thank you :D @Relativisticcucumber

perhaps one day i will be in germany and stumble upon a zwieback--for now i've only been at the frankfurt airport for a lay over :P @ACuriousMind

I feel like the linear combination of atomic orbitals method is quite suspect. How are we able to make absurd approximations yet get close to correct results.

@nickbros123 Roughly speaking, the non-linearity is governed by the fine structure constant, which is ~1/137.

> When perturbation theory is applied to quantum electrodynamics, the resulting perturbative expansions for physical results are expressed as sets of power series in α. Because α is much less than one, higher powers of α are soon unimportant, making the perturbation theory practical in this case.

Did anyone here like "Everything everywhere all at once"?

From ADVENTURES OF ISABEL BY OGDEN NASH rainydaypoems.com/poems-for-kids/cautionary-tales-for-children/…

Not everything Ogden Nash wrote was great, but some of his lines are absolute genius.

@RyderRude I thought it was pretty good

@SillyGoose oh, I don't recommend it - zwieback is what you eat when your stomach can't keep much else down :P

Is "consciousness causes collapse" a popular viewpoint among physicists?

I mean for physicists who care about the measurement problm

I see philosophers almost universally ridiculing this. I wonder what physicists think

MWI says that there is no collapse. But to connect MWI to experimental observations, I think one wud need an additional postulate : "There is a thing called consiousness that can perceive being in one of the worlds with probability given by Born rule"

Without this postulate, i think the wavefunction of mwi is just a mathematical object with no way to relate it to experiments

How does MWI tackle this problm?

@RyderRude what is consciousness

a hard problem

@Slereah idk :P

Tbf a measurement device is as poorly defined as consciousness

Tbf, if you consider measurement to be a poorly defined concept, any experimental results will suffer in their interpretation; correct?

@RyderRude MWI does not care about consciousness. When a measurement happens, it observes the branch of the universal wavefunction it is in. The proportion of branches that correspond to a specific result of that measurement is given by Born rule.

ncatlab.org/nlab/show/mind

nlab isn't much help

@ACuriousMind Is this something like relational QM then? I mean that every object can end up in one of the worlds upon getting entangled, by collapsing the wavefunction relative to its perspective. So we don't claim that only consciousness has this property.

And we don't treat consciousness as special

And we also claim that a single particle, that gets entangled from our perspective, actually ends up in one of the worlds from its own perspective. So we no longer draw any line between measurement devices and quantum particles, wrt collapse

Is this correct

Consciousness isn't related to measurement in QM any more than it is in any other theory

You might as well ask if a tree has really fallen if no one is there to hear it

What makes qm special is for me is that there r no hidden variables. But from our consciouss perspective, we DO observe only selected outcomes @Slereah

the same is true of non-conscious systems though, theoretically

We can't ask them but the same principles apply

@Slereah Do you buy the relational QM argument that a measurement device gets entangled frm our perspective, but collapses the wavefunction from its own perspective? This treats consciousness as just like everything else.

Maybe?

idk

It's one of those idea that's hard to really decide on

Yeah, this is a pretty neat way to begin to resolve the measurement problm. I run into a few big issues tho

which particle's perspective do we humans inherit? We do have a perspective as we observe selected outcomes. It is reasonable to assume that the particles in our lower body simply get entangled. So do we inherit the perspective of some particles in our brain

Another problm i find is that particles can get destroyed. So does the perspective gets destroyed?

And the last problm i find is that we can't verify that a measurement device has a perspective. A minimalist theory shud just assume what we can verify that is sufficient to explain experiments.

We can only verify our own perspective. And this is sufficient to explain experiments

These r the three problms i run into @ACuriousMind @Slereah

@PM2Ring is this true for molecular orbitals also?

Is there any proper accounts of the energy-time uncertainty relation in relation to virtual particles without any handwaving?

$|\tilde{k}\rangle=\hat{S}(t,-\infty) \hat{a}^\dagger (k) |0\rangle$

The above seems to be a perturbative expansion of a one particle state at time $t$

If I take the first order expansion, I will have $|\tilde{k}\rangle= |k\rangle - \imath g \int \mathrm{d}^4 X \hat{V} |k \rangle$

You know when you're so tired that you're not even convinced you're awake

Like it could just be a dream

What is wrong with my first order expansion above?

That integral will contain c/a operators such that the correction to $|k\rangle$ will be a multiparticle state

The other thing is that people say that only external lines correspond to real particles but they aren't even real particles since they aren't eigenstates of the interacting Hamiltonian

@DIRAC1930 why do you think that "real particle" means "eigenstate of the interacting Hamiltonian"?

the point of the asymptotic treatment is that we really only know how to define a particle in the non-interacting case

@DIRAC1930 see physics.stackexchange.com/a/691502/50583

But then how can we make the assertion that virtual particles aren't real if we don't know what the states are. If I do a standard perturbative expansion of the interacting states based on the free ones, I will get multiparticles states

@DIRAC1930 because the "virtual particles" aren't states

the lines in Feynman diagrams that we call "virtual particles" are, at best, a sum over intermediate states

So a standard perturbative expansion of the vacuum state has nothing to do with virtual particles

see this answer of mine for more on virtual particles

Hmm so I think I see my confusion

Is it correct to say that the perturbative expansion of the vacuum using QM perturbation theory is not describing virtual particles and those are actual states that are corrrections to the vacuum

what do you mean by "perturbative expansion of the vacuum"?

en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics) time independent pertubation theory

So the case where we are not adiabatically introducing the vacuum

on a technical level, I would not trust any methods that rely overly much on the idea that the free and interacting vacuum lie in the same Hilbert space

even the usual derivation of LSZ via the interaction picture is suspect due to Haag's theorem (but the result works nonetheless and can be made rigorous), I don't see anything good coming out of this kind of expansion

remember that in the usual approach the vacuum energy turns out divergent anyway and needs to be renormalized

so what are you doing computing "energy corrections" via perturbation theory? The result will just be infinity

while quantum field theory is a generalization of QM, many standard QM methods are just somewhere between too complicated and useless when you try to apply them to QFT

I'm not computing energy corrections, just the state $|n^{(1)}\rangle$ using the notation on that wikipedia page

why?

again, you already know from the standard approach the vacuum energy turns out infinite so this kind of expansion doesn't really make a lot of sense

if there existed a convergent expansion of one vacuum in terms of the other, then the vacuum energy would be finite

IIRC there is a model called the Lee model where you can renormalize based on the above QM perturbation expansion

Weinberg wrote a paper in 1947 or sometime around then I think

1956 maybe but the details aren't important

@DIRAC1930 ...and is this Lee model a relativistic QFT?

this sounds like another case where you pretend you're asking a general question about QFT but you actually have a very specific application in mind where the general answers I give may not necessarily hold

this isn't a very efficient way of communication if you only explain the specific application you're interested in at the end of the conversation :P

My background makes me extremely skeptical when we can't even write a state

a lot of QFT is indeed techniques developed to get meaningful answers from the theory even though we don't really know what the space of states of the interacting theory looks like

that's not something to fight against, that's the whole point: The interacting theory is hard, essentially intractable in the generic case

if you want to investigate the interacting theory in detail you need to use techniques adapted to the specific theory at hand, e.g. lattice models to study the realm of low-energy QCD

you don't need to like this, but the fact of the matter is that this standard approach to perturbative QFT works

if you want an expression of the interacting vacuum in terms of the free vacuum (however questionable due to Haag), the default derivation of the Feynman rules produces $\lvert \Omega\rangle = \lim_{t\to\infty} \frac{\mathrm{e}^{\mathrm{i}Ht}\lvert 0\rangle}{\mathrm{e}^{\mathrm{i}E_\Omega t}\langle \Omega \vert 0\rangle}$

this is a horrible expression

@nickbros123 Sure, it applies to QED in general. But please bear in mind that I said "roughly speaking". ;)

So just to recap, does the time energy uncertainty principle fit into all of this or is that another misunderstanding?

In some limited contexts, it can make sense to compute corrections over vacuum energy. e.g. - physics.stackexchange.com/questions/185171/…

But they r not computing corrrections due to another interacting field. It's smthing simpler

They hav also set the vacuum energy of the free theory to be 0

@DIRAC1930 the understanding that : "virtual particles exist for small periods of time" is false

So is there a correct interpretation of the energy-time uncertainty principle?

Yes. The answer here : physics.stackexchange.com/questions/53802/…

This answer on that page seems to be useful physics.stackexchange.com/q/222385

Yes. Also read the answer by Misha

So in perturbative treatments of QFT, this $\Delta t \rightarrow \infty$. Is that correct?

That formula is in Landau Lifshitz QM so I might look there

I dont understand. How r u interpreting this uncertainty principle in the context of perturbative qft?

@DIRAC1930 the interpretation in this answer is for a very specific application

Yes I was getting confused

Well in QFT, the interaction is introduced adiabatically from $t=-\infty$ to $t$

However I wonder if there is a general statement in terms of general QFT, i.e. starting with a fully interacting theory, assuming we can write particles states, and then deducing something from the the E-T uncertainty principle.

Well it would probably be the same for non-rel QM scattering so maybe there is something

The interpretation in Joshphysics's answer is general

It can be used to make quantitative deductions. But i wud say this uncertainty principle is not that fundamental. It's of limited importance

Doesn't this mean that the interaction must happen over some length of time for there to be a change in the state?

whatever happened to john duffield

was just wondering about the night sky

Does Wick rotation even make sense outside of an integral? I have a source claiming to Wick rotate $t=-i\tau$ in $\langle q_f,t_f\lvert\hat{A}\lvert q_i,t_i\rangle=\sum_{n,n'}\langle q_f,t_f\lvert n\rangle\langle n\lvert\hat{A}\lvert n'\rangle\psi_n(q_f)\psi^{\ast}_{n'}(q_i)e^{\frac{i}{\hbar}E_n't}e^{-\frac{i}{\hbar}E_nt}$ to get decreasing exponential and consider only the lowest energy exponential term $\propto e^{\frac{i}{\hbar}E_0(\tau_f-\tau_i)}$

@RyanUnger physicsdetective.com

Still kicking it seems

I think maybe in many non-rel systems, particle number is conserved therefore the field operator $\hat{\Phi}^\dagger$ acting on the ground state $|G \rangle$ (which is not the vacuum) has a physical interpretation of increasing particle number by $1$. Am I correct in saying that this is not true in rel systems i.e. it is difficult to give an interpretation to the Heisenberg field operators since they may create a superposition of states, some of which are many particle etc.

have any of ya'll read through moretti's spectral theory and quantum mechanics text? it looks quite nice :)--I am trying to decide what book to work through :P

@DIRAC1930 Did you read the first section of L&L4 on the relevance of time-energy or their section on virtual particles later on

trying to figure out if you can use mereotopology in physics

it is a challenging issue

Where is the section on time-energy uncertainty?

there seems to be mereology being used in tandem with geometry, and i naively assume where there's geometry there can probably be topology but idk

I mean it's trivial to do with geometry

The hard part is only doing it with mereology

But since there aren't points in mereology you can't really use $\mathbb{R}$

So most of the usual geometry goes out the window

Can't even easily define distances since there's no lines