When a particle is made to confine more and more to a particular position it breaks the energy barrier to get out because of the uncertainty principle. But, from where does the particle get the energy required to do so? Does its mass get converted into energy?

In that case consider a particle in a potential well equal to $mc^2$ where $m$ is the mass of the particle. Then to pass the energy barrier its whole mass is to be converted into energy so that the uncertainty principle is satisfied. So, a particle loosing all its mass and transforming into energy to get out of the potential well and again transforming into a particle.That's weird and I think it's not the case.

Does it undergo quantum tunneling during the escape process? In that case from where does that instantaneous increase in energy come from?

Note:I'm a high school student and I'm introduced only to basic quantum mechanics( just Bohr model and orbitals) in my school. Maybe the whole thing that I asked here is foolish. Sorry for that in that case.

  • $\begingroup$ A possible analogy, which is definitely not related to actual QM: suppose you have a particle surrounded with a wire mesh. Most of the time it bounces off the mesh but with some small probability it might move directly thru one of the holes in the mesh. Or, to get closer to QM, imagine the mesh's holes vary slightly in a random way (over time). Now the particle has to happen to meet up with a mesh hole just when it's large enough to allow the particle to pass thru. $\endgroup$ Jan 12 '14 at 13:37
  • $\begingroup$ Related: physics.stackexchange.com/q/54218/2451 $\endgroup$
    – Qmechanic
    Sep 12 '14 at 17:45

The energy does not change. The quantum mechanical solution of the potential problem with a barrier is such that it gives a probability for each specific energy level to go through the barrier .

In this explanation one sees a free particle reaching the barrier, and how the probabilities change, but not the energy:


Thus no extra energy is needed, but a lot of patience :) , because the probabilities are small usually and either one has to study a large number of the same conditions, or be in wait for the phenomenon.

Tunneling has been used to evaluate nuclear decays, for example the Polonium alpha decay lifetime :

enter image description here

  • $\begingroup$ Then, what determines whether a particle tunnel or not. Just probability? But, then that probability should have a physical meaning or interpretation. $\endgroup$ Jan 12 '14 at 12:46
  • $\begingroup$ Yea, just probability. That is why I added the alpha decay in, which is a completely random process on which particle decays. $\endgroup$
    – anna v
    Jan 12 '14 at 13:41
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    $\begingroup$ Note also that the barrier width can only be "a few" times the de Broglie wavelength of the particle before the barrier becomes to wide to allow tunneling. $\endgroup$
    – Kyle Kanos
    Jan 12 '14 at 17:47
  • $\begingroup$ @Kyle Kanos Didn't u mean to write 'too wide to allow tunneling'? $\endgroup$ May 3 '17 at 10:20
  • $\begingroup$ @Mockingbird of course , it is an easy error while tying $\endgroup$
    – anna v
    May 3 '17 at 10:32

There are more ways to view tunneling. One is that the energy of the microscopic particles is not constant, but fluctuates around some average value, for example due to fluctuations of the EM field (or other fields). Once in a while the fluctuation may supply enough energy to the particle to escape the potential barrier.

  • $\begingroup$ But, why doesn't the mass of a particle fluctuate the same way. If it does why do we observe the same mass for a particular fundamental particle anywhere in the universe. Does the observer effect come into play there? $\endgroup$ Jan 12 '14 at 12:34
  • $\begingroup$ Rest mass of the particle could be constant (by assumption), yet its kinetic energy may change due to external forces. $\endgroup$ Jan 12 '14 at 18:13
  • $\begingroup$ but will the energy fluctuation pose a problem for energy conservation? $\endgroup$
    – M. Zeng
    Apr 7 '15 at 4:52
  • $\begingroup$ @Timo, if the fluctuation is due to EM field, there is a work-energy theorem in the EM theory which supplants energy conservation (energy is conserved locally, but can flow in and out of any region of space). $\endgroup$ Apr 7 '15 at 23:18
  • $\begingroup$ @JánLalinský does that mean that regardless of the limitations of the measuring devices, we can never measure the energy of a particle precisely? $\endgroup$
    – M. Zeng
    Apr 8 '15 at 2:41

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