In our course on physical chemistry, which involves MOT, we have been taught that in the LCAO approach, the wave function for a molecule … say hydrogen ion ($\ce{H2+}$), can be approximated by a linear combination of two atomic orbitals.

From what I could gather from the very little I know about variational method is that this linear combination is going to be then optimised to give the minimum energy to find the ground state of the molecule in question. Is this correct?

Also, if we are simply using the linear combination as a test/guess function for a variational method, why should we restrict ourselves to only (say $+$ and $-$) two combinations? That is, how does the rule "Number of molecular orbitals = Number of atomic orbitals constituting them" arise?

Also, why is there the restriction in form of the rule that "Only atomic orbitals of close energies and proper symmetries combine to give MOs"?

  • $\begingroup$ I think the answer is great. The short answer to "proper symmetries" is that orbitals of different symmetry are orthogonal, so they can't combine to give MOs. $\endgroup$ Nov 10, 2014 at 19:43
  • $\begingroup$ By saying that we restrict to only $+$ and $-$ combinations, are you considering only the specific case of $H_{2}^{+}$? If so, the reason that we always have a symmetric or antisymmetric combination of the two $s$ orbitals is the mirror symmetry with respect to the plane bisecting the molecule. $\endgroup$
    – higgsss
    Dec 16, 2015 at 2:12
  • $\begingroup$ OP, I'd like to suggest you take a look at the first 8-9 lectures of Chemistry 107 (UCI Open Chem). It might prove helpful. $\endgroup$ Nov 22, 2019 at 15:46
  • $\begingroup$ Does LCAO form only sigma bond not pi bond? $\endgroup$
    – Apurvium
    Jul 23, 2021 at 9:45

1 Answer 1


You are right to be questioning the validity of this method, and I congratulate you for doing so. This is actually an extremely important skill, that differentiates the best students from the rest.

The variational method, like many methods from physical chemistry, is a method of approximation (a model) to what really happens. There exist a whole hierarchy of methods for computing the molecular orbitals of molecules (which are themselves models, being the stationary states of the Schrödinger equation), including at the top end post-Hartree-Fock theory and density functional theory. These methods provide quantitative information about molecular orbitals, but also require serious computer power (which is generally unavailable to undergraduate students). However, all of the basic physics can be explored and understood with simpler models that can be solved on a few sheets of paper, such as LCAO theory, hence why we teach them to undergraduates.

As regards to your question, we do indeed optimise a linear combination. We start off by assuming that the total molecular orbital wavefunction can be approximated using a linear set of atomic orbitals:

$$|\Psi\rangle = c_1|\phi_1\rangle + c_2|\phi_2\rangle + \cdots + c_n|\phi_n\rangle$$

We then need to find the coefficients. The variational principle (also known as the Rayleigh-Ritz method) states that the coefficients that give the best approximation to the wavefunction will minimize the energy, given by

$$\mathcal{E} = \frac{\langle\Psi|\hat{H}|\Psi\rangle}{\langle\Psi|\Psi\rangle}$$

Now, the second part of your question involves how to compute the two terms in this fraction. Without going into massive mathematical detail, the LCAO method can be recast into a matrix problem rather than an integral problem using the atomic orbitals as a form of basis vector. In the simplest case (Hückel theory) we assume that they are normalized, such that the denominator is always 1 or 0 (see later). The problem now is mostly how to determine the numerator.

In brief, each element of the Hamiltonian matrix is given by

$$H_{ij} = \langle\phi_i|\hat{H}|\phi_j\rangle$$

The variational principle applied to this matrix implies that the optimized energies are the eigenvalues of the Hamiltonian matrix, and the coefficients are given by the eigenvectors. Since an $n$ x $n$ matrix has precisely $n$ eigenvalues, this implies the "number of molecular orbitals" rule. Notice that the reason for the +/- combinations is by our own design (LCAO). We could have picked more complex trial functions by assumption, but this would make computation far more difficult.

It most certainly is true that in order for there to be a significant interaction, two orbitals must be close in energy. The detailed reasons are complex, but essentially it comes down to the size of the $\langle\phi_i|\hat{H}|\phi_j\rangle$. Orbitals that are far apart in energy have small values of this term hence interact weakly. This doesn't necessarily mean they can't form molecular orbitals, however, but these effects are negligible and are not important in understanding the chemistry, so are neglected in simple models (although are often included in some of the most complex modern methods).

The reason that only atomic orbitals of the same symmetry give molecular orbitals is the overlap integral $\langle\phi|\phi\rangle$. This is the total sum (integral) of the product of the AOs. This is zero for different symmetries. This must be non-zero to give an MO. Consider a $\mathrm{2p_z}$-orbital and a $\mathrm{1s}$-orbital for instance (for illustration purposes only!). The s-orbital is spherically symmetric, with all points the same sign. The $\mathrm{p_z}$-orbital has a "dumbbell" shape, with equal areas of different signs above and below the $xy$-plane. Therefore, the total sum of the product is zero. No overlap = no interaction.

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    $\begingroup$ Thank you very much for your answer! What I don't get is how does variational principle produce the energy of the higher states? After you answered, I searched for variation principle and the Hamiltonian matrix and I came across linear variation method, which minimises the energy wrt each coefficient and from condition for non trivial solutions we get the values of energy. How do we know that these correspond to the higher states of the molecule? $\endgroup$
    – transistor
    Nov 11, 2014 at 3:57
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    $\begingroup$ Whilst it is true that the variational principle is only rigorously true for the ground state, one can in principle use this to extract higher states based upon the orthogonality of the wavefunctions: a consequence of the Hermiticity of the Hamiltonian operator. This is possible if you know a quantum number (such as energy) that differentiates the states. For some trial wavefunctions, this involves complex projection operations, however due to the way the LCAO can be formulated using Hermitian matrices, we can get a bunch of orthogonal higher states "for free" $\endgroup$
    – DrHarps
    Nov 11, 2014 at 17:13
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    $\begingroup$ @DrHarps the variational principle is perfectly valid for excited states. The issue is that there is a difference between molecular orbitals $\psi$ and the state that they describe $\Psi$. This discussion has focused on the solution of one $\Psi$ (the ground state), but it is possible to solve for higher states $\Psi$ using techniques such as multiconfigurational HF theory. A somewhat crude description is that an excited state is the same MO set as the ground state, but with different occupations, orbital coefficients, and CI coefficients, cf. the State-Average CASSCF method. $\endgroup$
    – Eric Brown
    Nov 13, 2014 at 4:21

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