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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.
Mar
19
revised Hartree-Fock: Coulomb integral
added 905 characters in body
Mar
19
revised Hartree-Fock: Coulomb integral
Edited quote from Szabo & Ostlund.
Mar
19
suggested approved edit on Hartree-Fock: Coulomb integral
Mar
19
answered Hartree-Fock: Coulomb integral
Mar
19
comment Hartree-Fock: Coulomb integral
@Alejandro looks like you are badly and totally confused by the HF method. Basically, all you statements and formulas are "not quite right" to put it mildly. Szabo & Ostlund book is quite good at explaining the HF method, but you have to be careful and patient to follow through.
Jan
31
comment Are “uncertainties” in Heisenberg Uncertainity just standard deviations?
However, my point is that being armed with such an incredible device, we could estimate expectation values by sample averages as discussed in details in my answer, and consequently, could experimentally estimate uncertainties $\sigma_x$ and $\sigma_p$. Although the device is hypothetical, that is possible in principle.
Jan
31
comment Are “uncertainties” in Heisenberg Uncertainity just standard deviations?
So, in principle, we could imagine a hypothetical device which is infinitely accurate in measuring both position and momentum independently, but even with such device, if trying to measure position and momentum simultaneously, we will face the fundamental limit $\sigma_x \sigma_p \geq \hbar / 2$.
Jan
31
revised Are “uncertainties” in Heisenberg Uncertainity just standard deviations?
added 182 characters in body
Jan
31
comment Are “uncertainties” in Heisenberg Uncertainity just standard deviations?
true. Such accurate enough device is rather impossible. I have to rethink the matter and rewrite the answer...
Jan
31
awarded  Yearling
Jan
31
comment Are “uncertainties” in Heisenberg Uncertainity just standard deviations?
In other words, the fact that an accurate enough device is rather hypothetical does not invalidate the usage of sample averages as estimators for expectation values.
Jan
31
comment Are “uncertainties” in Heisenberg Uncertainity just standard deviations?
I think, accuracy is just an experimental issue. Once your measuring device is accurate enough, you actually can estimate expectation values by sample averages.