so here's a question my students have on their HW and i'm trying to see if there's an alternative calculation

you start with a particle in a box and add a square barrier in the center, one which is a lot higher than the lowest-lying energy levels. goal is to (approximately) compute the gap between the first excited state and the ground state

the direct way is to solve for these two states explicitly, then compute the energy levels and make careful approximations along the way

the splitting ends up being exponentially small, so perturbation theory seems like it can't be useful here

but the final (approximate) result is simple enough that it seems like there should be another approach, e.g., some kind of overlap integral

(i know how one would do it for a smooth double-well potential: there's a nice formula based on WKB in that case)

(said formula does give the correct exponential, but it can't reproduce the prefactor)

@ACuriousMind Sorry just to be sure, even though this isn't what you're doing it is equivalent to doing this and using $\mathcal F(0)=0$, right?

basically using the fact a linear map takes the additive identity to the identity for a vector space (of functions in this case)

@Charlie it is equivalent to doing this and using that the Fourier transform is injective, i.e. $\mathscr{F}(f) = 0$ implies $f=0$ (note that this is a stronger claim than $\mathscr{F}(0) = 0$)

I don't think I've ever seen any physicist prove injectivity, though :P

Is the injectivity dependant on which space we are defining it on?

sort of

what space other than $L^2$ did you have in mind?

I know the unitarity depends on how we define it, i.e. the factor of $2\pi$ out the front

Well basically just $L^n$, that's the only other case I could see it considered in my basic search of the internet

presumably mathematicians do all kinds of perverted things with Fourier transforms

yeah, so rigorously proving the Fourier transform is injective/bijective is actually really annoying depending on what you pick

e.g. on $L^2$ you can't use the normal integral representation for it because that integral isn't guaranteed to converge for arbitrary $L^2$ functions

I sort of remember reading that, but it seems like a detail unlikely to be relevant in physics :P

maybe the way to do what i'm saying above is to write $H\psi_L \approx E \psi_L+\Delta\, \psi_R, H\psi_R \approx \Delta\, \psi_L+E \psi_R$

for appropriately-chosen $\psi_R,\psi_L$

indeed, if you're happy to take injectivity on faith then sure, plugging in the transform into the x-space definition works, too

in which case $\Delta \approx \int \psi^*_R(x) (H\psi_L)(x)\,dx$ assuming normalized wavefunctions

@Charlie IIRC if $p \neq 2$ then the FT is not on $L^p$, but $L^p \to L^q$ with some relation between p and q

hmm

1/p+1/q=1, probably

great, now I have flashbacks to functional analysis, ugh

but we routinely take fourier transforms of functions on $\Bbb R^4$ in qft no?

sure but that's still L^2

or I guess classical field theory

you take transforms on $L^2(\mathbb{R}^4)$

yeah

oh

or, really, $C_c(\mathbb{R}^4)$, compactly supported functions

I see

I never followed up on why compact support was important beyond knowing the definition lol

something distribution related iirc

among other things

but it's also physically just saying "the fields go to zero at infinity" and it makes boundary issues with integrals go away :P

oh

in particular, integration by parts becomes trivial

$f(x)g'(x)$ becomes equivalent to $-f'(x)g(x)$ for any term within an integral

come to think of it, if the support of a function is compact this means that the boundary of $R^n$ is the only point at which $\phi(x)=0$, so the entire space except the points at inifinity are zero?

well, not the only point necessarily, but one of the points

the distribution thingy you remember is more commonly the Schwartz space of functions with bounded derivatives rather than compactly supported functions

I will admit I am again wildly stabbing at an intuitive answer

yeah something about the test function space of distributions or something

@Charlie compactly supported just means there's a compact set (which in $\mathbb{R}^n$ pretty much means "finite volume") outside which the function is zero

I was too busy not knowing what the Fourier transform was to learn about these things :P

we're saying "this function is only non-zero inside a finite volume"

(tbh I have no idea what you were saying about the boundary of $\mathbb{R}^n$, the boundary of that space is empty :P)

in physics terms it's like saying "i can make a box big enough that the field vanishes outside of it"

ok maybe I'm thinking about this wrong, but if $\lim_{x\rightarrow \infty}\phi(x)=0$, why is the support compact? Or are these not equivalent in the way I'm thinking

If the answer is it's complicated I will stop trying to guess it :P

@Charlie these are not equivalent at all

Ok I will cease and desist lol

(you're not alone though, it's a common point of questions on the site that they aren't, because people physicists often use the limit version to argue something that can really only be argued from compact support)

or, worse, they'll claim all square integrable functions have $\lim_{x\to\infty} f(x) = 0$ or some nonsense like that :P

D:

@Charlie the non-equivalence is not hard to see, just look at $f(x) = \mathrm{e}^{-x^2}$

it goes to zero at infinity, but since it's nowhere zero its support is the entire $\mathbb{R}$

oh

Is there a name for the domain on which they do vanish at infinity? or is this the Schwartz functions

I'm not sure what the question is

I've sory of seen this topic in more careful discussions of qft

or not specifically the domain sorry, the real functions on $\Bbb R^n$ that vanish at infinity

I don't think there's a name for them

although "vanish at infinity" probably requires more careful definition

ok fair enough, well my question is answered, and I appreciate you as always entertaining my further random guesses about things :P

the Schwartz functions are "nice functions that vanish at infinity" - not only do they go to zero, they go to zero fast enough, whatever that means :P

yeah I vaguely remember something about this

I'll go back to doing physics where we're not careful about the domain of functions we're talking about >:)

If I wanted to understand the difference between these "mode expansions" that we use in QFT and the Fourier transform, what would be the keywords?

like, the resemblance is there, but this clearly isn't just a fourier transform of $\phi(x)$ because you also have something which looks like the inverse fourier transofrm

Can anyone help me with the case where we consider a point charge that moves with a constant velocity and we need to find the electric flux ? Like any pdfs regarding this particular problem ?

@Charlie the mode expansion is "diagonalizing the Hamiltonian in Fourier space"

you have something like $a_p = \phi(p) + \mathrm{i}\pi(p)$ (some factors omitted :P), this is exactly analogous to $a = x + \mathrm{i}p$ for the single non-field oscillator

is there any proof that Gauss law hold for a charge that moves with constant velocity ?

I guess I'm still confused as to why if $\phi(x)=\int d^3p \phi(p)e^{ipx}$, it can also be equal to $\int d^3p (\phi(p)e^{ipx}+\phi^\dagger(p) e^{-ipx})$, in this case just letting $a_p$ be written as $\phi(p)$

@Charlie $a_p$ is not equal to $\phi(p)$, it's equal to $\phi(p)+\mathrm{i}\pi(p)$ (modulo prefactors), as I said

because it kind of just looks like the Fourier transform of $\phi(x)$ added to the inverse Fourier transform of $\phi(x)$

oh I can kind of see how they are different sure

I've just only ever seen this specific expansion used in qft and literally no where else

oh

the creation annihilation operators are functions of the FT of $\phi(x)$ and $\pi(x)$

It even has the factors of $2\pi$ which make me immediately think it's something to do with the Fourier transform

(the expansion not the a's)

@Semiclassical yes I thought so but I wasn't sure I remembered well and I did not want to check

I am trying to prove that Gauss's law is valid for charged particles with constant velocity, but I have one question about the electric field. If we consider 2 reference frames, the lab fram where an observers makes a measurement at an arbitrary point $\vec r =(x,y,z)$ and a the rest frame of the particle. The expression for the electric field given to me is some constant multiplying $\vec r'$ which would be the position vector in the rest fram of the particle.

I am trying to prove that Gauss's law is valid for charged particles with constant velocity, but I have one question about the electric field. If we consider 2 reference frames, the lab fram where an observers makes a measurement at an arbitrary point $\vec r =(x,y,z)$ and a the rest frame of the particle. The expression for the electric field given to me is some constant multiplying $\vec r'$ which would be the position vector in the rest fram of the particle.