1
$\begingroup$

A many-electron system with molecular Hamiltonian under the Born-Oppenheimer approximation has a finite ground state energy, which means its eigenenergy is bounded from below.

In my research, I need both the estimate of the maximum and minimum eigenenergy to set a parameter, so I was wondering whether such a system's maximum eigenenergy is also bounded from above. For hydrogen, whose eigenenergy is $E_n = -\frac{1}{2n^2}$, this conjecture holds since hydrogen's maximum eigenenergy is 0.

But I don't know how to prove or disprove this for all kinds of atoms and molecules. I guess this has something to do with the boundness of the Hamiltonian.

Improve this question
$\endgroup$
5
  • $\begingroup$ If a molecule is too energetic, bonds will break or it will ionize. $\endgroup$
    – mmesser314
    Commented Aug 6, 2023 at 1:32
  • 1
    $\begingroup$ "since hydrogen's maximum eigenenergy is 0..." No. There are also continuum states. $\endgroup$
    – hft
    Commented Aug 6, 2023 at 2:27
  • 1
    $\begingroup$ Most Hamiltonians have no upper bound. E.g., $p^2/2m$ can be a big as you would like. Just give the electron a lot of momentum. But the smallest $p^2/2m$ can be is zero. $\endgroup$
    – hft
    Commented Aug 6, 2023 at 2:29
  • $\begingroup$ @hft continuum states are not eigenstates of this Hamiltonian. $\endgroup$
    – M06-2x
    Commented Aug 6, 2023 at 11:26
  • $\begingroup$ @M06-2x Yes they are. demonstrations.wolfram.com/HydrogenLikeContinuumEigenstates $\endgroup$
    – hft
    Commented Aug 9, 2023 at 0:51

2 Answers 2

Reset to default
1
$\begingroup$

Is many-electron system's energy bounded from above?

No, it is not bounded from above.

For hydrogen... this conjecture holds since hydrogen's maximum eigenenergy is 0...

No, this is wrong. The Hydrogen Hamiltonian has continuum states, you are only considering bound states.

In Quantum Mechanics the Hamiltonian is generally not bounded from above. For example: $$ p^2/2m $$ can be as big as I would like, I just give the electron lots of momentum. But the smallest it can be is zero.

Improve this answer
$\endgroup$
4
  • $\begingroup$ If I only consider bound states, then is it safe to say the maximum energy of the bound states is bounded from above (by 0 since the potential energy of molecular Hamiltonian is 0 at infinity)? $\endgroup$
    – Andy Liu
    Commented Aug 6, 2023 at 3:52
  • $\begingroup$ Yes. But you originally asked about the Hamiltonian. The Hamiltonian is an operator with a fixed/known spectrum, and the bound states are only part of that spectrum. You are ignoring certain eigenvalues of the Hamiltonian and so your analysis will not strictly apply to that Hamiltonian. But, do what you will. $\endgroup$
    – hft
    Commented Aug 6, 2023 at 4:00
  • $\begingroup$ This answer is wrong, the kinetic the kinetic energy cannot be "can be as big as" we want in an atom and the continuum is without the atomic potential, this is trivial. $\endgroup$
    – M06-2x
    Commented Aug 7, 2023 at 13:15
  • $\begingroup$ @M06-2x the continuum is not "without the atomic potential." The Hamiltonian: $p^2/2 - Z/r$ has bound states and it also has continuum states. Just because you don't know about it doesn't mean it isn't a real solution. (see, e.g., demonstrations.wolfram.com/HydrogenLikeContinuumEigenstates) The kinetic energy is literally unbounded above and it can be as big as we want. $\endgroup$
    – hft
    Commented Aug 9, 2023 at 0:50
0
$\begingroup$

For a molecular system the upper bound is the state of one electron removed from the least bound conglomerate of electrons. As known from chemistry and linear algebra of quantum 1-particle states, the ionized molecule is a different object class. In many aspects the missing electron can be tensored as a positively charged pseudoparticle to the reduced n-1-particle state, either as a single particle surface state or as a volume hole state.

Since ionization energy depend strongly on the partitons of n, the particle numbers of fermionic n-particle correlation functions, nearly nothing con be predicted about magnitudes in general.

Improve this answer
$\endgroup$
1
  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Aug 6, 2023 at 7:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.