Why drop the vibrational ground state energy

  • #1
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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.
 

Answers and Replies

  • #2
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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.
 
  • #3
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I still don't get it. Why are we calculating the differences between energies instead of the absolute internal energies?
 
  • #4
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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.
 
  • #5
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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?
 
  • #6
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That’s how I understood what you quoted.
You can keep the constant, it just doesn’t matter.
 
  • #7
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I think I get it now. Thank you for your help!
 
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