# Quantum Tunneling and Conservation of Energy

According to my understanding quantum mechanics, the probability of any particular particle in the universe being at any specific location in the universe is very small but never actually becomes zero. Thus, a little bit of all of us is everywhere.

Let's assume I suddenly quantum tunneled from sea level to the top of Mount Everest. That jump represents a net increase in energy/mass for Earth, which would violate the first law of thermodynamics. My calculations indicate it would have increased the mass of Earth by around 0.0858657µg, unless it is somehow offset some other phenomenon.

It seems implausible to me that energy conservation for a closed system could actually be violated, so how would that net increase in energy/mass be accounted for?

Or does the fact that such a scenario by definition represents such a significant decrease in the entropy of the system mean that an increase in mass should be the expected result?

• Can you please elaborate why "That jump represents a net increase in energy/mass for Earth"? – Árpád Szendrei May 22 '19 at 4:10
• @Árpád Szendre it would have more potential energy. The subject could then fall again to sea level and acquire kinetic energy it did not previously have. – rghome May 22 '19 at 9:07
• In my example, I have "Earth" represent the combined system of the planet and everything on it's surface or in its atmosphere. Raising myself manually from sea level to 8000 meters requires energy—energy that would be released if I fell. If I were to climb that height, there would be no net effect on the total mass of the system "earth". However, in my example, I didn't climb: I tunneled. Where did that energy come from? If it came from nowhere, that would amount to an effective increase in the mass of the combined "Earth" system. – darco May 22 '19 at 22:29

There should exist wave functions $$Ψ$$ , i.e. solutions for the quantum mechanical equations applicable to the problem where the specific boundary conditions have been applied. Then the $$Ψ^*Ψ$$ computation will give the probability for the particle to be found outside the barrier.