We know that subatomic particles can and do tunnel through barriers, so it is theoretically "possible" somewhat that a grain of sand could tunnel through a paper, but Id like to get some perspective on it.

Can anybody give any sort of estimate of how long one would have to wait to expect to see a grain of sand tunnel through a sheet of paper? (for instance2 10 times the life of the universe")

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  • $\begingroup$ i would hope that such amplitude would depend strongly from the state of the sand grain: it should be different if the sand of grain is amorphous, random solid matter than say, actually a bose condensate $\endgroup$ – lurscher Oct 28 '11 at 20:00
  • $\begingroup$ This problem is not sufficiently defined in that it is not given how high an energy barier the piece of paper provides to the grain of sand. In other words: how much energy would be required for the grain of sand to go in classical fashion through the piece paper? (For the other unknowns one can make reasonable assumptions.) $\endgroup$ – Johannes Oct 28 '11 at 22:04
  • $\begingroup$ No, how long would you have to wait for the grain of sand to tunnel through the sheet of paper spontaneously (all particles quantum tunneling AT THE SAME TIME through the paper) I know it's infinitly (almost) unlikely, but I want a more elaborate answer than that. $\endgroup$ – SchroedingersGhost Oct 29 '11 at 0:45

That is hard even to estimate correctly. For a grain of sand to tunnel through a sheet of paper the probability is so small not because of the tunnel barrier but because it would require the whole grain to move in one direction spontaneously. Starting with the Transmission probability (from Wikipedia) $$ T = \frac{e^{-2\int_{x_1}^{x_2} dx \sqrt{\frac{2m}{\hbar^2} \left( V(x) - E \right)}}}{ \left( 1 + \frac{1}{4} e^{-2\int_{x_1}^{x_2} dx \sqrt{\frac{2m}{\hbar^2} \left( V(x) - E \right)}} \right)^2}$$

With the barrier height V and the energy of the incoming particle E you can plug in different numbers for the width of the barrier (x1-x2) and the height V which for realistic estimates will be tiny.

Then you have to estimate the likelihood of all $$\approx 10^{23}$$ atoms doing that at the same time, so in the order of $$T^{10^{23}}$$ with $$T \ll1 .$$

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  • $\begingroup$ Thanks for the answer. Unfortunately my advanced math comprehension is almost zero, so I didn't really understand what the probability came out as? 1 to ? $\endgroup$ – SchroedingersGhost Nov 1 '11 at 6:49
  • $\begingroup$ @SchroedingersGhost - I'm going to trust that Alexander's interpretation is good and say that the probability is arbitrary as T depends on unknown parameters. The important part is that the double exponent you need to raise T to to get the transmission probability of all of the particles will make any probability less than exactly 1 vanishingly small. If T were 1 in 10, T^10^23 would be 1 in 1 followed by 100 sextillion zeroes. That's a number so gobsmackingly large it gives me a headache to think about it. $\endgroup$ – Richard Terrett Nov 1 '11 at 9:18
  • $\begingroup$ An alternative interpretation would be to treat the entire grain as a single relatively massive particle with a tiny de Broglie wavelength $\endgroup$ – Richard Terrett Nov 1 '11 at 9:30
  • $\begingroup$ @RichardTerret: Yes, that would be a nice alternative approach and would also result in a tiny probability. SchroedingersGhost: An exact number is not really easy to give here, but you can try it with your calculator or Wolfram Alpha to estimate if T is 0.5 (like a coin flip) and see what happens. $\endgroup$ – Alexander Nov 1 '11 at 13:29
  • $\begingroup$ Thanks, and obviously I do not expect a exact answer... But if one could say for instance "atleast X times the age of the universe" or something to put it in perspective, I'd find that satisfactory. $\endgroup$ – SchroedingersGhost Nov 2 '11 at 11:31

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