2:49 AM
@JingleBells He didn't figured out, he just made it. (Branding >>> Product Market Fit)

1 hour later…
4:14 AM
@JohnRennie care for a chat in the hbar?

@Anthony hi :-)

Hello!

Thanks :-)

You're one of the names I scroll ahead to look for when I start reading a question.
That being said, I have a question that I've struggled to find a good answer to on the internet for a while. Maybe I'm just bad at searching.
Is there some obvious reason why Feynman diagrams are so popular?
I can't for the life of me understand. They are literally 1-1 with terms in the perturbative expansion, right?
So isn't the use of Feynman diagrams, at best, realizing you have some kind of graph-like structure in the terms? But this also doesn't seem to be incredibly illuminating - graphs are combinatorial objects!

People like them because they offer a way of visualising all the terms in the perturbation series, and lots of us find visual representations a great aid to understanding.

4:18 AM
Ugh.
Is that really the full story?

Yes

That's fine, and compelling enough.
Just, completely underwhelming.
Thank you!

Sorry if you expected something deeper!

Hahahaha.

I bet they were essential to Feynman when he was working out the theory in the first place.

4:20 AM
For convenience?

I think he was a visual learner

I don't know if he ever commented on this. It would be interesting to look at his publications and see if he said anything about it.

Oh, that's a good point.
Alright, I'm gonna get back to work.
Nice to e-meet you!

:-)

4:49 AM
0

Many sources assert that China appears to have eliminated COVID-19 within the country, except for the odd imported case or cluster, which is quickly isolated and eliminated. For instance, Worldometer's coronavirus tracker, based on data provided by each country's government, shows no new cases fo...

morning

4 hours later…
8:29 AM
Hey all!
Does this belong to MSE?
-1

Background So the aim is to find the value of the series (usually found in Feynman diagrams): $$S= a_0 + 1!a_1 \epsilon + 2!a_2 \epsilon^2 + \dots + n!a_n \epsilon^n$$ where $\epsilon \to 0$. We start by considering the function: $$f = \prod_{r=1}^N ( r\epsilon)$$ To make sense of we take log ...

Nevermind ... I realised I have to edit it

@MoreAnonymous I was just about to comment on the question that it makes no sense to me: 1. Your $S$ is just a generic polynomial of degree $n$. What does this have to do with "Feynman diagrams" and why do you think its value is hard to compute? 2. In what way do you think your terribly convoluted "result" with a limit and two integrations is simpler than that? 3. The value of $S$ as $\epsilon \to 0$ is obviously $a_0$. If you're trying to say that $\epsilon$ is small, then say that.

@ACuriousMind I think I did say $\epsilon \to 0$
Can't remember where I exactly saw the series
but this seems close
4

Suppose that $A(g)$ is given by some perturbative expansion around $g=0$: $$A(g) = \sum_n c_n g^n.$$ Then the statement that this expansion is asymptotic means that the radius of convergence is zero: for fixed $g$, no matter how small, the limit $$\lim_{N \to \infty} \; \sum_{n < N} c_n g^n$$ div...

2. One of the integrations is pretty easy solve. I think it would be easier that adding all the terms
I vaguely remember the factorial comes from the number of Feynman diagrms
@ACuriousMind Wait your right $\epsilon$ is small would have been more accurate

@MoreAnonymous Again, what you wrote down is just a generic polynomial. The problem is not in computing that once the $a_i$ are known.
The Feynman diagrams don't compute this series, they compute (parts of) the $a_i$.

8:44 AM
I guess I should have revised my Feynman diagrams in QFT ... Been forever I've had a look ... My bad

If you want this to be a question about physics you need to present the physical motivation in a much more coherent way

2 hours later…
10:43 AM
Let's make an AI model to simplify AI research papers.

2 hours later…
12:25 PM
The Fourier expansion of, say, the KG field is an integral (sum) of its general solution, the plane wave. When we quantise this field the quantisation of the field operators is equivalent to the quantisation of the $\hat a$ and $\hat a^\dagger$ Fourier coefficients in the expansion. Am I correct in saying that we are essentially turning the amplitudes of the individual plane waves in the mode expansion into operators? And this (sort of as a by-product) turns the field solution $\phi(x,t)$
itself into an operator?

@Charlie Yes

Awesome, thank you :)

3 hours later…
3:16 PM

3:40 PM
@JohnRennie It's not the worst example of questions like those. It does seem to be at least somewhat asking about the underlying physics principles in asking about turbulence, where the energy comes from, etc.
And of course catchy titles and odd scenarios always make it to HNQ. For better or for worse
I tend to get more annoyed with the irrelevant comments on questions like these haha

I think it’s something that’s currently being discussed about outside of PSE. This perhaps has inadvertently been attracting a lot of clicks here.

Hey, I have a question about the quantum harmonic oscillator, which I'm just learning about.
So the state vectors are eigenvectors of the Hamiltonian operator (right?). Does this mean that the state vectors form a basis for a Hilbert space? (I'm speaking in the context of the ladder operator method: https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator#Ladder_operator_method).

The state vectors can be expressed as superpositions of energy eigenvectors of the Hamiltonian

any vector in the Hilbert space is a "state vector"

But in general the state of the system does not need to be an eigenstate of the Hamiltonian
That was one of the most obvious revelations I eventually had when I first started learning QM. The books usually only focus on energy eigenstates, and so you start thinking in terms of only energy eigenstates. But really we study these states because they form the basis of any other state for that system.

3:52 PM
@ACuriousMind What about the zero vector

The zero vector can't be normalized

It's part of the Hilbert space though!

I thought it gave its letter of resignation a while ago

(I meant to ask whether the eigenvectors of the Hamiltonian are a basis)
Yeah that seems to be correct then from what you wrote

Yes, the eigenvectors of a Hermitian operator can form an orthonormal basis

3:54 PM
@Slereah we don't talk about the zero vector
it's a secret

"You wouldn't understand. It's a secret"

@BioPhysicist Ahh yeah that makes a lot of sense
@BioPhysicist So what you're saying is that a particle in a quantum harmonic oscillator can take on any state which is a linear combination of the eigenfunctions of the Hamiltonian?

@BioPhysicist Well, a generalized basis

Yes, unless I am missing something super obvious (something I tend to do with QM)

Thanks a lot :)

3:58 PM
Note that any such state will be time dependent i.e. not a solution of the time independent Schrodinger equation.

There's a bit of a chicken and egg problem here: The eigenvectors of a self-adjoint operator are a basis for the Hilbert space, that's why we want our operators to be self-adjoint in the first place so that we can formulate the Born rule for computing probabilities

I could eat some chicken and egg

@Slereah Also note that the actual chicken or egg question is not a metaphoric chicken or egg problem.

@ACuriousMind I'm guessing that's a reason why there is an "axiomatic approach" to quantum mechanics

4:18 PM
@BioPhysicist another important thing is that time evolution of energy eigenstates are trivial (phase) and they are stationary states. This is straightforward from the TISE

Yep :)

4:48 PM
Physics wisdom

5:37 PM
Do the creation annihilation operators acting on the free-vacuum in qft generate states that are eigenvectors of both the momentum and Hamiltonian operators? Are they also eigenstates of any other operators?
I assume they must at least be eigenstates of the Hamiltonian

The hamiltonian is roughly $$H = \int dk a_k^\dagger a_k$$

won't that always annihilate on the vacuum?

Not a big problem since we assume the vacuum has energy zero
It's the number operator
roughly speaking you just multiply the number of particle with momentum $k$ by the energy $\omega_k$
and sum them all up

oh I see ok I have seen this before
I'm still unclear on the difference between acting on the vacuum with the creation operator vs acting on it with the field operator. Both are often described as "creating a particle", but acting with the field operator will create a state that is an infinite linear combination of one particle states, each term having a different momentum
When we calculate scattering amplitudes we use the field operators, but we construct the initial and final states with the C/A operators

6:06 PM
you can try doing QFT with only the field operators if you want
it's not very fun :p

Hi guys could really use some help interpriting some physics math im attempting, this is aprt of a bigger issue im having but lets start witht he smaller/easier question
Looking at this page I see two equations that relate electric permitivity to tan(d):

tan(d) = (w*e''+c)/(e*e')
tan(d) = e''/e'
When i set these two equations equal to eachother, cross multiply and simplify i wind up with

e' * c = 0
Can that be right? must conductivity or the real part of permittivity always be zero or did i screw up?

6:31 PM
@Slereah Oh I didn't know that, that actually helps :P just knowing that it's possible

7:15 PM
0

Since the content on PSE is licensed under CC BY-SA $4.0$, I suppose the content that is once posted on PSE becomes part the free culture (correct me if I am wrong), and thus, even if the author decides to delete a post of theirs, the community can still decide to keep the content of the post up ...

7:41 PM
I posted a question instead and I got my answer, you guys can ignore my questions. Thanks though.

3 hours later…
10:35 PM
I'd really appreciate some help with this question, it has been plaguing me for a week now... physics.stackexchange.com/questions/581810/…

1 hour later…
11:52 PM
@jeff
@JeffreyPhillipsFreeman what is magnetic impedance?

ant
oops
@antimony same idea as electrical impedance just for magnetic fields rather than electric.. it impedes the propagation of a magnetic field though a material and describes the portion that is caused due to dissipation of energy vs stored energy (when complex)

Is it like the inductive component of reactance?

@antimony no thats actually different, that though i can see why you might get them confused
an

magnetic impedance can have a real and imaginary component?
whereas reactance can only be imaginary?