The h Bar

Slereah

04:05

Finite dimensional Hilbert spaces are isomorphic to $\mathbb{C}^n$

Nike Dattani

04:59

Does anyone here know who might be able to answer this question?

10

Q: How to find the projected Hamiltonian for lowest flat-band in general for the given system?

In [1], starting with the bosonic Hamiltonian (Eqn. 1) for the dice lattice model with half flux density (with Ahronov-Bohm phases incorporated), \begin{equation} H=-t\sum_{\langle j,\mu\rangle}(a^\dagger_\mu a_je^{iA_{\mu j}}+\text{h.c.})+\frac{U_\Delta}{2}\sum_{\mu \in \Delta,\nabla}a^\dagger_\...

JingleBells

07:22

👁👄👁

apt45

09:33

Peskin computes the transition amplitude for a free particle in at page 27 of his QFT book. At a given point, there is an integral to compute, which is

They claim that one can compute this integral by deforming the real integration path as in the following picture

to conclude that the integral is exponential suppress. How is that possible? The integral along the branch-cut is a different integral wrt the one on the real axis

are they regulating the integral? (like the i\epsilon prescription)

Slereah

@apt45 Residue theorem

apt45

@Slereah the residue theorem holds for closed paths...

Slereah

Hm

I think there's a trick you can use to make it work

Maybe integrating on the one-point compactification of the complex plane

so that the remaining path is pushed off to infinity

apt45

try to compute that integral numerically with Mathematica. The integral on the real axis does not converge

Slereah

Well yes, the integrals we deal with in QFT generally don't converge

But they weakly converge

ie in the sense of distributions

apt45

09:42

this integral oscillates

I just picked some random value for r and m

ACuriousMind

10:41

@apt45 Yes, the "contour trick" is the same as the $\mathrm{i}\epsilon$ prescription, see physics.stackexchange.com/a/138221/50583

It doesn't matter whether you shift the poles by adding the $\mathrm{i}\epsilon$ or shift the contour, either way, you're doing something to the integral. Trying to compute the integral as written numerically misses the point entirely :P

apt45

@ACuriousMind Thanks. The fact the numerical integral was oscillating lead me to think that the contour trick was a prescription

@ACuriousMind Is the "contour trick" unique? I mean, there could be any other choice of the integration path that makes the integral convergent but with a different result?

Slereah

10:57

@apt45 As long as the poles inside are the same, the contour doesn't matter

Although there are different functions for different contours

ie

apt45

but these are all closed contours

Slereah

Well, not all of them

But as said before, you can close the open contours by sending off part of it to infinity

ie this :

Sending $C_3$ to infinity

(by sending it to infinity, it converges to $0$ hence doesn't change the integral)

apt45

the integral on the two red curves is the same?

Slereah

Well I'm not enough into analysis to answer with firm convictions

So I'll leave it to @ACuriousMind

@apt45 www-thphys.physics.ox.ac.uk/people/FrancescoHautmann/…

Behold

Although... Hard to make a contour that doesn't run into the second branch

But otoh

Hm, no

apt45

11:47

@Slereah very nice slides! Thank you

Slereah

Complex

apt45

I am not sure about the integral in the lower half plane

The integrand grows exponentially for negative imaginary part of p

Slereah

Wait, it's from Peskin?

I'll check my notebook, see if I worked it out mb

Hm, no, guess I only worked it out with poles

Slereah

12:14

physicsforums.com/threads/…

try here also mb

apt45

yeah ok, but you are always assuming that your integral on the real p-axis converges

otherwise, it does not make sense, right?

Slereah

Well, if you want to do the proper derivation, you can do it, but i'm guessing you won't like it much!

It's probably in Reeds

The trick being that the propagator isn't a function $D(x,y)$, it's a distribution $D(\phi, \psi)$

Slereah

12:29

that sort of thing

apt45

yeah, sure

Slereah

hal.archives-ouvertes.fr/hal-01337433/file/…

The horror

Captain Bohemian

12:48

what on earth composes a black hole? Is it just a spacetime region with unusually big curvature? or there is really matter of unusually high density inside the event horizon of a black hole?

Slereah

Depends on your theory and model!

The Schwarzschild metric is a vacuum solution

But that's for general relativity and a specific kind of black hole

Captain Bohemian

but isn't a black hole the collapse of a celestial body due to gravitation? So if a black hole is just a strong-curvature spacetime region without matter, where does the matter composing the original celestial body go?

Slereah

As I said, it depends

A real black hole isn't like a Schwarzschild black hole

It's more akin to the Oppenheimer-Snyder spacetime

Charlie

13:16

@Slereah Ah I have heard something along those lines before thank you

B. Brekke

14:08

exp(a)exp(b) = exp(a+b). What is this property called?

Charlie

I've seen it referred to as the "exponent product rule".

B. Brekke

That works!

Captain Bohemian

14:51

does a macostate refer to an ensemble while a microstate refers to a quantum state?

ACuriousMind

@CaptainBohemian See physics.stackexchange.com/a/343457/50583

Charlie

There's an ACM answer somewhere on this site for anything :P

Charlie

15:13

This might be a really, really stupid question, but the Lorentz transformations can always be through of as going from a frame at rest to a frame moving relative to that frame, right? This is probably as trivial as it sounds

actually nvm

Slereah

@Charlie Yes

You can always define an observer at rest within that frame

ACuriousMind

15:40

Dammit, I wrote an entire answer and then realized right before posting it that the OP had asked a more boring question than I thought :P

@Charlie Not if the transformation is just a rotation, in that case, both frames are at rest relative to each other ;)

Charlie

And going from frame 1->2->3 is the same as going from 1->3?

I want to say that property has a name but I can't think of it

a kind of transitivity?

Slereah

Yes, that's the group property

Charlie

ah I see

Slereah

Well

ACuriousMind

@Charlie That's what makes it possible to represent it as a group (along with to every a->b transformation there being a b->a transformation)

Slereah

15:49

It depends what you mean by 1 -> 2 -> 3

Charlie

I rarely see the Lorentz and Poincare groups referenced in GR texts, are they considered more advanced or something? Would I encounter them first in QFT?

at least in introductory GR texts, I'm not reading anything advanced yet ofc

ACuriousMind

Physicists don't like to talk about groups, for some reason. You can even find them in classical mechanics if you read the more mathematically inclined :P

Slereah

The Poincaré group only pops up in GR in fiber bundle contexts, since it's only applicable locally

So it tends to be a bit more advanced, yes

apt45

Am I the only one feeling uncomfortable when I read sentences as “the well known result, as it is well known...” in physics notes/papers?

Charlie

I like when group theory is used it feels like it categorises things nicely, idk

B. Brekke

15:52

@apt45 Looks like a typo to me

Charlie

are you concerned about the grammar or the fact that they're assuming you know what they're referencing?

ACuriousMind

"The Minkowski metric is the metric invariant under the Lorentz group" did much more for my understanding of relativity than weird gedankenexperiments and talking about "the spacetime interval" :P

apt45

@Charlie the fact they assume I know what they are referencing.

Where is the typo? Ahaha

Charlie

not really a typo but "the well known result, as it is well known" is a bit of an odd sentence

unless you're referring to two separate statements

apt45

@Charlie ah sure, I am referring to two separate sentences, sorry :)

B. Brekke

15:56

@apt45 Ah, then I get it. It looks like something I could have written when I don't know how to start my sentence, and forget to delete my first attempt

Charlie

well my answer is the same either way, I don't personally like it when textbooks use the word easy, simple or trivially. Not because it's wrong but it's a bit off putting

I never feel like those words add anything to a text intended for teaching, other than making you feel bad if you have to take a second to think about it

apt45

@Charlie yes, that’s the same for me. I am trying to ban from my drafts/papers words like straightforward, easy, simple, well known etc...

B. Brekke

@Charlie I think it can be nice, if it is used correctly. I am currently using it a lot, but it needs to reflect on the nature of the approach/problem and not the skill of the reader

ACuriousMind

@Charlie I do think it's good if there is an indicator that this follows "easily" and that there aren't ten pages of horrible algebra hidden in that step. But often it's a legitimate question why it wasn't just written out explicitly if it's so easy...

apt45

@ACuriousMind the point is that You wouldn’t spend time and words to write something that is a triviality. We don’t write that a ball is a sphere, unless you are talking about rugby balls :))

So, if anyone states a result and then write that it is well known, then it means that the read could have followed the discussion in the text without referring explicitly to this result

I don’t see any way writing “it is well known” could be useful

ACuriousMind

16:07

I don't think it's that easy. For instance, Taylor's theorem is "well-known". But without the text explicitly stating the remainder term for me, I'm not going to recall it on my own. If the text doesn't say it's "well-known", then I might wonder where they got it from, while "well-known" means I likely simply need to look it up in the standard references for that field if I don't trust them.

Slereah

If you want a book that doesn't skirt proofs you can try Bourbaki :p

apt45

I am sorry, but I think that the authors should not decide what the reader should know or not

Slereah

Unfortunately it's hard to find the middle ground between doing that and writing a 2000 pages book

apt45

That’s also true

Slereah

A friend of mine actually wrote a GR book with most proofs written out, I think?

valek.webs.com//mp/gr_en.pdf

apt45

16:14

I don’t like papers/books written in Word :P

Shing

16:39

hey guys, I am reading lecture notes of Feynman's lectures on mathematical method, and got a bit confused here, would anyone be kind enough to explain this to me?

user434058

Yo! Just crossed (only) 100 reviews in all review queues! Noice!

bolbteppa

@Shing If we set $D^{-1} g(x) = \int^x g(u) du = h(x)$ then
\begin{align}
D^{-2} g(x) &= D^{-1} h(x) = \int^x h(v) dv = \int^x [\int^v g(u) du] dv \\
&= \int^x h(v) dv = \int^x d[h(v) v] - \int^x v dh(v) = [h(v)v]^x - \int^x v d \int^v g(u) du \\
&= h(x) x - \int^x v g(v) dv = x\int^x g(u) du - \int^x u g(u) du \\
&= \int^x (x - u)g(u) du
\end{align}

Cauchy formula for repeated integration

The Cauchy formula for repeated integration, named after Augustin Louis Cauchy, allows one to compress n antidifferentiations of a function into a single integral (cf. Cauchy's formula). == Scalar case == Let f be a continuous function on the real line. Then the nth repeated integral of f based at a, f ( − n ) ( x ) = ∫ a x ∫...

Shing

@bolbteppa thanks!

Charlie

17:42

How would I go about changing the placement of indices in mathjax? So objects like $\Lambda_\mu^\nu$ have the indices separated rather than on top of eachother?

ACuriousMind

@Charlie {\Lambda^\mu}_\nu

Charlie

Ah thank you

Charlie

20:35

I have encountered the phrase: "we define the four-velocity $U$ to be a vector tangent to the world line of the particle, and of such a length that it stretches one unit of time in that particle’s frame."

Does this statement about its length hold in any frame? Maybe I'm just getting hung up on the way it's phrased

I want to say that the four velocities length is an invariant since it's the the contraction of $U$ with itself

ACuriousMind

yes

Charlie

ok ty

Slereah

(it depends on the parametrization, though)

the vector, not the norm

Charlie

how can the vector be different other than changing it's norm?

the direction?

but that would mean the particle is taking a different path

Slereah

It does change the norm also, yeah

Although since this is only a real issue for null vectors, this isn't a big problem

Since the norm is always zero

Charlie

20:43

I can see how the tangent to the worldline would change based on parameterisation, since it's a derivative, but come to think of it I don't see how that wouldn't affect the norm

ah it can change the norm, that makes sense

Slereah

@Charlie It's basically the same as changing the unit

Charlie

so the statement that the norm of the 4-velocity is 1 only holds if I'm parameterising it with the proper time?

or -1 I guess depending on the metric

Slereah

Yes

Charlie

Ah I see

Slereah

otherwise you're changing the definition by $\dot{\alpha}^2$

for the parametrization $\alpha$

ACuriousMind

20:48

@Slereah He was asking about a particle and that "particle's frame", there are no null vectors here :P

Slereah

Well still applies to massive particles :p

ACuriousMind

If you read the passage he quotes, it doesn't depend on the parametrization. It doesn't say "the tangent vector" it says "a vector tangent to the world line, of such a length..."

The author wisely sidestepped the parametrization issue by fixing the length by hand :P

Slereah

Grave mistake!

You have to give every observer a parametrization and a tetrad frame!

It is even more fun

ACuriousMind

No, all the tetrads belong to me, I'm not giving them out

Slereah

I mean tetrads are basically three sticks

It's not hard to build!

ACuriousMind

20:51

Sticks don't grow on trees!

...wait.

Slereah

Measurement in GR is something you shouldn't think about too hard otherwise you start going insane

3

"Wait, how do I know that this rod will remain of fixed length!"

How can I be assured that they each move along geodesics!

Also does the FRWL metric mean that the hydrogen atom wavelength change???

Is it still a proper time clock

bolbteppa

Basically you just think of $d \tau^2 = dt^2 - dx^2 - dy^2 - dz^2 = \eta_{\mu \nu} dx^{\mu} dx^{\nu}$ so that $1 = \eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}$ if $\dot{x}^{\mu} = dx^{\mu}/d \tau$, and more generally you'd have $(d \tau/du)^2 = \eta_{\mu \nu} (dx^{\mu}/du)(dx^{\nu}/du)$

ACuriousMind

@Slereah come on, deep breaths. Think about something nice and Newtonian until the urge to question the foundations of reality goes away.

Slereah

@ACuriousMind This also applies to Newtonian mechanics!

Aaaah

Charlie

@bolbteppa ok thank you

I'm confusing myself a bit by comparing 4-velocity and 3-velocity, am I right in saying that in classical motion I can move between two points "more quickly" by just travelling faster through space, but in GR there is no such thing as travelling along a geodesic "more quickly"? Is this related to the fixed length of the 4-velocity?

Slereah

21:00

Technically this isn't a big issue as far as like

Epistemology goes

You're supposed to define the measurement of a theory from the tools you have

But you can't really do that in practice

Because similar tools will be different

It's hard to both have common physical assumptions for every tool and also estimate their different uncertainties

without going into insane details of their functioning

ACuriousMind

@Charlie Don't think about the 4-velocity as a velocity. The "velocity" you have intuition for is the spatial part of that (i.e. 3-velocity is the spatial part of 4-velocity), but the crux of relativity is that "spatial part" is a relative (=frame-dependent) notion.

Slereah

Synge had a whole paper about how to VIEW things in the context of GR

it's a tricky thing

ACuriousMind

The problem is that 4-velocity is a derivative w.r.t. the proper time along a path, while the 3-velocity you're used to is a derivative w.r.t. an absolute notion of time.

Whenever you use words like "faster" or "slower" in relativity, you must always say relative to what, otherwise it just doesn't mean anything.

Charlie

So it's better to think of it as simply a tangent vector to a worldline, rather than trying to draw some analogy to rate of movement

Slereah

Pretty much yes

Charlie

21:07

Ok I can work with that

Slereah

The fact that they all have norm 1 is a good indicator as to why

although the direction still contains the informations you'd like

Charlie

Ah so if we parameterise the worldline with a different parameter, we'll find the norm to be $\neq 1$, but we'll all still agree that it has the same value in every frame

Slereah

Yes

Charlie

Ok nice, ty

Slereah

Reparametrization basically just rescales the time unit

Charlie

23:49

On the topic of "superposition" in QM, does this just generally refer to the fact that the state vector can be written as a linear combination of eigenstates of any operator? For instance if my system happens to be in a state $|\psi\rangle$, I can write this as a linear combination of the eigenstates of some arbitrary Hermitian operator, is this what we mean when we say the system is "in a superposition" of states? Probably trivial question, but still.

This would also mean that even if the system has been forced into an eigenstate of some operator, it is still in a superposition of states of some other incompatible observable

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