Symmetry groups of molecule - Hamiltonian

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SUMMARY

The discussion centers on the relationship between molecular symmetry and the Hamiltonian of molecules belonging to specific point symmetry groups. It is established that the Hamiltonian of such molecules commutes with all symmetry elements of their point symmetry group. This is fundamentally due to the indistinguishability of atoms with the same number of protons and neutrons, particularly when using first-quantized formulations. The conversation highlights the necessity of enforcing identical coefficients in molecular force fields to maintain this symmetry.

PREREQUISITES
  • Understanding of molecular symmetry and point symmetry groups
  • Familiarity with Hamiltonian mechanics in quantum chemistry
  • Knowledge of first-quantized and second-quantized formulations
  • Basic principles of molecular force fields
NEXT STEPS
  • Research the principles of molecular symmetry in quantum chemistry
  • Study Hamiltonian mechanics and its applications in molecular systems
  • Explore first-quantized versus second-quantized formulations in quantum mechanics
  • Investigate methods for setting up molecular force fields with symmetry considerations
USEFUL FOR

Researchers in quantum chemistry, molecular physicists, and students studying molecular symmetry and Hamiltonian mechanics will benefit from this discussion.

Konte
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Hello everybody,

As I mentioned in the title, it is about molecular symmetry and its Hamiltonian.
My question is simple:
For any molecule that belong to a precise point symmetry group. Is the Hamiltonian of this molecule commute with all the symmetry element of its point symmetry group?

Thanks.
Konte
 
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A. Neumaier said:
Yes.
Thank you for your answer.
Could you please indicate how to demonstrate this (any link or book?)
 
Konte said:
Thank you for your answer.
Could you please indicate how to demonstrate this (any link or book?)
I don't know of a reference. This is more or less by definition, because atoms with the same number of protons and neutrons are indistinguishable. If you use a second-quantized formulation, you cannot create Hamiltonians where this fails.

In practice one uses first-quantized formulations only and enforces this through how the molecular force field is set up - by imposing identical coefficients on terms that differ only by a permutation.
 
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A. Neumaier said:
I don't know of a reference. This is more or less by definition, because atoms with the same number of protons and neutrons are indistinguishable. If you use a second-quantized formulation, you cannot create Hamiltonians where this fails.

In practice one uses first-quantized formulations only and enforces this through how the molecular force field is set up.

Many thanks anyway. In fact, I suspect that the answer is "yes" , so you reassure me largely.
I turn to somebody who can indicate any demonstration if possible please.

Thanks
 

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