- #1

- 36

- 2

## Main Question or Discussion Point

This is from *Statistical Physics An Introductory Course* by *Daniel J.Amit*

The text is calculating the energy of internal motions of a diatomic molecule.

The internal energies of a diatomic molecule, i.e. the vibrational energy and the rotational energy is given by

$$\begin{aligned}E_I(n,J) &= E_v + E_r \\ &= \epsilon _v n+\epsilon_r J(J+1), \qquad n,J=0,1,2,\dots ,\end{aligned}$$

where $\epsilon_v$ is the spacing between the vibrational levels:

$$\epsilon_v = \hbar \omega= \hbar \sqrt{\frac{K}{\mu}}$$

and $\epsilon_r$ is the difference between the rotational levels:

$$\epsilon_r=\frac{\hbar^2}{2I}$$

But the vibrational energy of the molecule is

$$E_v = (n+1/2)\hbar \omega$$

**Why do we ignore the ##\frac{1}{2} \hbar \omega## in the ##E_v## (ground state vibrational energy) when we calculate the total internal energy of the molecule?** The text refers back to another equation to explain why we dropped the ground state vibrational energy, but I still don't quite understand the reason behind it.

The text referred:

The ground state energy of the electronic system ##\epsilon_0## depends on the distance ##\rho## between the nuclei, and we can denote this as a potential energy ##U(\rho)##.

If ##\epsilon_0## has a sharp minimum at a distance ##\rho_0##, it is possible to approximate ##U(\rho)## by the harmonic approximation. Hence the energy of the molecule will then given by

$$E_{mol}=\frac{\boldsymbol{P}^2}{2M}+\frac{\boldsymbol{\pi}^2}{2 \mu}+\frac{1}{2}K(\rho-\rho_0)^2$$

where

##\bf{P}## is the momentum associated with center of mass

##\boldsymbol{\pi}=\boldsymbol{\pi}_{12}=\boldsymbol{\pi}_1 - \boldsymbol{\pi}_2## in which ##\boldsymbol{\pi}_\alpha## is the relative momentum of the atom ##\alpha##

##U''(\rho_0)=K## is the harmonic approximation of ##U##

In this case I get why they drop the ##\epsilon_0##, that's because ##\epsilon_0## is already included in the potential energy term. However for the first case I don't see where the ##\epsilon_0## goes.

The text is calculating the energy of internal motions of a diatomic molecule.

The internal energies of a diatomic molecule, i.e. the vibrational energy and the rotational energy is given by

$$\begin{aligned}E_I(n,J) &= E_v + E_r \\ &= \epsilon _v n+\epsilon_r J(J+1), \qquad n,J=0,1,2,\dots ,\end{aligned}$$

where $\epsilon_v$ is the spacing between the vibrational levels:

$$\epsilon_v = \hbar \omega= \hbar \sqrt{\frac{K}{\mu}}$$

and $\epsilon_r$ is the difference between the rotational levels:

$$\epsilon_r=\frac{\hbar^2}{2I}$$

But the vibrational energy of the molecule is

$$E_v = (n+1/2)\hbar \omega$$

**Why do we ignore the ##\frac{1}{2} \hbar \omega## in the ##E_v## (ground state vibrational energy) when we calculate the total internal energy of the molecule?** The text refers back to another equation to explain why we dropped the ground state vibrational energy, but I still don't quite understand the reason behind it.

The text referred:

The ground state energy of the electronic system ##\epsilon_0## depends on the distance ##\rho## between the nuclei, and we can denote this as a potential energy ##U(\rho)##.

If ##\epsilon_0## has a sharp minimum at a distance ##\rho_0##, it is possible to approximate ##U(\rho)## by the harmonic approximation. Hence the energy of the molecule will then given by

$$E_{mol}=\frac{\boldsymbol{P}^2}{2M}+\frac{\boldsymbol{\pi}^2}{2 \mu}+\frac{1}{2}K(\rho-\rho_0)^2$$

where

##\bf{P}## is the momentum associated with center of mass

##\boldsymbol{\pi}=\boldsymbol{\pi}_{12}=\boldsymbol{\pi}_1 - \boldsymbol{\pi}_2## in which ##\boldsymbol{\pi}_\alpha## is the relative momentum of the atom ##\alpha##

##U''(\rho_0)=K## is the harmonic approximation of ##U##

In this case I get why they drop the ##\epsilon_0##, that's because ##\epsilon_0## is already included in the potential energy term. However for the first case I don't see where the ##\epsilon_0## goes.