Furthermore, consideration of the tunneling in the framework of the Wigner theory assumes that the tunneling factor is small. However, the $\ce{O-H}$ frequency is very high ($\gg k_\mathrm{B}T/h$), and the H atom tunneling under a potential energy barrier when proceeding either from a ground or from other excited vibration levels may be significant.

Source: German, E. D.; Sheintuch, M. Kinetics of Catalytic OH Dissociation on Metal Surfaces. J. Phys. Chem. C 2012, 116 (9), 5700–5709. DOI:10.1021/jp2106499.

Why does a higher frequency influence the quantum tunneling effect?


1 Answer 1


It is not the frequency per se but the mass that is important,; the probability of tunnelling an energy E is given by

$$p(E)=\exp \left(-\frac{4\pi\sqrt{m}}{h^2}\int_{x_a}^{x_b}\sqrt{V(x)-E}\;dx \right)$$

where $x_a$ and $x_b$ are two points either side of the potential barrier through which tunnelling occurs. The integral is in effect a measure of the area of the potential $V$ above position $x_a$ to $x_b$ through which tunnelling can occur.

As m increases, for a given potential barrier and energy, the tunnelling probability falls.

  • $\begingroup$ But why does the author say it that way and compare it with kT/h? $\endgroup$
    – wangge
    Jun 25, 2017 at 19:47
  • $\begingroup$ @wangge Don't know other than $k_BT/h$ has units of $s^{-1}$ but in this case is not a frequency but units of a first order rate constant. $\endgroup$
    – porphyrin
    Jun 25, 2017 at 20:34
  • $\begingroup$ I believe the comparison to $k_BT/h$ is more commonly talked about as the dimensionless ratio $\hbar\omega/k_bT$. I have seen this comparison made quite a few times when discussing nuclear quantum effects, as it essentially measures whether or not we are in the classical regime where thermal effects will be most important. For something like $\ce{O-H}$, this ratio is quite large even at room temperature. Also note that this shows up in the tunneling rate given above by means of $E$ which will be quite close to the vibrational energy of the $\ce{O-H}$. Hence talk of the vibrational frequency. $\endgroup$
    – jheindel
    Dec 12, 2017 at 22:00
  • $\begingroup$ @jheindel good point. $\endgroup$
    – porphyrin
    Dec 15, 2017 at 23:49

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