Anharmonicity

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 $n$ 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
Zeroth-order wavefunction	$\psi^{(0)}_{nJM}=\frac{S_n(x)}{R}Y_J^M(\theta,\phi)$	70
Zeroth-order energy	$E_n^{(0)}=\biggr$n+\frac{1}{2}\biggr$\hbar\omega$	50
1^st-order perturbation Hamiltonian	$\hat{H}^{(1)}=\frac{1}{6}U'''(R_e)x^3+\frac{1}{24}U^{iv}(R_e)x^4$	71
1^st-order perturbation energy	$E_n^{(1)}=\langle\psi_{nJM}^{(0)}\vert\hat{H}^{(1)}\vert\psi_{nJM}^{(0)}\rangle$	72

where $n,K,M$ are quantum numbers and $x=R-R_e$ .

Substituting eq70 in eq72 and noting that $S(x)$ is real,

$E_n^{(1)}=\int_0^{\infty}\frac{S_n(x)^2}{R^2}\biggr\[\frac{1}{6}U'''(R_e)x^3+\frac{1}{24}U^{iv}(R_e)x^4\biggr\]R^2dR\int_0^{2\pi}\int_0^{\pi}[Y_J^M(\theta,\phi)]^2sin\theta d\theta d\phi$

For a set of orthonormal spherical harmonic functions, ${\int_0^{2\pi}\int_0^{\pi}[Y_J^M(\theta,\phi)]^2sin\theta d\theta d\phi=1}$ . Furthermore, ${\frac{dR}{dx}=\frac{d}{dx}(x+R_e)=1\;\;\Rightarrow\;\;dR=dx}$ and so,

$E_n^{(1)}=\int_{-R_e}^{\infty}S_n(x)^2\biggr\[\frac{1}{6}U'''(R_e)x^3+\frac{1}{24}U^{iv}(R_e)x^4\biggr\]dx\;\;\;\;\;\;\;\;73$

If we assume that the probability of a nucleus of a vibrating molecule undergoing a displacement of more than $\vert R_e\vert$ is very small, then $S(x)\approx 0$ for $x < -R_e$ , and we can express the lower limit of the integral in eq73 as $-\infty$ without any major error:

$E_n^{(1)}=\frac{U'''(R_e)}{6}\int_{-\infty}^{\infty}S_n(x)^2x^3dx+\frac{U^{iv}(R_e)}{24}\int_{-\infty}^{\infty}S_n(x)^2x^4dx\;\;\;\;\;\;\;\;74$

As mentioned in this article, $S(x)$ is either an even function or an odd funcion, which makes $S(x)^2$ an even function. Therefore, the integrand $S(x)^2x^3$ is an odd function, whose integral over all space is zero. Eq74 becomes

$E_n^{(1)}=\frac{U^{iv}(R_e)}{24}\langle S_n\vert x^4\vert S_n\rangle\;\;\;\;\;\;\;\;75$

Question

$R$ , $S$ , and $T$ are linear operators with matrix representations $\boldsymbol{\mathit{R}}$ , $\boldsymbol{\mathit{S}}$ and $\boldsymbol{\mathit{T}}$ respectively. The matrix elements of $\boldsymbol{\mathit{R}}$ , $\boldsymbol{\mathit{S}}$ and $\boldsymbol{\mathit{T}}$ are $R_{mn}=\langle f_m\vert R\vert f_n\rangle$ , $S_{mn}=\langle f_m\vert S\vert f_n\rangle$ and $T_{mn}=\langle f_m\vert T\vert f_n\rangle$ respectively, where $\{f_i\}$ is a complete orthonormal basis set. Show that $\boldsymbol{\mathit{R}}=\boldsymbol{\mathit{S}}\boldsymbol{\mathit{T}}$ if $R=ST$ .

Answer

Since $\{f_i\}$ is a complete orthonormal basis set, ${\sum_i\vert f_i\rangle\langle f_i\vert=I}$ , where $I$ is the identity matrix. So,

$R_{mn}=\langle f_m\vert R\vert f_n\rangle=\langle f_m\vert ST\vert f_n\rangle=\langle f_m\vert S\sum_i\vert f_i\rangle\langle f_i\vert T\vert f_n\rangle=\\\sum_i\langle f_m\vert S\vert f_i\rangle\langle f_i\vert T\vert f_n\rangle=\sum_iS_{mi}T_{in}$

which is the definition of the matrix multiplication of $\boldsymbol{\mathit{R}}=\boldsymbol{\mathit{S}}\boldsymbol{\mathit{T}}$ .

Consider the integral $\langle S_{n'}\vert x^2\vert S_n\rangle$ . As $x^2=xx$ , we have

$\langle S_{n'}\vert x^2\vert S_n\rangle=\sum_i\langle S_{n'}\vert x\vert S_i\rangle\langle S_{i}\vert x\vert S_n\rangle\;\;\;\;\;\;\;\;76$

Replacing $m$ in eq32a with $\mu$ and substituting the resultant equation in eq76

$\langle S_{n'}\vert x^2\vert S_n\rangle=\frac{\hbar}{2\mu\omega}\biggr\[\sum_i\sqrt{(i+1)(n+1)}\delta_{n',i+1}\delta_{i,n+1}+\sum_i\sqrt{n(i+1)}\delta_{n',i+1}\delta_{i,n-1}+\\\sum_i\sqrt{i(n+1)}\delta_{n',i-1}\delta_{i,n+1}+\sum_i\sqrt{in}\delta_{n',i-1}\delta_{i,n-1}\biggr\]\;\;\;\;\;\;\;\;77$

All the terms in the first summation vanish except for when $i=n'-1$ and $i=n+1$ . Using $i=n+1$ , the 1^st summation becomes $\sqrt{(n+2)(n+1)}\delta_{n',n+2}$ . Applying the same logic to the remaining summations, eq77 becomes

$\langle S_{n'}\vert x^2\vert S_n\rangle=\frac{\hbar}{2\mu\omega}\biggr\[\sqrt{(n+2)(n+1)}\delta_{n',n+2}+(2n+1)\delta_{n',n}+\sqrt{n(n-1)}\delta_{n',n-2}\biggr\]\;\;\;\;\;\;\;\;78$

Let’s now consider the integral $\langle S_{n'}\vert x^4\vert S_n\rangle$ . As $x^4=x^2x^2$ , we have

$\langle S_{n'}\vert x^4\vert S_n\rangle=\sum_i\langle S_{n'}\vert x^2\vert S_i\rangle\langle S_{i}\vert x^2\vert S_n\rangle\;\;\;\;\;\;\;\;79$

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

$\langle S_{n'}\vert x^4\vert S_n\rangle=\biggr$\frac{\hbar}{2\mu\omega}\biggr$^2\biggr\[\sqrt{(n+4)(n+3)(n+2)(n+1)}\delta_{n',n+4}+\\(2n+1)\sqrt{(n+2)(n+1)}\delta_{n',n+2}+n(n-1)\delta_{n',n}+\\(2n+5)\sqrt{(n+2)(n+1)}\delta_{n',n+2}+(2n+1)^2\delta_{n',n}+\\(2n-3)\sqrt{n(n-1)}\delta_{n',n-2}+(n+2)(n+1)\delta_{n',n}+\\(2n+1)\sqrt{n(n-1)}\delta_{n',n-2}+\sqrt{n(n-1)(n-2)(n-3)}\delta_{n',n-4}\biggr\]$

If $n'=n$ , the above equation becomes

$\langle S_{n}\vert x^4\vert S_n\rangle=\biggr$\frac{\hbar}{2\mu\omega}\biggr$^2(6n^2+6n+3)\;\;\;\;\;\;\;\;80$

Substitute eq80 in eq75,

$E_n^{(1)}=\frac{U^{iv}(R_e)}{8}\biggr$\frac{\hbar}{2\mu\omega}\biggr$^2(2n^2+2n+1)\;\;\;\;\;\;\;\;81$

With reference to eq74, the $x^3$ term disappeared for the computation of the first-order energy correction $E_n^{(1)}$ . So, we need to account for it by calculating the second-order energy correction $E_n^{(2)}$ . From eq270a,

$E_n^{(2)}=\sum_{m\neq n}\frac{\vert\langle S_m^{(0)}\vert\hat{H}^{(1)}\vert S_n^{(0)}\rangle\vert^2}{E_n^{(0)}-E_m^{(0)}}\;\;\;\;\;\;\;\;82$

Substituting eq50 and $H^{(1)}=\frac{1}{6}U'''(R_e)x^3$ in eq82

$E_n^{(2)}=\frac{U'''(R_e)}{6}\sum_{m\neq n}\frac{\vert\langle S_m^{(0)}\vert x^3\vert S_n^{(0)}\rangle\vert^2}{(n-m)\hbar\omega}$

As $x^3=x^2x$ , we have $\langle S_{m}^{(0)}\vert x^3\vert S_n^{(0)}\rangle=\sum_i\langle S_{m}\vert x^2\vert S_i\rangle\langle S_{i}\vert x\vert S_n\rangle$ and

$E_n^{(2)}=\frac{U'''(R_e)}{6}\sum_{m\neq n}\frac{\vert\sum_i\langle S_{m}\vert x^2\vert S_i\rangle\langle S_{i}\vert x\vert S_n\rangle\vert^2}{(n-m)\hbar\omega}\;\;\;\;\;\;\;\;83$

Substituting eq32a (where $m=\mu$ ) and eq78 in eq83, and using the same logic in deriving eq78,

$E_n^{(2)}=\frac{U'''(R_e)}{6}\sum_{m\neq n}\frac{\vert A\vert^2}{(n-m)\hbar\omega}\;\;\;\;\;\;\;\;83$

where

$A=\frac{\hbar}{2\mu\omega}\sqrt{\frac{(n+1)(n+2)(n+3)\hbar}{2\mu\omega}}\delta_{m,n+3}+3\biggr\[\frac{(n+1)\hbar}{2\mu\omega}\biggr\]^{\frac{3}{2}}\delta_{m,n+1}+3\biggr$\frac{n\hbar}{2\mu\omega}\biggr$^{\frac{3}{2}}\delta_{m,n-1}+\frac{\hbar}{2\mu\omega}\sqrt{\frac{n(n-1)(n-2)\hbar}{2\mu\omega}}\delta_{m,n-3}$

Since $\delta_{x,y}\delta_{x,y}=\delta_{x,y}$ and $\delta_{x,y}\delta_{x,z}=0$

$\vert A\vert^2=(n+1)(n+2)(n+3)\biggr$\frac{\hbar}{2\mu\omega}\biggr$^3\delta_{m,n+3}+9\biggr\[\frac{(n+1)\hbar}{2\mu\omega}\biggr\]^3\delta_{m,n+1}+9\biggr$\frac{n\hbar}{2\mu\omega}\biggr$^3\delta_{m,n-1}+n(n-1)(n-2)\biggr$\frac{\hbar}{2\mu\omega}\biggr$^3\delta_{m,n-3}$

Substituting the above equation in eq83 and expanding the summation gives

$E_n^{(2)}=-\frac{5\hbar^2U'''(R_e)}{8\mu^3\omega^3}\biggr$n^2+n+\frac{11}{30}\biggr$\;\;\;\;\;\;\;\;84$

With reference to eq266, $E_n\approx E_n^{(0)}+\lambda E_n^{(1)}+\lambda^2E_n^{(2)}$ . Since $E_n^{(2)}$ is negative, $\Delta E_n$ becomes smaller as $n$ increases.

Question

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