The h Bar

DIRAC1930

00:24

@ACuriousMind But when people were doing scattering calcs (before all of this mathematical analysis) in the 50s or 60s, there must have been a physical assumption why asymptotic states are free. Or was it moreso the case that it works?

ACuriousMind

@DIRAC1930 Compare this to "old quantum theory", like Bohr-Sommerfeld quantization

it also doesn't really "make sense"

but it worked in providing some sort of framework to explain atomic spectra

if you expect the initial versions of physical theories to make rigorous sense, you're not really engaging with how physics works in practice

we live in a sense in a golden age where we have decades old theories to explain every practical observation you might make

DIRAC1930

Yes but usually where there are inconsistencies, people don't ignore them. This seems like a massive issue that people just ignore

Obliv

Any chance the all knowing ACM knows the zero-input and zero-state response of a system $$Y(s) = \frac{Q(s)}{P(s)} + \frac{G(s)}{P(s)}$$ or equivalently $W(s) = \frac{1}{P(s)}$ the transfer function

ACuriousMind

@DIRAC1930 issues with QFT are not being "ignored" - that's why there is stuff like the Yang-Mills millenium problem!

it's just that physics, in practice, does not need completely rigorous foundations

you can dislike that attitude

Obliv

I know $G(s)$ is derived from $\mathcal{L}\{g(t)\}$ from the linear ODE with const. coefficients $a_n \frac{d^n y}{dt^n} + ... + a_0 y = g(t)$ I was hoping u could confirm that $P(s)$ is from the laplace transform $$\mathcal{L}\{\frac{d^n y}{dt^n}\}$$ and $Q(s)$ is from $$\mathcal{L}\{\frac{d^{n-1} y}{dt^{n-1}}\}$$

DIRAC1930

00:31

Why can't we use the Gellmann formula for localized wavepackets? I'm assuming there is a reason why

ACuriousMind

but in the current state of affairs that's much more a you problem than a problem most practicing physicists will care about :P

DIRAC1930

Lol

ACuriousMind

the physical argument for why asymptotic states are free is just physical: When particles are far apart, the effect of them interacting with far away stuff is negligible

DIRAC1930

But I thought we are dealing with fields and not particles

ACuriousMind

Haag's theorem has shown this to be rigorous nonsense for decades, for decades most physicists have regarded Haag's theorem as a mathematical curiosity rather than some sort of fundamental problem

DIRAC1930

00:36

I have a feeling this $\sin^2$ formula will solve everything

ACuriousMind

@Obliv I am afraid I don't really know anything about linear response theory beyond that it exists :P

Obliv

It's okay, I'm 99.9% confident it won't be on the test, just that I can use transforms to solve them

it was a remark in my book

DIRAC1930

If I start with a free particle and turn on the interaction immediately and I wait a long time, won't I get a dressed particle with the same energy if I wait

ACuriousMind

@DIRAC1930 I don't really know what that means, and I think that's at the heart of why most people don't think this is some sort of fundamental problem: In nature, interactions aren't "turned on"

the very act of thinking about this as an interaction being switched on is just a convenient fiction

i.e. it's an approximation anyway, but one that delivers experimentally very accurate results

so trying to put this on some sort of rigorous foundation is of questionable benefit

DIRAC1930

Sorry I meant if I view the problem using standard QM perturbation theory where V is the interaction operator

ACuriousMind

00:41

I mean isn't that what the usual LSZ derivation does when it switches to the interaction picture?

DIRAC1930

I haven’t looked at the LSZ derivation

ACuriousMind

...you've been trying to talk about the foundations of QFT without looking at the standard derivation of LSZ?

that's the central point of most intros to QFT

Obliv

Almost as important as the derivation of wumbo

ACuriousMind

how would you get to Feynman diagrams without doing either LSZ or functional derivative acrobatics in the path integral formalism?

DIRAC1930

I mean Feynman diagrams are used in calculating the 2pt functions. QFT isn’t just scattering. You don’t need LSZ to do the Lamb shift calc

DIRAC1930

01:02

There en.wikipedia.org/wiki/Old_quantum_theory#Limitations limitations seem just as bad as the ones in QFT

Leaky Nun

02:13

If the quantum wavefunction collapses simultaneously everywhere... but simultaneity is relative... then don't we have a paradox?

John Rennie

05:56

@LeakyNun See:

7

Q: Wavefunction collapse in relativity

It is well accepted that quantum theory has well adapted itself to the requirements of special relativity. Quantum field theories are perfect examples of this peaceful coexistence. However I sometimes tend to feel little uneasy about some aspects. Consider an EPR pair of particles light years apa...

Mr. Feynman

06:18

@ACuriousMind what really happened: I was going to bed and checked the hbar. I read your message only and without any context I decided to be furious :P

bolbteppa

06:42

@DIRAC1930 The result I keep referencing was found in the 1930's

Even in LSZ you assume at the beginning and end the particles are free and derive a formula from it, your stuck on why we make this assumption in the first place, the obvious thinking is that 'things far away don't interact', but the fundamental reason why this is the only thing we can really do in QFT consistently traces back to that first section

bolbteppa

07:24

you're*

Ryder Rude

12:19

Hello

@LeakyNun u shud think in terms of relational quantum mechanics. The wavefunction can only b collapsed relative 2 u. Any other observer goes in a superposition. They never collapse anything

nickbros123

Hello, I don't quite understand how, when converting the formula for work done to assemble a collection of discrete charges into a collection of continuous charges, the issue of self energy comes up (ie, the discrete sum has no inclusion of self energy whereas the integral form has the self energy inclusive)

Ryder Rude

And the wavefunction is not a physical wave @LeakyNun . It is only relational information. So, someone else can collapse the wavefunction simultaneously everywhere, but u wud never need 2 describe that collapse in ur reference frame. That collapse happened only relative 2 that observer

In ur description, that collapse nevr happened. The measurement device just went into a superposition. Collapse doesnt happen objectively 4 everyone

This is becuz the wavefunction isnt a physical wave or anything. So there is no issue with "simultaneity of collapse"

ACuriousMind

15:23

@nickbros123 I'm not quite sure what you mean, could you give some more details?

Mr. Feynman

15:40

@nickbros123 do you mean $\frac{1}{2}\sum_{i\neq j}\frac{q_i q_j}{|r_i-r_j|}\to\frac{1}{2}\int d^3\vec{x}\rho(\vec{x})\phi(\vec{x})$?

nickbros123

15:55

@ACuriousMind the equation @Mr.Feynman sent, this is the discrete sum that determines the work done to assemble a bunch of discrete charges. Whereas the integral written, is supposed to also represent this quantity, but when I extend this integral to give me $ W=\frac{\epsilon}{2} \int E^2 d\tau $, this new integral also includes the self energy of the charge distribution

Griffiths speaks about energy needed for "making" the point charge, which seems to have been included in the integral, but it's not obvious to me how it is

ACuriousMind

I'm not sure what exactly the question is

it is obvious to me that the discrete version explicitly excludes the term that would be the $i$-th potential acting on the $i$-th charge, but the continuous version has no "single charge" and hence no analogous omission of self-energy

this answer here shows that the discrete version essentially already subtracted out the "infinite self-energies" you'd get if you tried the continuous version for a point charge

maybe this is your confusion: These expressions are not equivalent! The continuous version computes a total energy, while the discrete version computes only an interaction energy (because the self-energy is divergent hence useless for point charges)

Mr. Feynman

16:11

It's worth adding that the infinity we are neglecting in the discrete formula is somehow trivial, in contrast to other divergent quantities arising when you consider a point charge and Maxwell equations

On a second thought, no. It's not trivial at all. It's precisely the same infinity that leads to the electromagnetic mass but tackled using a different path

Leaky Nun

17:00

@RyderRude "wavefunction is not physical" is not a universally accepted interpretation

it is the most widely accepted though

Obliv

I feel like I'm missing something. how do I do $$\mathcal{L}^{-1}\{\frac{1}{s^2 + 3s}\}$$ I tried factoring, partial fractions, completing the square

they only have linearity property afaik

Leaky Nun

partial fractions is linear

it's literally a sum

@JohnRennie hmm, that's interesting, thanks

Obliv

So like I did $\frac{1}{s(s+3)} = \frac{A}{s} + \frac{B}{s+3}$ ?

Leaky Nun

i guess the takeaway point is that the wavefunction collapse is instantaneous in the observer's reference frame

@Obliv yeah

Obliv

Oh

so A = 1/3 and B = -1/3 got it

Leaky Nun

17:08

I have a question about photon which is related to my confusion with the uncertainty principle; I'm aware that no actual contradiction is present here, but I'm struggling to understand the relation between position, velocity, and momentum; say we're measuring the speed of light, so we emit a photon and reflect it with a mirror and receive it, and measure the time difference;

but since the photon has a high uncertainty of position, we don't actually know where the photon is, so consequently we can't quite know when we measured it?

ACuriousMind

@LeakyNun a highly uncomfortable question, since photons don't actually have "position"

Leaky Nun

@ACuriousMind so what's the uncertainty principle?

ΔxΔp doesn't apply to photons?

ACuriousMind

true position operators only exist in the non-relativistic limit, so inherently relativistic object like massless particles never have good position operators

@LeakyNun the general uncertainty principle still holds - for any two operators $A$ and $B$ you have that $\sigma(A)\sigma(B) \geq \frac{1}{2}\langle [A,B]\rangle$

Leaky Nun

so that's what I don't understand about c; is c not the distance travelled divided by the time taken?

ACuriousMind

it just doesn't make sense to talk about "position"

@LeakyNun $c=1$ in my world, I don't understand the question ;)

Leaky Nun

17:12

sure, what does that mean then

Slereah

@ACuriousMind I guess technically classical theories have massless particles, but then they're confined to a single spatial slice, and therefore not very well localized either

Leaky Nun

if velocity is well-defined, and time is well-defined, then position must be well-defined right?

ACuriousMind

@LeakyNun sure, but what is "velocity"?

Leaky Nun

that's what I'm asking

ACuriousMind

I would argue it's just ill-defined when talking about a single photon

Leaky Nun

17:15

:o

ACuriousMind

remember: it's the speed of light that is "$c$", not "the speed of a single photon"

Leaky Nun

what does that mean?

it's like current vs drift velocity?

or water vs water molecule?

ACuriousMind

a classical light wave is a coherent state of indefinitely many photons, this wave has amplitudes (the expectation values for $\vec E$ and $\vec B$) and hence a velocity (defined as usual for classical waves)

it is that velocity that we mean by the speed of light

Leaky Nun

why doesn't that work for one photon?

ACuriousMind

the $\langle \vec E\rangle$ isn't the classical electromagnetic wave you want

or, rather, it's not clear that the "speed" of this wave has something to do with the "speed" of the photon

Leaky Nun

17:20

my classical brain is failing me; do you have a classical analogy?

ACuriousMind

that's not special to photons: you may try to associate some sort of speed to wavefunctions in position space if they look like waves, but there is no inherent guarantee that this has something to do with the "physical speed" $v = p/m$ of the particle described by that wave

Mr. Feynman

@ACuriousMind You once told me you don't like the way physicists deal with gauge theories. Is there any answer of yours discussing why?

Leaky Nun

@ACuriousMind oh no you're dividing by zero

ACuriousMind

@Mr.Feynman I probably was referring to this answer - most texts seem to think it's a good motivation for gauge theories that we want to "make symmetries local", I think that's nonsense and I find it surprising how widespread it is

@LeakyNun I'm saying this doesn't even work for a massive particle

you can't look at the particle's wavefunction and derive some sort of statement about its speed from the "speed" with which the wavefunction spreads

the "speed in position space" of the wavefunction and the "physical speed" are not necessarily correlated

but, well, for a free particle you can show that the group velocity of a free wavefunction is indeed $p/m$

Leaky Nun

is photon free?

ACuriousMind

17:25

so if you say that that velocity is the velocity of a particle in general, then sure, the photons velocity is $c$

but since the photon has $m=0$, there is no relation between this velocity and the momentum, so in fact your experiment has two problems: Not only is there no position operator, the velocity you're computing doesn't have anything to do with its momentum, so even if there was a position operator you wouldn't get a contradiction to the uncertainty principle

Leaky Nun

58

A: What equation describes the wavefunction of a single photon?

There is no quantum mechanics of a photon, only a quantum field theory of electromagnetic radiation. The reason is that photons are never non-relativistic and they can be freely emitted and absorbed, hence no photon number conservation. Still, there exists a direction of research where people t...

is this what you're talking about?

Mr. Feynman

@ACuriousMind oh, the "fetish" answer. I read it a bunch of days ago when Nihar Karve linked it. The way I see the passage from a global to a local symmetry is not as the purpose but as the tool by means of which we reach a theory that describes our world properly

Anyways, I thought your problem with Physics books was not mentioning associated bundles etc.

ACuriousMind

that's fine at an intro level

but they should talk about the Hamiltonian formulation more

because standard treatments leave people with this idea that gauge theories are these weird spacetime-dependent symmetries - I think the Hamiltonian version makes it much clearer what the gauge symmetry actually is: It's an artifact of us using more variables than the underlying physical system has degrees of freedom

Mr. Feynman

Oh, I see

Leaky Nun

@ACuriousMind this is very dank

ACuriousMind

17:31

@LeakyNun yes - non-existence of the photon wavefunction is a direct consequence of the non-existence of the relativistic position operator

Mr. Feynman

@ACuriousMind The books I've read up to this moment make this sufficiently clear or maybe that's the effect of courses and books and SE

Without mentioning the hamiltonian version and Dirac's theory, though

ACuriousMind

@LeakyNun and in fact there's a third problem with your experiment: If we're thinking about individual photons, you'd have to think about the reflection of the photon in terms of it being absorbed and then there's an emission of a photon at the reflected angle after that.

So some of the time while your clock is running there is no free photon at all, so even if the velocity had something to do with the momentum you'd be uncertain as to how your measured time actually relates to the velocity since the exact timing between absorption and emission is unknown to you

really the lesson here is: you can't think about photons as particles like you can with electrons

sometimes it works, but generally you're just producing confusion

Leaky Nun

so the speed of light is something that cannot be derived from each individual photon?

@ACuriousMind then I don't really understand how we can measure the speed of light without accounting for the "time it takes for the mirror to reflect the photons"

ACuriousMind

@LeakyNun you can't!

I don't think anyone would try to use your setup for precision measurement of the speed of light

Leaky Nun

what's your favourite way to measure the speed of light?

ACuriousMind

17:43

why would I have a favourite way :P

Leaky Nun

just any way then

because I don't see how you can measure the speed of light without a mirror

ACuriousMind

and actually "measuring the speed of light" is a somewhat weird notion ever since SI defined the meter via the speed of light

Leaky Nun

bruh

ACuriousMind

any "measurements of the speed of light" are now technically more like calibrations of our length scales

@LeakyNun why do you think we need a mirror?

Leaky Nun

@ACuriousMind because you can only ever measure the "two-way speed" of light, i.e. a round trip

ACuriousMind

17:47

oh, you want to be annoying about it :P

but I think the way with the least ambiguity is via interferometry

cf. en.wikipedia.org/wiki/Speed_of_light#Interferometry

Leaky Nun

bruh you got a mirror there

ACuriousMind

no, there's a beam splitter, not a mirror

Leaky Nun

there's clearly a mirror after the beam splitter

and also beam splitter works in a similar way right

I guess my question now is, since absorption and emission take time, and are necessary for any measurement, how can you have "the speed of light"?

ACuriousMind

oh, but we're not timing anything here

this is a measurement of the wavelength of light, we're computing the velocity by using light with a known frequency

so any absorption/re-emission shenanigans don't matter

Leaky Nun

you're just not timing it with a clock; you're still making an assumption that the light waves are reflected instantly, so that the waves can combine in or out of phase

ACuriousMind

17:55

@LeakyNun I don't think there's any such assumption here

the way this works is that we adjust the path length until we get maximally constructive interference, then measure by how much we have to adjust it until we get maximal destructive interference

whether the multiple reflections very slightly phase-shift one of the beams doesn't matter because this is a relative measurement - we don't care about the total path length, just about the difference between constructive and destructive interference

Leaky Nun

but the emission time is random so you need to take an "average" phase shift?

ACuriousMind

I mean that's another problem - you can't think about this experiment in terms of individual photons because they have no definite phase

the phase operator is definite on the coherent state associated with a classical EM wave, and there the number operator is indefinite

you either have "definitely a single photon" or you have "definitely a sharp phase"

which is consistent with my previous claim that the idea of associating a "velocity" with an individual photon is ill-defined

Leaky Nun

18:18

QM is really weird

ACuriousMind

I don't disagree :)

Obliv

19:12

Can some1 help me understand how $$f(t) = \begin{cases} g(t), & 0 \leq t < a \\ h(t), & t \geq a \end{cases}$$ is the same as $$f(t) = g(t) - g(t)\mathscr{U}(t-a) + h(t)\mathscr{U}(t-a)$$

$\mathscr{U}(t-a)$ defines $\begin{cases} 0, & 0 \leq t < a \\ 1, & t \geq a \end{cases}$ I think

Oh wait I get it nvm

becomes $g(t) - g(t) + h(t)$ and $g(t) - 0 + 0$

user 85795

19:52

(?)

DIRAC1930

21:12

@bolbteppa Thanks, I'll have to read and think about the first chapter of L&L 4 more carefully

Ryder Rude

21:44

@LeakyNun yes, i think the only r8 answer is that we do not understand quantum theory yet. Ultimately, a mathematical modification should come that would makes things clearer. Until then, we hav to go by one of the practical and mathematically consistent interpretations. Relational QM avoids issues with the "simultaneity of collapse" stuff, which is y it is elegant as a temporary understanding of quantum theory

DIRAC1930

22:22

I wonder if it's best just to accept in the L&L sense that wavefunctions become ill-defined in the high momentum limit and instead expand the free field operator in the position basis i.e. $\hat{\Psi}^\dagger(X)=\int \mathrm{d}X \psi^*(X) \hat{X}$ where $\hat{X}$ creates a one-particle state with the wavefunction $\psi^*(X)$.

On a similar note, I was thinking that if we put the QFT in a box with dimension $L^3$, we will have the wavenumber $k$ being discrete and quantized. Therefore, we could maybe invoke the adiabatic theorem since we have discrete eigenstates and perhaps then take the limit of $L\rightarrow \infty$

Obliv

Why in the very last part of the proof does the $\int_0^{\infty}e^{-sv}f(v)dv$ become $\mathcal{L}\{f(t)\}$ again

shouldn't u put $v = t - a$ back in to get $e^{-as}\mathcal{L}\{f(t-a)\}$

limits are defined in $v$

DIRAC1930

22:49

Landau writes something interesting
'Ths situation, however, becomes clear if, as is usual in theoretical physics, point interaction is regarded as the limit of some' distributed' interaction.'

This is in the Pauli Memorial Volume: Theoretical Physics in the 20th Century

Landau wrote an article that refers to L&L 4 chapter 1 titled 'Fundamental Problems'

I don't know if this is a good way of thinking about this. If energy conservation can only be verified in the limit of $\Delta t \rightarrow \infty$, then the Hamiltonian describing the experimental predictions can only be verified in the limit of $\Delta t \rightarrow \infty$ therefore only transitions from $t=-\infty$ to $t=+\infty$ can be reliably associated as arrising from a theory with a given $\hat{H}$

@bolbteppa What do you think of the above?

DIRAC1930

23:11

Then if energy conservation can only be verified in the limit of $\Delta t \rightarrow \infty$, then the same goes for symmetries

But this doesn't help the fundamental assumption that particles are free in the limit of $t\rightarrow \pm\infty$

Free particles are mathematical fiction. Dressed particles are what we observe in nature

There must be a reason why we can take this limit

I think Landau's statement 'Ths situation, however, becomes clear if, as is usual in theoretical physics, point interaction is regarded as the limit of some' distributed' interaction' may have something to do with it

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