meow

in an infinite crystal (let's say monatomic), there is an allowed state for every possible k-vector, correct?

(although any k-vector outside the first Brillouin zone is just a repeat)

@imbAF I'm afraid I don't. Maybe someone else can help you

I'm sorry but you can use GPT for math in some great ways to do menial stuff

hi

@bolbteppa what did u use it for

@TobiasFünke phew done physics.stackexchange.com/a/843373/92181

maybe there's some mistakes I'll have to go back and edit

but the gist of it should be hopefully ok

how did your talk go?

do u think $\partial _{\mu}(\frac{\partial L}{\partial(\partial _{\mu} \phi)})=\frac{\partial L}{\partial \phi}$ is an abuse of notation?

there is something fishy about this equation. it is supposed to be differential equation on $\text{configuration space}\times R$. but the tensor indices are those of spacetime

i think this is not supposed to be a tensor equation . it is just made to look like one. it is just an equation involving einstein summation

It's not on the configuration space, it's on the bundle

yes. thats what i meant. the tangent bundle of the config space

this eqn also gives the vibes of being a co ordinate independent equation. but that is also an illusion

this article has the co ordinate invariant version of the EL equation en.m.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation

by using the index notation, the previous equation gives the vibes of being a co ordinate independent equation

but since the indices are of spacetime instead of the config tangent bundle, it is an illusion of being co ordinate independent

but also, after one uses this equation to get the EoM on spacetime, the resulting EoM is a genuine tensor equation

there are some other problems, like how the canonical momentum is a dual vector, so this term feels fishy : $\partial _{\mu} (\frac{\partial L}{\partial (\partial _{\mu} \phi}))$

but it is a dual vector on $T^*C$ where $C$ is the config space, while the index $\mu$ is of spacetime

once one computes the canonical momentum, it turns out to be a vector on spacetime. like, the EoM reads $\partial _{\mu} (\partial ^{\mu} \phi)$

so the canonical momentum $\partial ^{\mu} \phi$ is a vector wrt spacetime indices, but a dual vector on the configuration space

also, the actual co ordinate independent EL eqn reads $L _{X}\theta = dL$. this is an eqn on the config space

and the tensors involved dont have spacetime indies

also, the only derivative involved is wrt time, cuz the config space formulation treats time differently

Me when the bundle theory hits 😐

🫣

Apparently the nlab defines what it means for a space to be linear in monadic terms

Ugh

can something be a vector wrt spacetime indices but a dual vector wrt configuration space indices?

if i write an EL eqn as d/dt (dL/dv)=dL/dx, then dL/dv is a dual vector

but once i actually compute this EL eqn to get the EoM, it becomes d/dt(mdx/dt)=-dV/dx. now, mdx/dt is a vector

it feels like dL/dv is a dual vector wrt configuration space indices, while mdx/dt is a vector wrt spacetime indices

what is the usual definition of taking the dagger of a vector of operators?

Say I have $\Psi = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$ as a column vector. What should $\Psi^\dagger$ be?

(Where $a_i$ are operators)

It should be the dual vector of the dagger of each

ie $\Psi^\dagger = \begin{pmatrix} a_1^\dagger & a_2^\dagger \end{pmatrix}$

Errr wait no

Wouldnt make sense for a self adjoint operator

Or would it

Oh okay

My understanding of daggers has gone downhill since I learned that daggers and adjoints are technically different

less monads more stabbing

Monads, mo' problems

stab them 🗡️

@Feynmate been active on main site

@Slereah what is the difference

Hi qwerty. The talk went quite well, I think. Thanks for asking.

congrats! :)

Nice answer! :)

@TobiasFünke I liked the part in which you said "quantum"

@SillyGoose this depends on what the "vector" Notation is supposed to mean to begin with. Do you consider a direct sum of Hilbert spaces? Or what?

@Feynmate haha I am waiting for the day some of us meet in real life without noticing but only then later come to the h bar and realize

@qwerty take myow upvote

there is so much to learn

Elections in a few days in Germany. I am scared :3

@TobiasFünke i am confused about that as well; i would like to write terms like $\Psi^\dagger A \Psi$ to represent a thing bilinear in c/a operators

i feel like it is more natural to just define an inner product $\langle \Psi, \Psi \rangle = \sum_i \Psi_i^\dagger \Gamma_i \Psi_i$ with some inner product form $\Gamma_i$...but i'm not sure

heuristically tho, i would think $\Psi \in \mathbb{C}^{2n} \otimes \mathcal{B}(\mathcal{H})$

where the latter tensor product is lienar operators over the hilbert space (including, say, c/a/ operators)

@RyderRude Do you prefer learning physics or researching physics?

Haven't watched the video but this may be interesting youtube.com/watch?v=Q4xCR20Dh1E

If it is correct then I suppose it is easy to translate into your preferred notation

@DIRAC1930 i prefer learning physics (and speculating about possible research)

@DIRAC1930 maybe they r hyping it for shareholders

they say it is potentially a million qubits

i think it is based on majorana spinors

i like speculating about possible research but i havent started any serious research yet

i have some ideas that i think r novel but they r too broad rn

it is a research direction. but it is not concrete enough to begin the research

@RyderRude You should try and see what happens

yeah... i think the thing i can do rn is to learn the material that is related to that research direction

Good idea

it is a broad bunch of topics... like i have to learn existing quantum gravity approaches, qft, gr, quantum measurement physics

i will see if i stumble across the right thing to realise that research direction

but i am open to it never being realised. nature doesnt have to conform to my philosophical expectations

i am primarily interested in learning (and speculating a bit)

@DIRAC1930 do u prefer learning or research

What is this equation called in the literature $\dot f = \{f,H\}$ ? Classical Heisenberg equation of motion? I couldn't find much sources that name it explicitly.

It's just the EoM for $f$

Written using the Poisson brackets

Man, it's crazy how that equation controls the time evolution of every classical object in the universe

i will write a paper on those ideas if i get any results

@User198 this doesnt have a name, which is strange

a special case of this is called Hamilton's equations

but u can also derive this from Hamilton's equations, so it is not that special of a case

@RyderRude What area is your research direction in?

@DIRAC1930 it is quantum gravity. i think one has to abandon both spacetime and hilbert spaces for this theory

abandon spacetime cuz the metric is no longer classical. abandon hilbert spaces cuz quantum mech requires time

This will take you like 200 years to finish lol

lol

it is about the journey

the real theories r the friends we make along the way

@SillyGoose unless you mean something non-standard by a vector operator, the adjoint on the vector operator is just the adjoint of each component. Think about the basic examples $\vec x$ and $\vec p$

In the tensor product language where these are elements of $V \otimes \mathrm{gl}(H) $, the adjoint just acts on the second factor.

@Feynmate Ok thanks

@RyderRude Hah ok thanks.

@User198 Hamilton-Jacobi equation

@Slereah ...I don't think that's right

HJ is an equation involving the action and the Hamiltonian, not arbitrary observables

meow

@ACuriousMind oh okay.

I want to learn more about density matrix, mixed states, phase space formalism; but I am looking for a gentle intro to those topics. Does such a book exist?

@ACuriousMind but then how do I actually turn a column vector of operators into a row vector of operators?

@SillyGoose I don't know what that means :P

well, actually, I guess you just mean the transpose/dual operation on $V$

I just don't know when that would come up

well i am trying to encode the canonical commutation relations for $n$ bosonic c/a operators into an "inner product"

@User198 np mate

@ACuriousMind I guess if you want to write something like $[x_i, p_j] = i \delta_{ij}$ as a matrix equation with the identity on the RHS

@ACuriousMind I don't think Slereah believes that :P

yes exactly actually

@fqq this is precisely what i am trying to do (but with c/a operators)

i mean i think this is the rough idea

(for bosons and two modes)

@fqq It's $\vec x\otimes p^T - \vec p \otimes \vec x^T = \mathrm{i}\mathrm{Id}$ with ${}^T$ the transpose/dual on $V$ (from my notation above) and $\otimes$ the Kronecker/outer/tensor product between $V$ and $V^\ast$. I don't really see what that would be useful for.

I mean I guess I should really just define hermitian conjugates on the respective spaces right?

@ACuriousMind I think in some cases it's a decent notation e.g. for the correlation functions of multi-component fields, and yes I agree with ${}^T$

someone just commented on a 10 year old answer of mine

lol

This only shows how old you have become in the meantime.

@RyanUnger I commented even older answers... :P

@Loong I'm aware of my Old Age

@SillyGoose What are you trying to do here?

What do I get when I measure spin along 2 different axis, but at the same time?

In theory at the exact same time*

If asking such question makes sense xD

@User198 No, the question does not make sense. What apparatus would you propose that performs that measurement?

@ACuriousMind It doesn't exist, yeah

Ok thanks

Can anyone help me with a suggestion. In our lecture, the lecturer started to show the quantization of the real scalar field as a primary example of the canonical quantization. In doing so, in an extraordinary lazy fashion, he talks about, poles, circumventing the poles , x_0>y_0 the negative imaginary plane x_0<y_0 the positive one, retarted grees function of the KG. operator (whatever that is), and other stuff that have to do with sketches and, something about 4 pole possibilities. But he

doesn't explain why things are the way they are, he just takes them as something very well known and goes along with it. He doesn't bother to explain i.e what is the retarded greens function or how we use the residue theorem and so on. Can anyone suggest me a rigorous derivation of the quantization of this field, with all the above mentioned elements?

@imbAF This might be useful for you imperial.ac.uk/media/imperial-college/…

@DIRAC1930 Thanks

May I ask you one thing

that has been bothering me for like 2 months now

in QFT

@imbAF As I have told you various times, there is no fully rigorous formulation of QFT. Proof of a fully rigorous version of the Standard Model is an open millenium problem. You have to be more specific. IF by "rigorous" you just mean "an explanation that satisfies me", you should stop using it in that respect, since without any further qualifiers the default assumption is that you mean mathematical rigor.

I mean a proof such that doesn;t make claims without proving them or at least explaining where is it coming from

For example

I will, copy past exactly what was written, simply as an example

Until the very moment that this was said

$\phi$ be schroedinger picture or Heisenberg one, $\phi$ has been a used to represent a state

And now, a claim is made, where $\phi$ is an operator, which I believe is because it can be expressed via ladder operators

But the problem is that in the notes, the state in qft is something like

$|\vec p_1,\vec p_1,....\vec p_N\rangle$

And at no point is the difference between $|\vec p_1,\vec p_1,....\vec p_N\rangle$ and $\phi$ explained, when $\phi$ was taken from QM, where it represented a state

Now it's a field operator, in QFT

@ACuriousMind What I want is to understand the quantization of the scalar field, without having concepts such retarded greens function of KG equation, consideration of the complex plane out of a sudden, clockwise and counter clockwise etc introduced and not explained why they were introduced. That is what I want

@User198 Poisson bracket formulation of Hamiltonian mechanics. @Feynmate @Slereah @ACuriousMind

@naturallyInconsistent That's not the "name of the equation". You're correct that's where that equation occurs, but there's not really the "name" for that equation...

@User198 You can do it if you assume that entangled stuff allows you to do that, but then the results are trivially given by the postulates of quantum theory. i.e. the experiments have already been done in that regard.

@ACuriousMind I think that's a name for that whole framework. It is manifestly a name, and a mouthful because it is a formal solution that is not really practical.

@imbAF I'm quite sure someone else had already told you about this tremendously huge chasm and how much of deception it actually is.

I don't remember tbh

Like, at least that version appears on Google search. Personally I would have called it Poisson bracket form of Hamiltonian EoM

I'm afraid I am not following you. What are you talking about?

@imbAF I recommend you go through second quantisation in L&L 3 as quickly as possible if you have enough time

It goes through 2nd quantisation in the non-rel scheme

I've read about 2nd quantization. But I haven't encountered an answer to my question

But perhaps what you suggest might have that

Well $\hat{\phi}$ is not to be confused with the wavefunction

@DIRAC1930 One thing. I believe in QFT we write $\pi=\frac{\partial\mathcal{L}}{\partial \dot \phi}$. In the topic of quantization for the scalar field, after giving a fourier decomposition of $\Phi(x)$ where ladder operators are present without an explanation, the lecturer says that $\Pi(x)=\dot \Phi(x)$

Is this true? the last part

@DIRAC1930 Yes, it was taken from QM where it was used for the wavefunction/state and it becomes field operator, somehow