The h Bar

5:01 AM

@bolbteppa the articles on Quanta are generally very good, in a field (popular science) where standards can be very poor indeed. In particular Natalie Wolchover's articles are excellent.

5:20 AM

Yeah I think some of the articles on there are a mixed bag but this one on the BMW stuff was very useful

Three different groups did the calculation independently and got the same answer so we can be confident the calculation has been done correctly. The problem is that we can't be sure it is the correct calculation.

antimony

Hi @JohnRennie just wanted to say thank you for giving some amazing explanations! You really help make clear what have been confusing topics

Assuming everything else is right, it maybe looks like the question is the values they used - my vague understanding of this is that the BMW people are basically saying one of the values they chose is not accurate enough, and if you account for it then the discrepancy narrows, and it wasn't included it because it needs to be evaluated more but they would in future work if it's analyzed to be worth including

@antimony you're welcome :-)

Q: How do I find the approximate surface area of a chicken?

6:27 AM

14

I'm working on building a chicken army and I'm trying to find out how much metal or kevlar (still deciding) I need to make armor for the chickens. this measurement does not need to be exact I'm just trying to get an estimate for how much I will need. You will be spared when my chickens take over ...

it seems as if physics has peaked.

Q: Folks, we need to talk about the surface-area-of-a-chicken question

37

This question How do I find the approximate surface area of a chicken? underwent a severe close/reopen yo-yo cycle. It was closed once by a moderator, then reopened by community members, then re-closed by community members, then re-opened again, and then re-closed by community vote again (as a ...

discussion scope specific-question vote-to-close

I find myself ambivalent about questions like that. I can see arguments both for closing them and for letting them stand.

8:49 AM

Why do people try to find the volume

You could have finite volume and infinite surface

there are better metrics to find surface

Find out the blackbody radiation of the chicken

total radiance should be proportional to the surface

9:13 AM

I think Prof. Steane's more serious answer does pretty much that

10:33 AM

hi, all

is there a way to summarize the exact framework of quantum mechanics? classical mechanics is the triple $(T^*\Bbb R^n = \Bbb R^{2n}, \omega, H)$ consisting of the standard symplectic 2-form $\omega$ and a Hamiltonian $H : \Bbb R^{2n} \to \Bbb R$

Does that simple description still work in constrained Hamiltonian dynamics

it seems there's too much going on in QM. there's the algebra $A = L^2(\Bbb R^{2n})$, and the operator algebra (allowing partially defined unbounded operators) $\mathcal{A} = "\text{End}(A)"$, a representation of the underlying phase space $\Bbb R^{2n}$ inside $\mathcal{A}$ (the position and momentum operators), a bracket on $\mathcal{A}$ (commutator). what exactly is the deal?

the symplectic structure on $\Bbb R^{2n}$ kind of gets encoded in the operator algebra as the commutator ($[x, x] = [p, p] = 0$, $[x_i, p_j] = \delta_{ij} ih$)

what the hell?

where can i find a precise context for QM

@bolbteppa i don't know physics terminology, so i don't know the significance of "constrained" there. but to my knowledge hamiltonian dynamics is an example of a hamiltonian flow on a symplectic manifold

@BalarkaSen well, a very generic way to formalize QM is just as a $C^\ast$-algebra with a distinguished self-adjoint element that's the Hamiltonian. You get the state space(s) via GNS construction

What you're looking for is very likely a 'geometric quantization' perspective

"Geometric quantization is an attempt at using the differential-geometric ingredients of classical phase spaces regarded as symplectic manifolds in order to define a corresponding quantum theory. Generally, the process of geometric quantization is applicable to other symplectic manifolds, not only cotangent spaces. "

ncatlab.org/nlab/show/geometric+quantization

@BalarkaSen Your classical mechanics definition is too weak, you need to allow symplectic spaces other than $T^\ast \mathbb{R}^n$ to accomodate gauge theories/constrained theories, where the "real" phase space is a submanifold inside $T^\ast\mathbb{R}^{2n}$, additionally quotiented by the action of gauge symmetry

10:45 AM

I already learned one thing from this nlab(!), that geometric quantization is apparently the Schrodinger picture version, and 'deformation quantization' is the Heisenberg picture version ncatlab.org/nlab/show/deformation+quantization

@ACuriousMind excellent! that seems like the way to go

mathematicians know the relevant procedures under "symplectic reduction"

@ACuriousMind fair, i was just thinking of simple minded classical hamiltonian mech in R^2n

I follow and agree with what you said and see bolbteppa's point

@bolbteppa i have heard this word but i felt like i should understand the story on a euclidean space before moving to manifolds

You see the quote above makes this point about other symplectic manifolds too

@BalarkaSen if one wants to be a bit more concrete, then one can switch this also to a tuple of a Hilbert space $\mathcal{H}$, a distinguished set of self-adjoint operators $\mathcal{O}$ on $\mathcal{H}$ called the "observables" and a distinguished observable $H$ as the Hamiltonian

10:49 AM

@BalarkaSen Depends what you mean by "Quantum mechanics"

It can be as simple as just the Dirac-von Neumann axioms

But those are super broad

They include a lot of "unrealistic" QMs

in case you obtained this via quantization of a classical system, you'll have $n$ copies of the algebra $[x,p] = \mathrm{i}$ in the set of observables, but this isn't necessarily a requirement - there are plenty of quantum systems that do not have a position or momentum degree of freedom

I don't think any of this is settled in the QM case but something like geometric quantization is probably a good place to aim for

there's no lack of quantization schemes

Or axiomatization of quantum mechanics

@bolbteppa trying to axiomatize QM is very different from looking at quantization procedures

Axiomatization of QM isn't done with your mind, it's done with your heart

10:52 AM

not all quantum systems are produced by quantization (e.g. qubits, and more generally any with finite-dimensional spaces of states)

You look at them and then you think "Yes, that's the axiomatization for me"

@ACuriousMind agreed. i like the C* algebra perspective

Yeah I have no idea about the qubit stuff

Also there's another aspect which is like

Most theoreticians when they say "axiomatization", they only mean the mathematical part

ncatlab.org/nlab/show/C%2A+algebraic+deformation+quantization

10:54 AM

But there are also axiomatizations which include the physical part

and these are much much worse

@BalarkaSen see physics.stackexchange.com/q/615696/50583 and its linked question for discussion of how to relate the $C^\ast$-algebra perspective to the "usual" physical viewpoint where you have a concrete space of states

"This is in contrast to formal deformation quantization, where one asks not for C*-algebras but just for formal power series algebras. Where formal deformation quantization is perturbative quantization (perturbation in Planck's constant, see Collini 16), strict deformation quantization is meant to reflect non-perturbative quantization."

@Slereah yeah i am not looking for a functorial description or whatever. what i want to see is exactly how much structure is going in the theory

i am currently obsessed with the idea that the symplectic camel theorem is a "classical" uncertainty principle

Well, at its core then QM is just a Hilbert space and some algebra of linear operators on it.

i would like to make this as precise as possible

10:55 AM

That is literally all you need

no my point is there's some underlying phase space

If you want a specific QM that's another matter

There is no underlying phase space in QM

I think this is an endless endless endless rabbit hole if you even try to apply it to conformal field theory

You can have a QM that's just the Hilbert space $\{ *\}$

the underlying phase space in QM is R^(2n). you have position and momentum coordinates. shrug.

10:56 AM

if you want to associate a space to your QM, that's a different matter

i dont know that there are many QMs haha im just a beginner reading a physics text

and trying to figure out what the hell is going on

Well then if you want to start, just start with the basics I suppose?

Which is just the theory of linear operators on a Hilbert space

that's the broadest QM there is

"A conformal field theory (CFT) is accordingly a functor on such a richer category of conformal cobordisms"

i understand what you mean, but your definition doesn't help illuminating any light on "position" and "momentum" operators, which are like Ch 1 material in Griffths, a physics basic text lol

if it doesn't help it doesn't help you know what i mean

i could go read some functional analysis text for unbounded operators but thats not what i want lol

He just throws the Schrodinger equation at you and tells you to use it

10:59 AM

Basically if you want your theory to mirror the underlying spacetime, what you need is to impose some conditions on your algebra of operators

if you have QM with position and momentum operators (which you pragmatically probably identify simply as the operators with $[x,p] = \mathrm{i}$), then you can do QM on phase space

ie impose the CCR, impose some symmetry operators, etc

this will structure your QM into something more realistic

@bolbteppa i am aware, im just saying your description has not helped

@ACuriousMind OK. I don't know terminology, but good to learn

if you're interested in symplectic non-squeezing and uncertainty, it would probably be a good idea to understand how the uncertainty principle manifests in this phase space formulation of QM

Like for your basic Schrodinger QM, you have your Hilbert space, you have your operators obeying the CCR $[x, p] = i$, $p$ generates space translations and $H$ generates time translation

That's about the structure you have for basic QM I think?

11:02 AM

@ACuriousMind yeah thats more or less what im trying to read, and see exactly how much goes into it

@Slereah yeah this is the context im familiar with

as mentioned, you can do this via some quantization procedure, or you can take these as axioms

or you can even not include all of that at all, for instance the Wightman axioms don't even mention the CCR at all IIRC

They just define an equation for the propagator

yeah im confused because there are a couple of different ways to contextualize QM; in another book i have seen it laid out probabilistically

Well it's the same thing as with all math theories

axiomatizations aren't fundamental

you can define them in a myriad of ways

The wiki on the phase space formalism shows some serious flirtation with violating basic quantum mechanics ("The concept of quantum trajectory is therefore a delicate issue here"), and even trying to treat position and momentum as in any sense equal is just a flagrant misunderstanding

Are integers a fundamental objects, sets, functions, natural number objects???

11:08 AM

@Slereah Of course I understand that, which is why I explained what I want to understand and what formalism is better for understanding it

I'm not trying to be Urs Schreiber haha

A good idea

although I think Urs Schreiber has a very definitive idea of what QM is

Only @Slereah will ever reach that goal

only way to explain anything ever is $\infty$-categories

I dunno, infinity might be too small

what if I need a theory of infinite categories

$\infty$-$\infty$-categories

11:09 AM

@bolbteppa with "equal footing" they just mean that the quantum state is represented as the q-probability function $f(q,p)$ that's both a function of position and momentum in contrast to Hilbert space formulations where the wavefunctions are only functions of one of them

But yeah if you want basic QM axiomatization as it is usually done, I think just do the usual quantization process?

OK, thanks! I will try to understand what that is

Yeah that phase space quantization is supposed to be legitimate

Thanks, all! Very helpful comments

which you can do via geometric quantization if you're feeling bold, or just define the algebra of observables via the phase space algebra

in which case Stone-von Neumann theorems gives you the Hilbert space etc etc

11:11 AM

Stone-von Neumann is an excellent theorem, I look forward to reading a proof soon

also make sure that your algebra of observable obeys the appropriate symmetries I guess?

I don't know all the details but classical symmetries are a little weird in QM because the Galilean group isn't good enough for those people

They want the Bargmann group or whatever it's called

Everything for QM on the phase space $\Bbb R^{2n}$ is in fact gauge-invariant under the symplectomorphisms though yes?

As in it is classically invariant

Coz preserves symplectic form => preserves the commutator

That's not "gauge", but yes it's invariant

Got it.

and also remember to have fun

2

11:14 AM

@Slereah important

Representation theory of the Galilean group

In nonrelativistic quantum mechanics, an account can be given of the existence of mass and spin (normally explained in Wigner's classification of relativistic mechanics) in terms of the representation theory of the Galilean group, which is the spacetime symmetry group of nonrelativistic quantum mechanics. In 3 + 1 dimensions, this is the subgroup of the affine group on (t, x, y, z), whose linear part leaves invariant both the metric (gμν = diag(1, 0, 0, 0)) and the (independent) dual metric (gμν = diag(0, 1, 1, 1)). A similar definition applies for n + 1 dimensions. We are interested in projective...

@Slereah it's just the usual business with projective representations

no different from having to use $\mathrm{SU}(2)$ instead of $\mathrm{SO}(3)$ or the Virasoro algebra instead of the Witt algebra

I can't find a good explanation of basic QM from this perspective, it's probably the best way to do QM in terms of QFT but the only accessible source is Ballentine which is iffy and doesn't really do it analogous to QFT

@ACuriousMind Does that mean that the Bargmann group has the spin group as a subgroup instead of the rotation group?

AFAIK the Bargmann group was just a central extension of the Galilean group?

@Slereah depends on how you define "Bargmann group". People usually just look at the algebra

11:18 AM

I don't know what that does, so could be the same idk

but if you look at the simply-connected group corresponding to the Bargmann algebra, it should contain spin and not just rotations, yes

Galilean transformation

In physics, a Galilean transformation is used to transform between the coordinates of two reference frames which differ only by constant relative motion within the constructs of Newtonian physics. These transformations together with spatial rotations and translations in space and time form the inhomogeneous Galilean group (assumed throughout below). Without the translations in space and time the group is the homogeneous Galilean group. The Galilean group is the group of motions of Galilean relativity acting on the four dimensions of space and time, forming the Galilean geometry. This is the passive...

The algebra is called the Bargmann algebra there

Why do we need the central extension, though

actually that's probably on PSE

@Slereah see physics.stackexchange.com/q/203944/50583

thx

If you enjoyed this question (and especially the answer), you will probably also enjoy the book on conformal field theory by Schottenloher (first linked book on the page), which covers this topic and much more. — Danu Oct 15 '15 at 21:20

Can't escape Amazon recommendation on PSE!

11:30 AM

IF you ENJOYED the viedo goys HIT that SUBSCRIBE button LIKE sobscroib

2

Danu the Youtuber

A: Representation of the Galileo Group and Central Charges

11

The boosts and translations do not commute in neither relativistic or nor relativistic systems, please see for example the case of the Poincare group. Since $K_i = M_{0i}$ , $(i=1,2,3)$, we get for the Poincare group: $[K_i, P_j] =i (\eta_{0j}P_i-\eta_{ij}P_0) = -i \delta_{ij}P_0$ Now, The Gal...

That's pretty good

12:01 PM

A pedagogical review of gravity as a gauge theory looks good

12:18 PM

Sorry this is a very weird noob question but if a TV or a computer is turned off, where did the electricity that turned the TV or computer go?

12:46 PM

usually devices are turned off by using a switch which breaks the electrical circuit

so that the charges can no longer move any more

and electric fields cannot propagate either

@Slereah so the electric fields just stay there?

the electric field drops off rapidly around the open switch

@Slereah so where did the electric fields go? Did they just disappear like that? I'm curious.

What do you mean, the tv etc requires a current (i.e. a flow of electrons) moving though it's wires to remain on, turning it off amounts to blocking the current flow

1:04 PM

@bolbteppa yes, exactly my question. So, the electric fields are still there but the fields got blocked?

Right, turning it off amounts to blocking the flow, but there are still electric charges all over the place, they are producing electric fields even when they're treated as stationary, but there's no closed circuit allowing them to flow through the tv to turn it on, any internal motion is random microscopic motion comparatively

@bolbteppa gotcha. Thank you for the detailed answer.

Ryan Unger

@ACuriousMind can you explain string theory

@RyanUnger tiny strings move

What more does one need

1:17 PM

@bolbteppa the equation, I assume

That's a 'detail'

@RyanUnger not in a coherent manner :P

I know a lot of bits and pieces about it, but it's never coalesced into some big picture, and I'm not sure whether that's my problem or there just isn't one :P

let's see what nlab says on the matter

Nothing specific for "string theory" itself

I guess you have to check like a specific string theory

That's too rudimentary for nlab

ncatlab.org/nlab/show/string+theory ?

1:28 PM

Even for "type IIA heterotic string theory" they don't show anything as vulgar as an equation

apparently there's something called "Little string theory", maybe it will be simpler

"A limit of the worldvolume theory of the NS5-brane, which is a string theory of “little strings” in six dimensions."

@RyanUnger you probably want ch. 14 of Nakahara or something, it throws the Riemann–Roch theorem at you pretty fast

1:40 PM

don't throw theorems at me please, I bruise easily

(Jason Bennett) Gravity as a gauge theory

►

To match those notes(!)

There better not be any plugs for SkillShare in there

1:56 PM

Completely coincidentally, those notes even have a discussion of Bargmann and basically the same central extension mass thing I linked to...

fqq

3:34 PM

@ACuriousMind obligatory xkcd xkcd.com/171

3:51 PM

also this (click image to progress)

4:04 PM

what about

Ryan Unger

4:25 PM

@bolbteppa something coherent

may I ask a Differential equations question here?

the math room is a bit dead right now

Mauro Giliberti

@satan29 if you have a proper question you should probably ask it in the site

doesnt seem very proper

@satan29 just ask. The worst that will happen is that we'll ignore you! :-)

XD

okay so its the use of the derivative as an operator.

for example, the equation

$$dy/dx + ay= f(x)$$

can be written as :

$$ (D+ a) y= f(x) $$

where D is the derivative operator d/dx ()

4:30 PM

@RyanUnger here is a good set of notes that has a ton of stuff

Since matrices are also operators in a sense, I kind of think of this as something similar to a matrix equation

$$ Ax=B $$

so, to solve for y, we need to invert the differential operator

i.e $$y= (D+a)^{-1} f(x)$$

Yes you can do that

You then have to figure out what $(D+a)^{-1}$ really means

now, if you look at the original equation, its a simple, linear ODE. (a is a constant).

we can easily see that $$y=e^{-ax}*\int e^{ax}f(x)dx $$

@ACuriousMind who learns K-theory before string theory

we can now conclude what $$ (D+a)^{-1}f(x) =e^{-ax}*\int e^{ax}f(x)dx$$

4:35 PM

@satan29 that's the Green's function approach, i.e. the inverse operator of a differential operator is the integral operator convolving with the Green's function of the diff. op.

as a sanity check, it makes sense for a=0. now, moving on:

The trouble is that my professor writes it in a different way

he writes

$$y= \dfrac{f(x)}{D+a} $$

@satan29 your notation is off - if you integrate over $x$, it cannot also be a free variable on the l.h.s.

@ACuriousMind didnt catch you?

In this example it's easiest to try to relate $\frac{dy}{dx} + ay$ to the product rule, thus trying to write $a$ a the derivative of something so the whole thing reduces to the product rule,
$$\frac{dy}{dx} + a y = \frac{dy}{dx} + [\frac{d}{dx} \int a dx] y = \frac{dy}{dx} + \frac{1}{e^{\int a dx}} [\frac{d}{dx} e^{\int a dx}] y = \frac{1}{e^{\int a dx}} \frac{d}{dx} [e^{\int a dx} y] $$
but the operator method generalizes e.g. to second order equations

@satan29 your $(D+a)^{-1}f(x) =e^{-ax}*\int e^{ax}f(x)dx$ doesn't formally make sense - you can't simultaneously integrate over $x$ on the r.h.s. and have it be the input variable to the function on the l.h.s

4:39 PM

i think it will be better with an $\equiv$ sign

anyways, let me quickly get to the point:

$$y= \dfrac{f(x)}{D+a}$$

he then proceeds to define this expression as

@NiharKarve I think we're not seeing someone learning it "before" so much as someone descending into madness trying to track down all the random math and physics that tends to pop up in string theory

(also it's a joke :P)

$$y=e^{-ax}*\int e^{ax}f(x)dx$$

so it seems as if it was just a notational convinience. We can live it with.

to quote Emilio, "my seriousmeter is always perfectly calibrated"

now, he takes the equation

$$( D^2-3D+2)y = e^x$$

and writes

$$y= e^x/ (D^2-3d+2)= e^x/ (D-1)(D-2) $$

$$y= e^x/(D-2) - e^x/(D-1) $$

and then evaluates these two expressions independently, based on the "definition"

and gets the correct answer

but that partial fraction decomposition really bothers me

why?

4:45 PM

@satan29 you may find Mathematical Methods in the Physical Sciences by Mary Boas helpful for this sort of thing in general

I took writing stuff like writing (D+a) in denominators just to be a notational convinience

apparently he is treating it exactly like regular fractions and its workking !?

@satan29 the partial fractions still work if you just write $(D-a)^{-1}$ in front instead :P

the rigorous way to write it seems to be $$(D-2)^{-1} (D-1)^{-1} f(x)$$

now how is that equivalent to $$( (D-2)^{-1} - (D-1)^{-1} ) f(x) $$

I think its just a quirk of the PFD in this case

there's nothing special about partial fraction decomposition that requires the notation$\frac{a}{b}$.

4:49 PM

try it with a different polynomial

@NiharKarve no, it works in general

$(D-3)^{-1}(D-15)^{-1}\ne((D-3)^{-1}-(D-15)^{-1})$ is what I'm saying

for example

i mean youll get 1/12 in the denominator for that..

I might be completely misinterpreting you

what i mean is

4:52 PM

Note
$$(D + a)^{-1} = \frac{1}{D+a} = \int_0^{\infty} e^{-s (D+a)} ds = \frac{1}{-(D+a)} e^{-s(D+a)}|^{s=\infty}_0 = \frac{1}{-(D+a)} e^{-\infty (D+a)} - \frac{1}{-(D+a)} e^{-0} = \frac{1}{D+a}$$
so that $(D + a)y = f(x)$ gives
$$y = (D+a)^{-1} f(x) = \int_0^{\infty} e^{-s (D+a)} ds f(x) = \int_0^{\infty} e^{-s (D+a)} f(x) ds = \int_0^{\infty} e^{-s a} e^{-sD} f(x) ds = \int_0^{\infty} e^{-s a} f(x - s) ds $$
but setting $z = x - s$ we see
$$y = - \int_0^{\infty} e^{(z-x)a} f(z) dz = - e^{-xa} \int_0^{\infty} e^{za} f(z) dz = - e^{-xa} \int_0^{\infty} e^{ax} f(x) dx $$

$$(D-a)^{-1} (D-b)^{-1} f(x)= 1/ (a-b) * (( (D-a)^{-1} - (D-b)^{-1} ) f(x)$$

@bolbteppa nice Schwinger parametrization

Also the limits of integration should probably be more general hmm (obviously flipping the integration limit fixes the sign but it looks weird)

@ACuriousMind you were mentioning some operator identity that seemed useful..

@satan29 it was just nonsense :P

you literally can do everything necessary without using the fraction notation

4:54 PM

It's literally just partial fractions there's no reason why it wont work

i mean I cant digest d/dx() as something being in the denominator

take the $(D-a)^{-1}f$ term and multiply by $(D-b)^{-1}(D-b)$, and the other term with $(D-a)^{-1}(D-a)$

and you arrive back at the term you started with

no need to write anything as fraction to do partial fractions

Charlie

what on earth is going on here

@ACuriousMind wait

what expression are you considering? initially

ohh the RHS of what I wrote?

@satan29 there is literally no conceptual difference between writing $(\mathrm{d}/\mathrm{d}x)^{-1}$ and $\frac{1}{\mathrm{d}/\mathrm{d}x}$

@satan29 yes, the $((D-a)^{-1} - (D-b)^{-1})f$

4:58 PM

@ACuriousMind it appears that this is indeed true, but I am having a hard time coming to terms with it XD

:57588730 $$ (D-b)^{-1}(D-b)(D-a)^{-1} f - (D-a)^{-1}(D-a)(D-b)^{-1} f $$

Since both $y = (\frac{1}{D-2} - \frac{1}{D-1})f$ and $y = \frac{1}{(D-2)(D-1)}f$ are such that $(D-2)(D-1)y = f$ e.g.
$$(D-2)(D-1)(\frac{1}{D-2} - \frac{1}{D-1})f=[(D-1) - (D-2)]f = [2 - 1]f = f$$
therefore since $y = y$ we have $ (\frac{1}{D-2} - \frac{1}{D-1})f = \frac{1}{(D-2)(D-1)}f$.

@ACuriousMind what next ? (so sorry for the repeated pings)

@satan29 yeah, now pull out the common factor $(D-b)^{-1}(D-a)^{-1}$, simplify and you get $(D-b)^{-1}(D-a)^{-1} (a-b) f$

@bolbteppa ohh interesting

@ACuriousMind OHH ofcourse, these commute unlike matrices

ayyy thats great, I have 2 proofs now

thanks a lot @ACuriousMind @bolbteppa

ah yes, the "treat like fractions" trick only works if everything involved commutes

5:05 PM

i should ask my math questions here more often it seems :p

I still have no idea why this inverse function stuff needs to be introduced at all, it's just a regular inhomogenous linear ODE:
$$
(D-1)(D+2)y=e^x
\\u=(D+2)y\implies u=xe^x+ ae^x
\\\Rightarrow (D+2)y=xe^x+ae^x
$$
and it's easy from there (and it's a lot quicker if you know the general-complemetary solution thing)

I suppose the fact that they can be decomposed like partial fractions is the big deal

so for eg if we had 3 factors on the LHS

a quick partial frac decomp, and if one has enough practise with evaluating f(x)/D+a , I suppose its faster

@NiharKarve you can use this idea to e.g. solve the Schrodinger equation and set up scattering etc

@satan29 the general-complementary solution is way faster, it's just find roots + ansatz

Hmm

5:11 PM

It also becomes more relevant when you consider higher order constant coefficient linear ode's which can be solved quickly by factoring (when one can even do the factoring)

idk, maybe there are other applications, plus its cool to have different weapons in your arsenal I suppose

I guess its useful in case of nonconstant coefficients?

you need to define $$f(x)/D+a$$ as $$ exp (-\int (adx)) * \int (f(x)*e^{\int adx}) $$

again, that's suspicious notation because the left hand side is a function of x while on the right side x is a dummy integration variable

ah i see the problem now, thats what ACM also meant

its just an "indefinite integral", if you may

yeah but you should still do something about naming everything $x$ :P

Interestingly the manner of parameterising the inverse of an operator/practically any object like that is quite useful in QFT integrals

Now I'm not an ODE guy

But I know this one really cool trick that works on most differential equations and even a lot of integrals

It's called Mathematica