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Depending on its strength, the attractive double dirac delta potential shown below

enter image description here

can support two bound states. They are called the bonding and the antibonding orbitals as shown in the figure below

enter image description here

It turns out that the antibonding orbital has higher energy than the bonding orbitals.

But from the figure above, it is cleat that the electronic charge density piles up more in between in case of bonding orbital and less in case of antibonding orbital. Does this not mean that the Coulomb repulsion between electrons be more in case of bonding orbital configuration than the antibonding case. If so, doesn't that mean the energy of the bonding orbital should be more than the antibonding orbital?

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    $\begingroup$ I seem to remember this being described differently. One would think that the anti-boding state has lower energy because the electron cloud is more localized at the positive potential. The potential energy in question is between the electron and the "positive charges". The fact that it's energy is actually higher is explained by an increase in kinetic energy which is associated with the slope of the wavefunction. But this may be a distorted version of an explanation I heard many years ago. $\endgroup$ – garyp Feb 15 at 12:33
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A single electron does not experience Coulomb repulsion with itself. The potential and wavefunctions you describe involve only one-electron, so the state energy cannot be associated (even as an approximation) with the Coulomb repulsion between two electrons.

On the other hand, the double delta potential can be seen as a very crude approximation of the Coulomb potential of two positive nuclei. The fact that for the bonding state, the electronic charge density piles up more in between merely reflects that the electron is more likely to be found between the two nuclei, justifying the name "bondig". The opposite holds for the anti-bonding state.

For the sake of argument, let us add another electron to your model and ask what is the ground state of the two-electron systems? Without explicitly stating other assumptions, you model again does not imply any interaction between the electrons beyond Pauli exclusion principle. If these electrons were of opposite spins, they would both occupy the lower-energy bonding state. If they were of the same spin, one would be in the bonding state and the other in the anti-bonding one.

Finally, let us ask if we have two electrons of opposite spins in the bonding state and we then "turn on" the Coulomb repulsion between the two, what would happen? It depends on whether the Coulomb energy between two bonding states is larger than the energy increase resulting from moving one or both electrons to the anti-bonding state. If it is not, the electrons stay in the bonding state and you get a covalent bond.

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  • $\begingroup$ Thank you for clearing my doubt. But for the single electron, what is the reason that the bonding orbital has lower energy? Why does the charge density piling up in between leads to lower energy? $\endgroup$ – mithusengupta123 Feb 14 at 3:44
  • $\begingroup$ Pieter has provided a good reason but would work if we had two delta "barriers" (instead of two delta "wells") in which case there exists potential minimum in between. $\endgroup$ – mithusengupta123 Feb 14 at 4:10
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    $\begingroup$ The fact that the bonding state has lower energy than the anti-bonding one has more to do with the kinetic energy than with the potential. A symmetric solution (i.e. the bonding state) will always have a lower kinetic energy than a similar but anti-symmetric one. One intuitive argument is using Heisenberg uncertainty principle: the anti-symmetric wavefunction has lower uncertainty about the position of the electron than the symmetric one, so it must have a larger momentum variance. Therefore, the kinetic energy (which is proportional to momentum squared) must also be larger. $\endgroup$ – Quantum-Collapse Feb 14 at 11:27
  • $\begingroup$ That is also true. One can approximate molecular orbitals in diatomics as those of particles in a box that is about three times as long as the internuclear distance. $\endgroup$ – user137289 Feb 14 at 11:32
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The $\delta$ potentials may be the main reason why this is a problem. In reality, the nuclei have $1/r$ potentials. The potential for a single electron is lower between the nuclei than on either side.

When the electron wave function has a node between the nuclei, the negative charge density is low. The positive nuclei also feel each other's $1/r$ potential and will then repel each other.

The equipotential surface can then be dissociative.

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  • $\begingroup$ Thanks. I can understand that the potential in the region between two delta "barriers" will small leading to charge density concentration in between, but here, I have two delta "wells". I think that the potential in the region between two delta "wells" is larger. $\endgroup$ – mithusengupta123 Feb 14 at 4:01
  • $\begingroup$ @mithusengupta123 The delta potentials are used in the Kronig-Penney model for the band structure of crystals. It makes it easy to calculate the band gap between the two locations of the standing waves with the same wavevector. For a molecule, I don't think the delta function potentials are helpful for intuition. $\endgroup$ – user137289 Feb 14 at 8:12
  • $\begingroup$ If two consecutive delta wells pose a problem, think of two consecutive wells of finite width and depth separated by a finite barrier. In that case too, the ground state is symmetric i.e. unlike to the antisymmetric wavefunction, the symmetric electron wavefunction is concentrated in a region where the barrier exists. This does not seem to me to be a lower potential energy configuration. $\endgroup$ – mithusengupta123 Feb 19 at 19:26
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As @Quantum-Collapse has noted in their answer, there is some confusion here between bonding and anti-bonding states which refer to the one electron problem, and the singlet and triplet states for two electrons. The former problem arises, e.g., when discussing the covalent bonds, in which case the bonding/symmetric state has lower energy, whereas the latter arises, e.g., when discussing the Helium atom where the synglet state, whose orbital wave function is symmetric in respect to exchanging the electrions, is the ground state. Note that teh symmetries are different here. The two problems get mixed when discussing the covalent bond in the hydrogen molecule.

To get closer to the question: the node theorem (see, e.g., here and here, but also in many textbooks) states that the ground state of a one-dimensional Schrödinger equation does not have nodes, whereas the n-th state has $n-1$ nodes. (Note that this is strictly true only in one dimension). The physical explanation is that nodes correspond to zeros of the probability density, i.e., the electron is confined to smaller regions in space, which means lower uncertainty in position and higehr uncertainty in momentum, hence the higher energy. This is admittedly a very hand-waving explanation, but the mathematical result is rigorous.

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    $\begingroup$ The theorem is commonly called Courant domain nodal theorem, and its proof (and a discussion of multidimensional case) can be seen in Courant, Hilbert "Methods of Mathematical Physics Vol. 1". $\endgroup$ – Ruslan Feb 15 at 11:36
  • $\begingroup$ Dear @Vadim, an increase in the number of nodes does not mean lower uncertainty in $x$. Think about harmonic oscillator. $\Delta x$ increases with increase in nodes. $\endgroup$ – mithusengupta123 Feb 19 at 19:30
  • $\begingroup$ @mathusengupta123 I stated that this is a hand-waving explanation - the electron is confined to smaller regions, in each of them the uncertainty is smaller. If you are familiar with the analysis of variance (ANOVA), we can talk about decomposing variance into intranode and internode components. But this would be taking this interpretation a bit too far. $\endgroup$ – Vadim Feb 19 at 20:16
  • $\begingroup$ @Vadim What do you think about Pieter's answer and my2cats's answers which offers explanations based on minimization of the potential energy and my comments to their answers? I don't think that that works here. $\endgroup$ – mithusengupta123 Feb 20 at 8:31
  • $\begingroup$ @mithusengupta123 Pieter's answer is completely off the point. The one by my2cts is not very different than mine - it is less general, but tries to work through specific examples... except that beyond 1D this is not generally true. Also, the true solution is not really a superposition. Note that the node theorem can also be proved from variational standpoint. $\endgroup$ – Vadim Feb 20 at 13:31
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There is a confusion between single electron and multi-electron states here.

First consider the single electron states. The bonding orbital has the lowest energy because it allows the electron to have the lowest potential energy. This is true as long as the bound orbital of V1, $\psi_1$, overlaps with V2 and vice versa. For the bonding orbital $$\Psi_B=\frac{\psi_1+\psi_2}{\sqrt{2+2S}}$$ the two orbitals constructively interfere at the potential positions, which gives it the lowest potential energy. For the antibonding orbital $$\Psi_A = \frac{\psi_1-\psi_2}{\sqrt{2-2S}}$$ the two orbitals interfere destructively. Here $S=\langle \psi_1 | \psi_2 \rangle$. The electron self interacts but this energy is independent of the orbital in the nonrelativistic limit and is considered absorbed in its rest energy.

For two electron states one can construct the bound state $$\Psi_0=|\Psi_B(1) \overline\Psi_B(2)|~,$$ where || indicates a Slater determinant. This state has the highest electron-electron repulsion but this is more than compensated by the lowest potential energy, which makes it the ground state. Note that $\Psi_0$ will have some admixture of $|\Psi_A(1) \overline\Psi_A(2)|$. The triplet $$\Psi_1 =\frac{|\Psi_B(1) \Psi_A(2)|}{\sqrt{1-S}}$$ is the first excited state, followed by the corresponding singlet $$\Psi_2 =\frac{|\Psi_B(1) \overline\Psi_A(2) - \overline \Psi_B(1) \Psi_A(2)|}{\sqrt{2+2S}} ~.$$

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  • $\begingroup$ Okay. Consider the single electron case. For the bonding case, in my example, the concentration of electron charge density more in the region in between two wells where the potential energy is larger compared to the values inside the well. How does this configuration reduce potential energy? $\endgroup$ – mithusengupta123 Feb 19 at 18:59
  • $\begingroup$ @mithusengupta123 As I said in my 2nd paragraph: 'the two orbitals constructively interfere at the potential positions'. $\endgroup$ – my2cts Feb 19 at 19:10
  • $\begingroup$ You said, "The bonding orbital has the lowest energy because it allows the electron to have the lowest potential energy." But in the bonding case, the electron wavefunction is dominantly concentrated in a region where the potential energy is larger. $\endgroup$ – mithusengupta123 Feb 19 at 19:14

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