I wonder if the bosses use smart reply. it's disgusting - to generate and show you what you "should" say before you've even read a personal message.

@qwerty To be fair, sometimes it's quite funny - Outlook consistently generates "only time will tell" as a quick reply for one of my coworkers and I wish they would just send that as a reply to everything

xD

@ACuriousMind funny ... sounds like a dilbert comic to me

@qwerty I feel you

@MoreAnonymous I don't know what "I then second quantize it" means.

@ACuriousMind When particle number is conserved I can go back and forth between first and second quantization description no?

@MoreAnonymous without a lot more context that doesn't really mean anything to me, no

I mean, I know what second quantization is, but the significance of the phrase "I second quantize the acceleration operator" escapes me.

well, actually there is well-defined meaning

@ACuriousMind I mean I express it as creation and annihilation operators

the second quantization of an operator $A$ on a Hilbert space $\mathfrak h$ is the operator $\mathrm d\Gamma(A)$ on the corresponding bosonic or fermionic Fock space $F(\mathfrak h)$ with $\mathrm d\Gamma(A)=\bigoplus\limits_{N=0}^\infty A_n$, where $A_n:=\sum\limits_{j=1}^n I\otimes I\otimes \cdots\otimes A\otimes I\otimes \cdots \otimes I$, with $A$ at the $j$-th position in the sum.

I hope I did not miss something, but that's essentially it.

Oh, and of course $A_0:=0$ and $A_1:=A$

@MoreAnonymous How are you proposing to do that for some random time-dependent operator like your $a(t)$?

the canonical c/a operators are defined in terms of the time-independent pair $x,p$ to be $a^\pm = x\pm\mathrm{i}p$

(or its generalization to fields in terms of the field and its canonical momentum)

there simply is no obvious meaning to just saying you do "the same" to $\ddot{x}$ in the Heisenberg picture

I'm not starting with the classical setup. I'm assuming I know the QM Hamiltonian of the physical system in a inertial frame

I've tried it for linear acceleration so far and it seemed fine

I don't see how that's an answer to the question of what you are actually doing

Let's say I have: $A = [[H,x],x]/\hbar^2$ Then I use this (adding below)

where:

Sorry it should be $A = - [[H,x],x]/\hbar^2$ where A is the acceleration operator not to be confused with the creation or annhilation

we're again at the point where you ask a one-liner question and have to explain it with a bunch of equally unclear equations; I'm afraid I'm not available for peer review of what seems to be an attempt at a paper

@ACuriousMind Ah ... my bad

@ACuriousMind classic magic 8 ball response xD

@qwerty lolz

@MoreAnonymous hi

i think one can use non rel QFT to define the second quantises acceleration operator

i think the position operator is $\int dx x \psi ^{\dagger} (x) \psi(x)$

@RyderRude hey

@qwerty only time will tell

could use "Concentrate and ask again" as an auto reply on the h bar, couldn't you? :p

@qwerty I think for me it would be "I don't know what that means"

holy moly haha

that's your signature phrase, it seems

i feel I will never get to know fundamental physics answers

they seem to be 300 years away

like the measurement problem

perhaps, yes

but who knows

yes

For the $-\frac{\lambda^2}{4!}$ interaction, in 2nd order one would have a term of the form $\frac{\lambda^2}{4!4!2}$. Should this term vanish?

@ACuriousMind "I'm not sure what you're asking" is more like you :P

Why is there none in the searchbar 💀

@RyderRude Any thoughts on the original question?