Can we accurately predict the properties of gold?

In summary: Thanks for these excellent replies. I did say QED, but in reality I meant QM in general including QCD. This is interesting. So even though they can see Hg, it still requires a complete and total recalculation of all the particles, which would be impossible in practice.That may have been a typo but it still works!Take the credit for inventing a new word. 😋Speaking of relativistic effects on QED, I don't know, I have to admit that for a long while I have found it very confusing because what is it that is moving fast with regard to the electron. I'm under the impression that if anything the electron in an atom is in the form of
  • #1
bland
146
43
TL;DR Summary
If gold never existed on Earth is QED understood enough to explain all there is to explain.
Let us assume magically that elements 77 to 79 do not exist on Earth, and we ignore any other consequences for the purposes of this query. Without actually making gold to see, is our understanding of QED sufficient to be able to look at Hg as a silvery metallic liquid and work out all the unique properties of gold, by supposing we remove a proton. That is, how many neutrons it would have to contain, it's gold colour (I know that we can explain the gold colour with our current knowledge) its ductility, malleability, resistance to corrosion, etc.

Hopefully I have explained my question clearly.
 
  • Like
Likes Jarvis323 and atyy
Physics news on Phys.org
  • #2
To determine the number of neutrons, QED is not enough; we need also nuclear physics, which in principle (but not in practice) can be reduced to QCD. For other questions QED is enough in principle, but in practice it's a question for quantum chemistry and solid state physics, which are applications of nonrelativistic QM. How powerful quantum chemistry and solid state predictions are in practice, I don't know.
 
  • Like
Likes gentzen, Jarvis323, bland and 3 others
  • #3
  • Like
Likes gentzen, Jarvis323, bland and 4 others
  • #4
vanhees71 said:
non-relativistic QM
The OP said QED, not non-relativistic QM.
 
  • #5
I very much doubt you can run a numeric simulation of 79 electrons and a nucleus and get anything but nonsense. There are 3200 charge pairs that need to be considered.

One can probably do okay with approximatios and interpolations, but that gets into the question of what "counts".
 
  • Like
Likes Jarvis323 and bland
  • #6
QED is not built to describe stationary states of atoms or molecules. It is particularly aimed at scattering events. There is, however, an approximation called the Bethe-Salpeter equation, but you are much better off with approximating the Dirac equation. Of course, simulating 79 electrons will never be achievable in practice by computer power, but there are parts of it which make sense, see the links in vanhees' post.
 
  • Like
Likes gentzen, Jarvis323 and bland
  • #7
Thanks for these excellent replies. I did say QED, but in reality I meant QM in general including QCD. This is interesting. So even though they can see Hg, it still requires a complete and total recalculation of all the particles, which would be impossible in practice.

What about something like if no one had ever made bronze, with our current knowledge would we be able to predict that tin and copper, would make a substance harder than either? Or is this just done by trial and error of metallurgists?
 
  • Like
Likes vanhees71
  • #8
Vanadium 50 said:
approximatios
That may have been a typo but it still works!
Take the credit for inventing a new word. 😋
 
  • #9
Speaking of relativistic effects on QED, I don't know, I have to admit that for a long while I have found it very confusing because what is it that is moving fast with regard to the electron. I'm under the impression that if anything the electron in an atom is in the form of a standing wave, yet relativistic effects imply to my limited understanding that a little solid object is moving very fast.
 
  • #10
PeterDonis said:
The OP said QED, not non-relativistic QM.
@vanhees71 responded to my post. When he does not quote the post he responds to, it means that he responds to the last post before his.
 
  • #11
PeterDonis said:
The OP said QED, not non-relativistic QM.
I was referring to @Demystifier 's claim that for atomic physics non-relativistic QM is enough in the sense that relativistic effects are only small perturbations, but that's not true for atoms with larger ##Z##.
 
  • Like
Likes Demystifier
  • #12
bland said:
Thanks for these excellent replies. I did say QED, but in reality I meant QM in general including QCD. This is interesting. So even though they can see Hg, it still requires a complete and total recalculation of all the particles, which would be impossible in practice.

What about something like if no one had ever made bronze, with our current knowledge would we be able to predict that tin and copper, would make a substance harder than either? Or is this just done by trial and error of metallurgists?
That seems to refer rather to the nuclear-physics and not so much to the atomic-physics aspect. To describe nucleons and nuclei "ab initio" with QCD is of course a very difficult task. For the description of some properties hadrons like their mass, spin, and parity there's of course lattice QCD, which is pretty successful nowadays to "predict" the hadron-mass spectrum as well as predicting higher resonances (particularly strange baryons) which have not yet been discovered by experiment.

The question of nuclear structure is of course even harder, but here there's been made much progress in using chiral effective models to describe the corresponding nuclear forces.
 
  • Like
Likes bland
  • #13
Since color ws mentioned, let me also point out that gold atoms are not yellow. Bulk gold is yellow. Simulating one atom is not enough. I don't know how many you need - I'd be surprised if 10,000 were enough. Perhaps 100,000.
 
  • Like
Likes vanhees71 and dextercioby
  • #14
Vanadium 50 said:
Since color ws mentioned, let me also point out that gold atoms are not yellow. Bulk gold is yellow. Simulating one atom is not enough. I don't know how many you need - I'd be surprised if 10,000 were enough. Perhaps 100,000.
Make it a two stage process: First determine the lattice constant(s) to appropriate accuracy, and then handle the lattice periodicity by suitable boundary conditions. You could also use the Born-Oppenheimer approximation, then you also have a prescribed periodicity of the lattice. (But then you should do simulations for many different positions of the atomic nuclei, so you better have some idea already of the approximate lattice constants.)
 
  • #15
gentzen said:
Make it a two stage process
Breaking an incalculable problem into two incalculable steps may not get you very far.
 
  • #16
Vanadium 50 said:
Breaking an incalculable problem into two incalculable steps may not get you very far.
The important point is less the two steps, but that you can model the periodic lattice via boundary conditions. This is what allows to avoid your issue: "I'd be surprised if 10,000 were enough. Perhaps 100,000."
 
  • #17
Yes, we can calculate the properties of gold, both atomic and bulk. In practice, you start with atomic gold. You have to include relativistic effects, so you'll likely use some implementation of the Dirac-Fock equations as your starting point. To include electron correlation, you'll probably use a multiconfigurational model. You could also use perturbation theory for that, but either way, it's a time-consuming calculation for high Z atoms.

Regardless, this has all been done years ago, and from those precise calculations, people have built approximations for the core electrons of the heavier elements known as pseudopotentials. These pseudopotentials are basically smoothed out mean fields of inner core electrons. This is because calculating the precise electron densities from a plane wave basis (the most convenient basis to use in solid state calculations with periodic boundary conditions) requires a very fine integration mesh, as the multielectron wavefunction is highly oscillatory near the nucleus. As it turns out, the smoothed out fields don't tend to make too much of a difference when calculating out valence electron properties such as band structure, color, tensile strength, phonon modes, etc. (in short, most of the properties you'd be interested in).

With a decent core pseudopotential in hand, all you have to worry about is the valence electrons and setting up the proper boundary conditions. DFT programs are good enough at this point to have implemented variable boundary conditions to optimize crystal structure fairly readily (although the calculations get a lot quicker if you know the symmetry of the crystal structure). After that, the valence electronic structure is a straightforward DFT calculation, and from there, you can calculate a whole host of other properties (photoresponse functions, phonon spectrum, electron-phonon coupling, etc).

Each one of these steps has undergone decades of research, development, and implementation, so it's pretty routine to calculate something like a band structure for gold. In terms of having to do a calculation on 1000 or 10 000 atoms, nanoparticles are in fact one area which is quite challenging for atomistic quantum simulations. Supercomputers can do it, but there's still a lot of research going on in this area. There's a lot of interest in nanoplasmonics and nanophotonics that is driving this work right now.
 
  • Like
Likes sysprog, mattt, bland and 6 others

FAQ: Can we accurately predict the properties of gold?

Can we predict the exact melting point of gold?

Yes, the melting point of gold is a well-known and highly accurate value. It is 1,064.18 degrees Celsius or 1,947.52 degrees Fahrenheit.

Is it possible to predict the density of gold?

Yes, the density of gold is a well-known and highly accurate value. It is 19.3 grams per cubic centimeter.

Can we accurately predict the conductivity of gold?

Yes, the conductivity of gold is a well-known and highly accurate value. It is a highly conductive metal, with a thermal conductivity of 310 W/mK and an electrical conductivity of 45.9 million siemens per meter.

Is it possible to predict the color of gold?

Yes, the color of gold is a well-known and highly accurate value. It is a bright, metallic yellow color.

Can we accurately predict the reactivity of gold?

Yes, the reactivity of gold is a well-known and highly accurate value. Gold is a noble metal, meaning it is unreactive and does not easily form compounds with other elements.

Similar threads

Replies
9
Views
3K
Replies
39
Views
6K
Replies
7
Views
3K
Replies
19
Views
6K
Replies
14
Views
4K
Replies
2
Views
4K
Back
Top