# Can a quark quantum tunnel outside a nucleus?

I read that a particle wavefunction can extend over a thin barrier and there is a probability of finding it over the barrier even though it doesn't have the necessary energy, what about quark?

So tunneling is when a particle goes from state $$A$$ to state $$B$$ through state $$C$$, generally with the following energy levels:

              __________
/     C    \
________/            \
A                  \____B____


The transition from A to B is energetically favorable but the transition to C is energetically forbidden. Quantum “noise” is perhaps the strangest form of noise you will ever encounter, because it can do things which violate conservation of energy, but it itself cannot violate energy conservation. Most other noise violates conservation of energy the whole way through; you can predict that eventually you will come to the same temperature as the noise source one way or the other, there is a “fluctuation-dissipation” theorem that makes sure of it.

Crucially, the space outside of the nucleus is not zero-energy for the quark. The color force is so strong that it would rather be balanced by tearing apart a quark vacuum into a quark and antiquark pair than leave a quark freestanding, so the free quark would immediately become a meson (probably a pion) and the nearby deficient nucleon would become a proton or neutron again (or maybe something more exotic like a Delta or Lambda particle, but that would likely decay into a proton or neutron again).

Since the free pion has a substantial mass there is just no improvement in the form of lowering the quark's energy state from the tunneling, so it just doesn't tunnel.

The same thing happens, by the way, if we're just talking about protons or neutrons tunneling outside of a normal nucleus. If the nucleus is stable it just sits there forever, because there is no added energy to be gained by a neutron leaving the nucleus. The nucleus has to be heavy with too many neutrons in order to trigger those sorts of spontaneous fissions. So it doesn't require a force that's as strong as the color-charge-unbalanced strong force -- it just is a problem with the vacuum energy state being higher than the bound state so that you would need a lasting energy violation in order to create a lasting free state.

No.

The trick to remember is that the strong force, unlike the forces you are likely more familiar with such as gravity and electromagnetism, does not get weaker with increasing distance. Indeed, it is quite unlike anything you have encountered in your ordinary life because it actually applies a constant (in magnitude) force regardless of the separation between particles, even down to zero separation, which technically means they cannot even "rest", as they are always subject to force at all times. Mathematically, this corresponds to a potential energy function of the form

$$U_\mbox{constant force}(x) := k \cdot |x|$$

which always increases with increasing distance $$x$$ of one quark from the other. So basically, your problem is that there isn't a "thin barrier", but instead an infinitely thick and tall one that has the form of an infinite "ramp", so to speak. If you set up and solve this situation for the wave function, you will find that while it does try to "penetrate" it, the amplitude dies off rapidly with increasing distance and never recovers. Thus no "tunneling", in the sense you are thinking, will occur.

(Of course, this is just a simple approximation: a full treatment requires the full machinery of QCD, and will say that eventually if you pull far enough, you will eventually have enough energy built up to form a new quark/anti-quark pair and at this point, they "snap", finally separating from each other but now bound to these new particles so still not isolated.)

(And it's also worth remarking on the strength of this force - i.e. the $$k$$ in the expression above: it is estimated to be around 10 kN, yes, that's kilo newtons, and thus is extremely macroscopic, for such a microscopic situation. In fact, this force is roughly the weight of a 1 Mg vehicle [about the size of a typical car] in Earth's gravity!)