The h Bar

Tobias Fünke

17:01

are A and B functions of p?

If so, then no; it has infinitely many possible solutions?! if they are constant, then it is not well-defined to begin with, I am afraid

imbAF

Honestly, I don't know what to say about that

Tobias Fünke

ehm ok

imbAF

Assume that I have this particular solution to K.G equation: $\phi(\vec x,t)=Ae^{-ipx}+Be^{+ipx}$ where p and x are in four vector notation.
Now for the general one I have:

$\Phi(x^\mu)=\frac{1}{(2\pi)^3}(Ae^{-ipx}+Be^{+ipx})d^3p$

Now I want to find the values of A and B, so I try to normalize the expression:

$\int \Phi(x^\mu)*\Phi(x)d^3x=1$

When substituting Phi* and Phi* I used p and p', so the integral will have d^3p'd^3pd^3x.

Assuming everything I wrong is correct. How would you proceed? My aim here is to get the $\frac{1}{2E}$ term

So this is where I stand

Tobias Fünke

I don't have MathJax atm, so it is not possible for me to decipher that, sorry. Perhaps someone else can help

imbAF

I think is best to make a thread for this, cuz I am not going nowhere for hours now

Tobias Fünke

17:05

but I told you today already that the $E$ term is due to the mass-shell condition

you want a Lorentz invariant measure

imbAF

That I understand

But as you can see

my calculations initiate from the wave function solution

and I want to reach the fourier decomposition in QM, not QFT

which means instead of the ladder operators, you have complex numbers a and a*

Tobias Fünke

I don't get what you want. But yes, you can solve the classical KG equation

and write down the general solution

but why do you want to normalize?

imbAF

because

trick to explain but I will try

In what I wrote above, you did notice how I expressed the general solution

oh wait you don't have MathJax

I will post a pic

This is how I start

with a very general expression of the decomposition

Feynmate

It's really just a choice of normalization. The expansion always holds, it's just about absorbing that term into the modes or not

imbAF

If you substitute the expression for $\phi$ in the integral, you will have an expression followed by d^3p' d^3p d^3x

I believe, that one should be able, through pure calculation

reach the following point

Feynmate

17:15

(and of course what Tobias said about invariance)

imbAF

While the way is different

namely here, we start with integration in spacetime d^4x\delta(...)\theta()

the end result must be possible from both angles

The reason why I am trying, what I am, is because I am starting from the wave function solution and then leveling up to this point. While in here and in many texts, we start from the fourier decomposition

Feynmate

It's the same idea; the first expansion is manifestly invariant; that's actually the way we prove that the measure with energy is invariant

imbAF

of the solution. And in no instance is made clear how that expression, which contains 1/2E was EVER derived

Tobias Fünke

then you follow the wrong resources

this should be explained in any reasonable text

...

Feynmate

Tobias entering ACM mode

imbAF

17:18

like L& L

who clearly do not derive the decomposition from the wavefunction solution of the eq.

Tobias Fünke

I don't know L&L; I just know that in any basic course this should be derived

imbAF

I believe that I gave a comprehensive explanation as to what the issue is here and why in no book I see the solution to my problem

@TobiasFünke the fourier decomposition?

Tobias Fünke

and I honestly do not understand why you spent hours with this instead of checking 1-2 notes and going through the relevant computation step by step

imbAF

@TobiasFünke I mean,.....

ok

Tobias Fünke

@Feynmate Please help me out :d I am no expert in relativistic QFT. But this should be derived in any intro course, no?

At least we did it in ours

and now after checking two books this was done there, too (although this are math books)

imbAF

17:21

You derived the decomposition starting from a particular solution, i.e wave fuction?

Tobias Fünke

I mean how the Lorentz invariant measure comes into play

imbAF

that comes into play

Feynmate

@TobiasFünke I'm a little tired so I'm not sure I'm any more helpful in this condition, but it would help me to really understand what's the problem with the normalization

@TobiasFünke This?

imbAF

if you START with decomposition in spacetime

Tobias Fünke

@Feynmate at least this is what I understood

imbAF

17:22

Ok, I will make it as clear as I possibly can

Feynmate

Well, if you are in a relativistic setting, you expect that the nornalization of states does not change with Lorentz transformations, so that's where the invariant measure comes into play

In the field expansion the energy factor is only there because of a normalization choice

Really, there is no reason in principle other than normalization, nothing to derive. The field is a function and you perform a Fourier expansion, then proceed to "divide" the Fourier transform into positive and negative modes, right?

16 mins ago, by imbAF

imbAF

I am working on posting soemthing. So by reading it, you'll be able to judge whether what I am trying to do is even possible or not. Because exchanging messages hasn't been working for the past 5 hours

Feynmate

I have just come into the discussion, so you could at least try to listen to someone who's trying to help you

This is the expansion. Understood, $A$ and $B$ are $A_\vec{p}$ and $B_\vec{p}$, functions of momentum

imbAF

Ok, but when I wrote them down I didn;t think of them as functions of momenta

I mean, I have to argue for it, no?

Feynmate

Then your integral is badly divergent

imbAF

17:31

correct

So how do you argue the momentum dependency ?

Feynmate

No, you don't those coefficients are just the Fourier transforms of the field

Nothing to argue

imbAF

So, when you have the K.G equation and you consider a wave function solution

you write

$\phi=A(\vec p)e^{ipx}$ ?

Feynmate

$\Phi(\vec{x},0)=\int\frac{d^3\vec{p}}{(2\pi)^3}\tilde{\Phi}(p)e^{i\vec{p}\cdot\vec{x}}$

That's the first step

imbAF

Ok, so that is the first mistake in what I wrote

I only have A or B

Feynmate

This is the Fourier transform, we haven't split into modes yet

imbAF

17:35

Ok

that is the generalized solution ?

could you say that?

since you are integrating over momenta

Feynmate

Again, I'm just Fourier transforming a field T_T

imbAF

I know that. I am asking if it is wrong to consider it as the general solution to the equation

But ok

Then what do we do, to avoid the divergence ?

Tobias Fünke

A FT is a FT, not a priori any solution

imbAF

ok

Feynmate

Now we use the fact that the field satisfies the KG equation and write the equation in momentum space i.e. the FT EoM

imbAF

17:39

hmm

Feynmate

$[\partial^2/\partial t^2+ \omega^2_\vec{p}]\tilde{\Phi}(p)=0$

Do you recognize this equation?

imbAF

atm I don't

FT of eom

Feynmate

That a HO EoM

With frequency $\omega_p$

imbAF

Ok, I recall the equation

Tobias Fünke

imbAF, since you speak german (?), you can check the book "Tutorium Quantenfeldtheorie"...perhaps it helps

imbAF

17:41

I am not sure how you got here

@TobiasFünke muss kaufen

Tobias Fünke

most German universities have contracts with Springer; it should be possible to get a PDF for free, at least with high probability ;)

anyway, I won't bother you guys anymore. see you later

Feynmate

*Oh no they are speaking a foreign language, I must say something to sound like I know what they are talking about. Come on, Feynmate, chill out. They won't notice a thing.*

Guten morgen

@TobiasFünke TOBIAAAAAAAAAS T_T

imbAF

I know that you can write in momentum space $(\partial^2_t + p^2+m^2)\phi$

so $\vec p^2+m^2=E^2$

right?

So, how do we continue ?

Feynmate

Now, for a QHO you know that $\tilde{\Phi}(p)=\frac{1}{\sqrt{2\omega_p}}(a_p+a^\dagger_p)$

I'm just writing the analogue of $\hat{x}\propto(a+a^\dagger)$ from QM, using the oscillators of our problem, $\tilde{\Phi}$

imbAF

yes \phi plays the role of \vec x

Feynmate

17:50

And of course, the respective annihilations/creations are labeled by the $p$ of the mode

Got this?

imbAF

meaning $a(\vec p), a(\vec p)^\dagger$

?

Feynmate

Just like $\tilde{\Phi}(p)$ plays the role of $x$, they play the roles of $a$ and $a^\dagger$

But of course you have a different mode for each $p$, so you label them with $p$

imbAF

yes

Feynmate

Okay, finally, we use the EoM of creation and annihilation of each mode, which are the same as QM

First replace this in the FT

9 mins ago, by Feynmate

Now, for a QHO you know that $\tilde{\Phi}(p)=\frac{1}{\sqrt{2\omega_p}}(a_p+a^\dagger_p)$

imbAF

I am not sure I am following

Feynmate

17:58

Just put that equation into the FT

imbAF

this one $\tilde{\Phi}(p)=\frac{1}{\sqrt{2\omega_p}}(a_p+a^\dagger_p)$ ?

in $\Phi(\vec{x},0)=\int\frac{d^3\vec{p}}{(2\pi)^3}\tilde{\Phi}(p)e^{i\vec{p}\cdot\vec{x}}$ ?

Feynmate

Indeed

imbAF

Ok

Feynmate

Then we want to write $\Phi(x,t)$, the one before was at zero. The time dependence is of course encoded in the coefficients, so we only have to write $a\to a(t)$

imbAF

with the subscript \vec p i assume

but ok

Feynmate

18:01

$\Phi(\vec{x},t)=\int\frac{d^3\vec{p}}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_p(t)+a^\dagger_p(t))e^{i\vec{p}\cdot\vec{x}}$

imbAF

I see

But one thing only. You have root of 2omega

I have seen notations where it;s simply 1/2\omega

Feynmate

For the field?

That happens in other cases

The field expansion usually has this normalization

imbAF

here

oh, this is another case

so you have other expression ?

@Feynmate physics.stackexchange.com/questions/442936/…

bolbteppa

@imbAF This is not a stationary state, the stationary state solutions are $e^{ipx}$, 10.6 of L&L vol 4, they are normalized as in 10.16.

Because the $p = (E,\mathbf{p})$ in this is only constrained to satisfy $p^2 = m^2$, the $E$ is not independent it depends on $\mathbf{p}$, so the stationary states are indexed by $\mathbf{p}$, however the sign of $E$ is also independent so we sum over the signs of it as well, and the general Fourier expansion as a sum over all possible stationary states is then 11.1.

You can then re-interpret the stationary states of this equation as describing a system of two 'species' of particles as explained after 11.1, where both species are described by the same equation in the same system, and interpret them as anti-particles

Feynmate

@imbAF You can do that, it just changes the coefficient of the modes

bolbteppa

18:13

The sign of the energy term in the exponential is just telling you whether the associated quantum field operator is going to be describing particles or anti-particles, as explained below 11.1

imbAF

Ok so it is vol 4 10.6 until 11,1

I was naïve into thinking that going as I went

one could get the same result

but of course, I ended up dealing with an integral that diverges

bolbteppa

(12.1 then discusses the more natural case where there is only one species of particles, which is what people would probably guess is going on at first)

Just read section 10, 11,12, skim 1 and 2, especially the last paragraphs of 2 for more on the sign of the exponential

Tobias Fünke

18:31

@Feynmate hehe

User198

18:54

Is there a generalization of the Moyal equation $\dot W= \{\{H,W\}\}$ ?

We can go Schrödinger equation $\rightarrow$ LvN equation $\rightarrow$ (Wigner transform to) $\rightarrow$ Moyal equation $\rightarrow$ ?

Now we are in phase space and practically made a full circle from the Hamilton's equation that are in phase space back to QM that is again in phase space. Did people feel the need to generalize further or is this practically it? The full circle?

I presume that it is (as always) possible to go even to greater generalizations and abstractions

ACuriousMind

19:53

@User198 it's not clear what you mean by a "generalization" here

imbAF

20:21

When we considered the canonical quantization of the real scalar field, one of the requirement was that the following commutation relations should be valid:

$[\phi(x),\Pi(y)]_{x_0=y_0}=i\delta^{(3)}(\vec x- \vec y)$
$[\phi(x),\phi(y)]_{x_0=y_0}=0$
$[\Pi(x),\Pi(y)]_{x_0=y_0}=0$

These were just given, no explanation how we arrive here, or why this is necessary or even what it means to consider the commutation of two field operators. At least in QM, the commutator was a way to tell if the physical quantities could be measured at the same time, or if the operators had a joint basis. But here …

ACuriousMind

if you understand why we do $[x,p] = \mathrm{i}$, you understand this. It's just the field version of the Poisson brackets turned into commutators.

imbAF

I understand the consequence of it in QM (not how we come about it), but in QFT is there an interpretation

ACuriousMind

I don't know how often I can say that the operators are still operators like in QM

so yes, operators that commute still have joint eigenbases and still can be measured simultaneously (if they're observables) etc.

imbAF

Ok but $\phi$ is an operator and it assigns an operator value at any spacetime point. So that is not an observable

At least that is what I believe

or suspect

ACuriousMind

I don't know what you're trying to say. An observable is a self-adjoint operator, my caveat is merely because e.g. a complex field has $\phi \neq \phi^\dagger$ and so is not self-adjoint.

imbAF

20:30

Ok, I'll be blunt. I don't know what evaluating $\phi(x)$ at a spacetime point should mean? Or should be? Is it a scalar, a complex value

I don't know

ACuriousMind

(At the physical level of rigor), the quantum field is simply an operator-valued function. That is, for each $x$, $\phi(x)$ is an operator.

imbAF

Ah, ok

And what would $\langle 0| \phi(x)|0 \rangle$ be?

Like what is meant with the expectation value of an operator-valued function?

Is there some physical significance to it ?

ACuriousMind

I'm not sure where the problem is. $\phi(x)$ is an operator for fixed $x$. So that's the expectation value of the operator $\phi(x)$ in the state $\lvert 0\rangle$.

imbAF

But isn't it different evaluating $\phi(x)$ at some point x^mu and calculating the VEV at some point x^\mu ?

ACuriousMind

The physical significance depends on what the physics significance of the field is to begin with. If it's the electric field $\vec E(x)$, the meaning should be rather obvious. If it's the Higgs field, it's less so :P

imbAF

20:34

They are the same thing?

ACuriousMind

@imbAF of course it's different. $\phi(x)$ is an operator, the VEV is a number.

expectation values of operators are numbers, again, it's just QM

imbAF

@ACuriousMind Yes but you say:

by "that"

you mean what?

ACuriousMind

I have no idea what you mean. Which 'that' are you referring to? Which message?

imbAF

@ACuriousMind This one

ACuriousMind

"that" refers to your $\langle 0\vert \phi(x)\rvert 0\rangle$

imbAF

20:38

Ok, but you did say that the physical significance of VEV depends on the field

i.e $\vec E(x)$ it would be the electric field strength at that point

if I am not wrong

ACuriousMind

I mean, it depends just like the significance of the expectation value of an operator in ordinary QM depends on the significance of the operator

there's literally no difference

imbAF

Yes, but of course I am assuming a case, where such a thing is possible

In QM i am not considering an operator, that doesn't represent a physical quantity

that ofc wouldn't make sense to talk about

ACuriousMind

you could

Tobias Fünke

oO

ACuriousMind

no one forbids you to e.g. talk about the expectation value of the self-adjoint operator $x+p$

imbAF

20:40

But because QFT is more abstract, and QM is like a reference point, I tried to give a meaning to expectation value of the field operator.

ACuriousMind

it's rarely useful, but the formalism allows it perfectly fine

imbAF

x+p ?

I have never encountered that

Tobias Fünke

@imbAF Roughly speaking, QFT is just QM with some additional postulates/an additional structure, namely to make QM compatible with SR

imbAF

what physical quantity would that be? sum of position and momentum O.o

Tobias Fünke

@imbAF it was meant as an example

ACuriousMind

20:42

@imbAF my entire point is that that is an example of a self-adjoint operator that usually has no physical meaning

yet it is formally an observable, and the formalism of QM allows you to talk about its expectation values, eigenvectors, etc.

imbAF

but for whom you can find an expectation value. I see

Ok I didn't know that

ACuriousMind

there's not really anything to "know" here. If you understand how QM works, you can understand that all those notions formally make sense for any operator

imbAF

but yeah, when I talk about significance of expectation values, I, ofc am cosidering something measurable, standard to say so,

@ACuriousMind True, but I make a distinction in QFT when it comes to operators. Namely, you can have all sorts of operators, who might represent a physical quantity i.e Hamilton operator in QFT or not, and field operators. This distinction doesn't exist in QM

I don't know if it is right to do

Or if it makes sense to do so

ACuriousMind

I have no idea what this "distinction" is supposed to be

I just gave you an example from plain QM of an operator without obvious physical significance

the difference is only that QFT forces you to perhaps consider such operators a little more often

imbAF

It is because, in my view, field operators, generate excitations of the field

And i consider them unique in that regard

ACuriousMind

20:46

we're just back at the point where I tell you to stop looking for direct physical meaning for every intermediate expression that may appear somewhere

the VEV of a scalar field is a really pointless example anyway because it will always be zero or a constant that you then set to zero by some tricks

imbAF

Ok, if I were to ask you what is the difference between the hamilton operator and field operator in QFT, for some theory, how would you word the distinction or difference between the two?

ACuriousMind

I don't understand the question. What's the difference between the Hamiltonian and position or momentum?

Allie

Hi all

I made some progress in understanding my resesrch :)

imbAF

If I were to classify operators, as caring meaning, i.e representing a physical quantity, then they do not differ in that regard.

If you were to list an operator, who represents no physical quantity, I would point that as a distinction between the two

ACuriousMind

I would suggest you stop trying to classify operators that way, at least at the beginning. The meaning of the operators becomes clear as you see what they are used for in the formalism

Tobias Fünke

20:49

@Allie nice :) you want/are allowed to share a bit?

imbAF

@ACuriousMind Then should I for the moment disregard their "meaning"?

ACuriousMind

at least in the sense you seem to use the term, yes :P

Allie

@TobiasFünke sure!

You know what OFDFT is?

imbAF

Is it because it's faulty ?

What is the issue with that?

Tobias Fünke

imbAF: The Hamiltonian is the Hamiltonian. The field is the field. The latter are used, for multiple reasons, to build observables, as the Hamiltonian. Roughly: The fields are the "building blocks" in QFT

imbAF

20:50

I am curious to know

Allie

DFT but not kohn sham basicallyb

Tobias Fünke

@Allie only superficially

yeah

ACuriousMind

@imbAF it's just not useful

Allie

So you would need a functional that is solely of density

KS ofc uses orbitals to get the non-interacting KE

imbAF

@ACuriousMind Can you elaborate ?

ACuriousMind

20:51

you are so far away from seeing how the entire actual machinery of QFT works that it's impossible to tell you the meaning of anything in it; you have to learn QFT to understand QFT

imbAF

@ACuriousMind And that would mean, a pure mathematical approach?

Allie

So one way to do OFDFT is to try to get approximations for the non-interacting Kohn-Sham KE

T_s[n] + Hartree[n] + XC[n] is ir total energy functional

That T_s would usually be the sum of the KEs of the KS orbitals right

DIRAC1930

@imbAF I think one of the only ways to learn QFT at least conceptually is to learn 2nd quantisation from L&L 3, and then follow L&L 4 to learn about the quantisation of the electromagnetic fields, Klein-Gordon fields, and then the Dirac fields. Ultimately, one has to make the leap that Pauli made in 1934 (which is eluded to in L&L 4) regarding the iterpretation of $j_\mu$.

ACuriousMind

@imbAF not exactly; but you are stuck at "what does the VEV of a field mean" when it will become perfectly evident what it means once you learn that a) only non-scalar field have non-zero VEVs anyway and b) non-zero scalar VEVs lead to symmetry breaking

but to understand the significance of that you have to learn all the steps in-between

Tobias Fünke

@Allie yes, I am listening :) I understand so far.

DIRAC1930

20:54

There is not much more conceptually to rel QFT compared to non-rel QFT. One has a field operator, that obeys certain field equations. In the non-rel case, that field equation is the Schrodinger equation, in the rel case, it is the Klein-Gordon, Dirac etc field equations

Allie

Okay, and so obviously getting T_s[n] is difficult just from n

DIRAC1930

The only difference is that one has both these positive and negative modes in the field expansion

Tobias Fünke

@DIRAC1930 Well, this is only partially true. Relativistic QFT is more complicated; for example, due to the lack of a well-defined position basis

@Allie yes

Allie

So one way you might think about it

DIRAC1930

However this is resolved by interpreting the coefficients correctly (as eluded to in L&L 4)

Allie

20:56

Is take the von Weiszaker functional

You aware of that is?

imbAF

@ACuriousMind The two things that you listed, I believe I don't understand them. But I want to. So idk, just reading different books and notations, will make me reach the point of understanding this? It feels like you need 1.0 across every BA physics, to be able to understand.

Tobias Fünke

@Allie yes

DIRAC1930

@TobiasFünke Okay then just use momentum basis in non-rel QFT

ACuriousMind

@Allie I assume you mean Weizsäcker ;P

Allie

It basicallt is the KE of a system of bosons with, of course, wave function sqrt(n(r)/N)

imbAF

20:57

@DIRAC1930 His way of treating 2nd Qunatization is messy or difficult to understand. But I get your point

Allie

He was a nazi ill spell his god damn name wrong if i want

DIRAC1930

@imbAF Follow a different source so you can at least follow the latter pages

You probably only need the "one-particle" operators

Allie

Anyways, so you can actually break up T_s[n] into T_vw[n] + V_P(r), where that v_P is the so called Pauli potential. It basically is the potential that will recreate the fermion system

So that you dont get KE of bosons you get KE of fermions

imbAF

@DIRAC1930 It's just the time constrain I have.

DIRAC1930

@imbAF I understand, I didn't have any time to do any of this when I was first learning QFT

Allie

21:00

So my research is all about trying to find (approximations of) v_P

Tobias Fünke

@imbAF my two cents: get one or two sources, study them step by step. You should be fine

@Allie interesting

imbAF

@DIRAC1930 I need to create time for 2nd quantization

Allie

The way Im looking at it is very different though

Im looking at it in terms of the so called exact elevtron factorization

Analogous to the nuclear electron factorization

Tobias Fünke

ah

yes

interesting! :)

ACuriousMind

@imbAF I have no idea what grades have to do with it, but yes, my position is that a lot of QFT becomes only clear in retrospect, once you have seen it at least once in its entirety until you arrive at the practical computation of scattering amplitudes for QED at least and then can appreciate different setups and approaches for it

Tobias Fünke

21:01

I am sure I will read some day a paper of you!

Allie

But instead you factor the N electron wf into a one electron wf and an N-1 electron wf that depends on what the one electron coord os

Maybe lol

Tobias Fünke

interesting

Allie

And it gives a really interesting new way of viewing the Pauli potential

Tobias Fünke

and regarding OFDFT: is it somewhat related with RDMFT?

because you could get the kinetic energy from the 1RDM

ACuriousMind

@Allie this is a horrible argument; do you really want to potentially confuse some guy named Weiszaker with a Nazi? :P

Allie

21:02

In the view of the EEF the pauli potential is not the correction from a boson system to a fermion system, but its a potential that actually captures the effects of the N-1 electrons around

imbAF

@ACuriousMind Well at least is good to know that If I keep blasting my head against the wall, one day I'll see the other side..hopefully. ACM I am banking a lot of trust at Weingand, when i start in Mars.

Tobias Fünke

@Allie hmmh interesting

Allie

Im really not sure, it honestly could be &tobias

Tobias Fünke

ok

Allie

So for example whaat i learned today is

In the view of the EEF, one of those terms in the Pauli potential is actually a correction to the one electron KE that accounts for the fact that if the environment (N-1 electron wf) needs to move in order for you to move the one electron, then there’s a “resistance” to movement

So the more the N-1 wf changes wrt changes in the 1 electron coordinate, the more reisstance you see, and that part of the potential is directly proportional to that resistanxe

You define the change in the N-1 electron wf by the fubini-study metric

Tobias Fünke

21:07

do you have a nice reference, by chance?

I think I want to learn more about the basics of OFDFT

Allie

Yes!

Tobias Fünke

hehe

Allie

OFDFT or EEF?

pubs.acs.org/doi/abs/10.1021/acs.chemrev.2c00758 Heres a good OFDFT paper

Tobias Fünke

both ;)

Allie

EEF: journals.aps.org/prresearch/abstract/10.1103/…

One day i will be as smart as u tobias

Tobias Fünke

21:10

thank you, much appreciated! I'll have a look tomorrow

@Allie haha, I bet you are much smarter already

Allie

Uhhh idk about that. Im still going through griffiths electrodynamics lolol

Tobias Fünke

I hate ED lol

Allie

Really

I dont mind it so far

Tobias Fünke

I am not good at it, no.

Allie

Not my favorite thing but

Tobias Fünke

21:12

Yes, it is for sure interesting!

Allie

I need to just catch up on my physics knowledge

I basicallt started learning physics like a year ago at this point

I have so much i need to cover and want to cover before i start my phd

Tobias Fünke

Today I dreamed that I met a former colleague, who recently got his PhD, and now he is offered, a few months after graduation, a professorship in NYC lol

@Allie yeah, you will do. You will also learn on the fly :) no worries

you will be fine

Allie

I wass about to say ong thats awesome

And then i read “I dreamed”

Tobias Fünke

xd

weird dream

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