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I understand that quantum tunneling is a pure example of the uncertainty principle but clearly transistor had to be powered to work properly, anyhow I like to know if it is true that particle must borrow energy to quantum tunnel then quickly pay it back? I thought this is supposed to be probabilistic phenomenon so how does energy come into play?


marked as duplicate by Aaron Stevens, John Rennie energy May 12 at 11:33

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  • $\begingroup$ there are solutions of the quantum mechanical equations for the system. see my answer here physics.stackexchange.com/q/479161 . The energy in tunneling does not change, and the uncertainty principle is not the explanation. $\endgroup$ – anna v May 11 at 3:07
  • $\begingroup$ @annav: I read the 2nd paragraph of this article and I am perfectly aware that quantum tunnel cannot beats light speed if we add in relativity. $\endgroup$ – user6760 May 11 at 3:14
  • $\begingroup$ did you look at the plot linked in my comment and the link to it $\endgroup$ – anna v May 11 at 3:19
  • $\begingroup$ @annav: yes it said reduced probability not reduced energy. $\endgroup$ – user6760 May 11 at 3:21
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    $\begingroup$ Unfortunately you are basing this off of horrible misuse of the uncertainty principle as well as pop-sci description of tunneling. $\endgroup$ – Aaron Stevens May 11 at 5:46

Quantum tunneling arises because of probabilities, and what you are suggesting is that the particle must borrow energy in order to tunnel. This is incorrect. Let us consider a one dimensional case where we have a particle of energy $E$ at a position $x$ and a potential barrier

$$V(x)=\begin{cases}V_0, \quad x=[0, 1]\\ 0,\quad \text{otherwise}\end{cases}$$

where $E<V_0$ and $x<0$ at a time $t=t_0$. For $t>t_0$, there is some probability that the particle will be at a position $x>0$. You are suggesting that the particle must borrow energy if $x>0$: you believe that it must be that $E\geq V_0$ if $x>0$. If this was the case, then the particle could exist within the barrier for an infinite amount of time. What we actually find in quantum mechanics is that the particle has a probability of existing within $[0,1]$ that is proportional to a decaying exponential (or something of the sort), so that the particle is not certainly going to pass through the barrier. So energy is not borrowed, the energy is always $E$, it just happens to be the case that there is still some probability that the particle exists within the potential barrier, or that it passes right through it.

  • $\begingroup$ Does this mean that if we were to detect the particle in the barrier that it would have a negative kinetic energy? $\endgroup$ – Aaron Stevens May 11 at 11:49
  • $\begingroup$ @AaronStevens No. I think the difficulty that both you and OP are having is that you are thinking in terms of Newtonian mechanics. Quantum mechanics reduces to Newtonian mechanics in limiting processes, but you cannot just use your intuitive sense to answer these problems. In Newtonian mechanics, the particle could never be inside of or pass through the well, so it makes sense that your Newtonian intuiation for this problem should not be applied, because we already know that Newtonian mechanics doesn't work here. $\endgroup$ – Kraig May 11 at 14:12
  • $\begingroup$ I'm not trying to use Newtonian mechanics at all here. In QM there is still a total energy $E$ and a potential energy $V$. If you knew the total energy at which the particle approaches the barrier and you find it in the barrier then you would find its potential energy to be larger than the total energy it started with, right? This is also commented by @DvijMankad on the main question: The reason tunneling is possible is because the kinetic energy can be negative in QM to balance the higher potential. Unless you mean we can't even talk about kinetic energy now? $\endgroup$ – Aaron Stevens May 11 at 14:21
  • $\begingroup$ @AaronStevens The question is discussing $E$ not the kinetic energy. What you're asking is beyond the scope of the question. I'd have to workout the actual value of the kinetic energy using the kinetic energy operator for this question to see if thats the case. And in fact, what you should do is just calculate the average kinetic energy within the barrier. Consider asking your own question to see if it indeed is negative. But I'm unsure if there's a correlation there, kinetic energy is meant to be a scalar so I suspect it will always be positive. $\endgroup$ – Kraig May 11 at 14:48
  • $\begingroup$ I know the question was just discussing $E$. Just wondering. Thanks. $\endgroup$ – Aaron Stevens May 11 at 16:04

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