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8

There is an updated list of errata* for the 3rd edition of Jensen's book at https://chem.au.dk/~frj/corrections_ICC3.html, which includes the mistake you pointed out (and more). It reads: page 138, (4.29), third line should read: [chiAchiA - chiBchiB](alpha.beta - beta.alpha) - [chiAchiB - chiBchiA](alpha.beta + beta.alpha) page 138, (4.29), fourth line ...


3

I'll specifically answer this part of your question: I would expect 50% of collisions in a gas of free hydrogen atoms to have up spins, and 50% to have down spins. So, on average, any given pair of hydrogen atoms are equally likely to either have parallel or antiparallel spins. Does this imply that only 50% of H–H collisions in that situation result in a ...


2

I think the premise of your question is flawed, which isn't your fault. It's the way students are introduced to quantum mechanics at the intro level. But if you could somehow generate a gas phase of free hydrogen atoms, it isn't correct to say that 50% would have "up" electron spin and 50% "down." What is correct is that if you measured ...


0

Yes -- if one or both electrons are in an excited state and they thereby occupy different orbitals. Of course, the excited state electron(s) would decay rapidly, so either the atoms blow apart from the resulting energy release or an electron spin-flips to preserve/strengthen the bond.


2

The short answer is that we'd like to do the first, but solving the multi-electron Schrodinger equation is impossible, so we use variations of your second option (ie considering many separate one-electron wavefunctions) to get something that matches experimental data as closely as possible, even though we don't ever get an explicit form of the true multi-...


7

The answer is a definitive affirmative yes. You don't need to reach out for proteins, organic dyes are known to suffer from solvatochromism in function of the polarity of the solvent they are dissolved. It either may be a bathochromic shift (or, red shift) of the UV-Vis absorption recorded, or a hypsochromic shift (or blue shift). In large proteins, you ...


1

MO theory fails pretty fantastically in a wide-variety of cases. While it works well near the equilibrium geometry of a molecule, it cannot dissociate even the simplest bonds correctly. It is well-known by computational chemists that any quantitative predictions of MO theory (ionization energies, HOMO-LUMO gaps, etc.) cannot be trusted, even for the ...


2

In a nutshell The tight-binding model relies on the assumption that we can forgo the GLOBAL antisymmetry of fermions and instead purely work with a LOCALLY antisymmetric basis. In that sense, the tight-binding model is an approximation to the total wavefunction that uses site-specific antisymmetric wavefunctions to form basis states for the total wave-...


1

The answer is yes and no. Electronic coordinates are needed when calculating the Fock matrix elements $F_{rs}$, which consist of one-electron integrals and two-electron integrals. Once you get the Fock matrix elements, then the Hatree-Fock-Roothaan equations are solved by matrix algebra through diagonalization without using electron coordinates. Therefore, ...


0

Think about the integration as solving the Schrödinger (or Fock, or whatever) equation for all coordinates in space at the same time. The whole point of this is to obtain a wavefunction which yields the energy eigenvalue to your operator no matter what electron coordinates you plug in. In other terms, $\hat H \Psi ( \mathbf{x} ) = E \Psi ( \mathbf{x} ) $ for ...


3

Spin multiplicity = $2S+1$, where $S$ is the total spin angular momentum. Now $S = \frac{n}{2}$ where $n$ represents total number of unpaired electrons. So now we can write spin multiplicity = $n+1$. Now coming to your question, $\ce{Mn}$ has 5 unpaired electrons in it. Therefore, its spin multiplicity $(S) = 5+1 = 6$. Similarly, $\ce{Mn^{2+}}$ has 5 ...


2

Because periodic waves can be represented using exponential functions using imaginary arguments I'm going to attempt a non-rigorous explanation that might give some intuition as to why imaginary numbers appear in the mathematics of quantum stuff. Most of this doesn't require a knowledge of what a Hermitian operator is or about vector spaces or eigenvalues. ...


9

Why is the momentum operator imaginary? The simplest explanation hinges on the fact that observables are represented by Hermitian operators in quantum mechanics. Once we accept this, then we can show that the momentum operator $\hat{p} = -\mathrm{i}\hbar\nabla$ is Hermitian precisely because of the factor of $\mathrm{i}$. We need to show that for any two ...


3

This problem can be thought of as a linear combination of atomic orbitals $\phi_-$ and $\phi_+$ to molecular orbital $\phi$ with broken symmetry (i.e. LCAO-MO and $c_1 \neq c_2$). The answer to it can be figured out as follows. As stated in the conditions, the normalized atomic orbitals are $\phi_-$ and $\phi_+$ for the left and right intervals centered at $-...


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