Dear Vladimir V. :

Your question is interesting, my field of research is n't Quantum Mechanics, nor Particle Physics, thus I'm sure I would n't be able to answer it fully. but I felt motivated to contribute anyways, since maybe it can help to give a logic direction to the ideas concerning your question.

An attraction between to quantum particles can be given just due two kind of fundamental interactions: by the Strong Nuclear Force and/or by the Electromagnetic Interaction.

The interaction between to quantum particles, by the Strong Force, will be always attractive, regardless of the Energy of each particle.

In the case of the Electromagnetic Force, we know -(I'm just telling things here I know we all already know)- the interaction between two Quantum Particles can be either attractive or repulsive, and we know that the observed case will depend solely on the Electric Charge of the particles themselves, regardless (again) of its energies.

Now, trying to get deeper on the interpretation of your question, I guess we can reformulate it as this: "Is it turns out that, always when the interaction between two Quantum Particles is attractive, it is due to the Negative Energy of the system ?".

Well, first of all, here I'm guessing that when we speak of a Negative Energy we are speaking of a Potential Energy, right ?, and in particular a Potential Energy from a positive or 0 point of reference, I guess.

Well, I think the answer is subtle.

Maybe for Elementary particles the answer is Yes: For an attraction between particles (elementary ones) (be Nuclear or Electromagnetic interaction) a drop of Potential Energy will always be involved, if you defined the total Potential Energy of the system at the initial time to be zero (which is the standar definition, since the Electrostatic Potential field is a conservative field, etc., etc.,) Please be aware that this is more a definition of reference than a piece of physics built into any specific law).

Now that I think about it well, for electrons, and atoms agains other atoms, the answer is also Yes.

An alternative representative view of this could be: the Potential Energy of the system always will drop, since you are trying to do "work" on the "environment" with this Energy of the "open" Quantum system. I know there is not such a thermodynamic definition of "work" at the Quantum level, yet.

What do you think Prof. Vladimir V. Lugovoi ? I guess my lack of explanation on the specific case of the contact interaction between atoms is evident.

This is the specific case of a material's surfaces characterization technique called Atomic Force Microscopy.

I found the attached graph, regarding on this topic and the definition of contact point. Maybe it can help to shed light upon the answer to the question somehow.

Best Regards !

It is more a convention to set zero energy meaning the two

particles become barely free of each other.

This is the case of the Hydrogen atom with an electron, or Classical gravitationally bound systems in the two body problem.

I would not read too much into this. It is just a convention.

Negative energy shows bound system and the two particles in bound system always exert attractive force on each other.

Dear Juan Weisz,

If I understand you correctly, since we have already introduced this convention, then whenever negative energy appears in the calculations, should we talk about the presence of a "not free", "bound" state of two quantum particles?

(For example, negative energy appears when calculating the molecular bond.)

Dear Franklin ,

Thank you for your reasoning. In addition to potential energy, in (non-relativistic quantum mechanics) there is the concept of calculating an averaged (in the quantum mechanical sense) operator corresponding to any physical quantity (for example, the Coulomb energy operator for the interaction of two electrons or for an electron and an ion). This averaging means that we will calculate the numerical values ​​for this operator, placed in a physical environment, described by the wave function that "participates" in the calculation of this quantum-mechanical averaging. So, if a system has two states, then it is a two-level system (see Feynman Lectures in Physics, Volume 8, "Quantum Mechanics".). In particular, it may turn out that one level is positive and the other is negative (but this is a very confusing task, albeit an interesting one).

Depends on what you call "attraction". If it is the possibility of a bound state, then yes. If it is an attractive force, then it also depends on the distance: attractive force implies positive derivative of the potential.