i cant find a resource/textbook recommendation post for anti-de sitter spacetime

is it even possible that its been not asked before?

Ah i found this one physics.stackexchange.com/questions/36303/…

I wish i had found this much earlier: physics.stackexchange.com/questions/12175/…

What is it like to be a retired physicist/mathematician? Mostly asking ex-researchers. Do you still study/research on your own?

I mostly just mess around and do whatever I find interesting at the time.

@ACuriousMind I have seen it in a Donald Duck comic

It had people living on a nucleus

Also one Disney character was actually an atom

this is what an atom looks like if you enlarge it

@JohnRennie Sounds very nice honestly

@ShikiRyougi It's not a bad lifestyle. And my ex-employer gives me money every month to do it :-)

I wish i could retire from now

but not possible XD

@JohnRennie I envy your position. I think it will feel so much better when you're not pressured to publish all the time. The pressure slowly deprives me of motivation and curiosity, since everything becomes career motivated, and not about physics itself... How did you keep motivated earlier in your career?

he just found toothpaste very interesting and so could work in industry instead of the academic treadmill you're describing ;P

@BlindMiner Yes, although Im not doing research yet (doing masters), the pressure of ''read this or else you are screwed'' a lot of times takes away all the excitement in reading the subject. I like to take my time reading a subject and wonder in which-ever direction I feel like at the moment.

The sad reality is, even after if you read all the stuff required of you, it will barely scratch the surface

Im aware of that, but I think thats expected due to how much physics and math have advanced

Its really sad that it will take years to understand something to a deep level, although somewhat exciting as well

those Aha! moments feel great

but the whole process is very hard, especially if you have depression and/or other mental issues

Yeah nobody said any of this was easy, going against every natural impulse in many ways

a lot of people around me have been telling me to give up because they see how much im struggling, but physics/math is one of the few things that give me some purpose

@bolbteppa what do you mean by going against every natural impulse?

Well there are plenty of free university outlets/services to talk to people if things are getting hard, and they are nearly always useful, worth considering

It took humanity hundreds and hundreds of years to figure out basic algebra after the thousands of years it took to get to geometry, etc... It's much easier to forget about all these numbers, or at most stick to the known stuff we have, let alone trying to push these subjects forward

@bolbteppa Dont worry, I have professional help and Im already in good hands :)

@bolbteppa Its so sad that I may die before I find out the ''truth''

That's what this stuff is for

very unfortunate that the alchemist dream didnt come true

@bolbteppa I'm not sure I'd use this as an argument for science/technology being "against every natural impulse": The exponential acceleration of technology/science/knowledge is just a very natural consequence of us inventing a series of technologies (e.g. language, writing, printing press, steam engine, railroads, airplanes, internet) that each make it possible for more people to communicate with one other quicker or have less of us fighting for mere subsistence.

There is a suspicion one could have that we didn't evolve our brains in an environment as fast-changing and information-laden as the (post)modern world and that that might cause problems, but that's one of these evo-psych takes I'm always very wary of because there's no way to test that

@BlindMiner After I finished my PhD I went into industry and that's very different. You tend to have shorter projects and you're learning different stuff all the time, which I found really interesting. You don't study things in as much depth as you would in academia though.

During the PhD I had so much cool stuff to do that I barely had time to eat and sleep, let alone worry about the future.

Does anyone knows where I can learn how to map the XXZ quantum chain to the Luttinger model? Basically I understand how to map the XXZ hamilotnian from spin operators to a model of interacting fermions with the Jordan-Wigner transformation.

What I miss is why the assumptions of Tomonaga should hold (but this is something I can take for granted for know) and most importantly how with the linear energy spectrum limit we can get the Luttinger model Hamiltonian from the one of the interacting fermions.

How do I know what form the density matrix takes in expressions such as $Tr (\rho A(t) B(t') )$ (in the Heisenberg picture)

As in how do I relate it to the Schrodinger or Interaction representation?

I don't know what you mean by "form"

Well people just write $\rho = \rho(t=0)$ but I'm not sure how this is justified

I mean, in the Heisenberg picture the operators are time-dependent and the state is not, so $\rho$ is also not time-dependent

So I can just arbitrarily take it to be the density matrix before interactions are switched on?

in the Heisenberg picture that sentence doesn't mean anything

there is no state "before" or "after", there's just a state

and that state is the time-independent $\rho$

Ah of course

In the density matrix formalism, the switch between Schrödinger and Heisenberg is just cyclicity of the trace: $\mathrm{tr}(\rho A(t)) = \mathrm{tr}(\rho U(t) A U^\dagger(t)) = \mathrm{tr}(U^\dagger(t) \rho U(t) A) = \mathrm{tr}(\rho(t)A)$

first expression is the "Heisenberg picture", last expression is the "Schrödinger picture"

(it's possible the daggers have to be the other way around, I didn't double check)

No that looks correct

If I have two operators $tr(\rho A(t)B(t')) = tr(\rho U_{t,0} A U_{t,t'} B U_{t',0}) = tr(U_{t',0} \rho U_{t,0} A U_{t,t'} B )$. Then from the invariance of the trace under unitary transformation $ tr(U_{t,t'} U_{t',0} \rho U_{t,0} A U_{t,t'} B U_{t't}) = tr(\rho(t) A U_{t,t'} B U_{t't})$

Is this correct for two operators?

seems legit - $\langle A(t)B(t')\rangle$ in the Schrödinger picture is what you get by evolving $\rho$ to $t$ and $B$ to the same time as $A$ (i.e. from $t'$ to $t$).

Do I then have to define interactions to be adiabatically switched on after reference time $t=0$ if I want to proceed with calculations?

I'm not sure what you mean - I don't really know what you want to compute here or what interactions you're talking about or why they'd need to be switched on adiabatically

if you're trying to do some kind of perturbation theory that relies on the adiabatic assumption, sure

but that's a tautology - if your approximative approach relies on an assumption, you need to actually assume it :P

Yes I want to do perturbation theory

@ACuriousMind lol

Hi guys, I have another stupid question about perturbation in qft. Say I have a vertex with 3 fields of the form H^2L, where H is some heavy field I'm trying to integrate out. In matching the 2-pt function of L, I encounter the diagram which looks like: -O O-, with the external lines capped off by tadpoles. It seems that this cannot be incorporated in a local effective action for L. I would need to multiply two spacetime integrals together like (\int L)(\int L).

Is there some reason for this diagram to vanish that I'm missing? A direct computation shows that it is not zero.

It's common lore that integrating out heavy fields produces a local, not bilocal, effective action