@user929304 they're there because the universe that shog 9 created doesn't have the processing power for exact numerical values for complementary variables, it raises the computation time by way too much
Phyiscal states are vectors in Hilbert space, and if it's not an eigenvector of an observable, there will be an uncertainty for that observable in that state. There is no because, this is a postulate about how the quantum world works
@ACuriousMind that part I get, indeed because of randomness involved in measurements of qm, I mean why their non commutability still relevant when I use diff copies
@user929304 It's not the non-commutability. You can compute the uncertainty of an observable without ever referring to a different observable, you just need to know the state
The non-commutabily is a red herring: Once you accept the notion of states and uncertainties for single observables , the uncertainty principle is an almost trivial consequence of the Cauchy-Schwarz inequality.
You can create an state which is simultaneously the eigenstate of several different observables, but once a pair of observables don't commute, you are forced to forever give up trying to construct a simultaneous eigenstate of both.
@user929304 So? $\Delta A(\psi)$ is a property of the state $\psi$, and nothing else. It also fulfills the HUP for any other observable $B$, but that's just by virtue of how operators on Hilbert spaces work, it is not any sort of extra constraint on $\Delta A$.
@ACuriousMind what originally confused me about that was the fact that even though I wasn't trying to measure a and b simultaneously, I was still getting a non zero uncertainty
@user929304 Yeah, the crucial thing to realize for that is that $\Delta A := \langle A^2\rangle - \langle A\rangle^2$ does not involve any other state or observable at all
you can think of them as simply vectors in the Hilbert space with the added condition that two vectors which differ only by an overall phase actually specify the same state
You shouldn't try to do quantum field theory without being very comfortable with the formalism of ordinary QM, both in the Hilbert space/operator and the path integral formalism.
The S-matrix is more like a bunch of <stuff|S|other stuff> where S is some scattering operator...but even the way you define the incoming and outgoing states is somewhat complicated lol
@Cows This normalization is not a physically meaningful operation, you just get rid of an annoying factor you'd need to otherwise keep track of when applying the Born rule.
I'm just making the statement that the ground state is the one for which the number <psi|H|psi> is the smallest number for any state in the Hilbert space.
Usually you are solving some sort of time-independent Schroedinger equation, and then you get the eigenstates of energy, and the lowest eigenstate is the ground state
only in QFT when you change from a particle-based view of the world to a field-based view of the world do you get stuff like "the ground state is the vacuum state"
My first encounter with actual QM was taught by an excellent prof who always made very clear when we were making assumptions that weren't true, or simplifying things that couldn't really be simplified.
the borne rule simply gives a method of interpreting the outputs you get when you perform mathematical operations on state vectors. It says, that the wave function psi(x) can be interpreted as a probability density function where int_a^b psi(x)*psi(x)dx gives you the probability that the particle in state psi will be found within the interval [a,b]
Filling some holes implies that you have a firm understanding of the overall structure with some pieces missing. But it appears that you have the inverse problems: Knowledge of many pieces without a good idea of what the overall structure is.
@ACuriousMind you are quite right. I was not as fond of physics back when I took QM the first time, and I used griffiths QM, took to semesters, but now I feel like I really want to get the full picture. But yeah i will definitely take QM again at some point. I though it was cool, but I did not really looooove it the way i do now
@ACuriousMind btw is there any significance to the hermite polynomial form for solution of harmonic oscillator? just curious since the other way with raising and lowering operators has an interpretation
@BernardoMeurer I really hope they return to the setting weirdness that defined Morrowind. The gameplay got better but the setting got a lot more boring
@ACuriousMind well can say H_0 or H_2 or something be mapped to some physical thing? the H here for hermite polynomials. I want to describe better but have some math language barrier here hehe
@Cows The Hermite polynomials times $\exp(-x^2/2)$ are the solutions to the time-independent Schrödinger equation, i.e. the wavefunctions of the energy eigenstates of the harmonic oscillator
There is no one single model that will teach you everything. The beauty and power of quantum mechanics is that it encompasses a wide range of phenomena.
My cousins are named Christian, Andreas, Johannes, Judith and Ronja, and you can't do anything with that information because none of them shares my last name
