|May17-12, 10:46 PM||#1|
Hartree Fock Symmetric Energy Expression
I just wonder why the energy expression of Hartree Fock method is symmetric. I tried to find out the reason on the Internet but I could only find that: since the Hartree Fock energy expression is symmetric, it is variational.
In Hartree Fock method, the repulsive energy between different electrons is averaged, so does that mean the amount of contribution of different electrons to the repulsive energy is indistinguishable which leads to the symmetric energy expression?
Thank you for your reply
physics news on PhysOrg.com
>> Iron-platinum alloys could be new-generation hard drives
>> Promising doped zirconia
>> Nanocrystals grow from liquid interface
|May18-12, 06:55 AM||#2|
What do you mean by "symmetric"? The HF energy equation is
with Phi being a Slater determinant. It is simply the expectation value of the trial wave function (if you evaluate this expression in terms of matrix elements, you end up with the usual sum over h_ii + 2<ij|ij>-<ii|jj> for a closed-shell determinant). HF is not variational because of the form of its energy expression, but because it is defined to be the method giving the wave function of this form (=determinant) which produces the lowest energy.
While there are plenty of many body methods with non-symmetric energy expressions of the form
(e.g., standard coupled cluster), and this form indeed prevents a variational solution of the equation (it is obviously not bounded for variations of Psi and fixed Phi), it is perfectly possible to make theories with symmetric expectation values which are *not* variational.
|energy, hartree fock, orbital, symmetry|
|Similar Threads for: Hartree Fock Symmetric Energy Expression|
|Hartree Fock Method Symmetric Energy Expression||Quantum Physics||0|
|Hartree Fock||Quantum Physics||2|
|Hartree Fock states.||Advanced Physics Homework||0|
|Hartree-Fock & post Hartree-Fock methods||Advanced Physics Homework||0|
|Hartree & Hartree-Fock||Quantum Physics||1|