My 2 cents to this: Not a particularly meaningful question.

The only meaningful thing about energies, as in eigenvalues of the Hamiltonian, is their DIFFERENCE. The evolution with time of the state of an isokated quantum system (Schrödinger picture) is completely governed by the differences of the eigenvalues of the Hamiltonian. The choice of where you put zero is up to you.

A common choice for zero is the ground state energy of the Hamiltonian. In this case clearly zero is zero, there is no effect of SO.

Other choices are more practical in various circumstances: The mean of the spektrum of the Hamiltonian, or the Fermi energy (definition discussed recently here), or some fictitious reference limit like the state with all charges at infinite distance ... The ground state energy then probably will depend on whether or not you put SO into your Hamiltonian. After all you consider SO because it changes energy differences and might split degeneracies.

Dr. Martin Pieper
Karl-Franzens University
Institute of Physics
Universitätsplatz 5
A-8010 Graz
Tel.: +43-(0)316-380-8564

Am 29.10.2016 23:20, schrieb Abderrahmane Reggad:
Hello again

I am waiting for an answer to my question .

My question is about the effect of the inclusion of the Spin-Orbit
Coupling on the ground state energy. 

I want to know if the SO affect the ground state energy also or It
only causes the splitting of the degenerate state energies.

Any information will be fruitful for me.

I hope find an answer to my question



Laboratoire de Génie Physique
Université Ibn Khaldoun - Tiaret


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