The h Bar

12:34 AM

@Mr.Feynman Wick rotation is a can of worms

the most general statement about it I know is the Osterwalder-Schrader reconstruction theorem showing the correspondence between Euclidean and Lorentzian n-point functions under $t\mapsto \mathrm{i}\tau$

and that's not really a statement about integrals

Silly Goose

1:14 AM

what is the deal with this so-called C*-algebras formalism of quantum. or hilbert space-less quantum

ACuriousMind

1:43 AM

@SillyGoose see e.g. this answer of mine but also this question of mine for parts that are (still) unclear to me

Mr. Feynman

7:44 AM

@ACuriousMind Starting my day with this message was hilarious

Silly Goose

man so hard to pick what to learn D:

Naveen V

does the h in h bar mean planck's constant ? 👀

Mr. Feynman

8:09 AM

@NaveenV $\hbar=\frac{h}{2\pi}$

Naveen V

ohhhh h-bar that is reduced planck's constant @Mr.Feynman?

Mr. Feynman

Yes, that's how you read it

Naveen V

i feel so dumb now 😆

Mr. Feynman

I found that out after a year of QM :P

Naveen V

i have a surface idea of QM and GR , what would you say is worth to study deeply first @Mr.Feynman

Mr. Feynman

8:14 AM

I'm not the right person to ask but I would say QM because introductory QM is far easier than introductory GR

Naveen V

ah thank you @Mr.Feynman

nickbros123

8:27 AM

@PM2Ring where can I read about this (1st year ug understandable level). LCAO was shoved down my throat by my Quantum chemistry prof.

PM 2Ring

9:00 AM

@nickbros123 Maybe you can discuss it with ACM...

2

A: Perturbation series in QED

Your latter option is what is meant - the perturbation series is an expansion in loop orders, and the power of $\alpha$ is what counts the loop order.

From en.wikipedia.org/wiki/Quantum_electrodynamics > if we want to calculate the probability amplitude for an electron to get from A to B, we must take into account all the possible ways: all possible Feynman diagrams with those endpoints. [...] we have a fractal-like situation

The fine structure constant is the scaling factor of that giant fractal Feynman diagram: as we zoom in, the contribution of a sub-diagram at zoom level n+1 is worth ~1/137 a sub-diagram at level n. So it doesn't take many levels before the sub-diagrams have negligible effect.

nickbros123

9:17 AM

Hmm, I dont think my math (or physics for this matter) is ripe yet, for this.

PM 2Ring

This makes perturbation theory in QED tractable. But figuring out the details was certainly worth a Nobel prize for Feynman, Schwinger, and Tomonaga.

In contrast, QCD perturbations don't work like that, which makes QCD much harder

@nickbros123 Ok. But even if you're not ready for all the details, it's good to get a rough qualitative feel for what's going on. So discussing it here in Chat is probably ideal.

But it's probably a Good Idea to discuss it with people like ACM, rather than me. ;)

nickbros123

9:32 AM

I know nothing about perturbation theory to be honest. my only exposure to quantum is through a 1 sem long introductory quantum chemistry course. The context of my question was purely based on the approach one takes to explain molecules of the kind of H2, where we add the wave functions of 2 H2+ in a linear combination.

PM 2Ring

9:43 AM

The fine structure constant got its name because it's related to how spectral lines split. en.wikipedia.org/wiki/Fine_structure

ACuriousMind

10:38 AM

@Mr.Feynman I meant it! I fear the day where someone asks me to explain how exactly spinors and Wick rotation interact in the cases where the Euclidean and Lorentzian spinor reps don't correspond to each other in obvious ways (e.g. one has Majoranas and the other doesn't)

Mr. Feynman

@ACuriousMind I'll wait until you forget this conversation (and I understand the context) and ask this :P

Slereah

@ACuriousMind also aren't you supposed to have some continuous path in between Lorentzian and Euclidian

What are the spin reps in between

ACuriousMind

@Slereah that's how Wick rotation is often presented - as a continuous rotation - but I don't think that's actually how one should think about it on a mathematical level

Slereah

is there a rigorous version of the Wick rotation

ACuriousMind

@Slereah I mentioned it above, Osterwalder-Schrader reconstruction

but I don't know how it deals with spinors/whether there is actually a spinor version of it

Slereah

10:52 AM

I'll look into it once my brain isn't fried anymore

I thought Wick rotation would be like some physics version of analytic continuation

ACuriousMind

it is

Slereah

doesn't that imply that there is some continuity?

is it like that awful twistor thing where you work on $\mathbb{C}^n$

ACuriousMind

well, the theorem just says "Under the following conditions, the analytic continuations of the Euclidean n-point functions are the n-point functions of a Lorentzian theory at pure imaginary $t$"

it doesn't assign any meaning to the value of these function "in between", i.e. at arbitary complex values

Slereah

fair enough

I do know that there's some notion of working on $\mathbb{C}^n$ and then euclidian and lorentzian metrics are special cases in twistor theory

idk if it relates somewhat

ACuriousMind

of course "intuition" wants to say that at 45° we have something like a 50-50 mixture of Euclidean and Lorentzian but I don't think that's what's going on

Mr. Feynman

10:56 AM

Anyways, I don't think my question needed complicated stuff about Wick rotations. Maybe showing the page will help

The blue part is clear

Slereah

What are spinors like for the $U(n)$ group 🤯

Mr. Feynman

The alternative way (red) is my question

ACuriousMind

@Mr.Feynman what is the question

everything that's written there is correct

it doesn't tell you why you'd want to do this analytic continuation but that doesn't mean you can't do it :P

just go with it

Mr. Feynman

I don't understand why I can just decide to analytically continue my function and then consider the imaginary axis

We get decreasing exponential this way, sure. It looks like some ad hoc manipulation to get what they want, though

@ACuriousMind If you mean that I could analytically continue that function, yes. The problem is that I don't see why it is allowed here - without changing the meaning of that expression - and it doesn't change anything

Slereah

Can you do analytic continuation to any signature

ACuriousMind

11:04 AM

@Mr.Feynman it's just formal manipulation

stop trying to assign meaning to every step of a computation, this generally sparks no joy in QFT :P

Mr. Feynman

@ACuriousMind Mhh... But at the beginning we have $t\in\mathbb{R}$, then by analytical continuation $\mathbb{R}\ni t\to t'\in\mathbb{C}$ and we now consider the line $t'=i\tau$ with $\tau\in\mathbb{R}$

ACuriousMind

in the end this weird continuation just disappears again because you end up with a contour integral that you can deform again to go along the real axis

Mr. Feynman

The thing about is not integrated though

ACuriousMind

not yet!

this is just an intermediate step

(I'm assuming this is part of a derivation of the Feynman rules)

e.g. Weigand's notes do this and explicitly tell you that the explanation of how this weirdness disappears again comes like 10 pages later

Mr. Feynman

ACuriousMind

11:08 AM

it's a typical "proof that falls from the sky" - you can't understand what's going on without knowing where it'll end up

Mr. Feynman

Then they do this for the N-point green function

And calculate the ratio to get rid of the normalization

And finally they relate this to a ratio of path integrals

ACuriousMind

sure - again, it's just formal manipulation, no one is asking you to interpret the meaning of the imaginary time here physically

Mr. Feynman

Maybe I'm not explaining well what I don't understand. My problem is that this formal manipulation is done to lead to decreasing exponential and take the dominant term, that is the least suppressed. It looks like something ad hoc to get what we want

We couldn't have excluded the other terms of the sum without that machinery

ACuriousMind

yes?

Mr. Feynman

But you are changing something making the real variable $t$ complex

ACuriousMind

11:14 AM

that is indeed the point :P

Mr. Feynman

The only thing that would look "legit" to me would be to write $\mathbb{R}\ni t=-i\tau$ without analytical continuation, which would not help because it would make $\tau$ imaginary and nothing would really change in my expression above (so I couldn't exclude the exponential). If I analytically continue, it works but the thing I'm left with after the analytical continuation is not (?) what I had in the first place

ACuriousMind

I don't really understand the problem - the claim is that these equations are true

whether the analytical continuation they involve is what you "started with" is not really relevant to that claim

Mr. Feynman

If it were just rewriting the real $t$ in another way I would understand we are just rewriting things differently, analytical continuation looks like more than just rewriting here

Unless you too mean that "formal manipulation" is more than just "rewriting in another form"

Relativisticcucumber

hello, i have a question about the following photo. In this equation, i am trying to figure out how the solution to the first component of 2.8 is 2.10. i see that we can write F in terms of A, which is by definition of the electromagnetic field tensor as i understand it? Then, i see that j can be written in terms of the wave function and gamma matrices, but i do not see how we unite these into one equation whose solution is 2.8?

ACuriousMind

@Mr.Feynman ah, indeed, we are not "rewriting" the equation you already have

we're just writing down another equation using the analytical continuation that is also true :P

Mr. Feynman

11:23 AM

Ok! That's what I needed

Then I can accept I will understand why later

@ACuriousMind And this equation is what we're using from now on

ACuriousMind

@Relativisticcucumber I'm not sure how what you say here relates to the text in the picture

the picture is just talking about (one half of) the classical Maxwell equations, there are no wavefunctions or $\gamma$s there

the claim that 2.10 solves the first component of 2.8 just follows from the general theory of Green's functions

Relativisticcucumber

sorry i was going off of this:

ACuriousMind

oh but what $j^0$ is doesn't matter for anything in the first quote, right?

that 2.10 is a solution doesn't depend on whether or not $j^0(x)$ is just a given function or given in terms of some $\psi(x)$

Relativisticcucumber

yes this is kind of what i am struggling with -- why we would want to represent it in this way? because if i write out this equation in component form, it naively to me at least looks like it would be very easy to solve? but i think i am missing something

ACuriousMind

@Relativisticcucumber the useful thing about the Green's function solution 2.10 is that you don't need to solve any equations for it

you can just compute the integral

it's a solution that's true regardless of how difficult the equation would be to solve "by hand" for some specific $j^0(x)$

Relativisticcucumber

11:34 AM

okay i see, i will look into how to solve this type of equation thank you

Slereah

I wonder if there's some structural way to understand Wick rotation

Both Lorentzian and complex structures are based on some vector field

Slereah

11:58 AM

> A more accurate way of saying what is in my 1988 paper is that I look at the twistor space of orthogonal complex structures over a 4d Riemannian manifold, and try and identify the electroweak U(2) as the subgroup of SO(4) at each point in twistor space that commutes with the complex structure.

🤔

> Analytic continuation between Minkowski and Euclidean space-time can be naturally performed, since twistor geometry provides their joint complexification.

closest I can find on the topic is Woit's stuff

DIRAC1930

For single particle QM, the Greens function (for any system with energy eigenstates $\psi_n(X)$) takes the form $$
G(X,Y) = \sum_n \frac{{\psi_n}(X){\psi_n^*}(Y)}{E-E_n}
$$ i.e. there is a pole at each excited energy level of the system.

The same thing appears to be happening in QFT except the poles from the excitations above the vacuum are either a multiparticle state (that contributes to a branch cut) or an isolated pole at $P^2=M^2$. As a function of $E$ (which is what we are interested in in non-rel QFT), this amounts to a continuum of poles for every allowed $\mathbf{k}$ value at $E =\pm \sqrt{ \mathbf{k}^2 - M^2}$ (i.e. the pole gives the dispersion relation in the language of non-rel QFT).

Ryder Rude

Do u like board games?

DIRAC1930

In the case of unstable particles, we will get a pole in the 2nd Reimann sheet (the unphysical sheet) which will affect the spectral function in the physical sheet. If we associate the pole with a particle, we will get a decomposition of the particle into propagators of free particles over a range of masses I think

I'm not sure the last but is correct. I need to go over it properly

Ryder Rude

@Mr.Feynman We will use the complex t to carry out computation and stuff, and in the end, we will restrict the results to real t

It works becuz complex is more general than real

Ryder Rude

12:33 PM

Tho is some cases, we leave the time as imaginary. e.g. in quantum statistical mechanics. The idea there is that the temperature stuff really behaves like the quantum theory of imaginary time

DIRAC1930

12:44 PM

@ACuriousMind What do you think?

DIRAC1930

2:47 PM

If the wavefunction of a free particle is $e^{\imath P^\mu X_\mu}$, then maybe the wavefunction for $1$ unstable particle could be $\int \mathrm{d} s \rho(s) e^{\imath P^\mu (s) X_\mu}$ where $\rho(s)$ is the spectral function on the physical sheet associated with the pole in the 2nd Reimannian sheet

If the pole is shifted slightly, then there will be an isolated spectral function away from the branch cut associated with the multiparticle states

It will be Lorentzian in shape

What do you think?

Obliv

Sorry for random math but can someone verify if this is correct, $y = ce^{-\int P(x)dx}, Dy = -cP(x)e^{-\int P(x)dx}$ ?

very basic chain rule I just want to make sure I did it right, I'm really iffy with calc 1 stuff

nvm im p sure that's correct

ACuriousMind

@DIRAC1930 relativistically it is not really clear what "the wavefunction" even means - remember there are no good relativistic position operators

@Obliv it's wrong/nonsense - what does $D$ mean, differentiation with respect to what variable? and something like $f(x)\int f(x)\mathrm{d}x$ is always suspect notation - a variable ($x$) should not occur both as a free variable and an integration variable

Obliv

3:05 PM

It's used in my book for differentiation w.r.t. x

Hmm I'm not sure what an integration variable is

or do you mean $e^y$ where $y=\int P(x) dx$ so treating $y$ as a free variable in the former but $y=x$ as an integration variable in the latter

it's still just a parameter in both cases though?

that's horrendous let me rewrite

DIRAC1930

What about eqn 23.1 in L&L vol 4

ACuriousMind

@Obliv I mean the variable behind the $\mathrm{d}$

@DIRAC1930 what about it?

there are contexts where it makes sense to talk about a particle wavefunction like that, particularly in the "hacky" viewpoints of "relativistic QM"

which ultimately turns out to be inconsistent in various places which is what we need QFT for

this progression non. rel. QM -> rel. QM -> QFT is pretty tricky and we need to be careful which descriptions are appropriate at what stage

I don't think it is useful to mix-and-match these various descriptions, in particular I almost never want to think about a "wavefunction" in a QFT context, except when I'm trying to match QFT with the QM description

DIRAC1930

Hmm but when we are describing asymptotic states, aren't we really talking about a wavefunction?

ACuriousMind

@Obliv maybe let's start a bit smaller: Of what variable is $y$ a function that you want to differentiation it w.r.t.?

or: what are the limits on that integration in the exponent?

Ryder Rude

We can talk about vectors in the hilbert space

ACuriousMind

3:18 PM

@DIRAC1930 In momentum space, perhaps

the momentum operators are fine, we can talk about the states $\lvert p\rangle$ and wavefunctions in that space

DIRAC1930

Ah okay, so the space I am in is momentum space

Obliv

@Acuriousmind they are of $x$ values, I'm trying to solve $a_1(x)\frac{dy}{dx} + a_0(x)y = g(x)$ but I solved the homogeneous case to get $y=ce^{-\int P(x)dx}$ where $P(x) = \frac{a_0(x)}{a_1(x)}$

I know you're just going to multiply by an integrating factor but I'm just working through it myself to make more sense of it

I just wanted to take the derivative of the soln. to the homogeneous case

DIRAC1930

My integration in my wavefunction above is just a superposition of momentum eigenstates

ACuriousMind

@Obliv if $y$ is suppoed to be a function $y(x)$ then it makes no sense that the exponential integrates over $x$

again: what are the limits on that integral?

Obliv

wait what

How does $y(x) = e^{\int f(x) dx}$ make no sense

ACuriousMind

3:24 PM

@Obliv because the r.h.s. is not a function of $x$

Obliv

assuming the interval is all of the reals and it's continuous

but it is.., $f(x)$ is a function of $x$

ACuriousMind

if the integral limits are just $\int_{-\infty}^\infty$ then the r.h.s. is just a number, not a function

Obliv

$y(x) = e^{\int x^2 dx}$ I swear that makes sense

ACuriousMind

no it doesn't

Obliv

but it reduces to $y(x) = Ce^{x^3\over3}$

ACuriousMind

3:26 PM

when you integrate with respect to a variable you "remove" it: $\int_{-\infty}^\infty f(x)\mathrm{d}x$ is not a function, it is just a number

I don't think you are using integral signs correctly

when you write $\int f(x)\mathrm{d}x$ you seem to mean "the antiderivative of $f$ w.r.t. $x$", but the standard way to write that with integral notation is $F(x) = \int_c^x f(z)\mathrm{d}z$ for some constant $c$

if the integral is supposed to be a function of $x$ the $x$ needs to appear in the limit, not as the integration variable

Ryder Rude

I think obliv means integral in the sense of "anti-derivative" @ACuriousMind

Slereah

Apparently Thales thought that both amber and magnets had a soul

Obliv

I am thoroughly confused, calc 1 was almost 8 years ago for me so I'm going off of old data

Slereah

This was the original unification of electricity and magnetism

ACuriousMind

@RyderRude I literally just wrote that :P

Ryder Rude

3:28 PM

So the integral can b wrt x, as well as the result can b a function of x

@ACuriousMind sorry then :P

Oh ye u literally wrote anti derivative

Obliv

@ACuriousMind to solve this integral you take some parent function $g(z)$ evaluated at $g(x) - g(c)$ correct?

ACuriousMind

"parent function"? Do you mean antiderivative?

i.e. a function whose derivative is $f$?

Ryder Rude

Maybe one shud just use another notation for anti-derivative to avoid this confusion. Something like $D^{-1}f(x) $

Obliv

yes, and I thought the point of an indefinite integral was to save time in writing what you did and the context would be sufficient?

so it should technically be $y = ce^{-\int_c^x P(z) dz}$ for some constant c

Oh..

Man it's been so long, so that's where the constant of integration comes from I completely forgot

Ryder Rude

In school, we took $\int f(x) dx$ to just mean $F(x) +C$. It's what that notation meant there

Obliv

3:38 PM

yep

Ryder Rude

But this notation maybe clashes a bit with the definite integral notation

Idk

I think one problm with this notation is that it doesn't really fix $F(x)$. U cud just as well choose $G(x) +C$ where $G(x) =F(x) +C'$

PM 2Ring

@RyderRude Sure, but that's not a function of x. It's a family of functions of x, parametrized by C.

Ryder Rude

The othr definition $F(x)=\int _c^ x f(z) dz computationally determines the LHS

@PM2Ring yeah. F can b defined as the family of functions satisfying $D^{-1}F=f$

PM 2Ring

Right

Obliv

$\int_c^x f(z)dz = F(x) + F(c) = \lim_{\Delta x \to 0}\sum_{i=1}^{\infty}f(c+\Delta x)\Delta x$ ?

PM 2Ring

3:47 PM

OTOH, I am sympathetic to Obliv's original notation. But you do need to be explicit about what happens with the constant of integration. Also, it's much more common to do stuff like $y(t)=\int_0^t f(x) dx$, where $x$ is a dummy variable.

Obliv

nvm I don't remember how to construct riemann sums lol

The integral notation is just shorthand for the associated infinite riemann sum anywho

so regardless, the idea is then you have nonhomogeneous case $\frac{dy}{dx} + P(x)y = g(x)$ and you divide both sides by $\frac{dy}{dx}$ the derivative of the soln. to the homogeneous case?

$1 + \frac{P(x)y}{-cP(x)e^{-\int P(x) dx}} = \frac{g(x)}{-cP(x)e^{-\int P(x)dx}}$

and solve for $y$ ?

$P(x)y = g(x) - \frac{1}{-cP(x)e^{-\int P(x) dx}}$

$y = \frac{g(x)}{P(x)} - \frac{1}{-c(P(x))^2e^{-\int P(x)dx}}$ seems wrong to me idk

ya nvm I don't think you can divide $y/y_c$ where $y_c$ is the complementary solution

$\frac{dy}{dx} = Dy_p$ , $Dy_p / Dy_c \ne 1$

Ryder Rude

4:11 PM

@Obliv u r suposd to multiply both sides by $e^{\int P dx}$

@Obliv both sides of this

Then undo the derivative product rule on the LHS. This will transform it to $D( y e^{\int Pdx}) $

Ryder Rude

4:26 PM

Why does nobody chat here!?

There r 11 people. We cud have a crowded conversation

Some people hav never said a word

Amit

Maybe it's the Quantum eraser again

Ryder Rude

Whats it about @Amit

Amit

A joke... ^_^ But quantum eraser is cool

Ryder Rude

It seems 2 b about decoherence

Amit

I'm not sure it's only that, there are several types of experiments showing that perhaps on the small scale causality is not really unidirectional

TSVF is an interesting attempt to formalize QM to explain that

Ryder Rude

4:34 PM

First time im hearing of this formalism

I heard of smthing related called the transactional qm

It was also bout retrocausality

Amit

Yes I think there are several people who hit upon this general idea at around the same time

Ryder Rude

Feynman and Wheeler had this crazy idea too. Tho i think this formalism is different from theirs mathematically

Amit

I remember Feynman telling this story of Wheeler calling him... "You know I think there's only one electron in the universe"

Ryder Rude

Yeah, thats what i read too. It was maybe part of Feynman's Nobel speech

Amit

Highly likely, he had a lot of interesting anecdotes in it

Ryder Rude

4:38 PM

It was rejected becuz the number of positrons is waaay lower than electrons

Or so we've estimated i guess

Transactional QM seems sci fi cool. But idk its math

TSVF is using two state vectors. One backward in time. Interesting

Amit

TSVF is motivated by some experiments too, like the EV bomb, the 3 boxes one...

Three box paradox

Ryder Rude

What peculiar thing happens that cant b explained by usual qm?

Idk about this experiment. I will read about it

Amit

Actually I know the EV bomb was done, not sure if the three box paradox was applied as an experiment

EV bomb without the bomb

The peculiar things have to do with weak values that apparently tell you something "paradoxical" about the state of the system in between two measurements

Ryder Rude

4:53 PM

When i first read about double slit experiment, i had a similar idea that the electron cud actually interact with the electrons in the future. Using this "interaction thru time", the pattern got made :P

It was becuz the book said "electron r fired one by one. So they cant interact"

But this idea was really trash, admittedly

geocalc33

I have a random question

DIRAC1930

I think the analogy between non-rel and rel is the same. In non-rel a particle is identified as an excitation that satisfies a dispersion relation $\tilde{E}(k)$ which is system dependent e.g. for a free particle we have $E=k^2/2m$ which is just the non rel limit of the dispersion relation $P^2=M^2$ minus the rest energy. In rel QFT, the renormalization condition is enforcing $P^2=M^2$ because that's all it can be from physical experiments

I'll see what happens when I have an unstable particle in it's rest frame and see it's modes in terms of the spectral function

Amit

5:14 PM

@RyderRude It's fun speculating about that. After studying a bit of GR I started thinking, maybe on such a small scale, the entire manifold is just waving around so the wave functions are just waves in the manifold or something

@geocalc33 It is customary to ask without prior notice

geocalc33

I was going to ask but forgot to actually follow up on that

I will ask momentarily

I'm wondering in which contexts does the generalised guassian function show up in physics

I'm referring to this function: $$ f(x)=e^{-x^p}$$

where for $p=2$ is the familiar gaussian

Amit

I think it often comes up in QM to model wave packets

Wave packet

In physics, a wave packet (or wave train) is a short "burst" or "envelope" of localized wave action that travels as a unit. A wave packet can be analyzed into, or can be synthesized from, an infinite set of component sinusoidal waves of different wavenumbers, with phases and amplitudes such that they interfere constructively only over a small region of space, and destructively elsewhere. Each component wave function, and hence the wave packet, are solutions of a wave equation. Depending on the wave equation, the wave packet's profile may remain constant (no dispersion, see figure) or it may change...

geocalc33

5:33 PM

Okay I will check out that link

Not trying to troll or anything but what if you permute the variables and sum

$$f(x)=\sum_{p=1}^\infty e^{-p^x}$$

Does this show up in physics anywhere where? Note that $p=1,2$ we have a geometric series, and a jacobi theta closed form respectively.

Amit

I don't know... if this construction has a name you can look it up in arxiv or something :)

geocalc33

I would venture to say it definitely does not appear anywhere in physics

because I haven't seen it anywhere lol

for $p>2$ there aren't easily attainable closed forms so it's kind of hard to understand the function.

Amit

Hmm it also diverges doesn't it?

geocalc33

it converges for real x>0

Amit

If it converges maybe it's useful for something lol

geocalc33

5:43 PM

but is only quasi analytic for real $x>1$

@amit yes lol!

Amit

If you can cook up $f(x)$ in a way that its integral is equal to $1$ using this monstrosity I guess you get a hypothetical wave function

no sorry, integral of square of itself

geocalc33

$$ \int_0^\infty f(x)^2 ~dx $$

is this what you mean?

Amit

should be from $-\infty$ to $+\infty$, you know a wave function

now where is that inf sign

lol

geocalc33

I believe you are talking about this $\int_{-\infty}^\infty f(x^2) ~dx$

Amit

No

geocalc33

5:53 PM

Can't integrate the f(x)^2 from -infinity to infinity

Amit

$\int_{-\infty}^\infty \mathrm{d}x \,\, ||{\psi(x)}||^2 = 1$

If you change $x$ to $|x|$ I think it converges

That's cheating I guess

More Anonymous

@JohnRennie i thought about your answer on the cosmological constant on a previous question. I think it's wrong?

1

Q: The energy conditions and cosmological constant?

So I thought it didn't matter which side of the equation the cosmological constant was one (did it emerge from geometry or the stress energy tensor). However, then I remembered the weak , strong, null, etc energy conditions. Now, if I presume the cosmological constant emerged from the stress ener...

Amit

What are you trying to achieve anyway?

geocalc33

@Amit Just wondering if there was an application that I was unaware of...

Amit

Okay, that's a very specific thing you're looking at though, I thought there may be a reason

Obliv

6:03 PM

@RyderRude so by multiplying a non-homogeneous linear D.E. of first order by the homogeneous solution, you can find the particular solution.

$y_c=ce^{-\int P(x) dx}$ is the homogeneous solution to $a_1(x)\frac{dy}{dx} + a_0(x)y = g(x)$

geocalc33

@Amit I mean there are some reasons I'm looking at it...they may not be very convincing reasons though lol.

Amit

Ah, fine :)

Obliv

I understand you get $D[\frac{dy}{dx}ce^{-\int P(x) dx} + y P(x) ce^{-\int P(x) dx}] = ce^{-\int P(x) dx}g(x)$ which you can rearrange and simplify to $D[yy_c]$ on the left hand side

geocalc33

it's a very very ugly function in many ways

1) admits a very complicated analytic continuation 2) no closed forms for x>2 3) loses analyticity on the real line for x>1

Obliv

how is the differential operator $D = \frac{d}{dx}$ defined ? It maps a set of functions to another set, with the notion of $D[xy] = x'y + y'x$, $D[x+y] = x' + y'$ etc

it's associative in function addition but not function multiplication?

geocalc33

6:12 PM

but it's a lesson in what happens when you "mix" incompatible spaces in some respect. i.e. you get a messy structure

case in point - just interchange the sum with product

Obliv

Is there a sequence of D.E. for each type,class,form, etc, that are "fundamental"

geocalc33

you obtain $f(x)=e^{\zeta(x)}$. Indicating the sum and the summand is a repulsive pair

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