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Jun
26
comment Energy of a system if the nuclear repulsion increases
And yeah, the question is perfectly fine here at Physics.SE, but it could as well be migrated to Chemistry.SE. Anyway, just for OP information: now when we (finally) have a rather vivid Chemistry.SE, I think it is more appropriate to discuss quantum chemistry there.
Jun
26
comment Energy of a system if the nuclear repulsion increases
@JonCuster, I'm not sure that it is necessarily the electron-nuclei attraction energy which will compensate this. In principle the quantity being minimized (the total energy with nuclei being fixed) has three more terms that could (over)compensate an increase in the nuclei repulsion energy. Besides, usually in quantum chemistry these three terms goes jointly as the electronic energy, so I wouldn't concentrate just on one of its parts.
Jun
26
answered Energy of a system if the nuclear repulsion increases
Jun
14
comment Really how can an observable quantity be equal to an operator?
@user36790, in fact, the author does not say "equal", he says "equivalent". Another (more usual way) to express the same idea is to say that observables in QM are represented by self-adjoint operators.
May
30
comment Why do momentum and position have the same direction in space in the Heisenberg Uncertainty Principle?
@SriramV, they were postulated to construct the theory which is consistent with observations. They mathematically express the way nature works at microscopic level.
May
30
comment Why do momentum and position have the same direction in space in the Heisenberg Uncertainty Principle?
@SriramV, yes we can measure both $y$ and $p_x$ simultaneously with arbitrary precision. But not $y$ and $p_y$ or $x$ and $p_x$.
May
30
comment Why do momentum and position have the same direction in space in the Heisenberg Uncertainty Principle?
@SriramV, yep. They are also usually (and a bit wrongly) called hermitian operators.
May
30
comment Why do momentum and position have the same direction in space in the Heisenberg Uncertainty Principle?
@NicolauSakerNeto, may be we can leave it as an excercise...
May
30
answered Why do momentum and position have the same direction in space in the Heisenberg Uncertainty Principle?
May
10
comment What is the quantum mechanical explanation of the octet rule?
Related question on Chemistry.SE.
May
10
comment What is the quantum mechanical explanation of the octet rule?
The binding energy of the electron, of course, decreases with $n$. ;)
May
2
awarded  Notable Question
Mar
19
comment Hartree-Fock: Coulomb integral
Let us continue this discussion in chat.
Mar
19
comment Hartree-Fock: Coulomb integral
@Alejandro, also note that I forgot to complex conjugate $\chi_{1}(1)$ to the left of $J_{2}(1)$ in few of my comments above and I couldn't edit them already.
Mar
19
revised Hartree-Fock: Coulomb integral
Correcting for complex-conjugates in the left parts of integrals.
Mar
19
comment Hartree-Fock: Coulomb integral
@Alejandro, oh, I think I got it. You have a problem with understanding this rather awkward physicists' notation for integrals which Szabo & Ostlund (and I following them) use in the book in which differentials come first. :D In this notation everything after differentials is integrand expression. So, yeah, you integrate the whole $\chi_{1}^{*}(1) J_{2}(1) \chi_{1}(1)$.
Mar
19
comment Hartree-Fock: Coulomb integral
@Alejandro, as I said, you can first calculate $J_{2}(1)$ using its definition and integrating over coordinates of e-2. You than can substitute the resulting potential $J_{2}(1)$ into $\int \mathrm{d} \vec{x}_{1} \chi_{1}^{*}(1) J_{2}(1) \chi_{1}(1)$ and integrate over coordinates of e-1. So, you multiply $J_{2}(1)$ by $\chi_{1}(1)$ from the right and by its complex-conjugate $\chi_{1}^{*}(1)$ from the lefts and then integrate.
Mar
19
comment Hartree-Fock: Coulomb integral
@Alejandro, now it looks like you are missing the second integration which enters an expression for a Coulomb integral. So $J_{2}(1) \chi_{1}(1)$ is not a Coulomb integral, it just an expression which defines the Coulomb operator $J_{2}(1)$ by its action on spin-orbital $\chi_{1}(1)$. The corresponding Coulomb integral is written as follows $\int \mathrm{d} \vec{x}_{1} \chi_{1}(1) J_{2}(1) \chi_{1}(1)$.
Mar
19
comment Hartree-Fock: Coulomb integral
@Alejandro, and yes, for an $N$-electron system you indeed do calculate all such Coulomb contributions for e-1 with all other $N-1$ electrons in the system. And you do the same of e-2, and e-3, etc. You then sum up all the Coulomb integrals (minus all the exchange ones) together with the sum of all core Hamiltonian energies to calculate the HF energy.
Mar
19
comment Hartree-Fock: Coulomb integral
@Alejandro I just wrote expression for Coulomb integrals explicitly. So, for instance, for the first Coulomb integral, you can think of it as follows. You first calculate the average Coulomb potential $J_{2}(1)$ which e-1 "feels" due to e-2 on $\chi_{2}$ by integrating $\chi_{2}^{*}(2) r_{12}^{-1} \chi_{2}(2)$ over coordinates of e-2. You then calculate the resulting Coulomb repulsion energy between e-1 on $\chi_{1}$ and e-2 on $\chi_{2}$ by integrating $\chi_{1}(1) J_{2}(1) \chi_{1}(1)$ over coordinates of e-1.