Schrodinger equation and atoms

[ralqs]

To what extent does the Schrodinger equation apply to an atom as a whole?

 

[dextercioby]

There's no such thing as <an atom as a whole>, but rather a multiparticle system: at least an electron and with it a positively charged nucleus. So SE will definitely apply, as it's the fundamental equation of QM.

 

[SpectraCat]

There's no such thing as <an atom as a whole>, but rather a multiparticle system: at least an electron and with it a positively charged nucleus. So SE will definitely apply, as it's the fundamental equation of QM.

Huh? Of course there is ... it all depends on the level at which you are working. Look for example at the Hamiltonians that are used to describe weak interactions between atoms that result in superfluidity, or BEC ... there is no explicit consideration of subatomic particles .. the atoms are treated as individual quantum particles.

 

[ralqs]

There's no such thing as <an atom as a whole>, but rather a multiparticle system: at least an electron and with it a positively charged nucleus. So SE will definitely apply, as it's the fundamental equation of QM.

So it applies exactly? How do you know? What's the proof?

 

[nickthrop101]

When you say an atom as a whole, i suppose you're talking about any multi-system of partticles that contains at least a proton in a nucleus and an "orbiting" electron ( i use the word "orbiting loosely, as it does not orbit as planets orbit a star, but appears to), you could add neutrons or more electrons and protons and have a full atom but a "full" or "real" atom only requires a proton in a nucleus. And if you use the schrodinger equation you can track any number of particles or groups of particles, although the maths gets harder as you use more and more particles it is still possible to track the future of a particle or a group of particles. This is easy as the equation is linear, meaning you can do it in any order, and add all the probability waves of the particles together :D

 

[espen180]

So it applies exactly? How do you know? What's the proof?

No physical equation is exact, and the Scrödinger equation is no exception. A better model is given by the Dirac equation, which also takes special relativity into account.