I'm confused as to how the above phenomena can take place since arent they breaking the law of conservation of energy (even, if temporarily)?
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$\begingroup$ Related: physics.stackexchange.com/q/111697/2451 $\endgroup$– Qmechanic ♦Commented Sep 12, 2014 at 17:38
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$\begingroup$ "Tunneling" is perfectly real, even in classical physics. The third law of thermodynamics states, that the temperature of a system has to be greater than 0K (we just keep neglecting that willfully in classical mechanics). Every real system makes small fluctuations around its classical phase space vector. The energy for those fluctuation comes from the temperature bath that couples to the system. For sufficiently large temperatures this can put the system above a hump in its potential energy. In quantum mechanics the mechanism that supplies these fluctuations is uncertainty. $\endgroup$– CuriousOneCommented Sep 12, 2014 at 17:46
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2$\begingroup$ @CuriousOne I wouldn't call what you describe as tunneling. Energy is conserved at all times in your picture. I don't think tunneling is allowed in classical mechanics. $\endgroup$– garypCommented Sep 12, 2014 at 18:10
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4$\begingroup$ I don't understand. That aside, the classical system never passes through "energetically forbidden" states to get across the barrier, while the quantum system does exist in forbidden states. That's why I call one tunneling and the other not. $\endgroup$– garypCommented Sep 12, 2014 at 18:39
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2$\begingroup$ Tell you what ... I'll make a question out of it, then you can explain. $\endgroup$– garypCommented Sep 12, 2014 at 18:44
4 Answers
Of course, one usual intepretation of quantum tunneling is that the particle will borrow some energy from the vaccum in order to pass an unsurmountable barrier otherwise and then restitute it asap after crossing the barrier.
As many others have said, this is a valid interpretation. I am not sure it is necessary though.
In fact, in quantum tunneling, what really differs from classical physics per se is not energy conservation (after all the energy is supposed to be conserved and quantum tunneling in the NH3 molecule is at fixed energy for instance) but barrier crossing.
The only thing quantum tunneling tells us is that it is not the case that a quantum particle with energy $E$ cannot overcome a barrier of height $V > E$, period. All the rest is a matter of interpretation and taste I think.
Now, how can a particle pass a barrier if it doesn't have enough energy to jump over it? The answer lies, in my opinion, in the fact that a quantum particle has rarely a definite position even when its energy is perfectly defined. Therefore, assuming that such an uncatchable object can remain always on one side of a finite barrier would be the strange thing to observe and not the other way around.
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1$\begingroup$ "is that the particle will borrow some energy from the vaccum", this is weird because even without vacuum quantization, tunneling happens. While for example spontaneous emission doesn't. $\endgroup$– agemOCommented Nov 24, 2014 at 15:58
First of all what you need to understand about quantum physics is that it's a theory of probability, not realist classical probability, but it is still a probability theory. The second thing you need to understand is that realism is wrong.
Conservation of energy in quantum physics simply means that the Hamiltonian is not time dependent. That's it. From this follows that the expectation value of the energy does not change in time. You shouldn't picture a particle moving around either and then some energy magically appears and the particle can jump across some potential. This is a wrong way of thinking about quantum physics and Bell's theorem is clear proof of that. Particles do not have well defined classical properties before measurement. The sooner you realize this the better.
Temporary violations of conservation of energy is allowed in quantum mechanics. A system can make a transition to a state that violates conservation energy by an amount $\Delta E$ as long as it stays in that state for a time shorter than $\Delta t$ where $\Delta E\Delta t \leq \hbar/2$. This is a form of Heisenberg's uncertainty principle.
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2$\begingroup$ An uncertainty in an energy measurement is not the same thing as a "short term" violation of energy conservation. You can't give a measured number "back". It stays in your lab book as long as you keep the book. $\endgroup$ Commented Sep 12, 2014 at 21:19
You are right to be confused because these processes do not conserve energy when taken individually. If you are lucky a couple of particles may appear close to you and you can collect their energy for free. However, the inverse is also equally likely. Imagine you spend some energy and store it in a couple of particles, then there is a chance that these will annihilate and you will loose your energy.
At the level of individual quantum processes energy is not conserved, but as soon as you average over multiple processes the energy gain and loss even out and all is right with the world again.
There is no experimental contradiction because the energy that you may gain by accidental quantum processes is tiny. If you try to do it again until you have accumulated a significant amount of energy, you will loose as much as you gain.
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$\begingroup$ One can't collect any energy from virtual particles, neither in the individual process, nor on average. Indeed, virtual particles arise only because we insist one describing a quantum system as localized during its evolution and the only reason we do that is because we haven't found a way, yet, to describe physics without a background metric. $\endgroup$ Commented Sep 12, 2014 at 18:31