When does the Hartree-Fock approximation fail?

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SUMMARY

The Hartree-Fock approximation fails when the system cannot be accurately described by a single Slater determinant, particularly in scenarios involving non-adiabatic processes or when considering the hard repulsive core of the nuclear force at short distances. In such cases, wave functions may require linear combinations of multiple Slater determinants to capture the interactions accurately. This indicates a breakdown of the mean field approximation, necessitating more complex models to describe the system's behavior.

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  • Understanding of the Hartree-Fock approximation
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  • Knowledge of nuclear forces and their implications
  • Concepts of mean field theory
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Homework Statement



Hi, I've read from Wikipedia that in the Hartree-Fock approximation, "Each energy eigenfunction is assumed to be describable by a single Slater determinant".

My question is... if the approximation fails and the system has to be described by linear combinations of more than one type of Slater determinants, what type of wave functions would they be determinants of? (besides the one-electron wave functions)
 
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You can't have Hartee Fock if the potential the particles move is is not the mean field. To answer the question "when does the Hartree-Fock approximation fail? "
 
Thanks for the answer.
actually... i decided to change the question to the one you see now (edited) on the first msg, a few minutes after writing it the first time...

i was able to change the body of the msg but not the title of the thread.
can this new question be answered here as well?

.. or is it a better idea to start a new thread with the right title?
 
just to ask - when would a mean field approximation fail?
Non-adiabatic processes?
 
Well I come from nuclear physics point of view, and the MFA fails when one takes into account the hard repulsive core of the Nuclear force at short distances.

But what it is called in general, I have no clue :)
 

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