Anharmonicity refers to the deviation of a system’s vibrational motion from the harmonic oscillator model, resulting in changes to the energy levels.

According to the harmonic oscillator model, the potential energy curve of a diatomic molecule is a quadratic function and there are infinite vibrational energy levels that are equally spaced. In reality, the potential energy curve resembles the one derived using the Born-Oppenheimer approximation (see diagram below). Furthermore, the energy levels are finite and the spacing between adjacent levels decreases as increases.

To understand how anharmonicity results in the converging energy levels, we refer to eq67, which was derived using the Born-Oppenheimer approximation by neglecting some terms in eq66. The inclusion of these terms via the perturbation theory will show the changes in the energy levels.

Consider the following:

Description |
Formula | Reference equation |

70 |
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71 |
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1^{st}-order perturbation energy |
72 |

where are quantum numbers and .

Substituting eq70 in eq72 and noting that is real,

For a set of orthonormal spherical harmonic functions, . Furthermore, and so,

If we assume that the probability of a nucleus of a vibrating molecule undergoing a displacement of more than is very small, then for , and we can express the lower limit of the integral in eq73 as without any major error:

As mentioned in this article, is either an even function or an odd funcion, which makes an even function. Therefore, the integrand is an odd function, whose integral over all space is zero. Eq74 becomes

###### Question

, , and are linear operators with matrix representations , and respectively. The matrix elements of , and are , and respectively, where is a complete orthonormal basis set. Show that if .

###### Answer

Since is a complete orthonormal basis set, , where is the identity matrix. So,

which is the definition of the matrix multiplication of .

Consider the integral . As , we have

Replacing in eq32a with and substituting the resultant equation in eq76

All the terms in the first summation vanish except for when and . Using , the 1^{st} summation becomes . Applying the same logic to the remaining summations, eq77 becomes

Let’s now consider the integral . As , we have

Substituting eq78 in eq79 and employing the same logic as before, we have

If , the above equation becomes

Substitute eq80 in eq75,

With reference to eq74, the term disappeared for the computation of the first-order energy correction . So, we need to account for it by calculating the second-order energy correction . From eq270a,

Substituting eq50 and in eq82

As , we have and

Substituting eq32a (where ) and eq78 in eq83, and using the same logic in deriving eq78,

where

Since and

Substituting the above equation in eq83 and expanding the summation gives

With reference to eq266, . Since is negative, becomes smaller as increases.