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The Morse potential can be used to describe the potential energy of the a diatomic bond:

Morse potential

This curve suggests that the lowest state $v = 0$ is the most stable, since it has the lowest potential energy. However, the energy of a system comprises both potential as well as kinetic energy, and stability should depend on the total energy, not just the potential energy. How does kinetic energy come into play here? Is it being neglected for some reason, or is this description of a diatomic molecule insufficient?

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    $\begingroup$ This is the potential energy plot; kinetic energy has to be accounted for in the usual manner, i.e., as given by the kinetic energy term in the {Hamiltonian,Lagrangian}. $\endgroup$ – Todd Minehardt May 1 at 19:03
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The curves labelled Harmonic/Morse show how the potential energy changes with internuclear separation, and the horizontal lines are the allowed total energies, potential plus kinetic. The potential energy is a maximum at the turning points (end of the horizontal lines in your picture) and zero at the $r_e$ value but because the total energy is constant the kinetic energy is the opposite and zero at the turning points and maximal at $r_e$. Its shape is just the inverse of that of the potential's at the energy for each quantum number.

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  • $\begingroup$ For a QM system (which the OPs diagram presumably describes) there are no discrete turning points, that is, the wavefunction "leaks" into classically forbidden regions of the potential. $\endgroup$ – Buck Thorn May 3 at 9:33
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    $\begingroup$ @Buck Thorn yes of course thats what you get from the wavefunction via expectation values but for the OP I have simplified things a little. However, the wavefunctions are for a time independent model you really need to use wavepackets to get any actual motion (bond vibration) but that is a whole other story. $\endgroup$ – porphyrin May 3 at 14:01
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The kinetic energy is not ignored. Each quantized state (labeled by the vibrational quantum number) has a total energy given by the expectation value of the Hamiltonian operator for the wavefunction. The contribution of the kinetic energy can be computed as the expectation value of the QM kinetic energy operator, or as the difference between the expectation values of the total energy and potential energy.

Because the math is simpler I will limit the discussion to the harmonic oscillator (ie quadratic potential). For the QM harmonic oscillator the expectation value of the potential energy turns out to be $$\left< E_\text{pot} \right> =\frac12 \left(v+\frac12\right)\hbar \omega=\frac12 E_\text{tot}$$ The expectation value of the kinetic energy is then $$\left< E_\text{KE}\right>=E_\text{tot}-\left< E_\text{pot} \right>= \left< E_\text{pot} \right>$$

So it turns out that the kinetic energy and potential values have identical expectation values, consistent with the prediction of the virial theorem for a quadratic potential. This is all outlined very nicely for instance in Atkins' physical chemistry textbook.

It is difficult to tell what energy levels are displayed on the diagram but they might very well correspond to the total energy given by $ E_\text{tot} = \left(v+\frac12\right)\hbar \omega$.

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  • $\begingroup$ As $\langle T\rangle =-E-RdE/dR$ and $\langle V\rangle =2E+RdE/dR$ the sum is still the total energy but kinetic and potential energy have opposite signs with harmonic potential as $E=kR^2/2$ $\endgroup$ – porphyrin May 3 at 14:16
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    $\begingroup$ @porphyrin For the harmonic oscillator $\left<V\right>=\left<T\right>$. The relation between kinetic and potential expectation values is straight from the virial theorem. See eg phys.libretexts.org/Bookshelves/Classical_Mechanics/… The link is to a classical treatment but the result holds for the QM case. $\endgroup$ – Buck Thorn May 3 at 14:47
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    $\begingroup$ Yes I am wrong, I picked up the equation for the electronic energy of a diatomic not the vibrational energy. $\endgroup$ – porphyrin May 3 at 15:23

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