0
$\begingroup$

Quantisation associated with translational modes can be modelled like a particle in a box (PiB).

$$E_n = \frac{n^2h^2}{8mL^2} \qquad n \in \mathbb{Z} \, | \, n \ge 1$$

By the equation above, the zero point energy associated with a PiB is $E_1 = \frac{h^2}{8mL^2}$. However, in my statistical mechanics lectures, we set this to zero by modifying the equation as below:

$$E_n = \frac{(n^2-1)h^2}{8mL^2}$$

Atkins does this too in Physical Chemistry, claiming to do so "for reasons that will soon become clear". But, alas, I'm still not sure why this is done.

Could somebody enlighten me?

| improve this question | |
$\endgroup$
  • 3
    $\begingroup$ Who cares about the zero point energy? It is always there, hence it will inevitably cancel out in the end. $\endgroup$ – Ivan Neretin Mar 29 '18 at 15:09
  • $\begingroup$ @IvanNeretin, someone cared enough to remove it. What advantages does that bring? $\endgroup$ – Jacob Mar 29 '18 at 15:17
  • 1
    $\begingroup$ OK, if you want a longer answer, here it goes. When Atkins and your lecturer mentioned these $E_n$, they most likely did that for some reason. They must have used these in some calculations later on. You have to read and actually repeat these calculations with pen and paper, like those ancient scientists you might have seen on YouTube. Do it all the way to the end, that is, until you derive some physically observable values. Then repeat everything again, but this time don't set zero energy to zero. See where this gets you. I claim that you'll get no difference in the end. $\endgroup$ – Ivan Neretin Mar 29 '18 at 15:33
  • 1
    $\begingroup$ The reason might be that adding a constant to the energy of every state does not not affect the probabilities for the system being in a certain microstate. $\endgroup$ – aventurin Mar 29 '18 at 18:22
  • $\begingroup$ The value of any energy such electronic energy in a PIB or vibrational energy in a harmonic oscillator will obviously depend on whether the reference zero of energy is added in or not, but heat capacity and entropy are not dependent on the zero reference. In reactions, to be consistent, the zero energy is taken as that of the molecule dissociated into atoms then the dissociation energy is $D_0$ if taken to the lowest level or $D_e$ if to the lowest point of the potential energy. $\endgroup$ – porphyrin Mar 29 '18 at 22:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.