When substituting the solution for ##x## in the other equation, I get:
$$y^2 \bigg(a'\cdot \frac{(-2h\pm s)^2}{4a^2}+2h'\cdot \frac{-2h\pm s}{2a}+b'\bigg)+c'=0$$
Where ##s=\sqrt{4h^2-4ab+\frac{4ac}{y^2}}##. Not sure how such an equation can be solved.
Also, shouldn't a quartic equation have a...
Why does this solution for ##x## not have the form of a fraction? It's missing a factor of ##\frac{1}{2a}## as mentioned in my opening post, isn't that how the solution of a quadratic function is defined according to the abc-formula?
Thanks, but wouldn't this lead to another method to solve these equations? I was wondering whether the quoted method can actually be used to solve this.
You mean as in dividing ##S_1## and ##S_2## by either ##x^2## or ##y^2##? I have tried that but I'm not sure how to continue to get ##x## or ##y## on one side.
Given are two equations:
$$S_1 = ax^2+2hxy+by^2 + c=0$$
$$S_2 = a'x^2+2h'xy+b'y^2 + c'=0$$
This source states that there are several methods to solve for ##x## and ##y##. One of them is the following quote:"Treat equation ##S_1## as a quadratic equation in ##x## and solve it for ##x## in terms...
"proton or neutron" as in exchange of a proton pair "or" neutron pair? If yes, could you please elaborate how this quote is compatible with this:
I am a bit confused.
I understood that specific phrase "won't have to be" as in "not obliged to be" which made me think that it is also possible for such a model to be antisymmetric under exchange of a proton and neutron as well, just as in the exchange of a proton pair or neutron pair.
But I assume you meant that...
Does "won't have to" imply that it is still possible to stay antisymmetric? Because in that case, the classical symmetry number would still not have a clear link (to me) since it relies on the physical indenticality of particles.
But whether the total wavefunction is symmetric/anti-symmetric depends on the spins of the nuclei, not on whether the nuclei are identical or not. It goes even further by the fact that the total wavefunction of e.g. dihydrogen can be symmetrical even if the two atoms have opposed spins.
So it...
Some rotational quantum states are not allowed for a rotating particle. At quantum level, these "forbidden" quantum states is based on the requirement of the total wavefunction being either symmetrical or anti-symmetrical, depending on whether the particle is a fermion or boson. The particle's...
I'm trying to understand the detailed concept of why the density of states formula is accurate enough to calculate the number of quantum states of an energy level, including degeneracy, within a small energy interval of ##dE##.
The discrete energie levels are calculated by
$$E = \frac{h^2 \cdot...
Never mind, I figured it out, thanks. My calcuations in post #30 are wrong, I didn't pay attention when doing them. I was overlooking the fact that the relation between ##dE## and ##dv## is a function of ##v## while the probability itself is also a function of ##v##. If the relation between...
@Ibix & @PeroK I think using your info's helped me a bit by explaining it as follows:
For a certain ##E_0## and a corresponding ##v_0##, the following must be valid:$$f(E_0)\cdot dE = f(v_0) \cdot dv$$
For ##f(E)##, the maximum number of particles lies within the range ##\frac{k_BT}{2} \geq...
You asked about what I based it on and I took the time to explain to you how I understood it so you could perhaps pinpoint how I should understand it with ##dE = mvdv##.
Merely substituting my whole explanation with "...the wrong thing" and that he meant something that I already mentioned...
The fact that he, as well as another user, liked my post #16 containing that very description of ##dE## as a statement I asked about would make me assume that my understanding about ##dE## was correct. Hence me asking about it after.
Perhaps I based my description on his mathematical fact...