The h Bar - 2016-04-10 (page 2 of 2)

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6:00 PM

I suppose, but make sure you do it in the right order.

Remember: murder-suicide, not suicide-murder. The order of operations is important.

It takes 24 hours to delete your account.

and you have to e-mail the mods.

xD

No use in wasting time, then, Ocelot.

@barrycarter we will see how my date tonight w/ the toothless alabaman goes

Ocelot, if you have the love of a good woman, that's all you need. However, I suspect this is not a good woman.

I would take offense to that but I'm the one who called her a toothless prostitute, so

6:03 PM

@JohnDuffield Whoa, you're in the UK?!

@barrycarter : yep, most people know I live in Poole.

user image

John Duffield - 1 OF 6

Watch them.

I'll wait for the book ;)

Did you actually order it?

That's one big pool.

6:06 PM

@0celo7 can you look over my linear algebra notes for the lesson i'm doing on friday?

That's Poole Harbour. On either side of the harbour mouth there's miles of golden sand. It's a great place to live.

I'm unhappy about simultaneity, and think @JohnDuffield's approach may have some merit.

@3075 Very busy.

@0celo7 i'll show you later then.

@barrycarter : IMHO you should be unhappy about the Andromeda paradox.

6:09 PM

@JohnDuffield That's exactly what started my unhappiness with simultaneity.

github.com/barrycarter/bcapps/blob/master/STACK/bc-andromeda.m has some of my thoughts

user image

That's my image of unhappiness.

6:39 PM

@barrycarter Mine too

@3075 you're got my attention for 20 minutes

user54412

6:55 PM

@FenderLesPaul Yes that's the conference.

7:10 PM

@0celo7 30 mins ago :'(

sorry I was on the bus.

7:57 PM

What's all this about ten stars?

@ChrisWhite cool

as for UCSB

It's really sweet, their gravity group is insanely big

and they're getting 3 new postdocs the coming fall, one of them is Strominger's student and another is Bousso's student

but the groups are pretty much all saturated as far as grad students go

which is why they accepted so few hep-theorists this year

only Gary is taking students

and I like what Bousso does quite a bit more than what Gary does

how was Gary's talk anyways?

@JohnDuffield It counts how many people openly dislove Ocelot.

Where?

@JohnDuffield On the stars thing to the right, but you have to click "show more" as its been pushed down by Ocelot's admission that he doesn't know what logarithms are.

@barrycarter : thanks, I see it. I clicked it, now it says 11.

8:04 PM

@JohnDuffield Awesome! At one point, Ocelot was going to commit suicide and stream it, but he's chickened out, @0celo7

@barrycarter : that doesn't sound too healthy Barry. I want no part of that.

@JohnDuffield Well, no, suicide is generally not considered too healthy. But it could be amusing ;)

@JohnDuffield if it was your message it would have 30 stars (or more).

In all seriousness, @0celo7 is just whining

@JohnDuffield why are you openly declaring your hate when it could be anonymous...

8:14 PM

@3075 : I want nothing more to do with this. Over and out.

k

@3075 You seem to have some anger over @JohnDuffield

idc really.

8:31 PM

@barrycarter What does that mean

11 stars!

Holy shit :(

@0celo7 Oh, that your "threat" to kill yourself is just attention-seeking :)

That makes me sound like some shitty child

@ACuriousMind From Wiki:

> The Hom-bundle Hom(E, F) is a vector bundle whose fiber at x is the space of linear maps from Ex to Fx (which is often denoted Hom(Ex, Fx) or L(Ex, Fx)).

It seems no one talks about the structure of the Hom bundle beyond its fibers.

@ACuriousMind But for the tangent bundle, can't we completely describe it by $TM=\sqcup_{p\in M}T_pM$, i.e. knowledge of the fibers at every point?

So $\mathrm{Hom}(E_1,E_2)=\sqcup_{p\in M}L(E_{1p},E_{2p})$?

Oh, I guess one has to construct a bundle atlas for $TM$.

The equation $E = \sqcup_{p\in M} E_p$ is true for every bundle, but it is an equation as sets, since the r.h.s. is not a bundle. To have a bundle you need to give the local trivializations.

Yeah, yeah

I'll ask an MSE question.

In all seriousness, @0celo7, most of the people who ticked that message probably don't hate you. They're just trying to let you know that you're sometime "that guy" that they hang out with: the one who doesn't seem to realize when he's being annoying.

8:46 PM

@0celo7 These notes, for example, do. (on the first page when googling "hom bundle" :P)

I thought about nuking it when I first saw it, but refrained because you put it up there yourself. I'm beginning to wonder if that was a mistake.

What would nuking it do?

@ACuriousMind :(

what page do they talk about it?

Starts on page 7

Ugh, fiber bundles are annoying

@0celo7 Rid the sidebar of an attractive nuisance.

8:50 PM

@dmckee Attractive nuisance?

@ACuriousMind I wonder if the following theorem "Let $N,N'$ be two submanifolds of $M$, and let $\gamma:[0,t]\to M$ be a geodesic s.t. $\gamma(0)\in N,\gamma(t)\in N'$ and $\gamma$ is the shortest curve from $N$ to $N'$. Then $\gamma'(0)\bot N$ and $\gamma'(t)\bot N'$" is equivalent to "take two parallel lines, then a third line intersects one of the others perpendicularly iff it also intersects the other perpendicularly"

I wonder if one can use this to prove the fifth axiom

From the assumption of vanishing curvature and trivial topology

Attractive nuisance doctrine

The attractive nuisance doctrine applies to the law of torts, in the United States. It states that a landowner may be held liable for injuries to children trespassing on the land if the injury is caused by an object on the land that is likely to attract children. The doctrine is designed to protect children who are unable to appreciate the risk posed by the object, by imposing a liability on the landowner. The doctrine has been applied to hold landowners liable for injuries caused by abandoned cars, piles of lumber or sand, trampolines, and swimming pools. However, it can be applied to virtually...

The phrase is also used by neighbors annoyed by the attracted nuisances.

@dmckee what's a good definition of parallel lines in the plane

Euclid went with their not having a point of intersection, didn't he? But of course that doesn't pass trivially into three dimensions.

The shortest distance between them is the same for all points on the lines, too, which does apply in higher dimensions.

@dmckee Really?

What is the shortest distance

In say 3 dimensions

@0celo7 In flat space.

9:03 PM

@dmckee I know, but I don't see that

> If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.

Why are old-timey theorems so hard to read

Take a point on one line. Construct a transverse plane at that point. Find the intersection between that plane and the other line. Draw the segment between original point and the intersection. It's length s the shorted distance between the lines at the original point.

Sure, my 9th grade geometry teacher wouldn't consider it finished, but I don't care.

You did 3D stuff in 9th grade?

Holy shit

i can't even do 2D geometry

@0celo7 Just the last week of class. We'd finished the rest of the book.

@dmckee Wait, we're defining parallel lines in 3-space how?

Lines that never intersect?

But she had a very rigid and formal expectation of what consituted a proof for the purposes of class.

@0celo7 Those might still be skew.

The distance thing is what makes them parallel.

9:08 PM

@dmckee Yeah, that's why I don't buy your claim.

@dmckee Oh, that's how you're defining them?

I think you actually have to run the construction above twice and get back to where you started to know they are parallel.

Mrs. Kirk would be proud of me: I learned something in her class.

I don't remember my geometry teacher's name.

Translating Riemannian geometry into Euclidean geometry is not as easy as you would think/hope

She lived in the same neighborhood. I watched her pets when she was on vacation a couple of times.

what is it with people having teacher who are actually human??

Mine were all freaking synths

@dmckee Aha! I think I can show the following using techniques from Riemannian geometry: Take two lines in the plane, A and B, and have line C intersect line A perpendicularly. Then, unless, C also intersects B perpendicularly, the distance between A and B is 0.

Which means they are not parallel.

Seems good.

9:17 PM

Now the hard part is where I actually use the fact that A and B are lines and that there is no curvature. Have to work on that part.

i.e. I'm doing something horribly wrong

@ACuriousMind I have a feeling you can help me with this one.

Actually let me make a broader appeal:

Dear everyone, the following question about elementary quantum mechanics has stumped myself and so far two other physicists, one of which is a good theorist :-)

0

Q: What do the wave functions associated to the fock states of a bound state system mean?

$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$ Consider a string of length $L$ under tension and clamped on each end. This system is described by the wave equation and has a set of modes. The $n^\text{th}$ mode has a spatial profile $$\phi_n(x) = \sin(n \pi x / L)$$ and frequency $\omeg...

quantum-mechanics second-quantization

Nov 7 '15 at 1:05, by 0celo7

@ACuriousMind Ahhhhhhhhhhhh

Shit shit shit

Why didn't past me write down my revelation anywhere

9:34 PM

@DanielSank I don't understand the question - the square well does not have a Fock space. It doesn't have creation and annihilation operators. Also the "wavefunctions" of n-particle states are not given by $\langle x \vert \psi \rangle$, but by $(\langle x_1\rvert\otimes\dots \langle x_n\rvert)\lvert \psi \rangle$.

I also don't understand why you're talking about Gaussians in the square well. The energy eigenstates of the square well are sines, not Gaussians

American Museum of Natural History: 2016 Isaac Asimov Memorial Debate: Is the Universe a Simulation?

Well, let me rephrase that. @DanielSank: Although you can build a Fock space just by taking the symmetric tensors on the 1-particle space, and define creation operators for the energy eigenstates, I don't see where you get the Gaussians from, or what you actually want to know.

9:56 PM

@ACuriousMind I'm dumb. I have a function $f:(-\epsilon,\epsilon)\to [0,\infty)$, and $f'(0)<0$. Then the book says: for small $s$, $f(s)<f(0)$, but shouldn't it be for small positive $s$?

Yes, should be positive.

@Slereah Let $M^{2n}$ be an orientable Riem. manifold with positive curvature. Let $\gamma$ be a closed geodesic. Then $\gamma$ is homotopic to a closed curve whose length is strictly less than that of $\gamma$.

Hello
It's me.

oooh, got the necromancer badge

@Slereah Apparently, you take a vector $V$ orthogonal to $\gamma$ at one point, then parallel transport it around the loop. You get a vector field along $\gamma$ like that, then vary $\gamma$ wrt. that field and check the second variation of arc length.

@ACuriousMind Do you know why parallel transport along a closed geodesic preserves a vector only if the manifold is orientable and even-dimensional?

@You who are "You"?

10:11 PM

@skillpatrol I am You.

You is me.

Another account @skillpatrol ?

We is one.

Don't you have like three?

@You And you need grammar lessons.

@0celo7 Hahaha. Who cares about grammar? This isn't the English Exchange. :)

@0celo7 Isn't a geodesic defined by preserving its tangent vector? Or are you talking about the parallel transport along it being the identity on all vectors?

10:13 PM

@DanielSank another random idea, brainstorm style: could weak measurement help answer this question?

@ACuriousMind The Fock state wave function is Gaussian, not the mode profile.

For a vibrating string's fundamental mode, the mode profile is sinusoidal in position, but in the ground state the amplitude of that mode is Gaussian distributed.

@ACuriousMind Oh yes, all vectors.

@ACuriousMind Fine, that's true but doesn't answer the question.

At least all vectors perpendicular to it.

(The general case then follows from preservation of perpendicular + tangent.)

@ACuriousMind Did you understand my description of the quantized vibrating string?

If so, the question is whether or not the same ideas apply to a square well:

1) Can I think of the single particle wave functions as modes, each of which can be in various Fock states?

10:16 PM

@DanielSank Actually, no. I don't know where that Gaussian comes from.

@DanielSank I don't know what you mean by "Fock state wave function" or "mode profile".

@ACuriousMind Let $V$ be a vector in $T_{\gamma(0)}M$. Then $|V(t)|$ is constant along $\gamma$ where $V(t)$ is the parallel transport of $V$

@ACuriousMind I explained this but maybe not clearly in the post.

At what point do you not follow?

So we should set up an orthogonal frame in the space orthogonal to $\gamma$

Then $V(t)$ is always orthogonal to the other vectors in the frame

@DanielSank I understand everything up to "For example, if a mode is in ∣0⟩, then that mode's displacement (i.e. amplitude) has a Gaussian probability distribution." and then again everything up to "For example, if we have the state ∣0000…⟩ in the square well, then we're saying that each mode of the square well has an amplitude which is Gaussian distributed."

Let's focus on the first part.

Suppose I have a vibrating string. Let us focus on the fundamental vibrational mode, which has a mode profile $\sin(\pi x / L)$.

Are you ok with this so far @ACuriousMind?

10:22 PM

Yes

Ok. This mode has the quantum mechanics of a simple harmonic oscillator.

For example, the energy in the mode is $n \hbar \omega$ where $\omega$ is the frequency of that mode.

and $n$ is the number of excitations in the mode.

This mode can be in any harmonic oscillator state, including but not limited to Fock states and coherent states.

These states can be expressed via wave functions.

If we do this, the two conjugate coordinates are the quadratures of the mode.

Do you know what I mean by quadratures, @ACuriousMind?

@DanielSank No

Ah ok then this is important.

If I have a classical oscillator, I can represent its motion in phase space. In particular, if I start the oscillator with a certain position and momentum, the motion becomes a circle with fixed energy.

Ok @ACuriousMind?

@DanielSank Actually, I'm not on board with this. If you quantize a string, you're essentially doing 1+1 QFT on a finite interval. QFT states are not wavefunctions, but functionals in the field.

@DanielSank Yes

@ACuriousMind Ok, so the position and momentum are often called "quadratures".

If this were an electromagnetic oscillation the quadratures could be the E and B field amplitudes.

In general, "quadratures" are the two 90-degrees-out-of-phase coordinates of an oscillator.

10:29 PM

Okay, they are the "natural" coordinates for an oscillator.

@ACuriousMind yeah

Okay, so what do you mean by "These states can be expressed via wave functions. If we do this, the two conjugate coordinates are the quadratures of the mode"?

So, once I take a mode of my vibrating string, in quantum mechanics that mode can have 0, 1, 2,... excitations. Each of these Fock states has an associated wave function where the coordinate of those wave functions are one or the other quadratures of the mode.

@DanielSank Sorry if I'm being dense, but I don't understand that statement. How do you associate the wavefunction to the Fock state? Are you saying "pretend it's a QHO state and write down the QHO wavefunction"?

@ACuriousMind You don't have to pretend.

Let's say it like this:

The Hamiltonian of a vibrating string can be written as the sum of Hamiltonians for each mode.

Each of these constituent Hamiltonians is exactly that of a quantum harmonic oscillator.

10:32 PM

Yes

We all know that the harmonic oscillator has wave functions.

In kindergarten we think of these wave functions as living in position space.

2

Obviously, the wave functions of a single mode of a vibrating string don't live in position space.

They live in quadrature space.

What kindergarten did you go to o.O

3

So if a mode is in the ground state, then that means the amplitude of that mode has a Gaussian probability distribution.

I can put this another way which might help.

I understand

My problem was with "wavefunction"

Suppose I have some electromagnetic object with a bunch of modes. I can pick a mode and say it's in Fock state 0, and that has some associated wave function, but that wave function has nothing to do with that mode's spatial profile.

@ACuriousMind What should I call it?

Suppose I pick a mode and that mode is in $|1\rangle$. Suppose the quadratures of this mode are $x$ and $p$.

10:35 PM

@DanielSank I would just call it the probability amplitude. "Wavefunction" is usually reserved for amplitudes in position or momentum

What would you have me call $\langle x | 1 \rangle$?

@ACuriousMind Fine, we can call it "probability amplitude" if you insist. I don't care.

ooooooo

he don't care

Anyway, doesn't matter what you call it, I get it now

So, you do the same construction for the square well

What exactly do you want to know about that case?

@0celo7 Go download war thunder

no!

doing Riemannian geometry now, doing QM later then watching movie with the woman and then more QM

10:38 PM

@ACuriousMind We usually learn about a single particle in the square well and solve for its "eigentstates" or "wave functions"

But then we learn about multiple particles.

So now we have Fock sates because each "single particle state" can be excited by 0, 1, 2,... excitations.

I want to know if there are Fock state probability amplitudes here.

Should I replace the term "single particle states" with "modes" and think about this system the same way I think about a quantized string.

Well, you can formally do the exact same thing here: Construct the quadratures as $a+a^\dagger$ and $a-a^\dagger$, and write down the amplitude as the projection onto the eigenstates of $a+a^\dagger$.

@0celo7 huh you're back to qm?

@3075 work

oh.

I don't think these carry any particular physical meaning, though.

10:41 PM

@ACuriousMind Well hold on just one moment.

::holds on::

@ACuriousMind I think one has to use Picardo-Lindyhopf.

Let's pick one particular "single particle state" of the square well. Let's choose the ground state, which has a spatial shape of $\sin(\pi x / L)$.

Remember the theorem

Suppose there are zero particles in this single particle state (we should say "mode" instead of "single particle state", but whatever).

The probability amplitude for $|0\rangle$ is a Gaussian.

So I put to you the question: What is this the probability amplitude of?

10:43 PM

Mar 29 at 23:44, by 0celo7

@ACuriousMind 9 times out of 10 questions about geodesics can be answered by appealing to PL :D

@DanielSank How do you know there is a $\lvert 0 \rangle$ state? Formally, the Fock space is constructed as the symmetric algebra on the one-particle space, that doesn't construct an "empty" state.

> How do you know there is a ∣0⟩ state?

Uh, because it's obvious that I can have zero particles in the well, in which case the system state is $|00000...\rangle$.

@DanielSank Not obvious to me. Given a collection of modes $\lvert i \rangle$, I define the Fock state $\lvert n_1,\dots, n_k\rangle$ to be given by $n_1$ copies of $\lvert 1 \rangle$ times $n_2$ copies of $\lvert 2 \rangle$ and so on.

Now, when I write $\lvert 0 ,\dots, 0\rangle$, that gives just 0 by that definition, not a state.

I'd say the empty well is just an empty well, not a system.

To me $|n_1,\ldots,n_k\rangle$ means "mode 1 is excited by $n_1$ quanta...".

That's not a definition

10:52 PM

I don't know what you mean by "$n_1$ copies of $|1\rangle$.

@ACuriousMind Neither is yours.

What's a "copy"?

What's $|1\rangle$?

It's a ket!

You'll all perish next to my knowledge

@ACuriousMind Please @ me if/when you're willing to talk about Riem. geo

@BernardMeurer Be silent, young one! :-)

@DanielSank I can try answering Mr University of Casual Sex and Beer (UCSB)

@BernardMeurer <3

10:55 PM

@DanielSank <3

@DanielSank Given a one-particle space $H$, the Fock space is given by $\bigoplus_{i\in\mathbb{N}} H^{\otimes i}$. Given a basis $\{ \lvert i \rangle \mid i\in\mathbb{N}_{>0}\}$ of $H$, a Fock state is defined for any collection of $\{n_i \in\mathbb{N} \mid i\in\mathbb{N}, n_i = 0 \text{ for almost all }i\}$ to be $n_1$-fold tensor product of $\lvert 1 \rangle$ times (as in "tensor product") the $n_2$-fold tensor product of $\lvert 2 \rangle$ and so on.

For example, if I have three modes $\lvert 1 \rangle,\lvert 2 \rangle,\lvert 3 \rangle$, then I write $\lvert 2,1,0\rangle$ for the state $\lvert 1\rangle\otimes\lvert 1 \rangle\otimes\lvert 2 \rangle$.

Why the hell would you label a mode with a ket?

That is unbelievably confusing.

Because it's a one-particle state

yeah but then you're mixing 1st quantized and second quantized notation.

let's at least not label the with integers.

we use integers to indicate the occupation of each mode in second quantization, so let's label the modes A, B, C or something.

Please.

Okay, label the modes alphabetically, then

So we have $\lvert 2,1,0\rangle = \lvert A\rangle\otimes \lvert A\rangle\otimes\lvert B\rangle$, yes?

11:01 PM

1

Q: Why is this question about measurement off-topic?

The question Reliable length scales in nature has been received very poorly, for reasons I don't really understand. In the help center, it is stated that questions concerning "Experimental designs and results" and "Experimental technology used in physics or astronomy" are on topic. I can reason...

(symmetrized, actually, but I don't want to bother with that)

@ACuriousMind Yes.

And yes, let us not write out symmetrized 1st quantized states.

Okay. What is your $\lvert 0,0,0\rangle$ supposed to be?

The whole reason we use second quantization is to avoid that.

@ACuriousMind That's a good point, but I'm not convinced.

In the case of a vibrating string, the state $|0000\rangle$ means that each mode is in its ground state.

@DanielSank Ah, but in the vibrating string case, you have the option to define that state as the ground state of the Hamiltonian

11:04 PM

I would assume then that $|000\rangle$ for the square well means that each single particle state of the well is unoccupied.

So there you have it.

You don't have that here

@ACuriousMind Oh goody, why not?

I think I'm about to learn a thing.

The ground state of the Hamiltonian here is just the lowest energy mode

Because here, the Hamiltonian is not the number operator + some constant.

Why can't I rewrite the hamiltonian of the square well as a sum of Hamiltonians for each single particle state?

@DanielSank Oh, it is. The Hamiltonian on n particles is just the sum of n 1-particle Hamiltonians

11:07 PM

The energy of the system is $\sum_{\text{mode }m} E_m (\text{number of particles in }m$).

@ACuriousMind Why are you talking about the Hamiltonian on $n$ particles?

I'm assuming the particle number is not fixed.

Maybe that's the key.

@ACuriousMind I don't know what a "one particle Hamiltonian" is.

Okay, let me explain what I meant:

I see what kind of physics you do @DanielSank

The Hamiltonian of 1 particle in a square well is $H_1 = \frac{p^2}{2m} + V(x)$ for $V(x)$ some rectangle function thingy. This Hamiltonian naturally extends to the whole Fock space just by letting it act on n-particle state $\lvert A_1\rangle\otimes\dots\otimes\vert A_n\rangle$ as $H(\lvert A_1\rangle\otimes\cdot\otimes\vert A_n\rangle) = (H_1\lvert A_1\rangle)\otimes \dots (H_1\lvert A_n\rangle)$

So your Hamiltonian just acts on particle states - but it is not defined on something like $\lvert 0,\dots,0\rangle$ because that doesn't lie in the Fock space!

I'm sorry, but what is $|A_1\rangle$?

The difference to the string is that with the string, you build the Fock space not from a 1-particle space, but from the ground state of the string by letting the mode operators act on it as creation/annihilation operators.

@DanielSank Some 1-particle state, call it however you like

11:16 PM

What's wrong with letting mode operators act on the ground state of the square well?

I can make creation operators $a^\dagger_A$, $a^\dagger_B$ etc. which add one particle to mode $A$, $B$,...

@DanielSank The ground state of the square well is itself a mode.

(Let's keep referring to the single particle states by A B C for consistency)

It's just one of the 1-particle states

@ACuriousMind Ok then I'll call it something else.

@DanielSank You have to define it first

11:18 PM

The "vacuum state"?

This "empty" state is not inside the Hilbert space we (or I) built

Ok, suppose I have a square well with no particles in it.

I call that the "vacuum state".

Now I start adding particles to various "single particle states" in the well.

@DanielSank How is that supposed to be a quantum system?

OR any physical system, for that matter

It has no d.o.f., it's not a physical system

@ACuriousMind Uh, I have a bucket in front of me. There are no balls in it. This is a physical system.

It's a bucket of zero balls.

Now I add one ball.

What's wrong with this?

You're being too intuitive. A classical system is define by a phase space and a Hamiltonian, a quantum systen is defined by a Hilbert space and a Hamiltonian

11:23 PM

@ACuriousMind I fail to see the difference between the square well and the vibrating string.

Look, suppose I tell you that I have a QFT defined over a finite interval with boundary conditions that the field is zero at the ends.

Doesn't that have a ground state with zero excitations?

It has

Ok, how is this different from the square well?

The square well is not a QFT

What would the field be, and what the Lagrangian/Hamiltonian?

I don't know enough QFT to answer that.

The string has a classical Lagrangian $L= \int\mathrm{d}x \frac{1}{2}\dot{\phi}^2 - \frac{1}{2}\phi'^2$, it is already a field theoretic system. When you quantize it, you find that the Fourier modes of $\phi$ become creation/annihilation operators on the ground state of the Hamiltonian

You don't have that for the square well. The classical square well is not a field theoretic system, it's just a bunch of particles in a box. You force the "second quantized description" on it by building the Fock space out of the 1-particle space, but it is not "natural" to it, in particular, you can't use the ground state of the Hamiltonian as the vacuum state, as it is just one of the 1-particle states.

11:35 PM

I read the words but I don't understand.

What is the difference between putting multiple particles in various single particle states in the square well, and putting multiple excitations in the various modes of a string?

That, in one case, you change your physical system (putting more particles into the box), while in the other case, you just change its state (exciting the string more or less).

Main site down?

@barrycarter Seems like it

@ACuriousMind I don't understand why adding particles is changing the system in a fundamental way.

Another try: The string is really special because it just looks like a bunch of uncoupled HOs in Fourier space

No other system looks like that, so you can't expect the Hilbert space of any other system to look like it

11:42 PM

But even interacting QFT's break down in this way, just with additional terms for the interaction.

well, assuming there's a free field part...

The Hilbert space of interacting QFTs is mostly unknown

The only Hilbert spaces we know are those of QM and that of the free field

o

oh

The crucial part of QFT is that you can compute the scattering amplitudes of an interacting theory in terms of the "asymptotic free states"

And only for the free theory we have a clue how to define a particle (state created by Fourier mode). There is no notion of particle state in a truly interacting theory

It's pretty amazing how well QFT computations work given how little we really understand about interacting theories ;)

Ok @ACuriousMind I need a break. My whole research group is now emailing each other trying to understand this. Thanks for your help.

I need to process what you've said.

@ACuriousMind Ok then, can you please help with my geodesics?

11:49 PM

hello @DanielSank I hope you got my emails?

hi guys, I have a question regarding number density and flux...

@0celo7 No, not in the mood for that

@TanMath Post the question on the site.

@ACuriousMind sigh

How can we derive Gauss's Law for E and M fields using the fact that the flux is equal to the number density in that surface?

@DanielSank I wasn't sure if it is fine to post it...it seems to be a little broad... So I thought I would discuss it here?

@DanielSank I'll stay tuned for the progress report ;)

11:55 PM

is this fine to ask on the main site?

@TanMath No, because it is not clear what your quesiton is

@ACuriousMind what part of the question?

@ACuriousMind What does the statement "parallel transport preserves orientation" mean?

@TanMath Gauß' law is usually the statement "electric flux through a closed surface is equal to enclosed charge", so what is there to derive?

@0celo7 That the parallel transpost of any right-handed system will be right-handed

@ACuriousMind I know that, I guess I'm asking how one proves that.

11:59 PM

@ACuriousMind i am talking about mathematically...

Since the manifold is Riemannian, we can probably use the volume form somehow

@TanMath What exactly is given?

And what exactly do you want to derive?

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