2:20 AM
2:54 AM
where, what?
Notes from Underground (pre-reform Russian: Записки изъ подполья; post-reform Russian: Записки из подполья, Zapíski iz podpólʹya; also translated as Notes from the Underground or Letters from the Underworld) is a novella by Fyodor Dostoevsky first published in the journal Epoch in 1864. It is a first-person narrative in the form of a "confession". The work was originally announced by Dostoevsky in Epoch under the title "A Confession". The novella presents itself as an excerpt from the memoirs of a bitter, isolated, unnamed narrator (generally referred to by critics as the Underground Man), who...
looks interesting
H O N K ~
3:14 AM
h o n nk

4 hours later…
7:09 AM
@SillyGoose amazon.in/…
7:22 AM
> He says that Eurytus made likenesses of the shapes of things in the natural world with pebbles and thus determined the number which belongs to each thing by the number of pebbles required.
Pythagoreanism was some weird shit
7:56 AM
@nickbros123 oh interesting
let $p(\vec{a} \lvert \vec{X})$ denote a condition probability of obtaining outcomes $\vec{a}$ upon measuring random variables $\vec{X}$.
What is the natural sense of turning the set of such "correlations" into a vector space?
In Griffiths Intro to QM, for the finite potential well he says:
"If Z_0 is very large, the intersection occurs just slightly below n_z=n\pi/2 with n odd. it follows $E_n+V_0\approx \frac{n^2\pi^2\hbar^2}{2m(2a)^2}$
Here (E+V_0) is the energy above the bottom of the well"
I mean several things are unclear here:
1. How to understand the intersections occur just slightly below...
2. Where does the index n comes from?
3.Where does the expression E+V_0 comes from?
4. E+ V_0 is not only above the bottom of the well, but above the well since -V_0<E<0 and thus V_0 + E >0
Can anyone help me understand what on earth is griffith saying here?
i think the "vector space" language is a bit sloppy? i think it means to only allow convex combinations of such $p(\vec{a} \lvert \vec{X})$. is that right?
Hello everyone! I was hoping to find help and get my doubt cleared.

Definition of Quasi-static: A quasi-static process is a thermodynamic or mechanical process that occurs very slowly, allowing the system to remain in a state of equilibrium at all times.

While explaining Potential Energy, we take an example by lifting a block of mass 'm' from the ground (H = 0, No kinetic energy and Potential Energy) to H = x, very slowly in "Quasi-static" manner. But according to the definition, if there state of equilibrium, how will the block even be lifted and taken up? After lifting, there can be equ
8:18 AM
hi
anyone want to play this lichess.org/ZYQt7EVP ?
8:39 AM
to what space do $H, V_k, \mathcal{L}$ belong? Concretely, they seem to belong to the same space $\mathcal{B}(\mathcal{H})$.
However, intuitively, they should belong to the dual space $\mathcal{B}(\mathcal{H})^*$
as in this context we are talking about them as linear operators over the space of density matrices (which is contained in $\mathcal{B}(\mathcal{H})$
perhaps there is just an isomorphism $\mathcal{B}(\mathcal{H}) \cong \mathcal{B}(\mathcal{H})^*$?
okay i think the answer is for finite dimensional $V$, $V \cong V^*$
9:19 AM
@SillyGoose no, they should not
the operation between $\rho$ and $V_k$ as $V_k \rho V_k^\dagger$ is just the multiplication in $B(H)$
of course you could phrase this as the operator $A \mapsto V_k A V_k^\dagger$ in $B(H)^\ast$, but the $V_k$ itself is just in $B(H)$.
Does anyone know books or articles that talk about light but more in a philosophical view? Because I'm interested in reading those. I'm looking for an answer if there is one of course to some things I've been asking myself, for example

If a photon experiences no time could it be in two places at the same time for someone that does experience time?
I've asked this to chatGPT and it gave me some satisfaction by saying some things that made sense, but than also some things that didn't made sense or just gave some info but not really related to or answered my question
9:42 AM
@imbAF On the graphs, the intersections are always slightly left of the vertical lines at exactly $n\frac\pi2$, where those are the solutions to the infinite square well problem that you MUST have seen right before because Griffiths covered it before this. And $n$ counts which vertical line is being considered. The text explicitly told you that $E+V_0$ is energy as measured from the bottom of the well; that is literally the lowest energy in the entire system, and you saw that the RHS > 0 too
@AkhileshG When a process happens quickly, you cannot define a single number as the pressure of the entire gas inside the system, and so nothing is well-defined. You must learn thermodynamics by considering the simplest things first, and that implies starting from quasistatic processes.
@naturallyInconsistent that response was brutal lol
10:33 AM
@Claudio which?
@naturallyInconsistent well I didn't accept its answers I just said it wasn't full of bullshit
@lynx_s philosophy is best for unsettled or unsolvable problems. i doubt there is any good philosophy on the nature of light
this wouldve been the topic of philosophy 1500 years ago maybe
also, no physicist would talk about "light experiencing zero time". light has no first person experience @lynx_s
Alright
Thanks

1 hour later…
11:45 AM
Ryder let’s play chess
11:55 AM
@Kenshin sure
Create a new one
Gg
12:15 PM
@Kenshin yeah
Not bad cool you’re interested in chess too
@Kenshin thanks for playing
ty
@Kenshin i think it's entertaining
i lost all today :P
You would have won the first game but for the timer
12:17 PM
yes.. i wasnt paying attention to the time
u r really good
cya
i should play longer maybe. i like some time to think
12:49 PM
@naturallyInconsistent the one about chatgpt. That being said I agree that quenching one's thirst for knowledge through AI is not something I have or will ever consider doing.
It does pretty bad job when it comes to any more specialized type of knowledge
@Jakobian It's rare seeing you here on the physics side hahaha.
yep you're right
@Slereah interesting. This is similar to formalism
I have a question, I've realized that when solving via contour integration, considering the trivial extension of a function to the complex plane might actually bring about wrong results ( namely $x \mapsto z$ )
that's something I was not aware of or that we ever discussed during our course
1:08 PM
@user85795 I am late to the discussion, but I think that people do math just to see what will happen. Its like a movie where you're one of the actors. They do it for entertainment
@Claudio possibly your function is not analytic/meromorphic after such "obvious extension"
So that methods of complex analysis are not applicable to it
@Jakobian yeah I never considered the possibility (understandably so, since up until now everything always worked fine :P) i admit I've been acting with little care
@ACuriousMind But a binary multiplication is not defined in a Banach space (without further specifications). Or is it?
Wouldn't we really be referring to an algebra structure
@SillyGoose Hm? $B(H)$ is the space of bounded operators on $H$, and you can multiply operators, no?
Yes, but as a Banach space it is just (to my understanding) a vector space with some additional properties
so?
it's a Banach space that's also an algebra
(i.e. a Banach algebra)
1:25 PM
@Claudio after looking at your post I think you made some other mistake, I suspect its because of how you set up your countour of integration
hm, well is there anything wrong with thinking of $\mathcal{L} \rho$ as $\mathcal{L}$ being a linear operator over the banach algebra $\rho$ belongs to
@Jakobian It's a standard key-hole path for multivalued functions tho
om nom
Although I must admit that recognizing that the contour I chose does require to slightly modify the complex extension of my function is probably a bit too advanced for the kind of course I followed
The problem is that my professor, when I sent him an email about this, just dismissed my question and told me "You should've used this contour instead to obtain the correct result" without trying to investigate a bit further why the key hole path failed :P
And that's why I asked the question
@Jakobian Anyways, please tell me any mistakes, either here or maybe make a comment
Complex analysis is not my strong suit so I won't try to find the error
1:39 PM
@SillyGoose well, the $\mathcal{L}$ is not like the $V_k$ and the $H$ - it's often called a "superoperator" (like the 'Liouvillian'), and it is indeed an operator on $B(H)$, i.e. in $B(B(H))$
but your text uses $H_S$ and $H$, I suspect that $H_S$ is the "state space", i.e. the usual Hilbert space, and the $H$ is the space of density matrices, so it's correct what's written there
@Claudio you probably ignored some poles - when physicists do this contour integration trick via analytic continuation, it's important that you don't move the contour "over" any poles your continuation has
1:53 PM
@ACuriousMind is $B(B(H)) = B(H)^*$?
@ACuriousMind I don't think the problem lies in the presence of unnoticed poles on the path of integration since the alternative function presented in the answer has the same poles and zeros of the one I originally used.
@ACuriousMind oh that was a mistake. i meant to type $\mathcal{L} \in \mathcal{B}(\mathcal{H}_S)$ (which I think based on what you said is incorrect).
@SillyGoose no, $B(H)^*$ inputs element of $B(H)$ and outputs a number, here you output element of $B(H)$
ohhh
oh my i see
what area of physics are you currently doing your research in if I may ask @SillyGoose
2:04 PM
How to explain brownian ratchet where ratchet is in a vacuum
@Claudio right now bell inequality related stuff, but it is not so interesting. hopefully soon quantum thermodynamic stuff
@SillyGoose setting up my mind on cooler stuff when I need to study something boring is indeed something I much relate to

1 hour later…
3:29 PM
we are organising a chess tournament with the math chat. everyone who wants to join, reply to this
4:11 PM
it will be physics team vs math team

2 hours later…
6:14 PM
Is this accurate:
$\int_{-\infty}^0 e^{kx}dx=-\int_{0}^{\infty}e^{-kx}dx$? ?
nah
6:31 PM
Hi everyone, I have a question regarding the formula of this exercise:
A stone is thrown from the top of a tower h = 50 m high with a velocity of magnitude equal to v0 18 m/s and an inclined direction of a $\theta$ = 50°, upward, from the horizontal plane. You get: 1) The time interval required for the stone to reach the ground and the distance it travels horizontally before hitting the ground; 2) The magnitude of the
speed with which the stone hits the ground; 3) The angle formed between the trajectory of the stone and the horizontal plane at the point of impact. Neglect air resistance.
Points 1) and 2) are clear, but point 3) I think we should use the formula:
$\tan (\theta) = \frac{v_y}{v_x}$, Could anyone explain to me why?
6:54 PM
If a system has bound and scattering states, which for simplicity we can consider both groups as eigenstates of the hamiltonian of the system. Does it make sense to consider the inner product between them and additionally that whether they form a basis, even though we are considering 2 completely different setups?
7:28 PM
in Mathematics, 26 mins ago, by Xander Henderson
You will never find a more wretched hive of scum and villainy than Physics chat. I'm not going over there.
8:05 PM
sorry, I wrote that text before without being very precise, if anyone wants to help me you can see this:
0

Hi everyone (sorry for the "unclear" title), I'm having trouble graphically representing the point of this: A stone is thrown from the top of a tower with $h = 50\text{m}$ and with a velocity of magnitude equal to $v_0 = 18 \text{m/s}$ and an inclined direction of $\theta = 50°$, upward, from the...

8:22 PM
@imbAF that's exactly the problem with the normalizability and discrete vs. continuous spectrum again - the scattering states will lie in the continuous spectrum of the Hamiltonian, and so these "eigenstates" won't be normalizable and won't have proper inner products with the discrete eigenstates
by proper you mean that the inner product gives zero ?
by proper I mean that it exists in a formal sense :P
I don't understand
like, think about how for position eigenstates, the $\langle x\vert x'\rangle = \delta(x-x')$ isn't really an inner product - the $\delta$ function is not a function, we're essentially saying $\langle x\vert x\rangle = \infty$, but by definition an inner product should have finite values
the "eigenstates" in the continuous spectrum are exactly like that
always ?
so not normalized
8:26 PM
@imbAF there should be no minus
@Jakobian Yeah, I realized
But thanks
@imbAF yes (again, proving and formalizing this requires a lot of functional analysis you simply don't have at an intro QM level)
I see
I have 2 more questions
Considering discrete and continues spectrum, the outer product of two arbitrary wavefunctions in each spectrum
1. If one were to calculate the outer product of some arbitrary wave function for the discrete case, for an arbitrary n, wouldn't one again use integral, to get the $\delta(x-x')$? If so, then what would be the differential term, the d(...)?
2. In both cases, when we talk about completeness we consider the same wavefunction at two different positions in space. What is the logic behind all of this? such that different locations in space are considered and if we get a delta function, that gives us completeness, and thus allows us to use these states that we are considering as a basis.
do you really mean outer product? I'm not sure what you mean
Yeah
$|\phi_n\rangle\langle\phi_n|$
which is what we use in calculating completness
But the first question I believe is clear
I only had to consider the simple case of a particle in a box
and the outer product of two such discrete eigenstates of the hamiltonian
8:32 PM
but for the inner product, the continuous $\delta$-function is simply the generalization of the discrete case - consider that a finite orthonormal basis $\lvert v_i\rangle$ has $\langle v_i\vert v_j\rangle = \delta_{ij}$ for the Kronecker delta, the continuous $\langle x\vert x'\rangle = \delta(x-x')$ is "just" the continuous version of this
I need to be a bit more clear
for the outer product, just consider again that you have a sum in the finite case: $\sum_i \lvert v_i\rangle\langle v_i\rvert = 1$ and a "continuous sum" is an integral: $\int_x \lvert x\rangle\langle x\rvert = 1$
I understand but I need to clarify what I want
I am considering this type of notation: $\phi(x)$ No bra-ket.
So for either case, discrete or continues you have, for completeness the following:
$\sum_n\phi(x)\phi^{*}(x')=\delta(x-x')$
or
$\int \phi(x)\phi^{*}(x')=\delta(x-x')$
What I wrote right now, is tied to my 2nd question
I mean you get that from my expressions by applying $\langle x\rvert$ from the right and $\lvert x'\rangle$ from the right, this is just the difference between abstract kets and the position repressentation
Ok
I thought there was more to it
When we want to know orthogonality we change the index n, meaning consider different eigenstates of H. Which makes sense
I mean, algebraically at least
But then when we want to check completeness we change the variable
again, in no bra-ket notation
It feels as if I am missing something
but I am not sure and maybe I am overthinking it
8:42 PM
you're making it more difficult for yourself by not doing bra-ket notation :P
lol
algebraically, as is clear in bra-ket notation, we're just summing the projectors onto the basis states, and they have to sum to the identity. Why this is "completeness" should be clear from elementary linear algebra
the position representation expressions come from this, as I said, by applying $\langle x\vert ...\rvert x'\rangle$ to it - this is the continuous version of taking matrix elements
the $\lvert v_i\rangle\langle v_i\rvert$ are operators, and you take matrix elements of an operator by sandwiching it in that way between two (a priori different) basis elements, i.e. the matrix elements of $A$ are $A_{ij} = \langle v_i\rvert A\rvert v_i\rangle$ in the discrete and $A(x,x') = \langle x\rvert A\lvert x'\rangle$ in the continuous case
now take $A=\lvert \phi\rangle\langle \phi\rvert$ and you get your position representation expressions
it's all just linear algebra generalized in a straightforward (but often confusingly presented) way to continuous "bases"
Weren't you taking about scattering states and bound states
how did it come to this
I keep telling imbAF that the bound vs. scattering business is a bit above the level they're currently at but it's taking a bit of time to sink in :P
It was that I was solving an exercise, of a delta like potential well for positive and negative energies
And for E<0 one gets only 1 bound state (somehow only 1 when the function depends on x) and infinitely many scattered states
And the exercise was asking about whether the states are orthonormal and after that whether they form a basis
8:50 PM
@imbAF yeah it's correct, it's a well known exercise
well
And I found weird that we are considering states of two different setups. Because the system, in one particular instance can either have E>0 or E<0, and for each case it exhibits a particular set of eigenvectors of H, which are a basis. If the particle has E>0, the basis is comprised of scattering states. There is no bound state present
so why bring it in the picture
when the system is unbound
And I found this merging of two different instances into one
weird
it's not two different setups
the 'E' is not part of the setup
case considered*
a system is given by a Hilbert space and a Hamiltonian on it, and your Hamiltonian here has a spectrum both above and below zero
and only all of the "eigenstates" for the full spectrum together form something you might call a 'basis' of the space
think of $H$ as $H_{discrete} \oplus H_{continuous}$
8:54 PM
$\oplus$, not $\otimes$
yep pardon
So a tensor product
Usually I associate tensor product when a system is comprised of subsystems
no, the tensor was a typo
it's a direct sum
the Hilbert space splits into the sum of two parts, one spanned by the discrete eigenstates, the other spanned by the continuous "eigenstates"
And why not Hdis +Hconti
I see
@imbAF I see what's the problem :P . See this reed.edu/physics/courses/Physics342/html/page2/files/…
last problem
that's probably what you need. What ACM is trying to tell you is very advanced stuff
8:57 PM
I did solve the problem. But looking at a clean derivation is also good
@Claudio Thanks
also there are many answers about bound and scattering states, just use the search bar and you'll find your way through
One last thing
lol
I don't know how to frame it
When considering the scattering states of the problem with the delta potential, they are continues because the variable considered, position, take continues values, is that right?
it's always 'one more question' with you - you''re physics Columbo
9:03 PM
lmao
ok I will leave it
wait
You can have a non degenerate discrete spectrum, non degenerate continuous spectrum, degenerate continuous spectrum
I know
My question was going to be
why does Griffith in his book, for the bound state of the above mentioned problem
says that we only have one bound state
Ofc there are no indicies
like for a particle in a box, which indicates only one
but the function depends on x
on position. Just like the scattered states, which also depend on position, and are infinite in amount
any time you consider a particle in QM you can express it as having a wavefunction that depends on position, that has nothing to do with discrete vs. continuous spectrum
consider the QHO, where the Hamiltonian has purely discrete spectrum but your states are still wavefunctions of continuous position
so is the energy
and the bound state in my problem, depends on the potential, which has a fixed value
while in the scattered state on momentum
is that right?
yeah that's just the x-space representation of your state
9:09 PM
I'm afraid I have no idea what you're trying to say
now you lost me
Why are the scattered states continues
?
Isn't it because the energy takes continues values?
continuous spectrum means exaclty that
the system, whose state is represented by the scattered states, takes continues energy values, something that the scattered states should reflect
hence continues spectrum,right?
They continuous because for every of these scattered states with energy/Hamiltonian eigenvalue $E$ you also have one with $E\pm\epsilon$ for arbitrarily small $\epsilon$
that's what we mean by continuous at this level
9:11 PM
That's what I also mean
why is this wrong
why is what wrong?
so is the energy
and the bound state in my problem, depends on the potential, which has a fixed value
while in the scattered state on momentum
is that right?
I'm sorry, but that still doesn't make any sense to me
how so?
like, it's not right, it's not wrong, I can't understand what you're trying to say
9:13 PM
@imbAF You seem very confused. Posing a well-thought question requires a lot of time and a good comprehension of the subject at hand. You've been jumping from question to another
From, what I know, from my lectures, for the particle in a box, the spectrum was classified as a discrete one, because the system in consideration could only take certain energy values
Again, that's the definition of discrete spectrum
For the free particle, the system in consideration, could take energy values from an interval, continues, or w/e the right way to frame it is
what does "the bound state in my problem, depends on the potential, which has a fixed value" mean? Like, what would be a potential that doesn't have a "fixed value"? What would be a bound state that doesn't depend on the potential? Literally all Hamiltonian eigenstates "depend on the potential" since different potentials produce different Hamiltonians
@Claudio I actually have been at a single theme, since I asked my last question
9:15 PM
and I can't understand what "in the scattered state on momentum" means, either - what's the verb here?
additionally it's throwing me off that the line breaks format the message like some kind of haiku gone wrong :P
the problem is the questions don't make sense, or at least I don't understand them
My piece of advice: re-read Griffiths, re-read the discussion we had above up until the direct sum part, and then formulate a 1 line question
For a particle in a box, we have more than one bound state. The spectrum is discrete because it reflects the fact that the system in consideration can only acquire certain energy values.
For a free particle, the spectrum is continues and it reflects the fact that the system can exhbit/manifest energy values from an arbitrary interval. I will not say from -infinity to +infinity because it's impossible for a system to have infnity energy. That's why I said interval.
Now in Griffiths solution to the delta potential he says the following:
9:34 PM
Yeah, in one of the books I used, it is in fact stated that the number of discrete eigenvalues depends upon the potential in question. I remember reading that under some specific circumstances (like V(x) having a lower bound and decaying at $+\infty$ in a certain way ), it can be shown that that number is finite or it can also be zero
but I haven't looked into these things, but I'm sure there are references and useful material online
Anyways, time to go to bed. Bye guys :)
@imbAF and sorry if I sounded rude, it was not my intention
Bye
no worries fra, gn
10:20 PM
@Pizza slope of the trajectory = derivative = $\frac{dy}{dx}$ = $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. but we also have slope = $\tan \theta$ where $\theta$ is the angle that the tangent to the curve makes with the $x$ axis
10:52 PM
Pythagoreanism originated in the 6th century BC, based on and around the teachings and beliefs held by Pythagoras and his followers, the Pythagoreans. Pythagoras established the first Pythagorean community in the ancient Greek colony of Kroton, in modern Calabria (Italy). Early Pythagorean communities spread throughout Magna Graecia. Pythagoras' death and disputes about his teachings led to the development of two philosophical traditions within Pythagoreanism. The akousmatikoi were superseded in the 4th century BC as a significant mendicant school of philosophy by the Cynics. The mathēmatik...
they believed that numbers could explain the nature of universe
> In Pythagoreanism numbers became related to intangible concepts. The one was related to the intellect and being, the two to thought, the number four was related to justice because 2 * 2 = 4 and equally even. A dominant symbolism was awarded to the number three, Pythagoreans believed that the whole world and all things in it are summed up in this number, because end, middle and beginning give the number of the whole.
chilling stuff