# Applications of Schrodingers Equation

by Winzer
Tags: applications, equation, schrodingers
 P: 605 If we can describe the quantum state through Schrödinger equation, we can calculate the observables of the hydrogen atom. Can we do this for other elements, like U238? Or does this have to do with quantum many body problem. I looked up analytical solutions for schrodingers equation and it is quite short compared to the number of elements that are out there. Can we use schrodingers to solve for lattice structures and free bodies?
 HW Helper Sci Advisor P: 4,739 yes we can use schrödinger for lattice. For atomic physics and nuclear physics, the number of equations becomes too many so effective methods are used instead, such as hartee fock and density functionals - which has their starting point at the schrödinger equation of course. The Shcrödinger equation is for non relativistic systems, so for high energy physics relativistic formulation must be used.
 P: 605 Are we able to model all of the elements? And when elements are mixed with each other? These other methods used like the hartee fock, are they accurate models/computations?
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P: 4,739

## Applications of Schrodingers Equation

I am not an atomic- molecular physicist, but I can tell you that they have achieved great success with that, also in nuclear structure physics which is even more complicated than atomic physics.
 Sci Advisor P: 1,865 Yes, the Schrödinger equation can be used for any element. But you can only use it successfully for the lighter elements. In heavy elements, relativistic effects become significant, and you need to use the Dirac equation instead. This is because the high nuclear charge leads to such a high velocity on the innermost ('core') electrons that they start picking up relativistic mass. (The relativistic effects can also be approximatively modeled by pseudopotentials such as RECPs - Relativistic effective core potentials) Most famously, a non-relativistic calculation of gold will show that it 'should' be silver-colored. The yellow color of gold is a relativistic effect. (originating with a 'p' band, shifted down from the UV to the blue range, giving blue absorption and a yellow color) Uranium, which you mentioned, is such a heavy element that non-relativistic calculations aren't really of much use. Hartree-Fock is essentially the simplest method that still gives reasonably accurate results. It essentially means solving the many-body problem by treating each electron as if it is moving in an averaged field of every other electron. Essentially you could think of it as you keep every other electron 'still' while you calculate the energy for one electron, then do the same for the other electrons, and repeat the procedure. (Self-consistent field method, SCF) The Schrödinger equation can be solved exactly (to within numerical precision) using other methods. In fact, the first calculations that were accurate enough to show that the 'new' quantum method should work for many-electron systems was such a method, the Hylleraas variational method, applied to Helium (1929). That method is almost never used today though, because it's difficult to expand to systems of more than two electrons. Also, it doesn't scale well. The most accurate method in common use is Configuration Interaction (CI), which if done over all possible configurations (full CI) is exact, within the limit of the basis set used and the nonrelativistic limit. The problem with full-CI is that it scales factorially with the number of electrons. So even with powerful computers, it can only be used for very small systems. Then there's also the Born-Oppenheimer approximation (assuming essentially infinite nuclear masses) which is generally used both with non-relativistic and relativistic calculations. It all comes down to how exact a solution you want and how much time you have to spend calculating it. But the consensus is that yes, quantum theory is exact in principle for every element and every chemical compound. (Edit:) I'll add a word on accuracy: It's not really that the methods are inaccurate themselves, not in absolute terms. You are calculating the total electronic energy of a system, and even Hartree-Fock will usually get more than 95% of the correct value for that. The issue is that you're usually not interested in that value. You're interested in the relatively small changes in that energy during chemical reactions and similar. And chemistry is in fact a very subtle thing. The total electronic energy for a typical molecule is on the order of hundreds, or even thousands, of Hartrees, whereas the energy involved in a chemical reaction is on the order of a tenth of a Hartree. Add to that the fact that chemical kinetics are exponentially related to energy, and you need very accurate methods to be able to make a reasonable prediction. The other thing that's often calculated are spectroscopic properties, and spectroscopical methods are very accurate. (1 cm-1 = 1.8 Hartrees)
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 Quote by malawi_glenn I am not an atomic- molecular physicist, but I can tell you that they have achieved great success with that, also in nuclear structure physics which is even more complicated than atomic physics.
Oh yeah? So that's why nuclear physicists are borrowing our methodologies, then?
 P: 605 Wow, thank alxm. All that sounds amazing--I had no idea. I know that we can model alpha decay of say U238(i think) and many other elements. Does schrodingers explain the beta(-) and beta(+) decay? Also how was it feasible to slip in those neutrino and antineutrino bit to conserve energy? Was this info found in schrodingers too? Conservation of angular momentum? Going even smaller, does schrodingers give accurate models for quark/s behavior? Or is that another beast?
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 Quote by alxm Oh yeah? So that's why nuclear physicists are borrowing our methodologies, then?
well yes LOL indeed those methods are supposed to be general, and should also provide good approximations for nuclear structure aswell
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