@SillyGoose definitely

@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...

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 $\phi(x)=\int \frac{1}{(2\pi)^3}e^{i\vec p \vec x}\tilde{\Phi(t,\vec p)}d^3p$ ?

@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.

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

someone needs to take you to fock space...

@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.

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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).

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?

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

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

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

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

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

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

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

it gives me "only taking and not giving" vibes

yeah that's true

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

be back later :) I have dance~

Okay, have fun! :)

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

hi

@TobiasFünke lets star this~

@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

@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.

@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

@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

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.

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

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.

It relates to the meta question you linked earlier

yes I see that

@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.

@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

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. ;)

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.

@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.

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

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

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

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

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

@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)

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.

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

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.

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

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

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.

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

@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.

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 $\Phi(\vec x,t)$

The Fourier expansion is an integral over 3-momenta right

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

Have you checked Weigand's notes, for example?

But would I also integrate over all time?

These have been recommended to you

@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

You do you

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

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

I am just curious

what can't you follow?

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

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

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

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

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

imbAF, you need the on-shell condition

@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

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

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

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

Keep those patent lawyers away

@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

@Slereah do you think this will happen for AI also

Just give me one minute

@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.

Ok

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

@skullpatrol It already has

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

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

@bolbteppa lmao

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

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

@PM2Ring hehe

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

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

But this isn't L&L

@bolbteppa which chapter ?

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

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

?

@bolbteppa oh

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

Ahaaa

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

the delta function factors are fixed

@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

@bolbteppa i will try..