The h Bar

Bohemian relativist

01:25

how different is the gauge theory based on the invariance of spacetime translation from the orthodox gauge theory like the Yang-Mills theory?

Marco

04:58

Hi, how can I plot a gradient field given the modules and phases of each gradient?

Nihar Karve

06:11

@Charlie do you even sleep

Charlie

10:02

@NiharKarve Huh?

Nihar Karve

@Charlie oh nothing lol I just see you commenting at like 2:00 AM or something

John Rennie

Moved

Jack Zimmerman

11:08

1

Q: Is $\zeta^{3,1}\simeq \Bbb M^{3,1}?$ Why or why not?

A semi-riemannian manifold $\zeta^{3,1}:=\zeta^{1,1}\times \zeta^{1,0} \times \zeta^{1,0}.$ I calculated the non degenerate product metric of it: $ds^2=\frac{drdt}{rt}+\frac{du}{u}+\frac{dv}{v}.$ Previously I had studied $\zeta^{1,1}$ and came to understand that it is diffeomorphic to the Minkows...

any help would be nice

Clara Diaz Sanchez

Love today's question about "drawing lines of worst fit". I'm grading student lab reports today, and I think that is one skill they exhibit to perfection ;)

Jack Zimmerman

:)

123

11:21

Hello World...

Can we define period of trig functions as integer value.

Sin(x + 2kpi) = sinx

2kpi is period. Is it correct use k as integer for period

Charlie

Yes

The "period" is not an integer though, you are multiplying $\pi$ by an integer and that is the period of the sine function

123

Thanks @Charlie

Nihar Karve

14:22

Did I just game the system? I got 201 rep today

I downvoted something then undownvoted it

bolbteppa

1

A: Wigner vs. BRST approach to Klein-Gordon

The faulty assumption is that the elements of the BRST cohomology can be represented by the elements of the unconstrained Hilbert space. They are really not, which means that one can't borrow the definition of the unconstrained inner product and use it. One has to invent another definition for th...

ACuriousMind

@bolbteppa ?

bolbteppa

The answer seems to be arguing that because the wave functions aren't normalizable i.e. plane waves don't have finite norm they're not in a Hilbert space so it means something, but continuous spectrum non-normalizable wave functions are everywhere e.g. the hyperbolic (scattering) orbits of the hydrogen atom, is the point just to say one has to use a different inner product due to this or something more

ACuriousMind

@NiharKarve yes, enjoy your free point :P

Nihar Karve

I will relish it

HNQ-baiting is so fun

Charlie

14:38

I think I've asked something similar before, but when we use the pole prescription in calculating QFT propagators, once we've pushed the poles out of the way with the $i\epsilon$ prescription, why can't we just do the integral along the real line, why do we have to close the contour around the top or bottom?

It is because once we take the $\epsilon\rightarrow0$ limit it will let us evaluate the integral through the poles?

bolbteppa

Pole prescription doesn't make sense if you're not doing a contour integral

Nihar Karve

@Charlie I might be wrong here, but the whole point of the contour integration in this case is that the real-line integral is hard. Instead, you perform the (easier) contour integral, notice that the curved bit has some known limit as you take the radius to infinity and end up with the integral along the real line (since the entire contour integral can be evaluated using the usual techniques of complex analysis, particularly simple if there are no poles enclosed)

Charlie

I don't see why the pole prescription doesn't make sense otherwise, I don't see why you can't just (as P&S put it) rotate the line along which we integrate slightly out of the real line into the complex plane, and evaluate that

We don't have to close a contour when we integrate along the real line, do we?

Nihar Karve

not usually, but often the integral along the real line resists the techniques of elementary calculus

Charlie

No surely not otherwise every integral of a holomorphic real function would vanish

so why can we not just use the pole prescription, move the poles off of the real line, and integrate along the real line then take the $\epsilon\rightarrow0$ limit?

without having to close our integral in the top or bottom half of $\Bbb C$ and pick up residues of poles

Nihar Karve

14:45

try doing it using elementary calculus techniques

it's pretty much impossible

Charlie

I could believe that, but still don't see why we must close the contour, surely we can just integrate along an infinite line rotated slightly into the complex plane, just like we do with the real line, and not bother closing the contour

I guess I don't know enough C-analysis, I hoped that wouldn't come back to screw me but apparently it has lol

Nihar Karve

I mean, closing the contour is the only way you get to use stuff like the residue theorem et al. which makes your life a whole lot easier

Charlie

Oh I guess so, maybe the slightly rotated integral is just unjustifiably hard to solve by itself

so we close the contour and just make it about knowing the residues of the poles

ok I guess i can live with that for now

Nihar Karve

I actually think $\int \frac{e^{itx}}{x^2+a^2} \mathrm{d}x$ is given as the prototypical example for a lot of complex analysis concepts

ACuriousMind

@Charlie ...how would you compute that integral without closing the contour?

bolbteppa

14:50

Yeah how would you just do it directly even if you could just insert a parameter to remove the problem with poles

Charlie

hmm

ACuriousMind

the trick is that the contribution of the path that "closes" the contour becomes zero if you push it out to infinity, so you can use the residue theorem to evaluate the original integral

Charlie

I guess I wouldn't personally know how, if doing so is a problem in principle then I guess that justifies closing the contour

ACuriousMind

there's no physical meaning to closing the contour, it's just how one computes that integral

the trick is purely mathematical

RewCie

What happens when you are overthinking and it's positive?

Sounds overthinking becomes addiction

Charlie

14:56

What do you mean "and it's positive"?

RewCie

The more you overthink, you feel motivated, better.

Charlie

Why would overthinking motivate you?

RewCie

No Idea...

It just does

Charlie

Then I don't know how to answer :P

RewCie

Cool

Charlie

14:58

People who overthink are often made very anxious by uncertainty

and relentless planning for the future is a sort of brute-force way of alleviating that

bolbteppa

@Charlie Feynman figured out how to deal with this in position space physically, see the argument around equation (17) here, then the momentum space pole prescription is configured to arrive at Feynman's position-space result (see around eq. (31) on)

Charlie

oh ty

ACuriousMind

15:53

@Aridhan Please do not advertise your questions here directly after they are posted; people that are interested in answering them will watch the main site anyway.

Ryan Unger

16:18

@ACuriousMind how do you know what has physical meaning and what doesn't?

ACuriousMind

@RyanUnger you never know for certain, but I've never seen any interpretation for that

@Charlie meanwhile, the shifting of the poles can be interpreted physically as introducing "damping", see physics.stackexchange.com/a/138221/50583

user 85795

Unrealistic results have no physical meaning :p

ACuriousMind

@user85795 ...and what result here is "unrealistic"?

user 85795

...in general

to answer:

11 mins ago, by Ryan Unger

@ACuriousMind how do you know what has physical meaning and what doesn't?

:-)

danielunderwood

I've always heard of wavefunction collapse from measurement introduced as "that's just the way it is" but I was thinking about it recently. How is wavefunction collapse not made necessary by classical/quantum correspondence (and potentially relativity)? If you're going to measure some observable and have it be consistent with a classical measurement, I don't see how you could get around the quantum state going there for at least some infinitesimal amount of time

Nihar Karve

16:32

@user85795 but it's not a realistic vs. unrealistic scenario, it's just whether a (correct) mathematical manipulation has an associated physical meaning

ACuriousMind

@danielunderwood wavefunction collapse is a property a quantum interpretation may or may not possess, it is not an objective feature of quantum mechanics

Wiki has a handy but potentially imperfect table here

2

danielunderwood

wow there are a lot more interpretations than I thought

I don't see how something like MWI could be consistent with a classical measurement, but perhaps I just need to read into the interpretations a bit more

user 85795

a curious mind is always on the look out for interpretations :-)

and, of course, its associated logical reasoning

Charlie

16:47

If MWI is consistent with quantum measurement it is automatically consistent with classical measurement no?

bolbteppa

It's not even clear (to me, anyway) if MWI is consistent

danielunderwood

@Charlie I think that's where my problem lies. In my head, it seems like you need wavefunction collapse for quantum measurement and classical measurement to be consistent. But if it's a property of the interpretation, then either that's not the case or I've ruled out a number of interpretations (unlikely)

@bolbteppa in terms of the interpretation of a whole or is a question of a specific part of the interpretation?

bolbteppa

The whole interpretation

ACuriousMind

17:03

@danielunderwood what do you mean by "quantum measurement" and "classical measurement"

the world is quantum, all measurements are quantum :P

@user85795 no, I find most of the discussion around interpretations rather pointless

danielunderwood

So part of my issue may be quantum measurements on a macroscopic scale, as we would have with a measurement device.
But my thought is that classically, measuring the position of something yields some number $x$ (within some error). If we make a subsequent measurement at time $t + dt$, the object has to be within $x \pm c dt$ to be consistent with relativity. If the wavefunction didn't collapse from this measurement, you could get something that is consistent with the Born rule, but not with the classical measurement

Though I suppose that could also be a problem that's fixed in relativistic quantum mechanics

ACuriousMind

@danielunderwood Collapsing the wavefunction is not the only way to get consistent results, see e.g. MWI

You're probably confusing the notion of wavefunction collapse with the technical statement that after a strong measurement the system is, for practical purposes, in an eigenstate corresponding to the measured value

danielunderwood

hmmm could be

Unfortunately all of my exposure to non-Copenhagen interpretations has been on the level of popsci

Before I go digging into random resources, are there any standard technical resources (lecture notes/review articles) on interpretations?

ACuriousMind

17:19

it's not really a large field, and I don't think there are many lectures on quantum interpretations apart from this one

Apr 3 '17 at 14:19, by Emilio Pisanty

Rob Spekkens - Foundations of quantum mechanics - QuICC Lecture 1

►

bolbteppa

There's plenty of perimeter lectures on interpretations as well

ACuriousMind

I would also wager that your exposure to Copenhagen has also been on the level of pop-sci because it's not really a codified interpretation :P

danielunderwood

Yeah I don't know that the word interpretation was ever mentioned in any of my QM classes

Perhaps I should check philosophy departments :D

ACuriousMind

since - apart from outliers like Bohmian mechanics that revise the whole formalism - they all share essentially the same technical underpinnings there isn't really a lot of "advanced" stuff you could do

I'd call quantum interpretations more metaphysics than physics, really, except for the few cases where they actually modify the predictions of standard QM

Charlie

17:52

It had never occurred to me that the rational numbers are a countable set that is also a field, my mental image of what a "field" is has been totally corrupted by physics

omg you can have fields consisting of as few as two elements, mind blown

bolbteppa

The reals are just one 'completion' of the field of rationals, by modifying the notion of closeness one can get p-adic numbers instead of real numbers

Arthur Morris

@Charlie About closing the closing the contour: the integral directly along the real line is not well defined, so you have to take the Cauchy principal value instead. It is possible to compute this without using complex analysis (see en.wikipedia.org/wiki/Cauchy_principal_value), but closing the contour in the complex plane provides a direct way to do it using the residue calculus which is much more elegant. That's all there is to it!

Not well defined due to the poles of course

Charlie

18:07

Ty, I see why it's advantageous to do things this way if the contribution of the arc closing the contour vanishes as it's radius becomes arbitrarily large

Ryan Unger

@ACuriousMind why aren't virtual particles physical?

ACuriousMind

@RyanUnger see physics.stackexchange.com/a/275099/50583

bolbteppa

19:20

It seems like wave function collapse is supposed to be an almost cartoon picture where one has $\psi \to \psi_i$, this seems to be the wiki claim and (as the link seems to say) it's apparently what von Neumann did, it really doesn't make sense to me though, and this is not what Copenhagen says:

Copenhagen says initially the combined wave function of the measuring apparatus (quasi-classical wave function) and quantum system is a product $\Psi^{sys}(x) \Phi^{app}_0(y)$ as they're initially independent, then the measurement takes place meaning they interact. Afterwards one gets a mess, but since the quasi-classical apparatus has a complete set of states we can expand the final wave function in this basis $\sum_n A_n(x) \Phi_n^{app}(y)$ where $A_n$ are to-be found.

Now the existence of classical mechanics is needed to argue that this sum is actually just one term since the classical apparatus is in a definite state, $A_m(x) \Phi_m^{app}(y)$ so we can say $A_m(x)$ has to BOTH encode the probability $a_m = \int \Psi_m^*(x) \Psi^{sys}(x)$ of the stationary state of the system that was measured AND the new wave function $\phi_m(x)$ after measurement, so $A_m(x) = a_m \phi_m(x)$, so $A_m(x)$ is just some integral equation evolving the initial wave function:

$A_m(x) = \int K_m(x,y) \Psi^{sys}(y) dy$ and it must reduce to the previous case fixing $K_m$. There's no witchcraft in this picture, no magic about the mystery of collapse, I'd really like to know why this is apparently so unseemly other interpretations are needed.

Apparently this simply picture was so objectionable that: "Everett's Ph.D. work provided such an interpretation. He argued that for a composite system—such as a subject (the "observer" or measuring apparatus) observing an object (the "observed" system, such as a particle)—the claim that either the observer or the observed has a well-defined state is meaningless;

in modern parlance, the observer and the observed have become entangled: we can only specify the state of one relative to the other, i.e., the state of the observer and the observed are correlated after the observation is made. This led Everett to derive from the unitary, deterministic dynamics alone (i.e., without assuming wavefunction collapse) the notion of a relativity of states."

It's hard not to say that this is just blatantly wrong

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