3:02 AM
@ACuriousMind In flat SR you can; in curved GR I really don't think this is possible.
@imbAF It is not at all weird because while each fundamental particle has a unique and specific rest mass, a system of particles will have its "rest mass" defined by this procedure, and thereby we see that all 4 numbers in the energy-momentum 4-vector are somewhat free to be any value it wants, and together they define the invariant rest mass/energy that is relevant to any situation of a system.
4:00 AM
ł

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6:22 AM
H O N K

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8:45 AM
@naturallyInconsistent ...which is why I said I don't want to launch into a diff. geo. lesson unless that's the point :P
9:00 AM
@ACuriousMind I have a hunch you're always lying when you say you don't want :P
9:22 AM
@Mr.Feynman I mean I enjoy talking about that but it would have completely derailed from the actual problem
@Sanjana hi. you can also use the more general eigenvectors $e^{ipx + i f(p)}$ where $f(p)$ is a real function of $p$, as the eigenbasis can have an arbitrary phase.
all these transform send derivative to multiplication, but not necessarily vice versa. The Fourier transform is a special case of this
but all these transforms qualify as a "transform to momentum basis", as -id/dx gets sent to multiplication by p

2 hours later…
11:32 AM
whats interesting is that, while the inverse transform obviously does map p back to -id/dx, the forward transform does not map x to id/dp. this is because the transform treats $x$ and $p$ asymmetrically because of the presence of $f(p)$
but the map drom -id/dx to p and the inverse work out. x gets mapped to $id/dp - f'(p)$ in the forward transform
at first glance, it may seem that this transform destroys all the phase information in the wavefunction because it is related to the Fourier transform by an arbitrary local phase $f(p)$
but that information is now stored in the $x$ operator which becomes id/dp - if'(p). This f'(p) has all the phase difference information, which can be recovered by a change of basis
12:05 PM
there is also some deep connection here with the SvN theorem
if $[X,P]=i$, then $[X,P+ig(X)]=i$ is satisfied even when $ig(X)$ is anti-hermitian. but because of this anti-hermiticity, P+ig(X) cannot be unitarily equivalent to $P$
this seems like it would violate the SvN theorem. i think SvN theorem needs to assume Hermitian representations of the CCR for the unitary equivalence to hold
in the above transformation, $id/dp - f'(p)$ turns out to be Hermitian. so that $[id/dp - f'(p), p]=i$ is a hermitian representation of the CCR, and it is unitarily equivalent to $[x, -id/dx]$
on wikipedia, it is mentioned that $Q$ and $P$ are required to be self adjoint in the SvN theorem

4 hours later…
4:13 PM
would it be a reasonable, beginner analogy to think about the sch eqn like newton's equation for the force exerted on a particle ?
and the wave function would be, i know it's very wrong or inexact, but it'd be like the standard solution, so naively the wavefunction is the equation of motion of an electron (for a single particle.)
The Schrodinger equation doesn't reference "force", but it is the equation of motion in quantum mechanics which gives you the time derivative of the wavefunction in the same way that Newton's equation gives you the time derivative of the momentum.
I'm not sure what you mean by "the standard solution". What determines the solution(s) of the Schrodinger equation is the Hamiltonian, $\hat H$ that you've chosen.
but isnt newton's law a differential equation of second order in terms of acceleration ?
Yes
so they are almost the same
Well they are similar in the sense that they both contain time derivatives, which is essentially what makes them an "equation of motion"
4:19 PM
they are different in the sense that SE is an eigenvalue equation ?
I should clarify, a time derivative of something describing the state of the system, not any old equation with a time derivative is called an equation of motion in general
See here for some clarification on the two usages of "the Schrodinger equation". Only one of which is actually an eigenvalue equation.
Also what is normally a potential in, say, the hamiltonian for a molecule
How would it appear in newton equations, as a summation of forces that give a total acceleration?
(i'm not a physicist)
Quantum mechanics is tough to use in the context of entire molecules, but for a single charged particle the potential is the Coulomb potential, have you seen that before?
What you're getting at is right in spirit - to answer your original question - the Schrodinger equation plays a similar role in QM as Newton's second law, $F=ma$, plays in classical mechanics
The reason I got here is, that I was wondering how to interpret the terms <ij || jk> for electrons in molecules
I'm not sure exactly what you're written there
4:25 PM
That is the physicist notation for two electron operators
It stands for something like this, I will get some wrong:
"Electron operators" isn't (at least a standard) term in physics
(in QM)
Sorry, yes, it's Coulomb operator the right term. Thanks for the correction.
@Minsky It sounds to me you've jumped in at the deep end - if you don't know how the Schrödinger equation works then you shouldn't be looking at stuff that involves QM operators
Well, i've read 70 pages and i understand how to solve the equation, but i haven't been to courses, so I don't know some of the interpretation
If you just want answers to questions you can get them here but as ACM says it sounds like you're getting ahead of yourself and any conversation is likely to only confuse you further
4:28 PM
For example the kinetic energy terms, are somehow easier to imagine
Also, you've read 70 pages of what exactly?
quantum chemistry by szabo and ostlund
Are you looking at Hartree-Fock theory?
solved all the exercises so far
yes
and roothaan equations
The problem you're going to run into is that you're trying to reconstruct the physics (which is more fundamental) from what you're read about in Chemistry. Which, as someone who has studied chemistry before is a difficult direction to go in.
If you want to understand the things you're referencing I'd recommend you start quantum mechanics from the physics side, from the start. Though I appreciate that is easier said than done.
4:31 PM
no, i'm trying to find analogies for fun
i think it's quite interesting, even if untrue
Well then you'll have to be happy accepting half-answers that are hand-wavey and don't really give you a satisfying answer :P
If that's all you're after then that's fine
i'm ok with that, this is my conceptual question, to go to the point, if I can
when you solve HF equations, you get or can construct many slater determinants
Mmm, ok, some of these words are familiar
and when combine them linearly, you get to the eigenvalue matrix equation with has combinations of those determinants with H
like <psi_0|H|psi_1>
I don't remember enough about Hartree-Fock theory to really know what you're referencing sorry. Maybe someone else knows more about it than me, but we're going to hit a language barrier of sorts quite quickly here I think.
4:35 PM
no problem, thanks
If you see a physics term that you're not sure about feel free to ask here, I'm sure you'll get lots of good answers. But I'm not aware of anyone who is regularly in this chat who is a quantum chemist
my question is not really about HF though, I was trying to show that I at least read the book
my question is about the interpretation of the integrals
but indeed, it uses H as expected for a molecule I guess
@ACuriousMind could be good to know what do you mean here, just to learn
my view of the equation is that for a molecular system I write the TISE which includes kinetic and coulomb operators and try to solve it
@Minsky why?
what do the solutions to the TISE tell you?
why what ? If you write a molecule, those operators make sense in terms of energy contributions
I mean why are you writing down and solving this equation?
what problem are you trying to solve here?
4:49 PM
the solutions are wavefunctions that satisfy how the electrons move in a field of fixed nuclei
no, not at all
well, you will have to explain that to the authors, it is almost a citation
the TISE is time independent, so it or its solutions do not describe any kind of "motion" at all
yes, I am aware of that
when I say motion, I mean that the operators correspond to that, even if we do not look at the time dependent
@Minsky but you just claimed the solutions describe "how the electrons move in a field of fixed nuclei"!
4:53 PM
yes, the solutions are just dependent on space coordinates, no time involved, and squaring them u get probabilities
if there is no time involved, how can you say "how the electrons move"
@Minsky so why are you looking at these solutions? why are they of interest to you?
To understand how each wave function contributes to the final energy, for example
Since you can get quite definite expressions for the fnial energy, without explitly defining the one-electron wavefunctions
0

What is wrong with this question. i have worked upon the suggestions provided by the comments.

I want to compare, or interpret, even vaguely, terms like <j | h | j> this being contributions to the slater determinants which in general look like < SD | H | SD>, so the first case I wrote is derived from this latter formula
4:57 PM
@Minsky what "final energy"? how do they "contribute"? There's typically infinitely many solutions to the TISE
what I meant by my earlier comment is that you seem uncertain about fundamental aspects such as what solutions to the TISE even mean or why they are relevant, so I don't think it is useful to try to interpret any specific results of HF - you're lacking the ground floor of the building but you're trying to build the third floor
what the solutions mean is rather simple though, I explain it:
from the linear combination of slater determinants which can be seen as N-electron wavefunctions, when you solve this system which ends up in the eigenvalue problem due to how you solve the linear variations approximation (in the book using lagrange multipliers), each of the energies can be interpreted as an excited state of the system
@Minsky if the $H$ in the terms there is the Hamiltonian, both of these expressions are just expectation values of energy for whatever states you're denoting by "j" or "SD", what is there to "interpret"?
yes, that looks to me quite similar to a newton equation for a single particle
but it cannot be
in what way???
5:04 PM
but in a conceptual sense there is now a probability involved
(and no time)
it is a second order derivative, just that it uses spatial coordinates
where is there a second order derivative here?
in h
that is a kinetic energy operator
oh you mean from the $p^2$ in the position basis
no I mean that the operator i wrote as h is the nabla squared operator
a 2nd order derivative in time is extremely different from a 2nd order derivative in space
5:06 PM
@Minsky yes, but it is a "nabla squared" because it's the momentum-squared operator acting on wavefunctions of position
yes
so what does this have to do with Newton's equation of motion except that there's a second order derivative here for a completely different variable :P
can't you get also some relatively similar newton equations that use space coordinates as well ?
i don't remember this exactly, but maybe a spring ?
physics rarely involves differentials of more than second order, so if you're going to try to relate every second-order equation in physics to Newton's laws you have a lot of disappointment ahead of you
@Minsky I have no idea what you mean. Newton's equation of motion is $F(x(t),\dot{x}(t),t) = m\ddot{x}(t)$, right?
there's no place for spatial derivatives in there at all
yes, i didn't worry much about that, because i thought there would be simple classic analogies of those terms
i.e second order derivatives, involving the space coordinates, not time derivatives.
5:12 PM
I don't understand what you mean by "analogies" or "those terms"
second order spatial derivatives of a function that describes a classic system
of course there are classical systems where you have spatial derivatives, e.g. the classical wave equation. That the free Schrödinger equation looks a lot like a classical wave equation is the whole reason we call the state functions of QM wave functions...
yes, but i don't want to make analogies with that equation only, since i'd like to force some particle analogies
I don't know what that means
ACM regarding our discussion yesterday. I have an additional question. We discussed my confusion regarding center of mass /frame/ momentum frame etc. I have a question regarding the statement that says that the s (in s channel) is the four momentum of the intermediate particle. Is it saying that the total four momentum of the center of mass is attributed to the intermediate particle?
Or that , at the same time, it can be considered the four momentum of the intermediate particle?
5:15 PM
the whole point of QM is that quantum objects display characteristics of both classical particles and classical waves, but they are neither and there are plenty of situations where their behaviour cannot be explained in either of these classical pictures
@imbAF You just stated that people say that the s is the 4-momentum of the intermediate particle and then asked me whether they mean that s is the 4-momentum of the intermediate particle. I'm confused what kind of answer you're looking for here :P
@imbAF that is just saying that since there is a short period of time in which the initial particles disappeared and before the formation of the final particles, the intermediate particle must thus carry the entire energy and momentum of the entire system, and that shows up as the s value
@ACuriousMind I didn't mean that
@naturallyInconsistent I thought of this, but the intermediate particle is observed in which frame?
the center of momentum, which was the frame we considered the center of mass of the system
What you wrote looks to me like this: "I have a question regarding a statement that says X. Is it saying X?"
@imbAF it does not matter, but yes, in the centre of momentum frame it would exist
then it shouldn't have momentum right?
5:19 PM
you apparently see important distinctions here between things I would consider trivial differences in phrasing
@ACuriousMind I am trying to understand in what frame of reference that statement was made
in no frame :P
you cannot observe a virtual particle
It is obvious that the energy and momenta of the initial particles need to be attributed to the intermediate particle.
and the whole interpretation of the diagram in terms of an actual process and actual particles is questionable to begin with
Ah ok
Out of curiousity, if the 2 particles are counter propagating in the z direction, the intermediate particle, I believe must have no momentum at all?
Or it's not given to be the case
5:21 PM
see physics.stackexchange.com/q/230113/50583 and its linked question for more discussion about the ontology of virtual particles than you probably want; but the upshot is that you 'really shouldn't think about these internal lines as concrete particles at all
internal lines?
if you try to think about them as actual particles you will derive all sorts of contradictions
you mean the lines in the feynmann diagrams?
@imbAF when you say "s-channel", you're looking at a Feynman diagram, no?
the phrase makes no sense without that
5:23 PM
yes
off-shell Fourier components of a field? xD what on earth is this
and when you say "intermediate particle", you mean some line in that diagram that goes from one vertex to another vertex
that's an "internal line"
Ok I will take it as it is that the statement regarding the four momentum of the intermediate particle isn't made in a particular frame
I mean the $s$ is the norm of a 4-momentum, it's a scalar
it's not frame-dependent to begin with
yeah
Just to clarify
What I thought, when I read the statement was the following: Since in the center of momentum frame the center of mass has no momentum and has the energy $E=\sum_i E_i$, physically what happens, in s channel, is that the intermediate particle is created, and I thought, one can say, that the four momentum of the center of mass (which is the total four momentum of the system) can be thought of as the four momentum of the intermediate particle.
And since the entire discussion up to the point of the intermediate particle, was considered in the center of momentum, than I can think that the virtual particle was created in the center of momentum frame, so it has no momentum. But I guess, it's simply to consider what you said
graphically that would be, two particles colliding, having an intermediate snapshot, of a virtual particle which is at rest, and than you have the products
I mean all of that works if we really insist on thinking about an actual particle here, but that's not what you initially asked - you asked something about the 's' variable I still haven't understood and then you started asking in what frame we're "observing" the intermediate particle. I don't see how you expected me or anyone else to get what you just wrote from what you initially asked.
5:31 PM
That's why I said "Just to clarify"
this is all fine, just don't take the particle picture too literally
Yeah, I think I got that from the discussion until now
Another thing is that I can't quite "read" the graphs, which show the dependency of the differential cross section from the energy,where we have peaks, which are, what we call in particle physics as resonances
My question would be:
Well honestly, first thing I want to know. If in a graph, such as the one i explained above there are resonances, which represent unstable particles, the same can be said when we consider the Breit wigner distribution, ofc there we don't have differential cross section but rather the value of the probability density function ranging from 0 to 1
What I can't understand is the importance of knowing the differential cross section, when the unstable particle is the result of the energy increase
dammit you are still on the cross sections business
I understood that
total and differential
My question is different than that
@imbAF How else are you going to predict how many clicks there are going to be in a detector placed at a certain position if not via the differential cross section?
5:41 PM
how does that help us with knowing / detencting the intermediate particle?
is it detected at that angle?
again, the intermediate particle is virtual
you can't detect it
we never detect the intermediate particle. it is not an observable
so what is the meaning of the peak than?
it's a peak in your cross section!
Ok and?
5:43 PM
the cross section for the process for which the diagram you're looking at is one contribution
@imbAF what do you mean "and?"
your detectors are going to be clicking a lot more at that peak than they would if the resonance didn't exist
Yes
how is that tied to the interemdiate particle?
I don't understand the question
you couldn't even draw the diagram if that intermediate particle didn't exist
My question has to do with the interpretation of the peak as the virtual particle. The peak, is the angle at which most of the scattered particles are scattered to. I understand that. Where does the interpretation of this peak as a virtual particle come from.
again, you're using the properties of the particle associated with the internal line to compute the diagram that yields the peak; different properties of that particle would yield different peaks, and non-existence of the particle would yield no peak at all (because the diagram wouldn't exist)
if your point is that this seems like a very "roundabout" way to detect the existence of a particle, then yes, it is
this is how particle physics has worked for over half a century now
But isn't the diagram a result of what you observe and not the reverse?
5:51 PM
no Higgs boson has ever hit a detector directly
@imbAF the diagram is the way in which the theory computes a prediction
we can postulate a particle with some variable properties exists and derive that depending on its mass and lifetime we should see certain peaks in certain cross sections
then you do the experiment and you either see the peak, which you then can fit to your theory to determine the actual mass and lifetime, or you don't, in which case this is evidence the particle doesn't exist
this is just the normal relation between experiment and theory, just that it's a lot more complicated than what we do in elementary lab courses :P
Ok, you postulate the existence of this particle with some properties (such as?), I fail to see how it's existence (again sorry if I am phrasing this incorrectly) is correlated to a differential cross section. But I will give it a thought
I really don't know how else to say it: The relation between the particle and the differential cross section is the Feynman diagram. When you postulate the existence then you can draw the diagram and it contributes to the cross section, when it doesn't exist you can't draw the diagram and don't get a peak in the cross section.
I would ask how one can show this " The relation between the particle and the differential cross section is the Feynman diagram." But i believe that represents a deep dive in QFT
why are you looking at these diagrams if you don't know how they relate to the cross section???
what do you think the diagrams do if not compute cross sections?
6:02 PM
oh, c'mon, you talked about the "s-channel" before and told me you know what Mandelstam variables are
so you know what a Feynman diagram is
I know what they are in the most simplistic sense
I know nothing about the rules of drawing them
and what it's deeper meaning is
then I'm even more confused what exactly you're trying to learn here
all this stuff about resonances and Feynman diagrams really only makes sense if you at least understand the most pragmatic rules of how to get them from a QFT
For example in the breit Wigner distribution, things are pretty clear, the intepretation of peak with the production of the intermediate particle
@ACuriousMind Ok than I believe I give up
6:06 PM
Hey everyone! Ive decided to do a 3rd masters from UK or canandmm
I need to learn the jargon used I can communicate my ideas better
Any suggestions?
@ACuriousMind So my questions are answered in QFT? What I am really asking is, whether they make sense as questions
@imbAF I mean, I haven't met a lot of people who are fully satisfied with the answers QFT provides, but yes, to the extent that your questions make sense QFT answers them :P
But I believe, the explanation as to how the peak is interpreted as the virtual particle, that must be answered
6:34 PM
The chat is basically a hive of people not fully satisfied with QFT :P
@ACuriousMind this message feel out of character... that "c'mon" rings american ACM vibes
7:13 PM
@MoreAnonymous Vaguely, what 2 did you do before, and what is the plan for after, is it just because you keep posting ill-defined questions and getting rebuked for it or more
For a system whose Hilbert space is $L^2(\mathbb{R})$ is it possible to have a Hermitian non-self adjoint momentum operator?

If I choose the domain to be $\mathcal{S}(\mathbb{R})$ I get self adjoint...If I use compactly supported $C^1$ functions, still I get self-adjoint...
@Sanjana What do you mean "If I choose the domain to be $S(\mathbb{R})$ I get self-adjoint"
certainly not all operators you can try to define on $S(\mathbb{R})$ are self-adjoint :P
@ACuriousMind But also the energy plays a role right? For example, the resonance might exist, but if the energy is not enough, we will not get the resonance at that particular angle. So one can't say that it doesn't exist. Energy should be taken into consideration, right?
@ACuriousMind momentum operator I meant (as in the first sentence)...or wait a sec...do you mean to say even the momentum operator defined on the Schwartz space is not self adjoint?
(Conditions on derivatives are implicit :p)
@imbAF Sure, since the cross sections depend on the 4-momenta of the scattered particles, they depend on energy. What's your point?
@Sanjana ah, I overlooked "momentum" but still, you really mean "are there dense domains of definition where the differentiation operator on $L^2(\mathbb{R})$ is not self-adjoint", right?
7:27 PM
Yes, but Hermitian
so you mean $p = p^\dagger$ but the domains of definition are different?
Yes
"Hermitian" is an annoying word because physicists usually use it as synonymous with self-adjoint
Symmetric is what the math people use, right?
@ACuriousMind It makes sense to say that the existence of the virtual particle will depend on the energy. I just I find it a bit deterministic, that the resonance is also dependent,( or characterized would be more appropriate) from the differential cross section. Can it happen that the same virtual particle in the same colliding experiment, while the energy is the same for it's existence, the differential cross section can be different in value for 2 different scattering events considered?
7:32 PM
@Sanjana they use both Hermitian and symmetric, it's just a matter of taste
Hmmmmmmmm
@imbAF I don't understand the question. I really think you need to learn this stuff properly instead of hoping that just one more question will somehow finally make it all make sense
ok
Forget about resonances until you get to the standard model, learn about cross sections in classical mechanics e.g. the Kepler problem
I have actually. I was able to understand the cross section, as a measure of probability, since it's proportional to the scattering rate, which is essentiall the probability of scattering per unit of time. Similarly to how the probability density function is a measure of the probability, by integrating over the corresponding interval, such is the cross section, a measure of it
while the differential cross section is proportional to the differential scattering rate
7:39 PM
@Sanjana wait a moment...how did you show self-adjointness for the compactly supported $C^1_c(\mathbb{R})$? I vaguely remember the domain of the adjoint should be larger there (some Sobolev space)
or rather I'm pretty sure you only get essential self-adjointness for the definition on $C_c^\infty$, meaning the operator with that domain exactly isn't self-adjoint
@ACuriousMind Yeah I said conditions on derivatives are implicit in another text :p
@ACuriousMind I showed essential self adjointness---yeah.
But don't we always include the closure...
okay got it...that would make it self adjoint but now i am having merely a hermitian one...ok ok
4

Total cross section $\sigma$ The cross-section of an event (e.g. collision) is the fraction of a total area that results in the event. This can be a true area (i.e. throwing darts at a dart board) or some analogous area ( i.e. meeting a person at a party). The probability of the event is then: ...

@Sanjana right - the closure is self-adjoint but the "unclosed" operator is symmetric but not self-adjoint
precisely because the adjoint has some annoying Sobolev space as its definition and not the same compactly supported functions
7:47 PM
Hmmm...but anyway "Essentially self adjoint" is not self adjoint at all...what a weird name!
it's not that weird
it means "it's not self-adjoint but it has a unique self-adjoint extension"
yes i know that...but doesn't it clash with the meaning of the word "essential" as used in daily english language (essential=necessary)....

Think...if "water is essentially needed" actually meant "water is not needed but it has a..."
@Sanjana in one of the quirks of the language, "essentially" is not just the adverb of "essential" :P
I don't think "water is essentially needed" is idiomatic English, you'd say "water is essential"
but when someone says "X is essentially Y", they mean X and Y are so alike in the aspects that matter right now that the differences are neglegible - but that there are differences, X is not actually Y.
oh
in that sense
sorry that was too many "h"s...wait a min

3 hours later…