The h Bar - 2015-10-20 (page 1 of 6)

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0celo7

00:12

@ACuriousMind I'm not in a hurry to learn everything. I'm in a hurry to learn something. Right now I feel like a complete idiot

DanielSank

The no-communication theorem states that no communication about the no-communication theorem can clear up the misunderstanding quickly enough to allow faster-than-light signaling.

0celo7

Too late

@dmckee posted that a day or so ago

DanielSank

w/e

Slereah

@DanielSank posted it before, actually

It just violated locality

So we only see it now

DanielSank

>.<

0celo7

00:19

Lol

Slereah

I once read a thing about experiments relating to quantum tunneling and FTL communication

And beyond the usual "It doesn't work" spiel, there was this bit about how

If you actually got it to work

The wavelength of the signal you could pass on interstellar levels would be so low as to be impossible to detect

(it was in a big list of FTL things for space exploration)

Engineering considerations are even less forgiving than physical ones

0celo7

00:50

@Slereah well those damn engineers

0celo7

01:44

@Slereah why $S$ for spacelike hypersurface and not $\Sigma$

0celo7

01:56

ffs my psychology prof did not enable the homework upload page

great now I get to be that A hole who emails him

0celo7

02:09

@ACuriousMind I forgot to mention: Denzler knew who Straumann was, he met him at ETH/Zurich U :D

tpg2114

02:31

@dmckee Seems your comment was on point... a deeper, non-particle and non-relativity question sits at +3 while a less-deep, non-relativistic analogue sits at +8. Although, neither are answered even though I know @ChrisWhite is working on the relativistic one ;) Maybe both are just hard to answer!

2 hours later…

David Z

04:54

7

Q: Will the box move?

Here is a man in a box with wheels on a frictionless surface. Case 1 - The person constantly pushes the box while standing near its wall. Why doesn’t it move? (I know it is because of the 3rd law, but I am not able to understand the “why” here, please explain.) Case 2 - If he runs inside the...

newtonian-mechanics

That seems like a no-effort question to me, probably homework-like... thoughts?

user685252

05:10

Almost 2,000 views in 22 hours?

:-/

DanielSank

Is it appropriate to flag answers to questions which violate the homework policy?

1 hour later…

user54412

06:20

4

Q: What is the way to prevent the cream from settling on the surface of hot milk with least manual labour?

I have seen Cooling a cup of coffee with help of a spoon but that question has a restriction of using only a spoon. My problem is that I have to boil the milk before pouring it in the bottle for the toddler and I don't want the cream to settle on the surface of the milk because then I have to pu...

thermodynamics experimental-physics home-experiment popular-science

user54412

Someone confirm for me: this doesn't happen with homogenized milk, right? But where does one get non-homogenized milk these days? Certainly not in the US.

user54412

Also, why does one boil milk for toddlers? Is the milk also not pasteurized?

user685252

Perhaps on the farm?

Galois in the Field

In quantum mechanics(?) we have the state of a physical system given by $\psi$ and the observable $H$ is the hamiltonian.

$H$ is a linear operator, is $\psi$?

I thought it was, but how does one derive it for Schroedinger's equation?

user54412

06:36

$\psi$ is an element of the Hilbert space upon which operators act. It is not an operator at all.

Galois in the Field

Oh gosh, I thought being an element of a hilbert space implied being a linear operator, but I suppose you are saying that a 1x1 matrix is essentially a scalar, so scalars exist in a hilbert space?

A hilbert space is a linear space that is complete under its inner product, so it consists of linear operators correct?

user54412

No, it consists of vectors. Operators map vectors into other vectors. Vectors at best can be cajoled into mapping vectors into scalars.

user54412

You can think of elements of a given Hilbert space as column vectors, all of the same length (possibly countably infinite).

user54412

The operators that act on the Hilbert space can then be thought of as square matrices of that length in each direction.

Galois in the Field

Okay, thank you, so a hilbert space $\mathcal H$ is a space consisting of column vectors all of the same possibly countibly infinite length, $n$ the dual space $\mathcal{H}^*$ consists of matrices(editted) of $n\times n$ size(except I am not sure what happens here in the countably infinite case)

user54412

06:45

$\mathcal{H}^*$ is the set of (linear -- everything here is linear) maps from $\mathcal{H}$ into $\mathbb{C}$. You can think of such maps as row vectors, not matrices.

Galois in the Field

Okay so the dual space consists of linear functionals, what is the space consisting of linear operators that act on $\mathcal {H}$?

user54412

When $n$ is finite, $\mathcal{H}^*$ is naturally isomorphic to $\mathcal{H}$, just as you can transpose column vectors to row vectors and vice versa.

user54412

When $n$ is infinite, $\mathcal{H}^*$ is larger than $\mathcal{H}$.

user54412

@GaloisintheField Um, just the space of operators on $\mathcal{H}$? I think I've seen it called $\Omega(\mathcal{H})$. Or sometimes $B(\mathcal{H})$ when considering only bounded operators.

Galois in the Field

Okay thanks, one last question. In Schroedinger's equation, we have $\frac{d\psi}{dt}$, where $\psi$ is the state of the system, so since $\psi\in \mathcal{H}$, it follows that $\psi$ is an $n$-length vector. What does this differentiation mean here?

Do we write $\psi$ out in terms of its basis, and then derive with respect to $t$?(if that makes sense)

Wait maybe I should say I have Schroedingers equation defined as $i\hbar \frac{d\psi}{dt} = H\psi$

Or is the point that we can just rewrite this as $\frac{d\psi}{dt} = -\frac{i}{\hbar} H \psi$

user54412

06:59

Basically yes. It's a common thing to write $\psi = \sum_n a_n \phi_n$ for some basis $\{\phi_n\}$ and then say $\dot\psi = \sum_n \dot a_n \phi_n$ by commuting the derivative and sum (generally works even in the infinite case thanks to nice properties of the Hilbert space). Then you plug these into the Schroedinger equation and solve for $a_n$ as a function of time, giving you the time evolution of $\psi$.

Galois in the Field

That's great, thanks so much, this clears up many of my confusions

user54412

If $H$ is diagonal in $\{\phi_n\}$ then you're left with $n$ trivial, independent ODEs to solve, which is why in QM you often hear "all we have to do is diagonalize the Hamiltonian."

Galois in the Field

Ahhh I see, that will probably come up very shortly, I am just going through some lectures notes on mathematical physics

2 hours later…

yuggib

08:34

@ChrisWhite for Hilbert spaces, $\mathcal{H}^*$ is isomorphic to $\mathcal{H}$ (Riesz's representation theorem)

Galois in the Field

My notes say an expectation value of physical observable on state $|\psi\rangle$ $\implies \langle A\rangle = \langle \psi |A |\psi \rangle$. They are saying that the expected value of observable $A$ on state $|\psi \rangle$ is found by calculating $\langle \psi |A |\psi\rangle$ ?

I don't know why they used the $\implies$ sign there

yuggib

It is the physical assumption/interpretation of the quantity, i.e. that of being the average value that the observable $A$ takes on the state

no implication needed, I agree

Galois in the Field

Okay thanks, that seemed a strange way to write it

yuggib

it is indeed

Galois in the Field

Doesn't $\langle \psi | A |\psi\rangle=\langle \psi | A\psi\rangle = \psi^\dagger A\psi$?

yuggib

08:44

it is the physicist's notation...anyways it is not equal to the last expression

Galois in the Field

Oh I thought $\langle v| = v^\dagger$ and $|v\rangle= v$

yuggib

it is the scalar product of $\psi$ and $A\psi$; i.e. $\langle \psi, A\psi\rangle$ (denoting the scalar product by $\langle\cdot,\cdot\rangle$)

what is your definition of $v^\dagger$??

Galois in the Field

conjugate transpose

yuggib

no, the scalar product is a scalar product, not a simple transposition

Galois in the Field

user image

Sorry his notes are a little strange to me

yuggib

08:47

do you know how a scalar (inner) product is defined?

Galois in the Field

I know how it is defined on a real space

yuggib

what is a real space?

a real general vector space?

Galois in the Field

real valued space I meant

It is just the classical dot product

yuggib

that is one possible scalar product on $\mathbb{R}^n$

Galois in the Field

Ahhh yes sorry

I know how inner products are defined

never read scalar product before

yuggib

08:49

ok, again physics terminology

Galois in the Field

The are sesquillinear

yuggib

exactly

so the Hilbert space is a complex vector space complete with some inner product

Galois in the Field

Yep

yuggib

and the notation $\langle\psi\lvert\phi\rangle$

used by physicists

stands for the inner product

an operator is often put as you wrote in the middle

Galois in the Field

And moving it to the left conjugate transposes it, and to the right leaves it unchanged I would now guess

yuggib

08:51

moving it to the left gives the adjoint operator

in infinite dimensions is a little bit more complicated than with matrices

anyways the idea is the same

Galois in the Field

Okay I see now

yuggib

the "bras" $\langle \psi\rvert$ can be seen as objects of the dual $\mathcal{H}^*$

that since it is isomorphic to $\mathcal{H}$ are identified with vectors of $\mathcal{H}$

Galois in the Field

So linear functions, and in the finite sense row vectors

yuggib

the vectors are denoted by $\lvert \psi\rangle$

yes

and the notation $\langle \phi\rvert \psi\rangle$ corresponds

to the linear functional $\langle \phi\lvert$ acting on the vector $\lvert\psi\rangle$

in finite dimension, the matrix row-column product

Galois in the Field

to $\begin{pmatrix}\phi_1&\phi_2&\cdots&\phi_n\end{pmatrix}\begin{pmatrix}\psi_1\\\ps‌i_2\\\vdots\\\psi_n\end{pmatrix}$

yuggib

08:55

roughly speaking

yes exactly

Galois in the Field

Awesome, thanks so much

yuggib

no prob ;)

Gennaro Marco Devincenzis

I find it funny that the most voted post in here is "I hate physicists".

(hello everybody)

Slereah

@0celo7 I am not greek

yuggib

technically, there is a more upvoted "I give up. Switching to astrology"

Gennaro Marco Devincenzis

09:08

which is almost equally funny.

yuggib

anyways, @Danu is not a physicist anymore :P

( neither am I :^) )

@GennaroMarcoDevincenzis yes, probably

Gennaro Marco Devincenzis

mathematical physicist?

yuggib

in some sense...

that is mathematics you know :P

technically, I did my PhD in analysis of PDEs

Gennaro Marco Devincenzis

interesting. what kind of PDEs?

yuggib

non-linear Schrodinger, non-linear Schrodinger-Klein Gordon, Maxwell-Newton

but it was just a disguise to study quantum field theories :D

it still is...

Gennaro Marco Devincenzis

09:14

heh, I feel you. the interrelationship between the two subjects is beautiful.

yuggib

yes, it's true

but when you go down the mathematical path, it is difficult to turn back :D

Gennaro Marco Devincenzis

I noticed that, and I agree, even though my mathematics essentially stops at an introductory course in PDEs and basic differential geometry

yuggib

you will eventually turn to mathematics as many many physicists before you did X-D

when you know better the community, you see that apart from France it is a very common trend

Gennaro Marco Devincenzis

uhm, should we open a personal chat?

yuggib

don't worry, the chat is deserted anyways :D

Gennaro Marco Devincenzis

09:20

fine. anyway, what do you mean by "apart from France"? what is the trend there?

yuggib

per questioni segrete possiamo sempre passare ad una lingua meno comprensibile per gli altri avventori ;)

Gennaro Marco Devincenzis

ahahahahah

yuggib

in France mathematical physics does not exist as a discipline

Slereah

(says the frenchman)

yuggib

so people that wants to do mathematical physics are actually in analysis

and mostly come from pure mathematics

you know...France is the country of mathematics

Slereah

09:22

It is Bourbaki country

yuggib

@Slereah knows something about that as a theoretical physicist :P

Gennaro Marco Devincenzis

yeah I do know. last year a field medalist came to Lucca (comics and games) and I believe he did some work regarding Landau damping

yuggib

who Villani?

Gennaro Marco Devincenzis

yep

yuggib

he wears always the same outfit also at conferences :P

with his green plastic spider

Gennaro Marco Devincenzis

09:23

yeah, I read about that. interesting character, I must say.

yuggib

now he is a superstar...not so active anymore

but I heard a talk of his, and he was quite good

obviously

they do not give Fields' medals away

Gennaro Marco Devincenzis

I thought so!

it wasn't a talk in Lucca though, he talked in two occasions: the first one was just an interview and in the second one he presented his book

yuggib

well, you have a very famous school of analysis applied to physics where you are as well...

Gennaro Marco Devincenzis

I just heard the first one because I couldn't reach the second one due to the city being overcrowded

Slereah

0

Q: Derivation of Dirac equation in curved spacetime

In all the Literature I have read, the covariant Dirac equation in curved spacetime is given as \begin{equation} \left(i\hbar\gamma^{\mu}(x)\left[\frac{\partial}{{\partial}x^{\mu}}-{\Gamma}_{\mu}(x)\right]-mc\right)\psi(x)=0 \end{equation} Where $\gamma^{\mu}(x)$ are the contravariant forms o...

dirac-equation spinors qft-in-curved-spacetime

Gennaro Marco Devincenzis

09:27

you mean Trieste?

Slereah

The answer has a link to vixra :D

yuggib

Pisa

Gennaro Marco Devincenzis

there indeed are several analysts here

the first ones that come to mind (because they held courses) are Spagnolo, Acquistapace and one whose name I forgot but he's probably the most famous one out of the three

Colombini

yuggib

colombini is quite known

but Ambrosio of SNS is even more known I think

Gennaro Marco Devincenzis

didn't know him, though I think I've heard his name before

there's also a Ricci

yuggib

09:30

transport equations in probability spaces

Gennaro Marco Devincenzis

is the PDEs field healthy nowadays?

yuggib

well, it is very active

Gennaro Marco Devincenzis

I always remember Arnold's introduction to his book on PDEs in which he says that most papers are "a specific solution to a specific equation" and he seemed annoyed at that

yuggib

well, it is not possible to do else (at least for now)

apart from some very general technique

you have to see how it works case by case

(especially in nonlinear equations)

Gennaro Marco Devincenzis

I figured as much, since in most physics textbooks non linearities are avoided like the plague

Slereah

09:35

Well

There are some general methods

Like inverse scattering

Gennaro Marco Devincenzis

is there an analogue of the fourier transform for non linear equations?

or, if not an analogue, a generalisation which might work in some cases

Slereah

Kinda?

Some nonlinear equations can be decomposed into solitons, I think?

yuggib

@Slereah Well, it is a strategy

but you have to see it works

there are many other strategies

but these are not theorems classifying existence or not of solutions for classes of PDEs

Slereah

Yeah

yuggib

@GennaroMarcoDevincenzis what do you mean by analogue of Fourier transform? in what sense?

Slereah

09:38

But it's still a pretty good recent advance

yuggib

@Slereah Dispersive estimates are even better IMHO :P

Gennaro Marco Devincenzis

I mean an analogue to the technique of transforming the equation, working in the transformed space and then antitransforming

yuggib

you use techniques based in Fourier transform (the so-called Fourier analysis) a lot also for non-linear PDEs

but in a different way

Slereah

@yuggib I will have to look it up!

@GennaroMarcoDevincenzis : Well IST is one of these

yuggib

like this book title "Fourier analysis and nonlinear partial differential equations " exemplifies ;)

Gennaro Marco Devincenzis

09:41

I'll give a look

yuggib

@Slereah Search dispersive estimates or "Strichartz estimates"

Slereah

My favorite method of solving non linear PDEs remain to look for a paper that already did it :p

Gennaro Marco Devincenzis

ahahahahah

sorry to bother you, but I always wondered: are techniques from perturbation theory are used in the analysis of PDEs as well?

David Z

@GennaroMarcoDevincenzis you can edit messages, y'know

Gennaro Marco Devincenzis

@DavidZ thanks, I didn't

David Z

09:51

just hit the up arrow (when the chat text box is empty)

Galois in the Field

@DavidZ Very cool. I always used the little down arrow on the left on my text

David Z

ah, yeah, that works too

yuggib

10:21

@GennaroMarcoDevincenzis Dyson series could be used for some application (not many)

Gennaro Marco Devincenzis

got it

Galois in the Field

10:44

In quantum mechanics we have energy levels given by a discrete set, what do these energy levels mean?

They are eigenvalues for an orthonormal basis of eigenstates?

Slereah

yes

yuggib

@GaloisintheField They are (if any) eigenvalues for the energy operator

that is a self-adjoint operator called Hamiltonian

Galois in the Field

The hamiltonian is the energy operator?

yuggib

yes

and it is also the generator of the dynamics

(i.e. Schrödinger equation)

note, however, that not every self-adjoint operator has eigenvalues

i.e. real numbers $\lambda$ such that the equation $H\psi=\lambda\psi$ is satisfied for some $\psi\in\mathcal{H}$

Galois in the Field

Fair enough

yuggib

10:49

if it has, those eigenvalues belong to the spectrum (i.e. the set of real values such that $H-\lambda I$, $I$ identity operator, is not invertible with bounded inverse)

however there are operators with non-empty spectrum and no eigenvalues

(purely continuous spectrum)

this is true only in infinite dimensions obviously

(also, the spectrum is never empty)

and it is a closed subset of the reals

Galois in the Field

And what's the meaning of energy levels? $\mathcal{E}_n = \{ \psi\in \mathcal{H}: H\psi = E_n\psi\}$? If we have an orthonormal basis of eigenstates, then we should only get $1$ element for each $n$ for $\mathcal{E}_n$

yuggib

the set you write is the set of eigenvectors

if you have an orthonormal basis of eigenvectors, then the set spans the whole Hilbert space

Galois in the Field

For each energy level though

yuggib

yes, there may be more than one

Gennaro Marco Devincenzis

has anyone tried to solve schroedinger's equation for the harmonic oscillator but with a quartic term (exactly, I mean)?

I'm currently trying with the same power series method used for the harmonic oscillator and I haven't got stuck yet

yuggib

10:56

$\psi(t)=e^{-it (-\Delta_x+x^2+x^4)}\psi_0$ :D

Gennaro Marco Devincenzis

heh!

yuggib

you asked for an exact solution...

Gennaro Marco Devincenzis

fair enough!

uhm, but aren't there conditions for defining the exponential of an operator? I recall reading that it must be bounded.

yuggib

alternatively, $\psi(t)=e^{-it(-\Delta_x+x^2)}\psi_0 +\int_0^t e^{-i(t-s)(-\Delta_x+x^2)}x^4\psi(s)ds$

if the operator $H$ is self-adjoint, you can always define $e^{-itH}$

(Stone's theorem, i.e. existence and uniqueness of the solution to the Schrödinger equation)

if you want to define the exponential by a power series, then it is only possible for bounded operators

Gennaro Marco Devincenzis

what is the alternative to the power series?

yuggib

11:02

spectral theorem

but it holds only for self-adjoint (and normal) operators

and it is much more complicated :P

Gennaro Marco Devincenzis

good to know!

yuggib

;)

anyways, the beauty of Stone's theorem is that it goes on the two ways

for every self-adjoint operator $H$ we know that $\psi(t)=e^{-itH}\psi_0$ is solution of the Schrödinger equation

and for any continuous (wrt $t$, in the Hilbert space norm) solution $\psi(t)$ of a Schrödinger-type equation there exists a self-adjoint operator $H$ such that $\psi(t)=e^{-itH}\psi_0$

better said, for each continuous map $t\mapsto \psi(t)$ such that $\psi(t)\psi(s)=\psi(t+s)$ for any $t,s\in \mathbb{R}$

then there exists a s-a $H$ such that $\psi_t = e^{-itH}\psi_0$

and solves the Schrödinger equation

the actual difficulty is to show that there are self-adjoint operators that describe well the physical systems

Gennaro Marco Devincenzis

11:24

now that's super cool. Is it obvious that every solution of the Schroedinger equation satisfies that property though?

oh ok, I misinterpreted it.

yuggib

@GennaroMarcoDevincenzis Of course you need also that the map has some form of linearity...i.e. $\psi(t)=U(t)\psi_0$, where $U(t)$ is a linear operator that encodes the time dependence

for there are also nonlinear equations of course ;)

0celo7

Bah, nonlinear crap

Just throw out the nonlinear terms, nothing much will change

yuggib

@0celo7 the world is linear (apart maybe from GR), but the effective world is non-linear

0celo7

String theory is nonlinear

So...

yuggib

the classical string theory maybe...the quantum one I doubt

not an expert

though

0celo7

11:36

Well it has self-interactions

Isn't that what nonlinearities mean in QFT?

yuggib

like any quantum theory (of fields)

the time evolution is always a linear operator in quantum theories

0celo7

The Lagrangian is a nonlinear sigma model for ST

Dunno if that means anything to you

yuggib

that is not the quantum evolution

this is the classical lagrangian :D

0celo7

I don't recall seeing the explicit time evolution operator in ST

But ST is just a fancy QFT so I'll trust you on that

yuggib

nevertheless, it would be a linear one

obviously no one knows how to find it

for now

neither for interacting QFTs :o

(in 3+1)

0celo7

11:58

>3+1

there'd be a lot of unemployed mathematical physicists if we were in 2+1

Jim

Success! Canada finally has a new leader! Can you say "Physics funding finally feels fantastically full"? That's a serious question. No matter how hard I try, I can't seem to say that and I'm wondering if you can

0celo7

Does the new research being funded raise GDP?

Jim

too early to tell

he was just voted in yesterday

actually, his party was voted in. In canadia, you don't vote for the prime minister

except in his riding

I'm still happy though! Our last PM smiled like a pedophile. This guy must be an improvement

Gennaro Marco Devincenzis

anybody knows this differential equation? $$-y''+2xy'+y(1+2x^4-2\alpha)=0$$

Jim

12:17

@GennaroMarcoDevincenzis what have you tried?

Gennaro Marco Devincenzis

power series solution

yuggib

@0celo7 Well...just the evolution for the $(\varphi^4)_3$ scalar theory is known...and for small coupling constants

so, we would have work even in 2+1 :P

in $0+1$, or maybe also $1+1$, we would have to think about something else

Gennaro Marco Devincenzis

which seems to work. I was wondering if the solutions are well known.

Jim

@GennaroMarcoDevincenzis can I assume x is independent of y?

and $\alpha$ is too

Gennaro Marco Devincenzis

$\alpha$ is a real parameter $\geq 0$. x is the variable in $y(x)$.

Jim

12:24

not well known as far as I know

have you asked a mathematician?

Gennaro Marco Devincenzis

I'm trying in the math.stackexchange chat.

skill patrol

12:37

I'm a doctor Jim, not a mathematician :P

Galois in the Field

If it isn't an annoyingly frequently asked question, can someone explain mathematically the idea behind schroedingers cat?

Jim

@skillpatrol sorry, Bones. I keep forgetting

skill patrol

:D

David Z

@GaloisintheField $\frac{1}{\sqrt{2}}(\lvert\text{dead}\rangle + \lvert\text{alive}\rangle)$, although I'm not sure how much that helps you

@GennaroMarcoDevincenzis FWIW Mathematica gives nothing useful

Jim

@GaloisintheField In quantum mechanics particles are described in states, let's call them $\Psi$ because I don't feel like doing the tex script for the proper state notation. An entangled particle is said to be in a state that is actually a combination of two or more independent eigenstates, $\Psi=\frac{1}{\sqrt2}(\Psi_a+\Psi_b)$. When you observe a particle, you can only observe it in one of the eigenstates. But which eigenstate will you observe? $\Psi_a$ or $\Psi_b$?

skill patrol

12:44

@GaloisintheField have you read this? There's a nice animation too :-)

Jim

We don't know until we make the observation, at which time it will change the particle's state to be exactly and only that eigenstate. Before the observation, the state is a combination of the two and behaves as such. It is both eigenstates simultaneously. Schroedinger's cat is jokingly said to be in the state David Z mentioned and so, until you look at it, it is literally both dead and alive

Galois in the Field

Okay so the Schroedingers cat analogy is bad then

obviously the cats heart is either beating or not

@Jim Observing it influences the particle in what way? mathematically if possible

David Z

@GaloisintheField Well, no, not necessarily. That's one way to take the analogy: that things which are seemingly obvious in classical mechanics just don't hold in quantum mechanics.

Then again the way Schroedinger intended it was more along the lines of what you're thinking

Jim

@GaloisintheField yes. On scales relevant to QM, there is no such thing as a passive observation. In everyday life, you can observe something from afar without interfering with it. But on QM scales, every observation requires interacting with the observed object

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