New answers tagged quantum-chemistry
4
Certainly experimental accuracy is a major factor. However, another argument for this rule of thumb is that at room temperature, a difference in free energy of 1.4 kca/mol results in a equilibrium/rate constant with a relative change of ~10. Therefore, in order to be within an order of magnitude of experiment, you need to have an accuracy of ~1 kcal/mol.
4
To study surface reactions I recommend the growing string method (GSM) for surfaces. Here is a nice quote from the paper which developed it (Ref. 1)
GSM’s efficacy was confirmed by comparison with CI-NEB on
an extensive set of reactions characteristic of modern surface chemistry studies. In these cases, GSM reduces the computational cost (in terms of ...
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 ...
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.
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 ...
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 ...
4
$\newcommand{\ket}[1]{\left|#1\right>}$
$\newcommand{\bra}[1]{\left<#1\right|}$
(1) Is this all correct so far?
Yep.
(2) If so, is it justified to use it in numerical derivations as an
eigenfunction of the Hamiltonian, i.e. can we write
$\hat{H}\ket\Phi\overset{?}{=}E_{0}\ket\Phi$ or are we restricted to
$\bra\Phi\hat{H}\ket\Phi=E_{0}$?
...
1
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$\newcommand{\Bra}[1]{\left<#1\right|}$
$\newcommand{\BraKet}[2]{{\left<#1}\left|#2\right>}$
It's an older question, but one worth answering!
For the uninitiated, second quantization is a bookkeeping technique that describes many-particle systems as excited states of a field, usually taken to be the ...
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 = ...
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 $3/2$...
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 \\...
0
Potentiology is a field that began in the early days of quantum mechanics:
1924: John Lennard-Jones publishes the Lennard-Jones pontial
1929: Philip M. Morse publishes the Morse potenital.
The most recent reference you gave was a potential from:
1940: by Hugh Hulburt and Joseph Hirschfelder
You seem to be hoping that in the last 80 years, someone might ...
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.
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.
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.
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).
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
\begin{equation}
S^{2}=S_{+}S_{-}-S_{z}+S_{z}^{2}
\end{equation}
(I'm using atomic units here)
and ...
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