The h Bar

qwerty

00:13

@SillyGoose definitely

00:28

@qwerty the "gets brighter" has nothing quantum to it. He is absolutely correct. There is nothing weird about it from a classical waves perspective. All the weirdness comes from when you realise that you must have an explanation for how photons go through them one by one.

@Feynmate miao miao rarely deals merely with memes...

imbAF

00:47

Is there a reason why sometimes in order to solve the K.G equation we consider $\Phi(x^\mu)=e^{-ix_\mup^\mu}$ and sometimes a fourier transform of it $\phi(x)=\int \frac{1}{(2\pi)^3}e^{i\vec p \vec x}\tilde{\Phi(t,\vec p)}d^3p$ ?

naturallyInconsistent

01:24

@User198 You are so close that Tobias did not even want to stress the small difference. You have a weighted sum with classical probabilities, so the equations must be $$\begin{align}\tag1\rho(t)&=\sum_jp_j\left|\psi_j(t)\right>\!\left<\psi_j(t)\right|\\\tag2&=\sum_jp_jU(t-t_0)\left|\psi_j(t_0)\right>\!\left<\psi_j(t_0)\right|U^\dagger(t,t_0)\\\tag3&=U(t,t_0)\left(\sum_jp_j\left|\psi_j(t_0)\right>\!\left<\psi_j(t_0)\right|\right)U^\dagger(t-t_0)\end {align}$$

$$\tag4\therefore\qquad\rho(t)=U(t,t_0)\rho(t_0)U^\dagger(t,t_0)$$ as wanted

Hence, if you postulated quantum dynamics for wavefunctions, then you will automatically obtain the correct dynamics for density operators by this derivation.

Silly Goose

02:54

another misleading statement: "the partition function for a system of non-interacting particles factors"

someone needs to take you to fock space...

Tobias Fünke

06:12

@SillyGoose who says that and where?

I've pointed out, again in a link I had posted here, that this is in general only true for the GCE, and the factorization is with respect to the single-particle states.

I really do not know where you got all the statements from ^^. I cannot remember reading such a single isolated statement in any book.

to add: what I refer to was for indistinguishable particles. for distinguishable ones it is indeed much easier, and can be generalized, I think.

bolbteppa

06:26

Theoretical Physics Benchmark (TPBench) -- a Dataset and Study of AI Reasoning Capabilities in Theoretical Physics

> While we find impressive progress in model performance with the most recent models, our research-level difficulty problems are mostly unsolved... While currently state-of-the art models are still of limited use for researchers, our results show that AI assisted theoretical physics research may become possible in the near future.

> The first iteration of our benchmark consists of 57 problems of varying difficulty, from undergraduate to research level. These problems are novel in the sense that they do not come from public problem collections.

Tobias Fünke

Why do so many people here have problems in accepting answers? For example, I know like 2-4 users who constantly fail to accept (very good) answers, for the vast majority of their questions (and they have many many, and even like around 10k rep).

2

I am then trying to point this out, and sometimes it works and sometimes not. I don't consider this fair, given how the site is intended to work. No?

qwerty

I don't know about the cases you're talking about, but it's possible that the OP might not understand the answer(s) fully and therefore delay or then forget endorsing them

Tobias Fünke

07:17

well, but wouldn't you write a comment and ask for clarification?

qwerty

speaking for myself only - there were a few very detailed answers on my energy torsor question

Tobias Fünke

And no, I don't think so. Again, with 8-12k rep you should know how this site works. I mean, don't get me wrong: It is obviously their decision wether or not to accept an answer... but I find it almost rude to not do, at least given they have hundreds of questions and almost no answers

qwerty

I haven't had the time to think about all of them in depth

Tobias Fünke

Yeah, there is no problem in not accepting an answer. But having like hundreds of questions and only, idk rn the exact number, 20% accepted is something different

qwerty

@TobiasFünke yeah in those cases, definitely it comes across differently

Tobias Fünke

07:19

it gives me "only taking and not giving" vibes

qwerty

yeah that's true

Tobias Fünke

they do not care for the site, they just care for their question. Personally, I will and did stopped trying to help those people, either in helping formatting (e.g. tags) or comments/answers

qwerty

be back later :) I have dance~

Tobias Fünke

Okay, have fun! :)

Tobias Fünke

07:31

10

Q: Problematic behaviour of user

Is the behaviour of a user who: in the majority of his (well) answered questions doesn't accept an answer and goes on asking further and further questions in the comments, almost always exceeding the scope of the original questions, always with a know-all attitude refuses to accept that an answ...

discussion

a meta post with an answer by ACM; but note also Emilio's comment

Ryder Rude

08:11

hi

naturallyInconsistent

@TobiasFünke lets star this~

Ryder Rude

@bolbteppa what kind of Ai assistance do these people expect in future research

i was thinking that u could give a specific kind of problems to the AI, just like u give specific problems to a calculator

so it is a time-saver, rather than something that humans couldnt do by themselves

naturallyInconsistent

@TobiasFünke Distinguishable -> factors essentially trivially. The case where we care a lot more, are indistinguishable particles, and in that case + non-interacting, we may elect to not use single-particle states, or at least choose orthonormal normal mode basis for them in such a way that the factorisation also happens. Especially in QFT; the 2nd quantisation scheme that is usually done in terms of 1-particle states is trying very hard to also look like it is factorising properly.

Some rare authors might also cover some states that are not reduced to 1-particle states thereof.

Needless to say, it is interacting particles that are incredibly difficult to handle.

qwerty

@TobiasFünke do you think the cases you've noticed are sufficiently different to require a different meta post? the question you linked was mostly about people trolling or sort of sealioning it seems

Tobias Fünke

08:52

@qwerty hmmh good question, I don't really know...mhmh

this btw partially overlaps with the non-use of mathjax

another annoying thing :d but enough rant, sorry guy

let me think about a meta post... perhaps ACM can already say if it will be useful

but yeah, they do not troll, and many questions are quite good

PM 2Ring

09:10

Here's a meta post related to accepting answers. Originally, the accepted answer was pinned to the top of the Answers list (when sorted by score), even if it wasn't the top scoring answer. But sites were given the opportunity to change that behaviour 3 years ago. The answers & comments on this question give a sample of our community's attitudes towards the Accepted mark.

20

Q: Do we want accepted answers to be pinned to the top?

SE is making the way accepted answers behave configurable per-site and is looking for input from our side what our preference is. Currently, accepted answers - answers that the asker of the question has explicitly marked as answering their question - always are shown at the top of the list of ans...

discussion accepted-answer

Tobias Fünke

Hi PM 2Ring. But does that relate to our discussion here?

PM 2Ring

Several members made the point that often the question author is the least qualified contributor on the page, so they may not have the skills necessary to evaluate all the answers. So their selection of an Accepted answer may not be a good indicator of which answer is best.

naturallyInconsistent

It relates to the meta question you linked earlier

Tobias Fünke

yes I see that

PM 2Ring

@TobiasFünke While I agree that it's satisfying to get the green checkmark, many community members (including several mods) claim that the Accept mark is over-rated, and people shouldn't be so concerned about it.

Tobias Fünke

09:16

@PM2Ring mhmh

I disagree.

If it is not accepted, the question will remain as "unanswered", no?

i.e. pop up from time to time at the front page

PM 2Ring

OTOH, even a newbie can accept the answer that they feel helps them best to understand some topic. But that can happen even if that answer is technically incorrect. ;)

Tobias Fünke

I don't see the point in leaving such questions open

I see your point(s)

But the cases I talk about have correct and often also several correct and good answers

They just don't do it, for whatever reason. Perhaps laziness. And as I've said, some of them even refuse to use MathJax regularly, and just copy/paste screenshots. There is also a meta thread about that.

PM 2Ring

@TobiasFünke No. But if a question has no positive scored answers it will get bumped to the front page from time to time by the Community bot.

Tobias Fünke

also if it has no accepted answer, no?

I might be wrong here, but I thought I saw posts with several answers and positive scores popping up (without an edit). Perhaps I misread or so

PM 2Ring

@TobiasFünke I don't think so. But I'd have to check to be certain

Tobias Fünke

09:23

and just to be clear: it is not about my own answers. As I said, I will stop and did so already to interact with these users. Of course, the "extreme cases" I mentioned are only a few members. But more generally, this and the MathJax thing annoy me, and even more so if it happens for users who are not new

Slereah

I'm just watching stuff about quantum computing and fuming about those suits talking about quantum mechanics

I'm just gonna retreat in my Grothendiecke wizard shed

PM 2Ring

In some communities, the Accept checkmark is important. Eg, on scifi & fantasy, it's used to mark the correct answer to a story ID.

Tobias Fünke

@PM2Ring Asking the other way around: What reason do you see in not accepting an answer (while e.g. not asking for clarification in comment)?

(and not only a few times, but a regular pattern)

PM 2Ring

Oh, it also annoys me when askers have a long track record of never or rarely accepting. But I try to not be as annoyed by it as much as I used to. OTOH, I write answers for all future readers, not just the OP.

Tobias Fünke

yes, agreed. But OTOH I do not want to reward "anti-social" behavior, roughly speaking

imbAF

09:29

Can someone help me with an issue I encounter in calculating the solutions to the K.G eq. I initially consider a wave function solution to the equation, which gives me two solutions. Normally I need normalize the solutions. After than I attempt at having a generalized solution, and that would be a fourier decomposition, an integration over all the the momenta values. And in the end I will substitute the complex numbers a and b* with operators.

The issue I have is with the normalization of the wave function. In order to do as such, do I integrate only over space or also for time.

bolbteppa

@RyderRude The paper gives examples of what they'd want to do

Tobias Fünke

The KG field is not a wave function. It is an operator valued function. It is not clear what you mean with normalization.

PM 2Ring

It's permitted to comment "Please consider accepting this answer if it has helped you", especially if the OP is a newbie, and you link to a relevant Help page &/or meta post.

It's even more diplomatic to comment on the question: "Please consider accepting one of the answers below". Some community members consider it bad etiquette to "beg" for acceptance. And of course it's definitely bad etiquette to beg for upvotes.

imbAF

Well, you can certainly do the following and consider the solutin as $\Phi(x^\mu)=e^{ip_\mux^\mu}$

Tobias Fünke

@PM2Ring yes. Regarding the users I am talking about, I kindly commented several times already to consider accepting an answer for their questions. To no avail, mostly.

imbAF

09:32

And in doing so you get two values for the energy

which means you have two such solutions. And you can, because of linearity, construct a solution from their sum. But what you have is an instance for an arbitrary p value. You can generalize by integrating over the values of momenta

What you'd get, would be a fourier decomposition of $\Phi(\vec x,t)$

bolbteppa

The Fourier expansion is an integral over 3-momenta right

imbAF

But what I need to do is: $\Phi(\vec x,t)=\Phi_1(\vec x,t)+\Phi_1(\vec x,t)=$

I need to initially normalize the solution to a specific p-value

In doing so I'd integrate over all space

Tobias Fünke

Have you checked Weigand's notes, for example?

imbAF

But would I also integrate over all time?

Tobias Fünke

These have been recommended to you

imbAF

09:36

@TobiasFünke My plan is such that I will start with Weigand notes in March and finish it by the end of April. What I am doing right now, is re-reading the notes that I took in the lecture. Going over them, and filling in everything that I read and is unclear

That way I will have the knoweldge from what my lecture gave

Tobias Fünke

You do you

bolbteppa

$\phi(x) = \int d^3 \mathbf{p} (A_{\mathbf{p}}(t) \phi_{\mathbf{p}}(\mathbf{x})+ B_{\mathbf{p}}(t) \phi_{\mathbf{p}}^*(\mathbf{r}))$ where the $\phi$'s have normalization factors in them

Tobias Fünke

I cannot even follow what you mean here. But perhaps this is my fault/problem

imbAF

I am just curious

what can't you follow?

Tobias Fünke

What you are trying to do here/mean. But as I said, it is probably my lack of knowledge to see it

imbAF

09:38

@bolbteppa yes but you need to somehow get the normalization factor which is $\frac{1}{2E}$

bolbteppa

The Klein-Gordon wave function has a continuous spectrum, so you need to know how to normalize wave functions in the continuous spectrum and the subtleties that come up with this

imbAF

Or perhaps is $\frac{1}{\sqrt{2E}}$

bolbteppa

The reason this is confusing you is because normalizing a stationary state in the continuous spectrum is not as straightforward as the discrete case, because you normalize against a delta function you can always sneak factors inside the delta function and change your definition so it can be very confusing and different books make different conventions

Tobias Fünke

imbAF, you need the on-shell condition

imbAF

@bolbteppa I don't see where the Delta function is coming from, but I will take your word for it. I want to just show you one thing. For clarity, so we both are on the same page. Or rather, for me to recollect my thoughts

bolbteppa

09:41

@imbAF You can choose your normalization factor to be whatever you want as long as you compensate for your choice somewhere else

Slereah

> Venegas-Gomez highlights that the quantum arena is very broad. “You can have a career in research, or work in industry, but there are so many different quantum technologies that are coming onto the market, at different stages of development. You can work on software or hardware or engineering; you can do communications; you can work on developing the business side; or perhaps even in patent law.”

imbAF

Ok, I need help just in this early step and I can continue on my own later.

bolbteppa

So you should follow what your book/notes does/do and stick with what they do carefully until you dig into this

Slereah

Keep those patent lawyers away

imbAF

@bolbteppa There is no book I am getting this. Only bits etc. Most of the books, start with the fourier decomposition expression but NEVER explain how they got here.

I will try to show you what I mean

skullpatrol

09:43

@Slereah do you think this will happen for AI also

imbAF

Just give me one minute

bolbteppa

@imbAF Section 5 of L&L vol. 3 explains this, without this starting point everything will look extremely confusing, then they scatter a few important comments about it later on in the book (e.g. when they discuss normalizing a plane wave around Sec. 16, here you see them changing the normalization by changing the scale inside the detla function) and when they discuss the current density a few sections later, and in volume 4.

imbAF

Ok

bolbteppa

The pain this has caused is why I know these section numbers :\

Slereah

@skullpatrol It already has

PM 2Ring

09:44

@TobiasFünke You may enjoy my math meta post about MathJax vs handwritten equations. ;) math.meta.stackexchange.com/a/33091/207316

bolbteppa

You really need to be careful with all this when defining scattering cross sections for example

imbAF

@bolbteppa lmao

Tobias Fünke

Why on earth do you refuse to check other books/notes but instead ask us what your notes could mean?

bolbteppa

Even in time-dependent perturbation theory this issue arises and causes problems

Tobias Fünke

@PM2Ring hehe

imbAF

09:45

@bolbteppa I see. Still just one more thing I have

bolbteppa

math.columbia.edu/~woit/QM/qmbook.pdf#page=473

Skim that chapter, you will see he derives the normalization factor in a clever way, I still think this misses the fundamental point but it's probably what you're looking for right now

imbAF

But this isn't L&L

@bolbteppa which chapter ?

bolbteppa

I linked to chapter 43 there, look at (43.3) - (43.9)

imbAF

I see it, that's quite the chapter

Is it a problem that sometimes people use (1/2pi)^3/2 and sometimes (1/2pi)^3

?

Ryder Rude

@bolbteppa oh

bolbteppa

09:52

No, that is a hint that some of this is arbitrary, you can change that factor as long as you compensate by changing your delta function

imbAF

Ahaaa

bolbteppa

o3-mini vs DeepSeek, explained by ex-Apple Engineer

►

Tobias Fünke

@bolbteppa it is not about the delta function; it is about the Fourier transform and its inverse

the delta function factors are fixed

bolbteppa

@RyderRude Go ask the 'reasoning' version something hard, like deriving the Einstein Field Equations from the EH action, ask it to explain it step by step

Ryder Rude

10:04

@bolbteppa i will try..

Silly Goose

13:57

@TobiasFünke im not sure what you mean it’s not true period for indistinguishable particles

Or okay it factors but not into single particle partition functions

Tobias Fünke

I mean that for (non-interacting) indistinguishable particles, the GCE PF factors into partition functions for the single-particle states

but that is not the case for e.g. the canonical ensemble

for distinguishable (non-interacting) particles everything is much easier

Do you agree with that? :)

Allie

Hi all

Meow

I have a question

Tobias Fünke

14:13

Hi Allie

Allie

So when I’m solving for V using Laplaces equation (Poissons? The one where theres no charge), it’s okay for me to have a discontinuity in E at the boundary of a conducting plate grounded at

V=0

Right

Like if the potential doesnt have a discontinuity and the laplacian of the potential is 0 on both sides, but the gradient of the potential has a discontinuity

Think y=sin x joined up with y=0

Silly Goose

@TobiasFünke i think this is only true if the hamiltonian is proportional to the number operator

which is not true generically for a system of non-interacting particles

e.g. Anisotropic harmonic oscillator

Or hm

@TobiasFünke i'm not sure i agree with this

Tobias Fünke

@SillyGoose It is true of $H$ is a one-body operator (in my field this exactly means that the particles are non-interacting); by this, I mean that it has the form of $H=\sum\limits_{ij} h_{ij}^a^*_ja_i$ (modulo typos)

Silly Goose

the GCE PF is $\text{tr} \circ \exp (-\beta \hat{H} + \mu \hat{N})$, yes?

Tobias Fünke

I've posted a link some days ago

It is true, at least if the single-particle space is finite-dimensional (although I guess physicists do not care too much)--I have not thought about the infinite-dimensional case too much to be honest. IIRC, the question was concerned with a statement in Coleman's book.

2

Q: Partition function for independent particles

I am trying to understand Section 3.8.3, "Independent particles", of Piers Coleman's Introduction to Many-Body Physics (self-study, mathematics background). He considers "a system of independent particles with many energy levels $E_\lambda$," with Hamiltonian $$H - \mu N = \sum_{\lambda} (E_\lamb...

quantum-mechanics statistical-mechanics condensed-matter many-body partition-function

I remembered their username :p

IIRC, also in Balian's book about stat.mech (either first or second volume) this is discussed

Silly Goose

14:27

@TobiasFünke I mean doesn't the first line in this assume $[H, N] = 0$?

they are labeling energy and number eigenvalues with number eigenvalues

Tobias Fünke

@SillyGoose Sure, that is always the case in GCE

Silly Goose

energies in an anisotropic oscillator are not labeled by a single number operator though

Tobias Fünke

All Hamiltonians in the form I've given can be diagonalized to be of the form $H=\sum\limits_j \epsilon_j \hat n_j$, and $N=\sum\limits_{j}\hat n_j$ anyway for any single-particle orthonormal basis!

I don't understand what you mean

You start by a single-particle Hamiltonian; you construct the second quantized form of it and end up with a Hamiltonian on the Fock space

naturallyInconsistent

@Allie When there is no charge, Poisson's equation reduces to Laplace's equation. V never has a discontinuity because V is only defined so that its derivative is E; its derivative exists, so it can only at most have kinks and not have gaps.

Silly Goose

$H = \hbar(\omega_1n_1 + \omega_2 n_2 + \omega_3 n_3)$ is the Hamiltonian for the anisotropic oscillator

Tobias Fünke

14:30

which by construction commutes with $N$

yes, that's fine

Madder

Hello guys, i really need some help
https://math.stackexchange.com/questions/3451152/reference-for-onsager-s-solution-of-the-2d-ising-model#comment10816316_3451152

i have exactly the same problem. i can not find any source that is dealing with onsagers original solution using quaternion algebras. The paper is very hard to read. I have searched very extensively, i have been unable to find anything replicating his work in more detail, i dont know what to do :D

Tobias Fünke

I think your misunderstanding roots in the formalism of second quantization (?!)

Silly Goose

@Madder are you particularly interested in onsager's solution or in any exact solution to 2d ising?

Madder

@SillyGoose I am writing my bachelor thesis regarding his solution. So i need to understand what he is doing in his paper.

Silly Goose

i see

he doesn't prove spontaneous magnetization though :P

he just writes it down

@TobiasFünke hmm

Tobias Fünke

14:32

let me summarize:

Madder

How can a paper with over 7000 citations have no explanation. This is just very bizarre.

Silly Goose

@Madder because no one really reads carefully :)

naturallyInconsistent

You should tell people right upfront that this is your situation so that they will not waste time wondering why you havent considered other solutions. Also, have you understood the easier solutions yet or not?

Tobias Fünke

Single particle Hilbert space $\mathfrak h$. Single-particle operator/Hamiltonian $h$ on this space. Construct the fermionic or bosonic Fock space $F(\mathfrak h)$ and the corresponding Hamiltonian $H=H(\mathfrak h)$. By construction, $H$ commutes with $N$ (on $F$)

Allie

@naturallyInconsistent right, my solutin had a kink, just wanted to make sure thats okay

Madder

14:34

@naturallyInconsistent I have read and understood some of them, but they are not interesting to me. as i said, my topic is exactly his original paper.

Tobias Fünke

no kink shaming

Silly Goose

@Madder more pessimistically, people cite what they think they should cite, not what they actually use or understand

Allie

But i checked the solution book and it was the right answer so :)

Silly Goose

@naturallyInconsistent i'm not sure any of the solutions are easy

naturallyInconsistent

@Madder the point is that you might be able to use the fact that you already understand some easier solutions, to reverse engineer Onsager's solution.

Allie

14:34

Only discontinuity shaming in here

Madder

@naturallyInconsistent No because his approach is very algebraic. The other solutions do not use his method at all!

naturallyInconsistent

ok

Silly Goose

schultz mattis and lieb's approach is purely algebraic..?

and way easier than onsager's solution

and if by algebra you include combinatorics, so is the combinatoric solution purely algebra

Madder

@SillyGoose but it does not use the quaternion algebra approach. It is not much of a help

Allie

I like getting problems right

Silly Goose

14:36

but why would you like to use quaternions

Allie

It makes me feel momentarily that im not an idiot

Silly Goose

or is that what you are interested in using i guess

Madder

Because it is the topic of my bachelorthesis.

Tobias Fünke

yeah hehe always feels good

Silly Goose

@TobiasFünke hmmm okay i see what you mean. then i wonder how to write the anisotropic energies in the $\hat{N}$ basis...

@Madder quaternions?

Madder

14:38

@SillyGoose The topic of my thesis is his original paper. In his original paper, he uses this approach. Thus i need to understand his exact approach.

Silly Goose

well good luck :P. i cannot imagine onsager's solution sheds much light on the underlying physics either...

Tobias Fünke

Goose: Your single-particle Hilbert space in this case is $L^2(\mathbb R)$. You construct the corresponding Fock space. You diagonalize the anisotropic oscillator and obtain an orthonormal basis $\varphi_j$ in $L^2$ with corresponding energies $\epsilon_j$ . Now the number operator on $F$ is $N=\sum\limits_j a_j^* a_j$, and similarly the Hamiltonian is $H=\sum\limits_j \epsilon_j a^*_j a_j$.

bolbteppa

@Madder Yes the paper is insane. Have you tried to study Feynman's version of it?

Tobias Fünke

How you do it in practice is, of course, another thing :p

I mean for generic cases it might not be easy to diagonalize the single-particle Hamiltonian

Silly Goose

@bolbteppa this pushes the problem into proving a combinatorial theorem, though. not very fun or enlightening

imbAF

14:42

@bolbteppa I read the chapter on Voigt's notes but I have a question, and perhaps you can clarify it.

Silly Goose

and feynman's own exposition of the solution is not very good (IMO)

SML's solution at least also teaches you about Jordan-Wigner and bogoliubov transformations. it also naturally shows you that gapped/gapless spectrums matter and some other more physically interesting ideas.

bolbteppa

But it is fundamentally a combinatorial problem, Feynman's version is a lot simpler like the original paper is insane comparatively

Madder

I am cooked.

Silly Goose

the most natural formulation of the problem is combinatorically, indeed. but i personally think combinatorics is just hard.

it is easier to map the problem into a theory of 1D free fermions lol...

Allie

Is bachelors thesis normal

I did NOT have to do that

Tobias Fünke

14:45

in Germany at least it is

bolbteppa

Yeah the combinatorics is very ugly, but this is just an ugly problem in general

Tobias Fünke

and similarly for a Master thesis

Silly Goose

at some unis in the states it is required

at others it is not

my uni requires a thesis

Tobias Fünke

we also had to defend it

Madder

so you are telling me there is not a single source on the internet doing his original work in more detailed fashion, this is honestly extremly bad news for me.

imbAF

14:45

@bolbteppa I was attempting something. Starting with the K.G equation, I consider a wave function solution to it $\Phi(x)=Ae^{ipx}$. And one gets two solutions. So the summation is a solution as well and it is of the form: $\Phi(x)=Ae^{ipx} + Be^{-ipx}$. If I'd want to generalize the solution I would, take my expression and just integrate over the values of momenta. But before doing that, I do need to normalize $\Phi(x)=Ae^{ipx} + Be^{-ipx}$. How to normalize plane waves, which is not possible

Tobias Fünke

which is quite normal here, I believe.

Silly Goose

@TobiasFünke is your $^*$ what physicsts use as $^\dagger$?

Tobias Fünke

yes

Silly Goose

you mathematician you

Tobias Fünke

ahah no, I am just too lazy to write the dagger

but yes, this is mostly used by mathematicians, and at some point I started to copy it

Silly Goose

14:46

@Madder i think his solution is so annoyingly complicated that people study other ones, perhaps

there is one professor who wrote several blog posts going in greater detail through the combinatoric solution and the SML solution

Tobias Fünke

can you link it, please?

I am interested too

bolbteppa

@imbAF This is a great example of the subtleties of normalization in the continuous spectrum that I was talking about. You can't naively normalize a free particle based on how it goes for the discrete spectrum, but there is a way to normalize based on the continuous spectrum, it's explained in Section 5 of L&L, there is even a physical interpretation of the fact that it doesn't naively normalize

imbAF

@bolbteppa Ok, I will have to read that because Voigt doesn't provide that

Also look at this here: physics.stackexchange.com/questions/442936/…

In this link, we start from a fourier transformation in spacetime

But that is one way to gain the result, the decomposition of the solution

But another one, is starting from the naive wave function solution.

Which as you say is in L&L 5

Btw which volume of L&L ?

bolbteppa

Vol 3

Madder

journals.aps.org/pr/abstract/10.1103/PhysRev.65.117

Honestly i am prepared to pay for someone to help me with interepating this paper. I am a broke student so please be merciful with your bids

imbAF

14:52

@bolbteppa Motion in a centrally symmetric field?

bolbteppa

The derivation in that link is good, but you can always change the normalization as long as you change something else somewhere else, and people sometimes do this, you have to ask what the most basic condition of normalization is in the continuous spectrum and sec. 5 is the ultimate starting point, look at the 'Momentum' section of Vol. 3 (Sec. 15?) there they explain how to do this for a free non-relativistic particle and you see what I mean about changing the normalization there explicitly

The condition by which people usually normalize Klein-Gordon is explained in Vol. 4 physically, we normalize the plane waves so that we can physically interpret a plane wave as 'one particle per volume V', or such that a particle has energy $E$ when you insert it into the KG energy-momentum tensor

imbAF

Yes, the derivation is good, but he is already starting from a fourier transform

I want to go on step back and reach the fourier transform from the wave function solution

Which is what L&L vol.3 chp.5 has as you said

Aha it's volume 4

Silly Goose

@TobiasFünke gandhiviswanathan.wordpress.com/2015/01/09/…

anyways...even the SML solution is not super satisfying

imbAF

Ok momentum @bolbteppa I found it. In vol 3. chapter 15, you are right. How you remember this lmaoo

Silly Goose

i also have my personal notes on the SML solution, which go in great detail about arriving at stated results

bolbteppa

14:57

Equation (1) in the Stack post is a linear PDE (KG equation), any linear PDE can be solved by a Fourier expansion, so equation (2) is just the abstract formula for a Fourier expansion, even that $(2\pi)^4$ is an example of what I meant about changing the normalization as long as something somewhere else balances it out. You see when they get to (9) that Woit does similar steps then changes things bringing in square roots

imbAF

Yes but I believe, and correct me if I am wrong that :
1. The (2) can also be written in d^3p, which would require the addition of 1/2E.
2. Considering only (2). One should be able to derive it from somewhere.

bolbteppa

Look at that blog post, it basically follows Feynman's solution, as hard as it is it's not insane like the Onsager paper

@imbAF Not sure what you mean, do you mean (2) in the Stack post?

imbAF

Yes, I am sorry

I also will take the (1/2pi)^3 notation

Silly Goose

i guess feynman's solution might be the most "sane" one. it might be possible to just prove to yourself linked-cluster theorem and then see if that result is enough to obtain the crucial relation $(30)$ (in the blog post).

imbAF

because I was able to explicitly derive it, hence argue its presence

Silly Goose

15:01

that is the hardest part of the solution IMO

bolbteppa

Physically, because $E^2 = \mathbf{p}^2 + m^2$, the sum over 4D should really reduce to a sum over 3D, so any 4D integral is artificial. You can derive the normalization factor by defining the general solution as a sum over $d^3 \mathbf{p}$ of the plane waves, and fixing the normalization factor so that $T^{00} = E$ for a Plane wave, which L&L vol. 4 do in the KG chapter

Silly Goose

i think SML is the most straightforward solution i have seen, though. no magic, just lots of algebra and transformations.

and some dubious limit taking as with all the solutions...

bolbteppa

15

Q: How many Onsager's solutions are there?

Update: I provided an answer of my own (reflecting the things I discovered since I asked the question). But there is still lot to be added. I'd love to hear about other people's opinions on the solutions and relations among them. In particular short, intuitive descriptions of the methods used. Co...

statistical-mechanics ising-model

Tobias Fünke

too many conversations for me. I cannot follow anymore :d

but thanks for the link, Goose

Silly Goose

no problem. it should be easy to find his other blog posts on the 2d ising model because i think he links to them within the blog post i linked

bolbteppa

15:03

@Madder you should ask the person in the post on that page to finish the 'to be continued'!

Madder

@bolbteppa will do

Silly Goose

feynman's own exposition of the combinatoric solution is given in his lecture notes on stat mech. he calls it the onsager problem i think

eduardo fradkin has really nice lecture notes on the SML solution as well, with a field theoretic emphasis; cf. quantum field theory and statistical mechanics: eduardo.physics.illinois.edu/phys583/physics583.html

bolbteppa

@imbAF Woit's (43.3) can be re-written in a crazy way using (43.7) by re-defining the coefficients, you won't 'naturally' derive it that way yet its equivalent, the way it appears involves a choice at this stage involving an arbitrary normalization factor that we fix later (by the usual continuous spectrum convention)

Tobias Fünke

Goose, is it also in his QFT book? do you know by chance?

Silly Goose

@TobiasFünke let me check

Madder

15:07

i will ask the mathematicians as well, maybe i find better luck there.

imbAF

@bolbteppa Honestly I just find everything confusing. The math is confusing, and on top of that I am jumping back and forth between different books who have different notations and different normalization factors. It's crazy to understand

bolbteppa

Welcome to the madness of QFT

Silly Goose

hm i thought i had fradkin's book totally legally downloaded...i guess i don't

@TobiasFünke sorry im not sure then

imbAF

@bolbteppa Why can't people write a coherent derivation. But only bits...

bolbteppa

Getting confused by all this means you're on the right track

imbAF

15:09

@bolbteppa And you know, to get coef. such as (1/2pi)^3 while easy, it is not that straightforward.

Tobias Fünke

Goose, I have the book

If you remind me later, I can check if you are interested :p but thank you!

@imbAF it is a convention. there is nothing to derive

every good book/lecture notes will state their conventions either at the beginning or at the end (appendix)

imbAF

Well, when I jumped from discrete momentum values to continuum I had to argue for the existence of this term

Silly Goose

fourier transform/discrete fourier transform/fourier series/fourier blahblahblah conventions indeed make me want to die

Tobias Fünke

hahah

imbAF

and I was able to consistently show that

Tobias Fünke

15:11

@imbAF ? where is the difference to e.g. solid state physics?

well, good

Silly Goose

@TobiasFünke to confirm, since $[H, N] = 0$ we can label energies using total number of particles $n = 1, 2, ..., \infty$, but generically we will have distinct energies labeled by the same total number of particles?

Tobias Fünke

yes

Silly Goose

so the energies might look like $\epsilon_{n,i}$ where $n = 1,2,...,\infty$ and $i$ is an index accounting for the fact that different energies might correspond to the same $n$.

Tobias Fünke

Well, as I've written, you can write $H=\sum\limits_j \epsilon_j \hat n_j$; so an eigenbasis is given by the occupation number basis (constructed from the single-particle eigenbasis) $|n_1,n_2,\ldots\rangle$

each of these states has a specific particle number, and a specific energy

imbAF

@TobiasFünke This was my argument for the term, in natural units

Tobias Fünke

15:17

and the energies are $E=\sum\limis_j \epsilon_j n_j$.

Does that help, Goose?

imbAF

The set up is a particle in 3D box

Tobias Fünke

as I said: it is a convention

whatever you do, do it consistently and it should be fine :)

Silly Goose

for some reason i am having trouble understanding why $[H, N] = 0$ should imply that we can actually just write $H$ in terms of single mode number operators $\hat{n}_i$

Tobias Fünke

no, you do not need that

I mean it does not matter a priori

What is important is that you can diagonalize the single-particle Hamiltonian on the single-particle Hilbert space

For any orthonormal basis (omitting discussions with domains), you have $H=\sum\limits_{ij} h_{ij} a_j^* a_i$, if I did not mess up the order of indices

now in an ONB where $h$ is diagonal you get the form I've written

And then you easily see that it commutes with $N$

Silly Goose

okay so you're saying just individually diagonalize each constituent of your many-body non-interacting system, which is permitted because they all act on independent hilbert spaces.

Tobias Fünke

15:26

ääähm

yes but no

It is just the math of second quantization

interpret it like you want, but the math is clear ^^. $H$ is one operator on $F$ , and so are the $\hat n_j$

Silly Goose

oh i think i see

Tobias Fünke

ok good

Silly Goose

$h: \mathcal{h} \to \mathcal{h}$ is the single particle hamiltonian. the one-body operator is $H: \mathcal{F} \to \mathcal{F}$ is $H := \sum_{ij} h_{ij}a^\dagger_j a_i$.

Tobias Fünke

yes

imbAF

In L & L vol. 3, when considering the normalization of eigenstates of the momentum operator it is written: $\int \psi_{\vec p}^* \psi_{\vec p'}dV$. Why is it considering two different momenta?

Silly Goose

15:34

if $h$ is diagonal, then the one-body operator is simply just written as $H := \sum_i \epsilon_i a_i^\dagger a_i$

Tobias Fünke

and actually it is just another representation of $H=0 \bigoplus h \bigoplus h_2 \bigoplus \ldots$, where $h_2= h\otimes I + I\otimes h$ and so on

yes

Silly Goose

right, all encoded in the c/a operators

Tobias Fünke

the above decomposition is with respect to the one of the Fock space: $F=\bigoplus\limits_{N=0}^\infty H_N$, where $H_0:=\mathbb C$, $H_1:=\mathfrak h$ and $H_N$ for $N\geq 2$ the (anti-)symmetrized $N$ particle Hilbert spaces

yes indeed

good ^^

and now back to the topic: Once you got this, you should be able to prove the assertion in the linked post. (again, this should work perfectly well for a finite-dimensional single-particle Hilbert space; for an infinite-dimensional, idk atm)

Silly Goose

because if one were to be pedantic, we could actually write $H:= 0 \oplus \biggl( \sum_i\epsilon_i (a_i^{(0 \to 1)})^\dagger a_i^{(1 \to 0)} \biggr) \oplus \biggl( \sum_i\epsilon_i (a_i^{(1 \to 2)})^\dagger a_i^{(2 \to 1)} \biggr) \oplus ...$

or something like this

Tobias Fünke

@SillyGoose depends on how you define everything, but I guess yes.

but this has nothing to do with being pedantic, it is IMHO just over complicating things ^^

actually, I don't know if what you write makes sense, but I do not have time to think about it further

I am procrastinating anyway too much haha

Silly Goose

15:41

lol yes i've got to do my stat mech homework...

okay well i think i understand better now thanks

DIRAC1930

16:28

@imbAF I don't think it's a good idea to learn QFT from Woit's book

imbAF

@DIRAC1930 you there?

DIRAC1930

Yes

imbAF

I am just trying to do some calculation for my own goal, idk if you read the thread above

the discussion

If you have like 10 minutes to discuss something

DIRAC1930

For this, I followed L&L 4's argument about normalisation

imbAF

which chapter

But still I want to discuss my idea of it

Just to test my understanding

So i'd like to discuss it a bit if you don't mind

DIRAC1930

16:37

I might not be much help about this

imbAF

ok

imbAF

16:50

what chapter in L& L vol 4?

Also would you be able to find A and B in this expression: $(\frac{1}{2\pi})^3 \int (|A|^2+|B|^2)d^3p=1$ ?

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