Chapter 18, Problem 18.33E

Interpretation Introduction

Interpretation:

The expected symmetry pairings of the nuclear and rotational wavefunctions is to be predicted.

Concept introduction:

The symmetry pairings of the nuclear and rotational wavefunctions is given as follows:

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym},J\text{even}\\ \text{degen:}\left(I+1\right)\left(2I+1\right)\text{degen:}\left(2J+1\right)\end{array}$

Answer to Problem 18.33E

The expected symmetry pairings of the nuclear and rotational wavefunctions are,

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym}\\ \text{degen:}1\text{degen}:5\end{array}$

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym}\\ \text{degen:}1\text{degen}:9\end{array}$

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym}\\ \text{degen:}1\text{degen}:13\end{array}$

Explanation of Solution

It is given that the diatomic oxygen has an antisymmetric ground electronic state. The oxygen nuclei are bosons $\left(I=0\right)$.

The symmetry pairings of the nuclear and rotational wavefunctions is given as follows:

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym},J\text{even}\\ \text{degen:}\left(I+1\right)\left(2I+1\right)\text{degen:}\left(2J+1\right)\end{array}$…(1)

Substitute $\left(I=0\right)$ in equation (1).

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym},J\text{even}\\ \text{degen:}1\text{degen:}\left(2J+1\right)\end{array}$…(2)

For even values of $J$, the system will be symmetric.

Substitute $J=2$ in equation (2).

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym}\\ \text{degen:}1\text{degen:}\left(2\left(2\right)+1\right):5\end{array}$

Substitute $J=4$ in equation (2).

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym}\\ \text{degen:}1\text{degen:}\left(2\left(4\right)+1\right):9\end{array}$

Substitute $J=6$ in equation (2).

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym}\\ \text{degen:}1\text{degen:}\left(2\left(6\right)+1\right):13\end{array}$

Hence, the expected symmetry pairings of the nuclear and rotational wavefunctions are,

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym}\\ \text{degen:}1\text{degen}:5\end{array}$

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym}\\ \text{degen:}1\text{degen}:9\end{array}$

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym}\\ \text{degen:}1\text{degen}:13\end{array}$

Conclusion

The expected symmetry pairings of the nuclear and rotational wavefunctions are,

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym}\\ \text{degen:}1\text{degen}:5\end{array}$

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym}\\ \text{degen:}1\text{degen}:9\end{array}$

$\begin{array}{l}{\Psi }_{\text{tot}}={\Psi }_{\text{nuc}}\times {\Psi }_{\text{rot}}\\ \text{sym}\text{sym}\text{sym}\\ \text{degen:}1\text{degen}:13\end{array}$