most of discussion is about dependence of some wave packet properties on particle mass. But this doesn't have anything to do with weak equivalence principle: we have mass-dependent wave packet broadening even without any gravitation — even in free space!

Maybe there's some inequivalent formulation of weak equivalence principle, which speaks about mass-dependent effects in cases where classical mechanics doesn't have them, but then it shou
@bolbteppa I like the original more. But the departed is not bad either
@ACuriousMind why see a movie about Belgium and not an actual Belgian movie
Man Bites Dog (French: C'est arrivé près de chez vous, literally "It Happened Near Your Home") is a 1992 French-language Belgian black comedy crime mockumentary film written, produced and directed by Rémy Belvaux, André Bonzel and Benoît Poelvoorde, who are also the film's co-editor, cinematographer and lead actor respectively. The film follows a crew of filmmakers following a serial killer, recording his horrific crimes for a documentary they are producing. At first dispassionate observers, they find themselves increasingly caught up in the chaotic and nihilistic violence, eventually becoming...
Greatest Belgian movie
@MoreAnonymous yes. but even this answer doesn't bother to point that this is not a quantum specific phenomenon
i have read articles from professional physicists and all of them miss the point I have made in my post
I cannot log in into my account on a movie database right now, I could suggest another thousand movies, which do not come into my mind rn... hehe
15:45
@Slereah My favorite French movie
The Dinner Game (French: Le Dîner de Cons, pronounced [lə dine d(ə) kɔ̃]; literally Dinner of Fools) is a 1998 French comedy film written and directed by Francis Veber, adapted from his play Le Dîner de Cons. It became that year's top-grossing French film at the French box office (second overall behind Titanic). == Plot == Pierre Brochant, a Parisian publisher, attends a weekly "idiots' dinner", where guests, who are modish, prominent Parisian businessmen, must bring along an oblivious "idiot." The ideal "idiot" is usually one who is obsessed by a ridiculous hobby or theme, whom the other guests...
@bolbteppa Terrible taste
So my textbook says
but the two Korean movies I mentioned are classics :)
these mass dependent effects also show in classical mech. the quantum effect is an exact analogue of that @MoreAnonymous
@RyderRude I don't think Im disagreeing
15:47
it's because the wavefunction encodes momentum. "Same wavefunction" means "same momentum distribution"
@MoreAnonymous great :)
you can see who downvoted it?
@Allie ? :d I am waiting hehe but idk if I can help you. depends on the topic
@MoreAnonymous No, it's pure speculation. @RyderRude you've been told before not to do this, stop it.
Reading a bit more into this "we have chosen to have the same initial momentum for both the particles, instead of the same initial velocity"
15:50
...
Why wont the same inital velocity be the same?
@Slereah Either that or
The Fifth Element is a 1997 English-language French science-fiction action film conceived and directed by Luc Besson, and co-written by Besson and Robert Mark Kamen. It stars Bruce Willis, Milla Jovovich, Gary Oldman, Ian Holm, and Chris Tucker. Primarily set in the 23rd century, the film's central plot involves the survival of planet Earth, which becomes the responsibility of Korben Dallas (Willis), a taxicab driver and former special forces major, after a young woman (Jovovich) falls into his cab. To accomplish this, Dallas joins forces with her to recover four mystical stones essential for the...
I can define a velocity operator and apply it to the wavefunction no? @RyderRude
@MoreAnonymous indeed. but if u initialise two different wavefunctions with the same velocity instead of the same momentum, the mass dependent effect should disappear
Can one actually ping two users in chat? (In comments this does not work)
15:52
@TobiasFünke yes
ok thanks
@MoreAnonymous these calculations show that if two initial wavefunctions are the same, then the evolution is mass dependent. But if the wavefunctions are chosen to be identical, we have chosen the momenta to be identical
and ofc particles of different masses with same momenta behave differently
this is exactly what happens in classical mech
i think I tried to vaguely work it out in the Heisenberg picture. if we choose the initial velocity to be the same (which means the wavefunctions are different), the mass dependent effects disappear
but i haven't worked it out fully
i will maybe write this in my paper
@TobiasFünke you can only reply to one message, though - the rest of the pings have to be generic, not message specific
is the 2D ising model's free energy singular at the critical temperature due to the partition function being zero at the critical temperature?
I compute rather that the partition function blows up to infinity at the critical temperature
$Z = \text{tr}(V^N)$ where $V$ is the usual transfer matrix for 2D ising. then at the critical temperature there is an infinity-fold degeneracy of the (nonzero) "maximum eigenvalue" so, I get that $Z \to \infty$
16:00
A photon collides with a crystal elastically
The photon changes wave vector by deltaK
My book says that the whole crystal recoils with momentum -deltaK
@ACuriousMind Thank you.
That makes sense from a conservation of momentum
But im a little confused about where the energy for the crystal’s momentum comes from
Since the photons energy will be the same?
Is there something sillly im missing
doesn't elastic collision mean that the overall energy is conserved?
kinetic*
Exactly
But doesnt the crystal now have extra energy?
Because it has a momentum
where is the problem then?
16:09
The photons energy doesnt change, the crystals energy does
Isnt that violation of conservation of energy?
Why do you say the photon energy does not change?
Because its an elastic scattering
but this means that the total kinetic energy is conserved
There has to be something stupid im missing
The photon doesnt change energy, right? For the elastic scattering
Sorry, I cannot follow anymore :s
maybe I misunderstand what you mean
@Allie as I said, idk if I understood your question... but of course the energy of the photon changes
16:17
Are you in position or momentum space? @RyderRude
maybe someone else can help
@bolbteppa And the only book I can find that claims that the cooper pair density is not the same as the condensate density. L&L and those cursed footnotes of theirs
@MoreAnonymous i want to imagine two wavefunctions corresponding to particles of masses $m_1$ and $m_2$ such that $|\psi _1 (x)|^2= |\psi _2 (x)|^2$ and $=|\psi _1 (\frac {p}{m_1})|^2 =|\psi _2 (\frac {p}{m_2})|^2$
and then I want to show that they produce the same interference pattern, i.e. $|\psi _1 (x,t)|^2=|\psi _2(x,t)|^2$
@MoreAnonymous both spaces are relevant. we have to impose that the initial wavefunctions have the same position and velocity distribution
Okay ...
so if we start with the same position and velocity, the mass dependent effects should disappear. i will try to prove this
16:22
Also I must say I don't have access to the paper ... Im downloading Sakurai
imagine a free particle for simplicity. now, the time evolution in the Heisenberg picture is $X(t)= X- Vt$ where $V$ is the velocity operator. Now it is clear that $\langle \psi _1|X(t)|\psi_1\rangle = \langle \psi _2|X(t)|\psi _2\rangle$
because we have chosen psi_1 and psi_2 such that they have identical positive and velocity distributions
this means that the expected value of position in the time evolution is the same for both particles
i think this means mass dependent effects disappear
in the hesienberg picture: you have
$\hat v =\frac{d \hat x}{dt} = \frac{[\hat x, \hat H]}{i\hbar} = \hat p /m$
@Allie Overall kinetic energy is conserved. So to do this properly, the photon energy can change if the crystal picks up the corresponding kinetic energy (by changing its momentum by $\hbar G$). But often this $\hbar G$ would be so miniscule (because the mass of the crystal is large) that you can't really measure it, so it's neglected in an approximation. In that approximation, of course, energy is no longer conserved - but only by a small amount you've explicitly chosen to neglect.
16:27
I think?
I dont think you get $\int \hat V dt = V t $
Ahhhh!
because there are unitary operators there @RyderRude
@MoreAnonymous the position operator at t=0 is the X (like, say the multiplication by x operator)
and dX/dt=V
so when you do conservation of momentum, the kinetic energy imparted on the crystal is so miniscule that the change in photon energy can be neglected in an approximation?
@RyderRude yes
16:30
@ACuriousMind
where V is the operator ihd/dx /m
so u get X(t)= X+Vt
@Allie yes (at least that's how in interpret the excerpt you posted)
No ... you are in the Heisenberg picture $V = U^dag \frac{p}{m} U$
but the excerpt does not talk about the energy of the photon at all, no? am I missing something?
now, choose psi 1 and psi 2 such that they have identical initial position distribution and velocity distribution.
@MoreAnonymous i don't get it
16:31
where U is the Unitary operator
@ACuriousMind thanks! thats very clear! i didnt realize that that momentum on the scale of a crystal would be miniscule
@TobiasFünke No, but the energy conservation is what Allie is asking about - the apparent contradiction between saying the collision is "elastic" and then neglecting the crystal recoil momentum
@MoreAnonymous are u computing V(t)
i knew it was something silly lol
I missed that last part, it seems
16:32
but i think this stuff is really cool
basically what we are doing is (in explict notation):
$ d ( U^\dag x U )/ dt = U^\dag V(t) U = U^\dag \frac{p}{m} U $
learning about phonons. guys i think im turning towards the dark side
i had inorganic chemistry this morning and it HURT ME
i was like THERES PHYSICS BEHIND THIS
im really really loving physics even tho i ultimately still want to apply it to chemistry
Sorry it should be:


$ d ( U^\dagger x U )/ dt = U^\dagger V(t) U = U^\dagger \frac{p}{m} U $
16:34
i am trying to approach it differently
That's Heisenbergs equation of motion
at t=0, choose some representation of the CCR. Call it X and P
I need to find a way to put "unitary" in my next username
so X(0)=X, P(0)=P
Or "coherent"
16:34
CCR = ??
now dX/dt=P/m
Canonical commutation relations
so X(t)= X+ P/m t
@MoreAnonymous i mean a representation of [X,P]=i
@MoreAnonymous creedence clearwater revival
@RyderRude I dont think you can write this.
16:37
we have an ODE. X(0)=X , P(0)=P, dX/dt=P/m , dP/dt =0
first we get P(t)=P
and next we get X(t)=X+P/m t
Whit when you write X you mean X the operator in the Heisenberg picture right?
it is just the explicit solution of Hamilton's equations @MoreAnonymous
Or are you doing classical mechanics?
@MoreAnonymous no. That is the X operator in the Schrodinger oicture
X(t) is the X operator in the Heisenberg picture
16:39
at t=0, X(t)=X
I implicitly mean dX(t)/dt=P/m
so we have this X(t)= X+P/m t
now imagine two wavefunctions which are identical wrt their expectation value of X and their expectation value of P/m
$X(t)= U^\dagger X U \approx X +P/m dt $
You can write upto first order dt
@RyderRude @MoreAnonymous if either of you understood the mathematics of QM to the point you could write a paper about it, you would immediately see that both your expressions yield the same for the free particle by the BCH formula (factors of $\mathrm{i}$ supressed because I can't be bothered): $U^\dagger x U = x + \frac{t}{m}[p^2,x] + \frac{t^2}{2}[p^2,[p^2,x]] + \dots$. All terms except the first commutator vanish, giving $x(t) = X + \frac{p}{m}t$, similarly $p(t) = p$.
@MoreAnonymous yes. but I am writing it for finite times
@ACuriousMind we understand this???
i am not saying that I contradict @MoreAnonymous 's approach
i am just taking a simpler approach
@RyderRude Then why this back-and-forth instead of you simply telling MoreAnonymous that this is how to see that the two ways to compute the evolution yield the same result?
Ah I see ... my bad
Thanks @ACuriousMind ... was a bit confused :P
16:43
@ACuriousMind i wrote above that "I am taking a simpler approach"
i understand that Hamilton's equations follow directly from the U^{\dagger} O U formula
but I just think that directly solving the Hamilton's equations is easier than bothering with U dagger A U
@ACuriousMind modulo $\mathrm{i}$ ACM is back :P
@RyderRude I gtg. But do let me know if you complete the calculation. I'm curious about ti
yes. i will write it
ok so we have X_1(t)= X + P/m1 t and X_2 (t)= X+ P/m2 t
This reminds of when my prof's approach of deriving the GL equations of motion from the GL action by explicit variation with the subsequent claim "you can see that also the EL equations will work"...
@SignorFeynman I've always been puzzled by the people who think that somehow doing the variation explicitly is easier or preferable to just plugging stuff into the EL equations
16:49
now imagine $\psi _1$ and $\psi _2$ are chosen such that $\langle \psi _1 |X|\psi _1\rangle = \langle \psi _2|X|\psi _2\rangle$. And $\langle \psi _1|\frac{P}{m_1}|\psi _1\rangle = \langle \psi _2|\frac{P}{m_2}|\psi _2\rangle$
Like, they're not wrong that that works, too, but why did we derive the EL equations in the first place if you're going to repeat the derivation every time :P
then it follows that $\langle \psi _1|X(t)|\psi_1\rangle =\langle \psi _2|X(t)|\psi _2\rangle$
this means that, if initial position and velocity are same, then time evolution is same. Mass dependent effects disappear
but we need to improve on this
Oh, me too. I've always found it so cumbersome. The EL are soooo much easier. :P

To make things worse, in that case it was said in such a way that I suspect that the Prof legitimately thought it was a different thing :P
first we need to show how we can choose two wavefunctions such that their initial position and velocity distributions are the same
and second, we need to generalise the argument beyond just expectation values. i have only shown that the expectation values of position are identical
i haven't shown that the interference patterns are identical
in classical mech, we can always choose the same initial position and velocity for two particles cuz they r independent variables. but in QM, they r related by a Fourier transform, so we need to prove it
i have written the formulation of the problem here chat.stackexchange.com/transcript/message/67117985#67117985
 
1 hour later…
18:25
The
18:46
The indeed
 
1 hour later…
19:58
the???
Isn't the vacuum after "the" rendering for you people?
20:54
@RyderRude no, arXiv does not quite let anyone post. you have to have to be approved by someone already on arXiv personally
21:30
For a particle that moves with relativistic speed, the distance that it covers during its lifetime (assuming it decays) is larger than what we would expect, because of the time dilation from particles POV?
21:49
@imbAF It travels farther, because it survives for a longer time because of the time dilatation.
From its own POV right?
The clock that is in the particles frame would tick slower than a clock that is stationary in a lab.
22:09
Can someone give me some cool examples of canonical transformations, but that are not too advanced; only in CM still.
I found out about: $Q=\sqrt{m\omega}q$ , $P= \frac{1}{\sqrt{m\omega}}p$
It brings the Hamiltonian into a more geometric preetty form ok.
That example is for a harmonic oscilator.
What are some else? And why is the usage? Just to make the Hamiltonian more preety? I heard you can solve diff. equations just via a canonical transformation, no integration needed.
Is that true?
 
1 hour later…
23:15
@imbAF : The more careful way to say this is that in our frame, the distance the particle travels is greater than we would expect if we didn't know about relativity. If we do know about relativity, the distance is exactly what we'd expect. And in the particle's frame, the distance is the same as we would expect whether or not we knew about relativity.

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