# Why do solutions of electron in a box (and in a ring) predict coefficients for LCAO (linear combination of atomic orbitals) in 1D systems?

The solutions to the 1D particle in a box quantum mechanic system are standing waves (zero at both ends of the box) with 0,1,2... nodes for increasing energy (zero for the ground state).

If I look at the LCAO solutions for a linear system, combining AOs of the same shape and energy (e.g. conjugated double bonds, linear array of sodium atoms), the coefficients follow the pattern of a standing wave with 0,1,2... nodes. For example, when I combine 3 AO's in a row, the solutions are +++, +0-, +-+, i.e. zero, one, and two switches of signs as I go to higher energy states. For the state with one node, the node is in the center of the box, corresponding to the middle AO, which has a coefficient of zero. (More examples, i.e. hexatriene and pentadienyl cation, are on slide 126 and 128 of this document.)

The solutions to the particle in a ring system are standing circular waves (same value at 0 and 360 degrees) with 0,2,4... nodes for increasing energy. The animation shows a standing circular wave with 8 nodes:

If I look at the LCAO solutions for a circular system, again combining AOs of the same shape and energy (e.g. p-orbitals in an aromatic system), the coefficents follow the pattern of a circular wave with 0,2,4... nodes. For benzene, the solutions are ++++++, +++--- and +0--0+, +-++-+ and -0+-0+, +-+-+- (see picture):

What underlies the connection between the unbound electron in a box or corral with the formation of covalent bonds of bound electrons?

• You mean why are they qualitatively similar, with more nodes corresponding to higher energy? – Buck Thorn Jun 14 at 14:15
• @BuckThorn Yes, but it also predicts the positions of the nodes, i. e. which AO will get a negative or positive coefficient or be zero. Maybe it also predicts the relative size of the coefficients, but I’m not sure about that. – Karsten Theis Jun 14 at 14:19
• When you use Hückel theory to solve for the π MOs you will get coefficients that have some sin/cos dependency, which reproduces the form of the standing wave. It's easy enough to show this for small systems but I think the maths to prove it more generally is non-trivial. Also, I've never heard of it as "particle in a corral" - usually "particle on a ring". – orthocresol Jun 14 at 15:35
• @orthocresol I changed it (if you google it outside of Texas, ring is more popular than corral). – Karsten Theis Jun 14 at 15:45
• The lack of symmetry of degenerate states such as $\psi_2$ makes me wonder to what extent these drawings are just "cartoons" drawn to illustrate or fit the expected trend of energy being proportional to the number of nodes. Or, more to the point, to what extent the solutions are unique, as opposed to chosen to represent what we would like to emphasize. AFAIK any linear combination of the two benzene $\psi_2$ MOs drawn here will have the same E ie be an eigenfunction of the problem as written. – Buck Thorn Jun 14 at 20:41

A short answer is that the functions that satisfy the Schrodinger equation for a system are largely determined by the potential energy part of the function, since that's the part of the Hamiltonian that varies with different scenarios. The particle-in-a-well (= 1D box) is defined by having infinite potential energy outside the walls and a constant finite energy within. That turns out to be a very good approximation of the potential energy function of a pi system or a line of identical metal atoms.

At the end of the line of atoms, an electron faces a steep potential energy curve. Not quite the vertical wall of the 1D box, but pretty close to it. The same is true on the sides, so the system can be treated as 1D. Within the system, the electron moves essentially freely (or exists in a broadly delocalized way) with very little change in potential across the system.

Since the potential functions are nearly identical, the solutions to the Schrodinger equation take the same mathematical form.

• "with very little change in potential across the system"... I agree with you up to this point. The position of the nodes in the delocalized-pi ring is determined by the position of the nuclei. The nuclei are what makes a comparison between particle-in-a-box and any real system non-trivial. I would not say that the potential functions are identical! – Buck Thorn Jun 15 at 7:19
• @BuckThorn - The potential functions are very different, but they are taken out of the equation (the AOs already take account of them). When making MO's from the AO's, all that happens is to delocalize the electrons without changing the electron density, so the Coulomb interactions are the same (= alpha in Hueckel theory). – Karsten Theis Jun 15 at 21:49

I want to add a slightly different but related perspective to the excellent answers already posted: Comparing the Hamiltonian of particle in a box and for Hückel theory. First though, I want to compare boundary conditions and potential (Coulomb) energy.

Boundary conditions

For the particle in a box, the boundary conditions are explicit: The particle has one degree of freedom (1D) and has to be inside the box. For LCAO, there are implicit boundary conditions: The electrons can only be where they would be in the atomic orbitals.

Potential energy

For the particle in a box, there are no Coulomb interaction within the box; this is a free electron in that region. For the Hückel theory in its simplest version (all p-orbitals, all carbon atoms) of LCAO, the Coulomb interaction is encoded in the $$\alpha$$ parameter (pre-computed, or just given as a conceptual parameter). The two scenarios are very similar: there is a region of low energy (inside the box or near the atoms) and a region of high energy (outside the box or away from the atoms). You can either describe this as "confining potential" (as in the answer by Buck Thorn), or you can describe this as absence of Coulomb term in the Hamiltonian because the boundary conditions (and the choice of basis functions for Hückel theory) take care of the "confining".

The Hamiltonian

For the particle in a box, the Hamiltonian is just about the kinetic energy (no electrostatic interactions), so it is simply the second derivative with some constants:

$$H = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2}$$

For Hückel theory, the Hamiltonian is not explicitly shown, but the result of $$⟨ϕ_i|H|ϕ_j⟩$$ are given as $$\alpha$$ for i=j, as $$\beta$$ if i and j are neighbors, and are neglected (set to zero) otherwise. The $$\alpha$$ corresponds to the Coulomb interaction, and its contribution is constant. With the atomic orbitals as basis functions, you might say the electrons are free otherwise (this amounts to saying that the Coulomb interaction is the same around each atom). In a way, the Coulomb interaction is taken out of the equation because all the AO's have the same energy, and the MO's are just linear combinations of them (so there will not be more electron density between atoms for the MO's compared to the AO's - still, the electrons are more delocalized and therefore have lower kinetic energy).

The $$\beta$$ term roughly corresponds to the kinetic energy. If two adjacent AOs are combined with equal signs of coefficients, the resulting wave function is smoother (has a smaller second derivative) than when the signs of coefficients are opposite. Another way of saying that is that if the coefficients "have a node", the kinetic energy is higher. So this models the kinetic energy part of the Hamiltonian.

The solutions

If the Hamiltonian (or the implicit Hamiltonian) are similar because the only relevant term for finding a solution is the kinetic one, then the resulting wave functions are expected to be similar. One difference, though, is that the particle in a box can have arbitrarily high energies (number of nodes) while the LCAO result has a finite number of energies constrained by the available basis functions.

• Can we address your last point by considering the limit as the number of atoms in the line or ring goes to infinity rather than just a small system like benzene? – Andrew Jun 16 at 13:20
• @Andrew I suppose so, for example for the sodium example. As you go from a couple of atoms to the bulk system, the discreet energy levels turn into bands, and the electrons are delocalized over the entire volume (not free, though, but more like hopping from nucleus to nucleus). – Karsten Theis Jun 16 at 14:11

Attractive Coulombic interactions impose a confining potential, and single electrons, being wave-like in character, will display standing-wave probability densities within such a potential. In a 1-D system, lower energy corresponds to probability density distributions with less nodes, because highly spatially-varying wavefunctions correspond to a higher electron kinetic energy. All of this is explained here in an excellent answer to another post. In the case of an aromatic ring, the location of the nodes can be inferred from the fact that the approximate MOs are constructed from AOs consisting of p-orbitals, which have density centered above the nuclei on the plane above the ring.

In summary,

• the wave character of electrons accounts for the "standing wave" appearance of the density
• the wavefunctions are sensitive to the potential, specifically the location of nuclei, but symmetrical distribution of nuclei (as in a ring or line) will result in nice symmetrical MOs that might make important details less obvious
• the 1D character of the potential accounts for the proportionality of number of nodes and energy

Finally, I should point out that the MO's constructed to generate the solution to the aromatic ring problem are obviously finite in number, as the AOs used to construct them are finite in number (#MOs=#p-orbitals). The same is true in a sense of the sodium row problem (#MOs=#AOs=#1s orbitals). This is absolutely not true for the particle in a box. There you don't need more AOs (or atoms) to construct a new wavefunction, and the position of the nodes is determined by the number of nodes and size of the box, not by the position of any nuclei. The analogy breaks down because you are not constrained by any AOs (or nuclei).

• I'm sorry, I'm still confused what point you are making about the linear combinations. No matter how the orbitals are recombined or rotated, I believe the result will always resemble a circular standing wave with two nodes (=single nodal plane). It is this resemblance that is the motivation for the question. – Andrew Jun 16 at 13:22
• @Andrew Upon further inspection it seems my "MO uniqueness" comment was vague at best so I dispensed with it. – Buck Thorn Jun 17 at 7:59