Hartree Fock Symmetric Energy Expression

Click For Summary
SUMMARY

The Hartree Fock (HF) method's energy expression is symmetric due to its reliance on the Slater determinant, represented as E=\langle\Phi|H|\Phi\rangle. This symmetry arises because the repulsive energy between electrons is averaged, leading to indistinguishable contributions from different electrons. While the HF method is variational in nature, it is not solely due to the symmetry of the energy expression; rather, it is defined by the determinant form that yields the lowest energy. Other many-body methods, such as standard coupled cluster, can exhibit non-symmetric energy expressions that do not allow for variational solutions.

PREREQUISITES
  • Understanding of Hartree Fock method and its principles
  • Familiarity with Slater determinants in quantum mechanics
  • Knowledge of variational principles in quantum chemistry
  • Basic grasp of many-body quantum theory and coupled cluster methods
NEXT STEPS
  • Study the derivation of the Hartree Fock energy expression and its implications
  • Explore the concept of Slater determinants and their role in quantum mechanics
  • Investigate variational methods in quantum chemistry beyond Hartree Fock
  • Learn about coupled cluster methods and their applications in many-body systems
USEFUL FOR

Quantum chemists, theoretical physicists, and students studying computational methods in quantum mechanics will benefit from this discussion, particularly those interested in the Hartree Fock method and its implications in many-body theory.

arsc
Messages
2
Reaction score
0
Hello.

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 Phys.org
What do you mean by "symmetric"? The HF energy equation is
E=\langle\Phi|H|\Phi\rangle
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=\langle\Phi|H|\Psi\rangle
(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.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 24 ·
Replies
24
Views
8K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K