# Tag Info

19

A few years ago the XUV physics group at the AMOLF Institute in Amsterdam were (to my knowledge the first to be) able to directly image the orbitals of excited hydrogen atoms using photoionization microscopy. For more details see the paper, Hydrogen Atoms under Magnification: Direct Observation of the Nodal Structure of Stark States. A.S. Stolodna et al. ...

19

Generally speaking, atomic and molecular orbitals are not physical quantities, and generally they cannot be connected directly to any physical observable. (Indirect connections, however, do exist, and they do permit a window that helps validate much of the geometry we use.) There are several reasons for this. Some of them are relatively fuzzy: they present ...

15

The first images of hydrogen s orbitals were obtained in 2013 by physicists in the Netherlands.

11

For the first rows in the periodic table, this can easily be explained by the fact that electrons possess four quantum numbers (usually $n$, $l$, $m_l$ and $m_s$). These numbers are restricted as such: $$n = 1, 2, 3, ...$$ $$l = 0, 1, 2, ..., n - 1$$ $$m_l = -l, -l + 1, ..., 0, ..., l - 1, l$$ $$m_s = -\frac{1}{2}, \frac{1}{2}$$ By Pauli's ...

10

I remember doing SCF calculations back in the early 80s, and it was by no means guaranteed that the calculation would converge or that it would give you the ground state. Several of my calculations diverged at the first attempt, though putting a bit more thought into the starting point would usually produce convergence. I don't think I ever accidentally ...

10

By definition, "hydrogen atom" refers to the neutral system with one proton and one electron, so it cannot hold any extra electrons. However, protons can hold more than one electron, in which case the system is termed a hydrogen anion. This is a stable, bound system, and the reaction $$\mathrm{H}+e^- \to \mathrm{H}^- \tag 1$$ releases about $0.75\:\rm eV$,...

9

...it makes no sense to calculate any property of the particle with the "initial" wave-function, since this is simply the incorrect wave-function for the new well? The wavefunction can't be "incorrect for the well". Your wavefunction is just an initial condition for time-dependent Schrödinger equation. Here's how it would evolve if you solve the time-...

9

Your intuition is spot on. If we have two atoms approaching each other with a large kinetic energy then they will have too much energy to form a stable molecule. Their electrons will interact as they approach, but the two atoms will simply whizz past each other and head off into the distance. In many cases the reaction is more like: $$\mathrm{AB + CD \to ... 7 Assumptions: I will be talking about Hermitian (more generally self-adjoint) operators only. This means that I will assume that the operators in question have a set of eigenvectors that span the Hilbert space. As mentioned by tomasz in a comment, this is not exactly necessary, since more general statements can be made, but since we are dealing with basic QM, ... 7 This is well explained on the basis of sub shell electronic configuration. But first, let's look it by the concept of shells alone. See in the example of calcium, it has 20 electrons. Of course the outermost shell can accommodate 18 electrons. If it goes like 2,8,10 then the outermost shell contains 10 electrons. The stable state is either having an ... 6 The mathematician John Baez recently wrote a long series of blog posts about using quantum techniques for non-quantum stochastic systems, in which chemical reaction networks played a central role as an important special case. This culminated in a paper entitled Quantum Techniques for Reaction Networks, which might be something close to what you're looking ... 6 The right formula is very similar to yours,$$ \frac{4!}{8}{100 \choose 4}+3!{100 \choose 3}+2\times 2!{100 \choose 2}+1!{100 \choose 1} = 12,753,775 $$I think that by comparing the coefficients in front of the (correct) binomial numbers, you may determine how you need to fix your calculation. 6 You have one 1s atomic orbital for each H atom, and one 1s, 2s and three 2p for each C. This makes a total of 36 atomic orbital in the whole molecule, and so you have 72 spin states. The 2s and 2p's orbitals are going to hibridate giving three 2sp^2 orbitals and one 2p orbital, so it has its characteristic \pi-delocalized electronic estructure on the ... 6 To avoid re-treading old ground, this answer contains some previous literature that has been mentioned on this thread, as well as the surface layer obtainable via naive google searches: A. H. Wilson. The Ionised Hydrogen Molecule. Proc. Roy. Soc. Lond. Ser. A, Math. Phys. 118 no. 780, pp. 635-647 (1928). These lecture notes for CHEM-UA 127: Advanced ... 6 I would suggest that you try to write a small code that can perform a Hartree-Fock calculation. Specify molecule and basis set. Get the integrals. The integrals can be hard to calculate on your own so I would suggest to obtain those from different sources, see Programming Project #3: The Hartree-Fock Self-Consistent Field Method Make a diagonalization of ... 6 As part of an undergraduate project I calculated the electron density for various small molecules such as water and ammonia, and the disappointing result is that they are all basically formless blobs with only small bumps where the hydrogen atoms are. They ended up looking much like your last picture of water: though even that picture exaggerates the bump ... 6 A reasonable answer is given in A.R.P. Rau, "The negative ion of hydrogen". J. Astroph. Astron. 17, 113 (1996) where Rau explains as follows: Of particular interest among the Ν = 2 states is the lowest one of ^3 P^e symmetry, described in independent-electron terms as 2p^2. This is bound below the \mathrm H(N = 2) threshold with about 9.6 \:\rm ... 5 I'm not sure how easy it would be to be to give a rigorous explanation of this. Here's an explanation based on the atomic orbital approach to the electronic structure of atoms. This is only an approximation, but I think it gives a good flavour of what is going on. To clarify Terry's question: if you take the atomic orbital approximation to electronic ... 5 In principle, yes, you can compute all chemical reactions by Feynman diagrams, since the underlying theory is QED. In practice, to a reasonable accuracy, the mechanism of a chemical reaction can be describe by the transition state theory, ref. Atkins, "Physical Chemistry". The theory uses many assumptions of the transition state related to the equilibrium ... 5 If you take the reaction$$\ |\text{CH}_4,2 \text O_2 \rangle\ \ \overset{\text {burn}}\longrightarrow \ \ |\text{CO}_2, 2 \text H_2\text O\rangle\ "$$There are 7 nuclei and 42 electrons. In nonrelativistic quantum mechanics, a state of this system is a function on a ~150-dimensional space. It's essentially impossible to do any calculations on such a ... 5 First, there is no classical explanation for chemical bonding, so any explanation has to include some quantum mechanical ideas. Second, perhaps the simplest quantum idea here is that electrons exist in energy levels. Continuing the hydrogen example the lowest electron energy level in H_2 has less energy than the lowest electron energy level in atomic ... 5 I would forget about the movement of the wall. The potential is the infinite square well of width 2L (potential is \infty aside from the region 0 < x < 2L, where it is 0), and the wavefunction is$$ \Psi\left(x,t\right) = \sum_{n=1}^\infty c_n \psi_n\left(x\right) \exp\left(-\frac{iE_n t}{\hbar}\right),  where $\psi_n\left(x\right) = \sqrt{1/... 5 Another way to look at spin, complementary to the other ways, which I find helpful is look at an abstract generalisation of the concept of angular momentum and forget about things like classical tops. This generalisation begins in something called Noether's Theorem which you probably haven't met yet. You need some background but the idea is essentially ... 5 To understand what these orbitals are, you first have to understand the notion of superposition in quantum mechanics. In regular classical physics, a particle or a system must be in a definite state. A car is at a particular mile marker on a highway, moving at a particular speed. The Moon orbits around the Earth with a particular velocity at a particular ... 5$\hat{O}$will have a set of eigenvectors$\phi_i$and eigenvalues$o_i$. You can expand your wave-function$\psi$in terms of the basis set$\phi_i$, i.e.$\psi=\sum_i c_i\phi_i$. If you make a single measurement then your wave-function$\psi$will collapse to one of the eigenvectors$\phi_i$with probability$|c_i|^2\$ in which case your measurement is ...

5

A basic difference between quantum mechanics and classical mechanics is that the potentials do not act on masses in quantum mechanics. Instead they are part of the differential equation that has to be solved to give the wavefunction for the system under consideration. In the case of a single atom, lets take the hydrogen atom, the differential equation is ...

4

This is really just a footnote to Chris' answer, but any change in the bond angle of water would change its rotational and vibrational spectra. We can detect the spectrum of water molecules in the Orion nebula, which is 1,344 light years away, so we know the bond angle hasn't changed in the last 1,344 years. Presumably we can detect water in more distant ...

4

There is no guaranteed solution to get convergence on the ground state. But there are really good algorithms used like DIIS. And as always you need a good starting point to not get stuck at a local minima. And this is e.g. Hückel operator or INDO guess. As the Hartree-Fock method relys on the variational principle you find a lower energy with a better trial ...

Only top voted, non community-wiki answers of a minimum length are eligible