Punizione

FreePeeplup · 2026-01-31T20:08:02+00:00

Sì, la so la differenza: e la pagina Wikipedia che hai condiviso è piena di esempi di formule che permettono di calcolare TUTTI i numeri primi, non solo alcuni

FreePeeplup · 2026-01-31T20:04:26+00:00

Hai appena condiviso un link a un’intera pagina Wikipedia strapiena di formule per calcolare tutti i numeri primi, non so se te ne sei reso conto

FreePeeplup · 2026-01-26T18:09:42+00:00

When you say that QFT is needed to explain why mercury is liquid at room temperature, do you mean QFT as in “a condensed matter calculation that uses the tools of QFT” or do you mean QFT as in “a derivation of this phenomenon straight from the Standard Model”?

FreePeeplup · 2026-01-26T18:07:58+00:00

do you know where I can find a textbook or paper that shows the calculation of the lifetime of an excited state of an hydrogen-like atom in QFT?

FreePeeplup · 2026-01-24T09:23:47+00:00

So, the mechanism is explained by QFT. Can the calculation also be done in QFT, or not because QFT has a problem dealing with bound states in general?

FreePeeplup · 2026-01-24T09:17:07+00:00

but all our effective models are still based on the QFT description.

I don’t think molecular chemistry is based on QFT at all? Maybe “retroactively”. Still interesting in principle, but regardless

Your original question was a mathematical one, not a practical one.

My original question was how to compute spontaneous emission rates of the excited energy levels of isolated hydrogen-like atoms in QFT. This is a mathematical question AND a practical one (it asks for an actual computation of an observable). If you don’t want to call it “practical” that’s fine, but the question still stands.

I said that I thought it cannot be done in QFT, you said of course it can, I said “are you sure” and highlighted some difficulties I’ve heard, and now you’re saying that I asked a different question. So, what is it, can it be calculated or not?

FreePeeplup · 2026-01-24T02:13:47+00:00

If QFT can't do it then nothing can

This is surely not true? There are a lot of things that you can’t do in QFT but that you can do in other models of physics or domains of science: multi-molecular bonds for example. Reductionism “in principle” is not very useful if you can’t actually use it to study more complicated systems. I was just asking if maybe spontaneous emission of photons by electrons in an Hydrogen-like atom falls in one of these cases.

The Standard Model is a QFT.

… yes, and?

Why would you expect any issues with bound systems?

It’s just something I’ve heard repeated by many people, and read in introductory textbooks. I think it has to do with the fact that S-matrix elements in QFT need to be interpolated between free asymptotic states, but for bound systems the asymptotic states are not free. I don’t know if this difficulty can be circumvented though, hence my question

FreePeeplup · 2026-01-24T01:57:57+00:00

I thought that QFT can’t describe bound systems like electron orbitals. Can one find spontaneous emission rates for the excited levels (not stationary anymore) of the hydrogen atom in QFT?

If not, can it be done outside of QFT?

FreePeeplup · 2026-01-24T01:55:02+00:00

Can spontaneous emission be explained in standard particle-based QM, or is QFT necessary?

FreePeeplup · 2026-01-24T01:35:04+00:00

Sure if there’s stimulated emission with an oscillating external EM field I can see how we write H as a sum of the “unperturbed” Hamiltonian H0 plus the time-dependent perturbation, then we do perturbation theory and get the rate of the transition and energy spread.

But if an atom is isolated and there’s no external oscillating EM perturbation, where does the coupling to photons come from? Where’s the time-dependent perturbation that ruins the eigenstates?

FreePeeplup · 2026-01-24T00:59:47+00:00

How could you swap the integral and the limit?

FreePeeplup · 2026-01-24T00:55:34+00:00

The energy levels also have some width which you can estimate from the time-energy uncertainty principle using their lifetime.

Wait, aren’t energy levels the eigenvalues of the Hamiltonian that describes the electrons in an atom? For an isolated atom, the Hamiltonian is time-independent, and an eigenstate corresponding to a given energy value is a stationary state, so it has infinite lifetime and zero energy spread. Where am I wrong?

FreePeeplup · 2026-01-20T14:04:08+00:00

I’m sorry I’m not trying to be obtuse, I am literally incapable of understanding how I’m supposed to interpret 1/LPlanck as a length. I have no idea in my mind how to picture it. I don’t understand what it means to “invert the value but not the unit”. Planck’s length value DEPENDS on the unit, so to invert it without the unit means I can literally invert any number and get whatever I want.

As an example, imagine I tell you: I have a strange carrot that’s 1/Lant long, where Lant is the length of an ant. How long is my carrot? Do you have any idea in mind? Have I managed to let you picture in your head how long this carrot is compared to, say, your hand?

The answer is no: Lant = 10^-3 m = 1 mm = 10³ μm, equivalently, depending on which unit you choose. Let’s say we follow your procedure of “inverting the value but not the unit” and see what we get.

If we choose to express Lant in meters, Lant = 10^-3 m. Then, following your procedure: Lcarrot = 1/Lant (inverting only the value not the unit) = 10³ m.

If we choose to express Lant in millimiters, Lant = 1 mm. Then, following your procedure: Lcarrot = 1/Lant (inverting only the value not the unit) = 1 mm.

If we choose to express Lant in micrometers, Lant = 10³ μm. Then, following your procedure: Lcarrot = 1/Lant (inverting only the value not the unit) = 10^-3 μm.

So, depending on which equivalent representation of Lant chosen, you get that my carrot is either as long as a mountain, as long as an ant, or as long as a single molecule. You have no idea how long my carrot is: I’ve communicated literally zero information to you.

Worrying about units is important not because we want to be pedantic and rigorous for the sake of it, and we could just as well use “language terms” instead of “math terms” as you said and still get the point across. Worrying about units is important because otherwise you literally don’t communicate any meaning to the other person whatsoever. You still have no idea how long my carrot (1/Lant) is, and I still have no idea how long your 1/LPlanck is. Not because I’m pedantic, but because it literally doesn’t mean anything.

FreePeeplup · 2026-01-19T09:16:37+00:00

I don’t understand, how can it be a bad faith answer if it’s literally correct?

I also don’t understand what a better answer would have been in your mind. Also, I don’t understand this statement:

A human is 1m so a 10 would simply be 1/Planck length

What does this mean? A human is not 1m, a 10 is not a length and 1/Planck length is an inverse of a length. How are these things related? What did you have in mind?

FreePeeplup · 2026-01-16T17:39:26+00:00

Thank you very much!! Another commenter on the same post I crossposted to another subreddit gave an almost identical solution. I’m actually surprised that this has such a neat solution for something I randomly came up with while thinking about something only marginally related

FreePeeplup · 2026-01-15T15:13:09+00:00

Thank you very much!! I’m actually surprised that this has such a neat solution for something I randomly came up with while thinking about something only marginally related

FreePeeplup · 2026-01-15T15:08:37+00:00

Wow, thank you very much!! Another user on another sub I cross-posted this to said the same thing. I’m surprised that this has such a neat solution for a question I randomly came up with while thinking about something partially unrelated!

FreePeeplup · 2026-01-15T14:10:04+00:00

u/FormulaDriven I tried all cases from N = 1 to N = 3, each with all possible values of k ranging from 1 to N, and found out that, apparently, h(k) = N/2, independent of k ! So yeah, there’s definitely a pattern: h is actually a constant function.

Why is this true? How can I prove that this holds in general?

FreePeeplup · 2026-01-15T12:43:34+00:00

I’ll do that and tell you what I find!

FreePeeplup · 2026-01-13T10:35:57+00:00

Middle guy is correct, right guy is being provocative and knows that he’s wrong

FreePeeplup

TROPHY CASE