« first day (5197 days earlier)      last day (17 days later) » 

00:01
yes...
00:12
i am reading L&L vol. 5 on quantum (bose) liquids.
they have a dispersion relation expanded around a local minimum $\epsilon(\vec{p}) \approx \epsilon(\vec{p_0}) + \frac{1}{2} \epsilon''(\vec{p_0}) (\vec{p}-\vec{p_0})^2$
then we want to compute the integral $\int d^3p e^{-\epsilon(\vec{p})/T}$ in the limit $T \to 0$.
naively, i can apply Laplace's method to approximate this integral. I get the result $e^{-\epsilon(\vec{p}_0)/T}(2\pi T/\lvert \epsilon''(\vec{p}_0)\lvert)^{3/2}$
But L&L obtain the result $e^{-\epsilon(\vec{p}_0)/T}\vec{p}_0^2 (2\pi)^{3/2} 2(T/\lvert \epsilon''(\vec{p}_0)\lvert)^{1/2}$
how can this possibly be?
00:35
i can also just exactly compute the integral in mathematica....
01:23
hiya tobias
I have a pregunta!
I'm finally reading about the time dependence of the Schrodinger eqn
So if we're looking at some eigenket $\hat{H} |\psi \rangle = E|\psi \rangle$ of the Hamiltonian
the TDSE tells us $i\hbar |\dot{\psi} \rangle = \hat{H} |\psi \rangle = E|\psi \rangle$
Now, this tells us that the rate of change of the eigenket over time is proportional to the eigenket itself but multiplied by a purely imaginary number. Is that basically telling us that only the phase is changing? Since your rate of change is basically always perpendicular to the wave function's current phase at any given point. Idk if im explaining this clearly but
Something like this in the complex plane, where at any value of the wave function (shown in the complex plane as the endpoint of r) , the rate of change is i * that value, which is just a rotation of 90 degrees
thats the best explanation i can give rn im lazy. but hopefully if im correct that wil make sense
almost like youre rotating the eigenkets, but you have to do them separately because they rotate at different rates depending on the energy/eigenvalue
 
1 hour later…
02:46
Consider a diffusion problem where I want to say that there's an absorber at some point. Why is it that the correct boundary condition is to set the probability density to zero at that point?
I guess the idea is that, if there's an absorber, then that means we're sucking up any particle that hits that point, and if the probability density is continuous then something something it has to go continuously to zero...?
03:21
@ACuriousMind would have been a fun time to be hypnotised by the stroboscope though miehehehe
@SignorFeynman circuit QED and cavity QED are broken extracts from full QED just for the sake of those special circumstances. They are even less of a complete theory than QED proper.
 
5 hours later…
08:07
morning
@DanielSank hmmh sounds reasonable... could you provide more context? (just out of interest)
Hi @Allie --no, not only the phase changes. but what is true is that the norm of the vector is constant in time, i.e. the "length" is constant. This is shown in several posts on the main site, or should be content of a QM book. It basically follows from the fact that the solution to the problem is $\psi(t)=U(t)\psi_0$, where $U$ is the time evolution operator and $\psi_0$ the initial condition; $U$ is unitary and the claim follows
Well, but okay: If $\psi_0$ is an eigenstate of $H$, i.e. $H\psi_0=E\psi_0$, then yes, the time evolution is just a change of the phase and for all times the vector $\psi(t)$ is indeed an eigenvector--all of this assuming that the Hamiltonian is time-independent! You can show this either via direct computation, or arguing that $U$ commutes with $H$.
Yes thats what I meant :3
That in general its not just a phase change, but for energy eigenstates it is
yes, sorry, I missed that you are considering only eigenvectors of the (time-independent) Hamiltonian
yes
are you starting with TDDFT or just general QM?
And for a linear combo of energy eigenststes the phase factor doesnt factor out bc they phasr at different rates according to the eigenvalues
General QM
Started reading Shankar a bit
Im starting to understanf Lagrangian and Hamiltonian a bit better
@Allie indeed
nice and nice :)
Yay how awesome
I have a lot of physics to catch up on
Gn
08:33
I think I figured it all out.
08:49
thanks
it was on my to-read list, actually :p
09:21
@naturallyInconsistent Indeed. It's just that the name sounds like an oxymoron. A circuit is the most practical thing I can think of, quantum field theory is not exactly "practical" :P
Do you know any ways to find a book/paper just having an extract of 5 lines?
I wish books were Lie groups
09:59
@SignorFeynman lol
mhmhm no, I don't know any way, except of course google scholar/books (or google itself)
you could try an AI :p?
I tried D:
10:20
Apparently the proof of incommensurability of lines of Hippasus was probably more of that kind than square root of 2 : atractor.pt/mat/incomensurabilidade/pentagono-_en.html
10:43
i have a really nice proof that the trace of an operator is basis independent. u won't find this on wiki or in books
Mh? Do you mean in regular linear algebra? It's a pretty straightforward result
if u think about it, the trace of an operator $T^a_b$ is equal to $T^a_b \delta ^b _a$, where $\delta ^b _a$ is the Kronecker delta
now the Kronecker delta remains Kronecker delta in any basis. So this is basis independent
one can easily generalise this to show that the contraction of a general tensor is a tensor
@SignorFeynman yes. but I have an easy way to look at it
any contraction of any tensor can be written in terms of Kronecker deltas
i didnt see this, e.g. in Wald's book
also on wiki
it is an easy way to look at it
@SignorFeynman what do u think
i shared this on the math chat but they didn't reply :(
but they helped me arrive at this proof
11:04
isn't this, if I understand correctly, just the usual proof in disguise?
I mean, it's not really different from the standard proof even in index notation @RyderRude
$T^{a}{}_{b}\delta^{b}_a$ is $T^a{}_{a}$
And if you write the transformation of law of $T$ you have something like $T'=STS^{-1}$. Write this in index form and then contract the two indices. The trasnformation matrices will yield a delta
@SignorFeynman yes. but I hadn't seen it written the former way. i thought $T^a_a$ was an elementary expression
@TobiasFünke the proof I saw used the cyclicity of trace. i think this proof is just a one line proof
My proof is one line too
@SignorFeynman maybe..
@SignorFeynman yes
i think starting with the index notation is a lot clear
@RyderRude $T'^{a}{}_{b}=S^{a}{}_{c}(S^{-1})^{d}{}_{b}T^{c}{}_{d}\implies T'^{a}{}_{a}=\underbrace{S^{a}{}_{c}(S^{-1})^{d}{}_{a}}_{\delta^{d}_{c}}T^{c}{}_{d}=\delta^{d}_{c}T^{c}{}_{d}=T^{c}_c$
Not maybe, it is
11:17
@SignorFeynman but u can write this in multiple lines :P
but it is the same proof, yes
i think I hadn't come across this proof. i saw the one with cyclicity of trace
That's definitely a faster one, if you have already proved cyclicity
If you haven't it's about the same length
yes.. but i think this proof is easier to understand
also, this one generalises to tensor contraction
the cyclicity one only works for matrices
It works for linear operators
The way you can make it work fot matrices is if you have a metric structure to raise indices and associate a $(1,1)$ tensor to every rank $2$ tensor
In that case, the trace of the tensor is just the trace of the associated operator which is well defined (note that the mixed components of the metric are delta)
@SignorFeynman but i think matrices r already (1,1) tensors?
matrices are usually used for functions $f: V\to V$ @SignorFeynman
it is after u pick a basis that this becomes a matrix
but it depends on conventions. one can define a matrix as a (0,2) tensor or (2,0) maybe
11:34
@RyderRude matrices are matrices
yeah ...
youtu.be/71q2q9sGkX8?si=HWQk8QmqombvFwrg this physicist thinks QM is probably fundamental and not something that requires future correction
he keeps saying "we have found no problems in QM", while acknowledging the measurement problem several times. it is like QM has held physicists hostage :P
12:03
@RyderRude all these notions are isomorphic to each other
12:27
Infinite series are underrated
12:39
@TobiasFünke yes
@DIRAC1930 they r interesting. math allows people to compute them without taking infinite time. i wonder how this is related to computability theory
perhaps computability theory would define them as a result of a finite computation
infinite series are defined to be limits as opposed to "result of an infinite computation"
@Allie Yes, that's exactly what's going on, and you can just see it in the full solution of the SE: The time evolution for an energy eigenstate $\psi_E$ of energy $E$ is just $\psi_E(t) = \mathrm{e}^{-\mathrm{i}Et}\psi_E$.
no need to guess, you can just solve the equation :P
@SignorFeynman you can almost always find it by putting the excerpt you have into google, surrounded by quotation marks
the quotation marks mean you only search for exact matches
13:11
In philosophy and mathematics, Newcomb's paradox, also known as Newcomb's problem, is a thought experiment involving a game between two players, one of whom is able to predict the future. Newcomb's paradox was created by William Newcomb of the University of California's Lawrence Livermore Laboratory. However, it was first analyzed in a philosophy paper by Robert Nozick in 1969 and appeared in the March 1973 issue of Scientific American, in Martin Gardner's "Mathematical Games". Today it is a much debated problem in the philosophical branch of decision theory. == The problem == There is a reliable...
this paradox divides people 39% vs 32 % and each side is absolutely sure about their answer
i think the correct answer is one box and i too can't fathom why anyone would think two box
> To almost everyone, it is perfectly clear and obvious what should be done. The difficulty is that these people seem to divide almost evenly on the problem, with large numbers thinking that the opposing half is just being silly
@ACuriousMind Oh, I knew quotation marks did something on google. Unfortunately it doesn't work because it's an old text (no digital version) and I don't know how to type equations in google search engine :P
just omit the equations, usually one or two sentences work fine, but sure, if there is no good OCR'd version of it you're out of luck
I think I'll accept for now that qubits are quite technical to get too invested for a single exam
13:38
oh, just found out that the "wrong Englert" that is now with NUS, wrote some QM texts that are very kind to students.
13:53
> Computer science is no more about computers than astronomy is about telescopes - Dijkstra
14:47
The quantum course i am ta-ing for covers quantum harmonic oscillator before covering formalism of qm…why doe
15:21
@SillyGoose Ehm, what? D:
16:02
what do you all think about the idea where a computer can encode a human's present state and predicts all their actuons
Landau's book does the Harmonic oscilator differently
I think he does matrix mechanics or something
16:22
Has anyone studied the spin-statistics theorem? I'm thinking about studying it
@DIRAC1930 The proper proof is rather technical, but if you want to look into it, Streater and Wightman's book is still good
Is the original proof by Pauli or did someone else do it aswell?
Fierz and Pauli wrote the first version of it, I haven't read it
16:50
@SillyGoose is the "quantum formalism" perhaps assumed to be known? I.e. is it an advanced course?
17:17
@TobiasFünke it is the "quantum II" undergrad course, but the prof also told me that people still had trouble with bra-ket notation by the end of the semester lol.
also the topics are what i would consider a "quantum I" level course, im not actually sure what is covered in the actual "quantum I" at this uni
yeah, this can differ from uni to uni or even prof to prof
^^
TA means you offer exercise classes? Or what?
for my uni it means that we run workshops (the students do problems and can ask questions and etc.)
and also host office hours and grade
maybe elsewhere they are called recitations?
I see. I guess in Germany we'd call it "Tutorium", although I think there is a different mostly between eastern and western German unis on what this means exactly
nice. I always enjoyed such things, because teaching is one of the best ways to learn
may I ask: do you get payed for that? or is it mandatory that you TA X courses to earn your PhD?
i thought i would enjoy this semester (since quantum is very fun to teach), but i think the material that i find most interesting is not always the priority of the course. i also often disagree with the professor's organization of the content/choice of topics.
@TobiasFünke for my program it is mandatory to TA for 1 year. we also get paid for TAing. but later on you get paid through (usually) grant money from your supervisor. some people do alternate between getting funded by PI and TAing though (if there isn't that much grant money to go around).
this course follows griffiths which i think has a supremely poor organization of quantum content
it also dumbs down the mathematical structures at play to the extent that primary points are obfuscated in my opinion
17:33
@SillyGoose yeah, ikr... ^^
I see, interesting
@SillyGoose I just skimmed through the book a few times, I really did not like it too much... but it is the standard book in the US, no?
@TobiasFünke it seems standard-ish. I know my uni and NYU use it. Caltech uses Shankar. My undergrad uses Townsend. So I'm not sure what is the most common
i feel like they should just use sakurai or cohen or what not
17:53
Hiya tobias
18:09
is the discussion of quantum liquids in L&L vol.5 outdated?
Isn't vol 9 devoted like nearly entirely to Fermi-Liquid theory?
18:30
Oh i didnt realize i should look at that
The Goldstone theorem says: " For each spontaneously broken continuous symmetry there is a massless particle in the theory".
Is the massless particle a new excitation of the field? I mean, if we consider some sort of field theory, whose excitations are particles of a certain type. And then we investigate symmetry breaking and we find out that this is the case. Does a new type of particle belong the the theory ?
It is very terse however
Subir Sachdev has some nice video lectures
@imbAF of course there is a corresponding scalar field to the Goldstone boson, which you would see if you read more than that one-line summary of the theorem :P
That was the def. of the theory
Additionally we worked to prove it and I read other definitions but they are nearly the same
But the way it is framed
it makes it as if, one needs to detect symmetry breaking to make a claim about the existence of a boson
massless one
You replied to me about the VEV of Lorentz invariant vacuum state. To tell you the truth, lorentz invariance was something we talked in two particular cases. Invariance of scalars
So to talk about lorentz invariance or trasnformations of a state
is not something I am familiar with
You can argue that I should have deduced that, but I didn't
18:58
meow
thanks ACM
19:25
is the klein gordan field the complex scalar field?
I have seen in many textbooks the term klein gordan field
But isn't the KG equation, the eom that results from using the euler lagrange equation for the given lagrangian density of the complex scalar field?
19:40
@imbAF the KG equation can also be applied to a real field. See this physics.stackexchange.com/q/442936
@imbAF I've rarely seen this label, scalar field is more typical. As RR told you, it may also be a real scalar field. That being said, the Dirac field components too satisfy the KG equation, so it's really a bad name :P
And it's "Gordon"
Not Gordan
19:55
there was a debate between Einstein and Bergson and no one even has a transcript
does anyone know where to buy hardcovers of L&L volumes from?
 
1 hour later…
21:06
hello
21:19
Is it assumed always in CMT that dispersion for small momenta is linear?
or does anyone have a counter example with a real material
21:49
@SillyGoose Are you still doing Fermi-Liquid theory?
@DIRAC1930 yesh
Btw L&L 3 has a discussion on these complex energy states that are used in Fermi-Liquid theory
Section 132
It is however impossible to comprehend
22:22
Does the area in the phase space (q,p) have a meaning?
Like say in thermodynamics, there is the (V,p) diagram, and the area in that diagram is the work done; or heat exchanged if you look at (S,T) diagrams. $dU=TdS-pdV$ so $TdS=pdV$, $\delta Q=\delta W$ so the area is preserved.
22:37
@User198 (Gibbs) entropy - which is a statistical notion of entropy - is well approximated by the logarithm of the volume of the phase space occupied by the system. Well, in the microcanonical ensemble it is exactly the logarithm, which is the so-called Boltzmann entropy
Roughly speaking, the volume in phase space is just the "number" of states. If you choose an appropriate unit, it's a the number of states in a more sensible way. The appropriate unit is $h$, the Planck constant and it can be justified with quantum mechanical arguments
@SillyGoose back in the day I bought some secondhand from AbeBooks.
@SignorFeynman Yes it confused me because that area (volume) has units of J s. So i tought I can somehow think of that area as kind of energy, didn't occur to me that it could be number of states. Thank you.
But it is still weird, that "number of states" (whatever that is) has units of J s.
I just stumbled on this question of yours: https://physics.stackexchange.com/questions/715474/on-shell-poisson-brackets-and-time-derivative

Your first equation (Liouville's equation) is: $\frac{\partial\rho}{\partial t}=-\{H,\rho\}\tag{1}$

But on the Wiki they define it as ${\displaystyle {\frac {\partial \rho }{\partial t}}=\{\,H,\rho \,\}}$

Why is the sign difference?
22:59
@SignorFeynman they usually do not.
What? Using the phrase "KG field"?
no. The Dirac field components dont usually satisfy KG equations
@User198 because as I said, it's just proportional to the number. You usually choose a unit with respect to which you measure e.g. $h$, so you make it dimensionless
@naturallyInconsistent the free Dirac field components?
@SignorFeynman that's the special case in which it does. Hence, usually
Well, you are right. Strictly speaking interacting ones don't satisfy the (free) Dirac equation either, but it's still a coupled Dirac equation
Oh probably you are thinking of the fact that the minimal coupling in the squared Dirac equation is not the same as the coupled KG equation
@SignorFeynman Ah, ok thanks.
Oh, well, you refreshed my memory, nI :P
It's been a while since I played with spinors
The important part is to realise that KG is only a crutch to help get the shape of the Dirac equation. The consequences of the Dirac equation being a different equation altogether shows up in the energy eigenvalues being different from KG equation when you introduce any interaction, even just with a potential.
Indeed, just the example I mentioned above is an evidence of the fact that the Dirac equation includes spin

« first day (5197 days earlier)      last day (17 days later) »