I'm proud of you!

It did require asking DS & a friend for help

my friend walked me through the phone and, most importantly, made me give up on trying to flip the egg like a chef with just one hand on the pan

...what exactly about "crack open egg, put into ban, stop before it's burnt" did require two people to assist you? :D

@ACuriousMind It's not that simple

lol, just don't try flipping stuff like that

@Slereah I hath not...

the free vacuum in the wavefunctional rep is just $1$

(seen as a constant functional, square integrable wrt the gaussian measure)

@ACuriousMind Why is there no [magic] tag for questions?

@yuggib Well 1 up to a phase, yes

the ladder operators I don't know exactly...probably they are derived by linear combination from the fields

@Slereah no no, just one

@ACuriousMind please edit kthxkss

Why no phase?

Aren't all wavefunctions defined up to a phase

@BernardMeurer [There is].

So the normalization is just $\int d\mu(\varphi) = 1$, right?

not really...they are the square integrable functions wrt the gaussian measure

With $\mu$ the Gaussian measure

exactly

Hm

So $\langle 0 | \varphi | 0 \rangle = \int \varphi d\mu = 0$ makes enough sense, I guess?

It's the mean of the gaussian measure

yes, that's it

I have this Feeling that maybe $a^\dagger_k \approx \varphi_k$, with $\varphi_k$ the one particle wavefunction

And I guess maybe $a_k = \frac{\delta}{\delta \varphi_k}$, if that makes sense?

@BernardMeurer I have nothing to edit there

@ACuriousMind I'm glad :)

@Slereah not exactly

the field is the multiplication by the variable, and the momentum is the derivation plus the variable

Yes, but isn't that what we are supposed to get

yes, more or less

$\hat a^\dagger_k | 0 \rangle = a^\dagger_k 1 = \varphi_k(x)$

Although...

I suppose the two operators wouldn't be complex conjugates then

Hm

the cre ann ops can be derived by the. field and momentum by linearity

yeah

Something like $$a_k(x) = \frac{1}{\sqrt{(2\pi)^n 2 \omega_k}} \int d^n x e^{ikx} (\omega_k \varphi(x) + \frac{\delta}{\delta \varphi(x)} - \varphi(x))$$

Not quite sure how to simplify it, though

well, something like that

but maybe there is a more explicit form somewhere in the literature

probably

I tried looking up Rovelli but then again it has $\pi = \frac{\delta}{\delta \varphi}$

Might not be too exact

Also is $\pi = -i(\frac{\delta}{\delta \varphi} - \varphi)$?

Seems a bit weird, unit-wise

yeah it's something like that

It ends up giving a term like $\varphi (\omega_k - 1)$ for $a$

@BernardMeurer Uh, do you not know what he looks like?

@ACuriousMind ...what

image of what

Image of the point

under?

@0celo7 ACM you mean?

yes

@0celo7 I know how he looks. Even tried adding him on facebook but he said he hated me and wished I was dead :(

savage

That may or may not be an accurate representation of events

@ACuriousMind I still don't get what "fiber preserving" means

@ACuriousMind We'll see about that in court

I'm bad at algebra

I'm the victim, you can't disagree with me. Stop victim shaming

@ACuriousMind Argh, what the heck does fiber preserving mean :/

I don't know what your problem is

Everyone says it maps fibers into fibers or something

Which fiber gets mapped into which fiber?

$f(E_x)\subset E'_{\pi(f(x))} = E'_x$.

Aha, that's my issue

What is $f(x)$

what the heck

what is $\pi(f(x))$

What is $x$?

A point in the base.

$f$ is defined on $E$, not the base.

Assuming $f:E_1\to E_2$

Okay, yeah

Fiber preserving just means $f(E_x)\subset E'_x$.

???

What if the bases are different

And what is $E'_x$

Then it means that the diagram on that Wiki pages commutes

lol

@0celo7 I'm calling the bundles E and E' because I don't want to type double indices

@ACuriousMind oh

@ACuriousMind which one

@0celo7 ...there is only one that applies for the case where the bases are different.

The fiber-preserving aspect in that case is that if one point of $E_x$ lands in $E'_y$, then already $f(E_x)\subset E'_y$.

How do you define the map between the bases

You don't

Wiki's $f$ IIRC

It is a requirement that it exists.

But you can recover it from $f(E_x)\subset E'_y$ in an obvious way.

@ACuriousMind ...

how

I'm not doing too hot today

No, I'm not writing that down

@ACuriousMind do you just project the mapped fiber onto the second base

yes

@ACuriousMind I would ask why you didn't write that but you'd get annoyed

@ACuriousMind what is the tensor product of matrices?