# Why drop the vibrational ground state energy

• I

## 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.

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mfb
Mentor
It doesn't matter if you only look at differences between energies. If you take relativity into account you have much more energy via the mass of the particles - but you don't have to care about that.

I still don't get it. Why are we calculating the differences between energies instead of the absolute internal energies?

mfb
Mentor
The absolute value is meaningless. Changing it doesn't change anything. If you add a fixed amount of energy in a consistent way, no physical observation changes.

So you saying what we calculated is not the actual internal energy of the molecule, but rather it is the energy relative to the ##\epsilon_0##? Just like how we define the gravitational potential energy to be zero at infinite distance? We set the zero point ourselves?

mfb
Mentor
That’s how I understood what you quoted.
You can keep the constant, it just doesn’t matter.

I think I get it now. Thank you for your help!

mfb