The claim is that this wavepacket will move to the right with velocity $k_0$ (in units where hbar=m=1) and will approximately retain its original envelope

To justify this, I take the Fourier transform of the initial wavepacket to get $\tilde{\psi}(k) = \tilde{\phi}(k-k_0)$. The subsequent time evolution is $\tilde{\psi}(k,t)=e^{-i k^2 t/2}\tilde{\phi}(k-k_0)$

@Semiclassical this isn't a claim about Fourier transforms

it's (presumably) a claim about the Schrödinger equation

well, I think the claim about FT is with regard to $\tilde{\phi}(k-k_0)$

namely, that it should also have support in the nbhd of $k=k_0$

@Semiclassical as in, multiplying by $e^{ik_0x}$ is a shift in momentum space?

yes

@Semiclassical if $\phi(x)$ is flat then yes

mostly I'm trying to see where the approximations come in

@Semiclassical well, in saying "flat"

namely, why this doesn't work if $\sigma=0$ and $k_0$ is finite

hmm, maybe I see it

@Semiclassical if $\sigma=0$ then it works in that it shifts the center of the momentum distribution, but its tail becomes extremely heavy

yeah

The logic I had in mind was that 'smoothing' at the edge of the flat $\phi(x)$ is equivalent to a Gaussian convolution

which in turn would mean that the Fourier transform would amount to multiplying the $\sigma=0$ FT by a Gaussian, thereby smoothing it out

^ the $\sigma=0$ case

which in turn justifies $\tilde{\phi}(k-k_0)$ having most of its support near $k=k_0$

... is ugly

ew

well, what'd you expect

lol

starting off a Schrödinger equation with a discontinuous function

just because it's expected doesn't mean it's not ugly

of course it's "ew"

@Semiclassical true that

@Semiclassical if it is a gaussian convolution then obviously yes

well, I think one can approximate it as such

and once one does that approximation, then that logic is fine

how good of an approximation that is...hmm

@Semiclassical ... depends on the actual shape of the smoothing

true

but the bottom line is, yes, don't worry about it

Nice.

for "smooth enough" shifts from constant to zero, the support of your shifted wavepacket is arbitrarily tightly localized around $k_0$

the relevant picture from the paper is this one (though there's one unrealistic aspect of it which annoys me):

where obviously the tightness is controlled by the requirement that $L\gg \sigma \gg 1/k_0$ and that the function be smooth

the second one is the relevant one. I should probably only have included it, come to think of it

the unrealistic bit is that they have $L\gg \lambda_0 >\sigma$ in the above picture when they really need the last inequality to be reversed (and state as much in the paper itself)

@Semiclassical the higher amplitude on the right?

@Semiclassical ah

yeah, well

i get why it's like that

it'd be hard to draw it with the inequality reversed

but at a glance it gives the wrong impression

@Semiclassical is it really important?

if I'm trying to simulate it in MMA, yeah

I guess it is, the time behaviour would be dominated by the boxcar's diffraction more than the movement to the right

Right

if that inequality is reversed, it'll lose its shape

well, it'll always lose its shape

sure

it's always just a matter of how quickly

the question is whether it'll do so before moving one wavelength to the right, or something like that

right.

I think that that's the right benchmark

i guess the issue is that, if you look at it too quickly, you might assume that you could take $\sigma\to 0$ without issue

fair

anyways

the project I'm making for myself is ultimately to simulate the scenario they give there

but I figured starting from the free evolution was a good first step

@Semiclassical indeed

actually, my endpoint is a bit more ambitious than that. where I want to get to eventually is a spin-1/2 particle scenario. but getting this version to work is a necessary warm-up

@EmilioPisanty one thing that's bugging me there is that i know i've seen formulae pertaining to $\text{erf}(z)$ along lines like $\arg z =\pi/4$

but I can never track those down when I need them

@Semiclassical ugh

dubious

well, that's what sqrt(i) is

if they're not in the DLMF then I wouldn't know where to look for them

yeah

yeah, no, exact expressions are definitely out

not sure what I'm thinking of then

the issue, I think, is that erf has stokes lines along arg(z)=pi/4 and arg(z)=3pi/4

(maybe anti-stokes? I never remember the terminology right)

reference: books.google.com/…

oh frabjous day, they actually have the relevant equation available: books.google.com/…

@Semiclassical Is this some WKB stuff?

not explicitly, but it's in the same ballpark

Ok, just wondering.

it's all asymptotic stuff

Dangit, @ACuriousMind where's that answer where you explained why density matrices have to work they way they do?

I thought I favorited it, but I can't find it in my favorites list.

@DanielSank here

@EmilioPisanty I think what i have in mind is the relation of the error function to the fresnel integrals

$${\displaystyle {\begin{aligned}C(z)+iS(z)&={\sqrt {\frac {\pi }{2}}}{\frac {1+i}{2}}\operatorname {erf} \left({\frac {1-i}{\sqrt {2}}}z\right),\\ S(z)+iC(z)&={\sqrt {\frac {\pi }{2}}}{\frac {1+i}{2}}\operatorname {erf} \left({\frac {1+i}{\sqrt {2}}}z\right).\end{aligned}}}$$

@ACuriousMind I must not understand this website.

Thank you.

@DanielSank I just search for answers of mine with "partial trace" in them whenever you ask me for that answer - since I haven't written about partial traces all that much, it's easy to find in the results ;)

I see.

ignore the amp; in there

@EmilioPisanty What i find weird in there is how having small but finite $t$ is apparently analogous to Gibbs phenomenon

I'm sure it can be explained in a sensible way, but it just weirds me out

@Semiclassical it's a good question

Probably for the maths site

You should ask it

@DanielSank if you do favorite it, you know you can search with infavorites:mine, right?

Or some very similar syntax

@EmilioPisanty I may, yeah. It should come down to a statement about the large-z asymptotics

@Semiclassical hmmmmmm

Probably not, I think

Well, it sorta has to. The arguments of your error functions are of the form z~x/sqrt(t)

But you will probably lose out on the high-k components first and thus mimic an incomplete convergence

On a separate track though

So small-t means large z

@EmilioPisanty hmm, nice

Of course, this is all irrelevant to my original problem since the whole point of smoothing out the rectangular function is to ensure none of the crazy stuff shows up