The Diatomic Molecule: Exercise 80 discusses the idea of reduced mass. Classically or quantum mechanically, we can digest the behavior of a two-particle system into motion of the center of mass and motion relative to the center of mass. Our interest here is the relative motion, which becomes a one-particle problem if we merely use $\mathbf{\mu }$for the mass for that particle. Given this simplification, the quantum-mechanical results we have learned go a long way toward describing the diatomic molecule. To a good approximation, the force between the bound atoms is like an ideal spring whose potential energy is $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{kx}}^{\mathbf{2}}$, where x is the deviation of the atomic separation from its equilibrium value, which we designate with an a. Thus,x=r-a . Because the force is always along the line connecting the two atoms, it is a central force, so the angular parts of the Schrödinger equation are exactly as for hydrogen, (a) In the remaining radial equation (7- 30), insert the potential energy $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{kx}}^{\mathbf{2}}$and replace the electron massm with $\mathbf{\mu }$. Then, with the definition.$\mathbf{f}\left(r\right)\mathbf{=}\mathbf{rR}\left(r\right)$, show that it can be rewritten as

$\frac{\mathbf{-}{\mathbf{ħ}}^{\mathbf{2}}}{\mathbf{2}\mathbf{\mu }}\frac{{\mathbf{d}}^{\mathbf{2}}}{\mathbf{d}{\mathbf{r}}^{\mathbf{2}}}\mathit{f}\left(r\right)\mathbf{+}\frac{{\mathbf{ħ}}^{\mathbf{2}}\mathbf{I}\left(I+1\right)}{\mathbf{2}\mathbf{\mu }{\mathbf{r}}^{\mathbf{2}}}\mathit{f}\left(r\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}\mathit{f}\left(r\right)\mathbf{=}\mathit{E}\mathit{f}\left(r\right)$

With the further definition show that this becomes

$\frac{\mathbf{-}{\mathbf{ħ}}^{\mathbf{2}}}{\mathbf{2}\mathbf{\mu }}\frac{{\mathbf{d}}^{\mathbf{2}}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\frac{{\mathbf{ħ}}^{\mathbf{2}}\mathbf{I}\mathbf{(}\mathbf{I}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{2}\mathbf{\mu }\left(x+a\right)}\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{E}\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

(b) Assume, as is quite often the case, that the deviation of the atoms from their equilibrium separation is very small compared to that separation—that is,x<<a. Show that your result from part (a) can be rearranged into a rather familiar- form, from which it follows that
$\mathit{E}\mathbf{=}\left(n+\frac{1}{2}\right)\mathit{ħ}\sqrt{\frac{\mathbf{k}}{\mathbf{\mu }}}\mathbf{+}\frac{{\mathbf{ħ}}^{\mathbf{2}}\mathbf{I}\left(I+1\right)}{\mathbf{2}\mathbf{\mu }{\mathbf{a}}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\begin{array}{c}\mathbf{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\\ \mathbf{I}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\end{array}$

(c)

Identify what each of the two terms represents physically.

The potential energy between the bound atoms,$U\left(r\right)=\frac{1}{2}k{x}^2$.

x is the deviation of the atomic separation from its equilibrium value,

The reduce mass is $\mu$.

The equation 7.30 is given by,

$-\frac{{ħ}^2}{2m}\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)R\left(r\right)+\frac{{ħ}^{2}I\left(I+1\right)}{2m{r}^2}R\left(r\right)+U\left(r\right)R\left(r\right)-ER\left(r\right)=0$

(a)

Consider the equation 7.30 as,

$-\frac{{ħ}^2}{2m}\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)R\left(r\right)+\frac{{ħ}^{2}I\left(I+1\right)}{2m{r}^2}R\left(r\right)+U\left(r\right)R\left(r\right)-ER\left(r\right)=0$

Substitute$\frac{1}{2}k{x}^2$for U(r) and$\mu$ form into above equation.

$-\frac{{\hslash }^2}{2\mu }\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)R\left(r\right)+\frac{{\hslash }^{2}l(l+1)}{2\mu r^2}R\left(r\right)+\frac{1}{2}k{x}^{2}R\left(r\right)-ER\left(r\right)=0$ …… (1)

According to the definition,

$f\left(r\right)=rR\left(r\right)$

Differentiate the above equation with respect to r.

$\frac{d}{dr}f\left(r\right)=R\left(r\right)+r\frac{dR\left(r\right)}{dr}$

Differentiate the above equation with respect to r.

$\begin{array}{r}\frac{d^{2}f\left(r\right)}{d{r}^2}=\frac{dR\left(r\right)}{dr}+r\frac{d^{2}R\left(r\right)}{d{r}^2}+\frac{dR\left(r\right)}{dr}\\ \frac{d^{2}f\left(r\right)}{d{r}^2}=2\frac{dR\left(r\right)}{dr}+r\frac{d^{2}R\left(r\right)}{d{r}^2}\\ \frac{d^{2}f\left(r\right)}{d{r}^2}=\frac{1}{r}\left(2r\frac{dR\left(r\right)}{dr}+r^2\frac{d^{2}R\left(r\right)}{d{r}^2}\right)\end{array}$

The term$\frac{1}{r}\left(2r\frac{dR\left(r\right)}{dr}+r^2\frac{d^{2}R\left(r\right)}{d{r}^2}\right)$in the above equation can be written as $\frac{d}{dr}\left(r^2\frac{d^{2}R\left(r\right)}{d{r}^2}\right)$.

$$\begin{gathered} \frac{d^{2}f\left(r\right)}{d{r}^2}=\frac{d}{dr}\left(r^2\frac{d^{2}R\left(r\right)}{d{r}^2}\right) \\ \frac{d^{2}f\left(r\right)}{d{r}^2}=\frac{1}{r}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)R\left(r\right) \end{gathered}$$

Substitute$\frac{d^{2}f\left(r\right)}{d{r}^2}$for$\frac{1}{r}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)R\left(r\right)$into equation (1).

$-\frac{{\hslash }^2}{2\mu }\frac{1}{r}\frac{d^{2}f\left(r\right)}{d{r}^2}\frac{{\hslash }^{2}l(l+1)}{2\mu r^2}R\left(r\right)+\frac{1}{2}k{x}^{2}R\left(r\right)-ER\left(r\right)=0$

Substitute $f\left(r\right)/r$ for R(r) into above equation.

role="math" localid="1660026902814" $$\begin{gathered} -\frac{{ħ}^2}{2\mu }\frac{1}{r}\frac{d^{2}f\left(r\right)}{d{r}^2}+\frac{{ħ}^{2}I(I+1)}{2\mu r^2}\frac{f\left(r\right)}{r}+\frac{1}{2}k{x}^2\frac{f\left(r\right)}{r}-E\frac{f\left(r\right)}{r}=0 \\ -\frac{{ħ}^2}{2\mu }\frac{1}{r}\frac{d^{2}f\left(r\right)}{d{r}^2}+\frac{{ħ}^{2}I(I+1)}{2\mu r^2}\frac{f\left(r\right)}{r}+\frac{1}{2}k{x}^2\frac{f\left(r\right)}{r}=E\frac{f\left(r\right)}{r}=0 \\ -\frac{{ħ}^2}{2\mu }\frac{d^{2}f\left(r\right)}{d{r}^2}+\frac{{ħ}^{2}I(I+1)}{2\mu r^2}f\left(r\right)+\frac{1}{2}k{x}^{2}f\left(r\right)=Ef\left(r\right).......\left(2\right) \end{gathered}$$

Hence the equation$-\frac{{ħ}^2}{2\mu }\frac{d^{2}f\left(r\right)}{d{r}^2}+\frac{{ħ}^{2}I(I+1)}{2\mu r^2}f\left(r\right)+\frac{1}{2}k{x}^{2}f\left(r\right)=Ef\left(r\right)$is obtained for $f\left(r\right)=rR\left(r\right)$.

Now consider the coordinate transformation $x=r-a$.

$$\begin{gathered} x=r-a \\ r=x+a \end{gathered}$$

Substitute$x+a$ forr and$g\left(x\right)$ for$f\left(r\right)$ into equation (2).

$-\frac{{\hslash }^2}{2\mu }\frac{d^{2}g\left(x\right)}{d{x}^2}+\frac{{\hslash }^{2}l(l+1)}{2\mu (x+a{)}^2}g\left(x\right)+\frac{1}{2}k{x}^{2}g\left(x\right)=Eg\left(x\right)$ …… (3)

Hence the required equation$-\frac{{\hslash }^2}{2\mu }\frac{d^{2}g\left(x\right)}{d{x}^2}+\frac{{\hslash }^{2}l(l+1)}{2\mu (x+a{)}^2}g\left(x\right)+\frac{1}{2}k{x}^{2}g\left(x\right)=Eg\left(x\right)$ is obtained for $f\left(r\right)=g\left(x\right)$.

(b)

It is given that x<<a. Therefore,${\left(x+a\right)}^2\approx a^2$.

Substitute$a^2$for${\left(x+a\right)}^2$into equation (3).

$$\begin{gathered} -\frac{{ħ}^2}{2\mu }\frac{d^{2}g\left(x\right)}{d{x}^2}+\frac{{ħ}^{2}I(I+1)}{2\mu a^2}g\left(x\right)+\frac{1}{2}k{x}^{2}g\left(x\right)=Eg\left(x\right) \\ -\frac{{ħ}^2}{2\mu }\frac{d^{2}g\left(x\right)}{d{x}^2}+\frac{1}{2}k{x}^{2}g\left(x\right)=Eg\left(x\right)-\frac{{ħ}^{2}I(I+1)}{2\mu a^2}g\left(x\right) \\ -\frac{{ħ}^2}{2\mu }\frac{d^{2}g\left(x\right)}{d{x}^2}+\frac{1}{2}k{x}^{2}g\left(x\right)=\left[E-\frac{{ħ}^{2}I(I+1)}{2\mu a^2}\right]g\left(x\right) \end{gathered}$$ …… (4)

Compare the equation (5) with 5.25 of the textbook equation, which describe the simple harmonic motion with mass $\mu$, oscillation frequency ${\omega }_0=\sqrt{\frac{k}{\mu }}$and effective energy $E^{\text{'}}=E-\frac{{ħ}^{2}l\left(l+1\right)}{2\mu a^2}$.

According to the simple harmonic oscillation given in section 5.7 of the textbook, the equation (4) can be written as,

$E^{\text{'}}=\left(n+\frac{1}{2}\right)ħ{\omega }_0$

Substitute $\sqrt{\frac{k}{\mu }}$for ${\omega }_0$into above equation.

$E^{\text{'}}=\left(n+\frac{1}{2}\right)ħ\sqrt{\frac{k}{\mu }}$

Therefore, energy for the diatomic molecules is given by,

$E=(\mathrm{n}+\frac{1}{2})ħ\sqrt{\frac{\mathrm{k}}{\mathrm{\mu }}}+\frac{{\mathrm{ħ}}^2\mathrm{I}(\mathrm{I}+1)}{2{\mathrm{\mu a}}^2}\begin{array}{c}\mathrm{n}=0,1,2,...\\ \mathrm{I}=0,1,2,...\end{array}$

Hence, the required equation is obtained for x<<a.

(c)

Therefore, energy for the diatomic molecules is given by,

$E=(\mathrm{n}+\frac{1}{2})ħ\sqrt{\frac{\mathrm{k}}{\mathrm{\mu }}}+\frac{{\mathrm{ħ}}^2\mathrm{I}(\mathrm{I}+1)}{2{\mathrm{\mu a}}^2}\begin{array}{c}\mathrm{n}=0,1,2,...\\ \mathrm{I}=0,1,2,...\end{array}$

The above equation contains the two terms, first term represents the vibrational energy at the frequency ${\omega }_0$. This term is associated with the harmonic oscillation during the separation of the diatomic molecules. The second term represent the rotational energy with angular momentum$L=ħ\sqrt{I\left(I+1\right)}$ and moment of inertia $\mu a^2$.This energy is associated with the rotation of the mass about an axis of the center of the mass at the distance .