this might be naive, but it seems like it must work. if it didn't work, then we'd just come up with another lagrangian which did produce the EoM of interest and explain away the arbitrariness as we do
@SillyGoose Is there a way to show the shape of the BZ for the fcc lattice, without drawing it, but considering the Brillouin condition $\vec k \vec G=\frac{1}{2}|\vec G|^2$
a new confusion has arisen for me. so in condensed matter, we have a unique form of the hamiltonian. that is, the hamiltonian is not translation invariant generally speaking, but it is invariant under the period of the lattice. what can i say about the conservation laws associated with this case?
sorry the wording is not quite precise, but hopefully the question is clear.
seemingly, we can define some linear operator $P$ which acts on $L^2(\mathbb{R})$ like $P(a\psi(x) + b\phi(x)) = a\psi(-x) + b\phi(-x) = aP\psi + b P \phi$.
sure, in the Hamiltonian formulation everything that commutes with the Hamiltonian is both "a symmetry" and a "conserved quantity", in the sense that it's a constant of motion (classically)
But the usual Noetherian correspondence between "a symmetry" and "a conserved quantity" is that you have a continuous symmetry $U(t) = \mathrm{e}^{\mathrm{i}Tt}$ and the $T$ is the conserved quantity. Of course, you also always have that $[H,T] = 0$ and so you might call the discrete operation induced by $T$ itself "a symmetry" but that's not how we use the words - $T$ itself induces the infinitesimal symmetry
take for instance the case of the lattice Hamiltonian @Relativisticcucumber started this discussion with - sure, you have that the translation operators $T(a_i)$ commute with the Hamiltonian for lattice vectors $a_i$ and so all the $T(a_i)$ are also "conserved quantities". But we don't talk usually about it in those terms because the $T(a_i)$ are only unitary operators and not self-adjoint operators, i.e. they are not observables in general
@RyderRude Of course there is, parity is in all the kinematic groups $\mathrm{O}(3), \mathrm{O}(1,3),\dots$ and hence can act on the classical configuration and phase space.
@ACuriousMind yes. we can talk about the action of parity transform in classical mech. but there's no observable P on phase space whose Poisson bracket with Hamiltonian is zero
but in quantum mech, we have [H,P]=0, which leads to the eigenvalues of P being conserved
@RyderRude That's just the result of it not being a continuous symmetry, hence there being no corresponding generator of parity in a Lie algebra that would be mapped to a function on phase space. It's not strange, it's exactly how this works.
Note that the parity operator is a priori only unitary, not Hermitian, it's not an observable either.
of course it just so happens that in most representations you end up with a boring parity operator that is both unitary and self-adjoint, but it does not really belong to the quantum algebra of observables in the abstract
but any treatment of QM must in the end admit that not in every case every self-adjoint operator is an observable because otherwise it cannot explain what superselection is, see this answer of mine
(and also Valter Moretti's answer to the same question)
Molality = No. moles of solute (Here $Cu^2+$) / Volume of Solvent
For volume of the solvent do we take everything except of $Cu^2+$ or just water in this case (200g of water + Water of crystallisation that has been dissociated) ??
For this problem you need to find the molar mass of CuSO4 · 5H2O somewhere on the Internet (it is 249.6854 g/mol)
@AkhileshG I think molality is number of moles divided by mass, not by volume
Because if you got 200g of water and then dissolve 50g of CuSO4 in it, then the volume will change, and it is not that obvious what the resulting volume will be
I believe it may be confusing because those words differ just by one letter (molaRity vs molaLity), and molarity is indeed number of moles divided by volume, but the question you are trying to solve most probably asks for number of moles divided by mass
Can one show the shape of the BZ of a certain type of lattice from the consideration of the Brillouin condition $\vec k \vec G=\frac{1}{2}|\vec G|^2$ ?
Yes we need to take the whole mass, you just need to add 200g of water to 50g of CuSO4, and the whole mass is 250g
Now we need to calculate how many moles of $Cu^2+$ are there. Molar mass of copper is 63.546 g/mol, while molar mass of CuSO4 · 5H2O is 249.6854 g/mol (you can find those numbers on the Internet or calculate them on your own using the periodic table)
So just divide 63.546 by 249.6854 is around 0.2545, so in other words 0.2545 of the CuSO4's mass is copper. You have 50g of CuSO4, so you have 0.2545*50 = 12.726 grams of copper
The substance that is present in the largest amount is called the solvent and the one present in the smaller amount is called a solute. There can only be one solvent in a solution, but there can be many solutes. http://www.ilpi.com/msds/ref/solution.html#:~:text=Definition,there%20can%20be%20many%20solutes. [This site is not secure, so be careful]
So since a Solution can have not more than one solvent, which the question has already decided it to be water, rest act like solutes. That means we cannot take SO4 in mass of solvent.
Do you have the answer for this problem? I took the whole mass (including SO4) and what is got was 0.2 mol/kg, if the answer is different I'm sorry for confusing you because it will mean we indeed cannot take SO4's mass into account
i think it's fascinating that the poisson bracket does not have the structure to be able to determine discrete symmetries, but the commutator does have that structure
Which means we need to find out how much water is there is 50g of CuSO4 · 5H2O: water's molar mass is 18.01528 g/mol and there are 5 molecules of water, so water constitutes 5*18.01528/249.6854 = 0.3608 of CuSO4 · 5H2O's mass. Therefore, 50g of CuSO4 · 5H2O contains 0.3608*50 = 18.04g of water, which makes the solvent mass 200g + 18.04g = 218.04 g
12.726 grams of copper is 12.726/63.546 = 0.2 moles of copper, which means molality is 0.2/0.21804 = 0.9185 mol/kg, I hope this is correct :P
True, physics is strict and formalized, that doesn't go too well with lots of philosophy's subjects, I believe sometimes you need to mix both physics and philosophy to really get the full picture about what is going on
i still think that physics tells us a lot more about the ontology than any other science. e.g. physics says, with precise evidence, that the flow of time is relative. philosophy could never have expected to produce a result like that
so physics is the best estimator of ontology that we have, but it's just not sufficient
@ACuriousMind Incidentally, concerning that question of mine about Green's functions, vacua (the one which you answered talking about Haag, among other things), I've found a book doing things like Maggiore, so I may shed some light on the obscure notation :P
@NairitSahoo No. Hint for proof: This is a linear algebra question about tensor products.
@MoreAnonymous I will never tire of recommending Pratchett's Discworld novels to people, they're not only my favourite fantasy novels, they're my favourite books, period.
In that dream there was someone purchasing books using my Amazon account and I couldn't find out their identity. Then I noticed the books were in German and about philosophy, so I opened the chat and tagged @ACuriousMind to know if he was using my account to buy books, but he denied
@SillyGoose this is kinda an interesting take to me, silly goose. I think arbitrary is fine for all the reasons ACM and slereah gave but I never considered the choices as inconsistent. to be clear, are you saying that the resulting theories as also subtly inconsistent or is that to you different to ambiguous/logically dubious. also did you have any other examples, or just those two?
@SillyGoose I object to both your examples. Some presentations may be ambiguous, but those can be fixed by people who read it and obviously see how to present the same information in a properly unambiguous way. It is also not dubious when any theory asserts that some assumption leads to infinite energy; it is just immediately implying that such a thing is unphysical.
@imbAF You are never asked to do that. The real lattice is measured in units of length and the reciprocal lattice is measured in units of radians per length. Having different units makes it clear that they are entirely different things requiring different pictures to show. In which case, you can pick $2\pi/a$ as the unit length for the reciprocal lattice.
@Relativisticcucumber crystal momentum is conserved. It is a concept all unto itself and quite intricate.
@qwerty cc @SillyGoose the complaint about the "measurement postulate" is legitimate insofar as this is essentially what the measurement problem is all about; the complaint about the infinite-energy plane waves is not since the solutions to Maxwell's equations become unique only upon imposing physically meaningful boundary conditions and we often choose "should go to zero at infinity" as our boundary conditions precisely to exclude these unphysical solutions
@ACuriousMind I'll even object to your characterisation that the wacky fowl's complaints against measurement postulate has anything to do with the measurement problem at all!
I wouldn't call it inconsistent, rather underspecified
@naturallyInconsistent I read the complaint as being that "measurement" is left undefined in the standard presentations of the Born rule beyond "I know a measurement when I see it"; this is (one aspect of) the measurement problem
Like, every text I know that defines the measurement postulate the usual standard way, defines states as living in Hilbert space beforehand, and implicitly imports linear algebra and Sturm-Liouville so that eigenfunctions and eigenvalues are well-defined. The measurement postulate thus simply asserts that the act of measurement corresponds to a specific mathematical operation of projection. How is any of these underspecified?
@naturallyInconsistent The problem is that there is no clear definition of which physical things constitute a "measurement". Yes, once you know you're doing a measurement, the mathematical apparatus tells you what the result is, but there's no clear-cut rule for looking at some setup and telling whether it's "doing a measurement" or not
@ACuriousMind even that is objectionable: for most of the history of quantum theory, people sufficed with the measurement problem being left alone. Clearly the inadequacies of that standard Copenhagen presentation of Born rule lies in the æsthetics, not in the logical underpinning of the theory. I mean, I also hate how that looks, but the theory is manifestly functional.
@SillyGoose I'm still not sure what you're claiming is inconsistent, precisely, and if this is all just a qm measurement thing, or if you had other examples
@ACuriousMind There is no clear-cut rule because there are two completely different uses that has to have different results. Sure, when we do a diffraction grating experiment, the different momentum values move to different positions on the screen and so are manifestly separated out and follow the measurement postulate. However, the p^2/2m in the Hamiltonian is not separating out the wavefunction's components for separate study, and so it aint a measurement as is.
@SillyGoose nobody is claiming that you are "claiming [that] the theory was not operational"; we are much more interested in what you would think is the ambiguity at all.
@naturallyInconsistent yeah actually me asking the q stemmed from me trying to figure out how i should think about the crystal momentum
condensed matter has so many topics that are so hard for me to wrap my mind around -- just when i think i am starting to get band structure we move onto fermi surfaces
