# Tag Info

8

Probably the easier way to explain it is to rely on the fact that if Li indeed had a ground state with $S = 3/2$, this would necessarily comprise four levels with $M_S = +3/2, +1/2, -1/2, -3/2$. This sort of splitting can be observed under a magnetic field, for example (the Zeeman effect). Note that it's not possible for the Pauli principle to forbid $M_S = ... 8 You can use the "compare" command in the Jmol program, either online or as free-standing program. A tutorial on how to use the program for this purpose is here, http://proteopedia.org/wiki/index.php/Superposition_with_jmol. It outputs the root mean square of the pairwise differences (RMSD) as well as the superposed coordinates. 6 Start from the AO Hartree-Fock equation and its adjoint $$\mathbf{F}^{AO}\mathbf{T}=\mathbf{S}\mathbf{T}\epsilon \text{ and } \mathbf{T^\dagger}\mathbf{F}^{AO}=\epsilon\mathbf{T^\dagger}\mathbf{S}$$ where$\mathbf{T}$is an$N\times n$matrix that is essentially the occupied block of$\mathbf{C}$($n$is occupied,$N$is total orbitals). We use this$\...

6

You can do a structural comparison with VMD, using the RMSD trajectory tool. The RMSDTT is a default plugin for VMD (it comes with it, no need to do any fancy installation, so should help students access it).

6

The implementations of MNDO, AM1, and PM3 along with PM6 and PM7 in MOPAC include Sr and Ba, but not Ra. MOPAC is free for academic use, but modern versions of it are not open source. Most of these models have an open-source implementation in SPARROW (GitHub). An entirely different class of semiempirical models (GFNx) is available in the open-source XTB code ...

5

In terms of how the electrons are arranged*, the reason you get only spin $1/2$ is because the most stable configuration has two of the electrons in the same 1s orbital, which by the Pauli Exclusion pile forces them to have opposite spins. This leaves only the third electron, sitting in the 2s orbital, to contribute a net spin. You could get a spin of $3/2$...

5

Starting from statement that the Fock matrix and the density matrix commute in an orthonormal basis. $$[\mathbf{F}, \mathbf{D}] = \mathbf{FD} - \mathbf{DF} = \mathbf{0}$$ The orthonormal basis matrices can be substituted for their equivalents in an atomic orbital basis \begin{align} \mathbf{F} = {} & \mathbf{X}^\dagger \mathbf{F}^{AO} \mathbf{X} \\ \...

4

There's a "pair-fitting" function in pymol that will align molecules given some user-defined reference atoms, and output an RMSD across those points of comparison. You can get pymol educational version for free, so students would be able to install it themselves.

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3

SDD is a basis set description, which uses Stuttgart/Dresden ECPs. They are readily available for download at the BSE, but also via Univ. Köln. And they are pretty easy to pick out. See the Pseudo keyword for what is Gaussian 16 is using currently, which should be the same for 09. At least in the BSE there hasn't been an update to the S/D ECPs since their ...

3

Solution $S^2\alpha = \left(S_x^2 + S_y^2 +S_z^2 \right)\alpha= S_x(S_x(\alpha)) + S_y(S_y(\alpha)) +S_z(S_z(\alpha))$ from the ''rules", where for instance $S_x$ operating on $\alpha$ gives $\frac{1}{2}\hbar\beta$, we make the three replacements to get $=S_x(\frac{1}{2}\hbar\beta) +S_y(\frac{1}{2}i\hbar\beta) +S_z(\frac{1}{2}\hbar\alpha)$ Now take ...

3

One possible way: As you say $${\bf{S}^2} = S_x^2+S_y^2+S_z^2$$ Thus \begin{align} {\bf{S}^2} &= \left( \frac{\hbar}{2}\right)^2 \left[\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} + \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \\...

3

I would say Chimera, no question. It has a good GUI with a lot of tools, including ensemble comparison. I find it both much easier and much prettier than VMD which has been recommended here.

3

You can reorder the Z-matrix with a free program like Molden. For example open the molecule in Molden and press the ZMat editor. From there you should see a button that says reorder Z-matrix. Click the atoms one by one, making sure to select the HCH atoms sequentially so they have an explicit angle. Dummy atoms are another good option.

2

Rotational frequencies are not close to zero. They are only smaller than vibration frequencies. You may know that frequencies and energies are proportional at the atomic scale. It does not require much energy to make a molecule move or rotate. More energy is needed if you want to deform it and try to stretch its bonds.

2

The complete, formal mathematical answer is a bit complex. It might be better to get that discussion from Math Stack Exchange. An answer to the world of Chemistry is more straightforward: Yes, for any physically relevant operator and any relevant function, you always can. That’s a fundamental property of quantum mechanics: (functional) states are ...

1

Semiempirical quantum chemistry models do carry out proper Hartree-Fock calculations, including both Hartree (electron-electron electrostatics) and Fock exchange contributions to the total electronic energy. As noted in the question, the tensor of 4-center Coulomb integrals is heavily sparsified, approximated, and parameterized to simplify and accelerate ...

1

Let us begin with the following principle: the CC ansatz always returns an appropriate electronic wavefunction (meaning a combination of Slater determinants). Still, at least in a non-relativistic approach, the Hamiltonian does commute with the Spin operator $$S^{2}=S_{+}S_{-}-S_{z}+S_{z}^{2}$$ (I'm using atomic units here) and ...

1


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