0
$\begingroup$

I was wondering if I can say to a layman that "upon throwing the ball on a wall an enormously large number of times, there is a small probability that the ball will go through the wall", while explaining quantum tunneling (alpha decay example is abstract and artificial for a layman).

My doubt if whether the wall region can be modeled as a finite potential barrier (infinite potential barrier - which is not of Dirac delta form - will not allow tunneling).

Also, the wall seems to have all the other characteristics of the artificial barrier potential we set up in quantum mechanics, am I missing anything?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

infinite potential barrier – which is not of Dirac delta form – will not allow tunneling

isn't quite right. A barrier where there's a finite region of infinite potential will not allow tunneling, nor will potentials with singularities going suffiently fast $\to \infty$. But it's easy to construct a non-dirac potential with singularity that still permits tunneling; in particular the one-dimensional singularities of the $\tfrac{1}{|r|}$ peaks as which you might model the nuclei's coulomb potential aren't much of a problem. You can basically model the ball's CM amplitude as a Bloch wave there.

So, yes, a macroscopic ball can in fact tunnel through a brick wall. Of course, the probability is exponentially small in the thickness, so indeed an enourmously large number of throws is required. More problematically, it is far more likely for the ball to, say, spontaneously disintegrate into two identically-shaped halfs, to develop a tight chemical connection to the wall, or perhaps catch fire.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Can you elaborate how you deduced that "it is far more likely for the ball to, say, spontaneously disintegrate into two identically-shaped halfs" $\endgroup$
    – Man
    Commented May 27, 2013 at 12:15
  • $\begingroup$ That's more of a rough estimate, the idea being that for the ball to split only $\mathcal{O}(n^{2/3})$ atoms are involved in a "tunneling" process for which there is easily enough energy available from thermal excitations, at only the length scale of molecular bounds. Through the wall, we have $\mathcal{O}(n)$ atoms tunneling a macroscopic distance. $\endgroup$
    – leftaroundabout
    Commented May 27, 2013 at 12:28
1
$\begingroup$

In order to have quantum tunneling one needs a specific wavefunction which is the solution of the boundary conditions of the problem. Specific means all energies and phases are known.

The wall, as a potential, cannot enter into a simple quantum mechanical equation because it is composed of an enormous number of molecules each in its wavefunction according to the boundary conditions the neighbors impose.

Since there exist about 10^23 molecules per mole the number of variables entering the quantum mechanical equation is enormous both as far as the wall is concerned and as far as the ball is concerned, and there is no way coherency could be found between these two systems to allow for a consistent quantum mechanical wavefunction describing the ball that would give a small probability of tunneling.

Both objects are in decoherence and can only be meaningfully described classically.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Realistically, you're right of course. But in principle, tunneling isn't really concerned with decoherence: since the amplitude decays exponentially rather than periodically, the pure constituents of a mixed ensemble state will have their tunneling amplitudes all interfer constructively, regardless of random phase differences. $\endgroup$
    – leftaroundabout
    Commented May 27, 2013 at 12:26
  • $\begingroup$ @leftaroundabout The exponential solution comes from simple quantum mechanical problems. In crystals, for example, there is periodicity in the channeling of particles along the symmetry axis or planes, (example: arxiv.org/ftp/arxiv/papers/0704/0704.0031.pdf but c )rystals can be considered coherent quantum mechanically. $\endgroup$
    – anna v
    Commented May 27, 2013 at 18:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.