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Quantum harmonic oscillator tunneling puzzle

  1. Jan 17, 2015 #1
    My problem is described in the animation that I posted on Youtube:



    For the sake of convenience I am copying here the text that follows the animation:

    I have made this animation in order to present my little puzzle with the quantum harmonic oscillator. Think about a classical oscillator, a swing, a weight on a spring, a pendulum in a clock .... It has a certain characteristic spring constant, and a mass. For simplicity let us choose units so that these constants are numerically equal to 1. Let us also assume that the time scale is choosen so that the period is equal 2 pi. Given energy E, the classical oscillator will vibrate with an amplitude A. Its deviation x(t) from the equilibrium position x=0 is given by the formula

    x(t) = A cos t

    The relation between energy E and amplitude A is simple: square of A is 2E. For a classical oscillator the energy E can be any positive number.

    For a quantum oscillator, assuming units in which the Planck constant hbar = 1, the energy can not be arbitrary, it is quantized: 2E(n) = 2n+1. To each energy level there corresponds a "quantum eigenstate", the "wave function". Using Mathematica notation it is given by the formula

    psi[n_,x_] = 1/Sqrt[Sqrt[Pi] 2^n n!] Exp[-x^2/2] HermiteH[n, x]

    This wave function (notice that it is real-valued) is normalized so that its square gives the probability density of finding the oscillating point (with energy E(n)) at the point x. Classically we will never find x beyond the turning ponts xmin = -A and xmax = A. But for our quantum oscillator there is always a nonzero probability of finding the point in a classically forbidden region. We say that that there is a non-zero tunneling probability.

    Now the puzzle:

    Here is a relevant sentence from one of the popular textbooks:

    "The probability of being found in classically forbidden regions decreases quickly with increasing n, and vanishes entirely as n approaches infinity, as we would expect from the correspondence principle."

    "Quanta, Matter and Change. A molecular approach to physical chemistry", P.W. Atkins, J. de
    Paula and R.S. Friedman, Oxford University Press, Oxford/New York, 2009, p. 66

    So I used Wolfram's Mathematica to calculate these tunnelling probabilities for n = 0,...,512. The calculation have been done symbolically in order to avoid as much as possible numerical errors.
    The animation shows the sequence of plots of probability densities, the classical limits, and also, in a sub-window, the tunelling probability for each n. There is also an U shaped curve representing the classical probability density of finding the swing at a given position given only its energy, but not knowing its phase. In the animation each graph is scaled so that the classical turning points are always at x=-1 and x=1. The vertical axis is also scaled so that the total probability (the area under the probability densities) is always 1.

    It can be seen that indeed, the tunneling probability, at first, decreases rather quicly, but then its decrease rate is essentially slowing down with the increase of the quantum number n. This is puzzling. It is a surprise. From my correspondence with one of the authors of the quoted monograph I know that it is also a surprise for him. For instance, I could not find any estimate of what n is needed to have the tunneling probability smaller than, say, 1/1000000? Is it known?
     
  2. jcsd
  3. Jan 17, 2015 #2

    dextercioby

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    Hi Arkadiusz, and welcome back to PF. I don't there's a 'hidden' explanation to this issue. It's just (simple) mathematics and a particular behavior of a 1-variable function which is found starting with the known and accepted principles of quantum mechanics, the same principles that led us (through QFT) to the most successful (perturbative & not mathematically rigorous) theory we still have nowadays. QED and its famous g-2 accuracy.

    As for your particular finding here, it's for me just another proof that making unsubstantiated claims in textbooks is a pretty intelectually dangerous activity.
     
  4. Jan 17, 2015 #3
    What's missing is perspective. Consider diatomic oxygen, it has a vibrational zero point energy of [itex]790cm^{-1}[/itex] which is roughly [itex]1\times 10^{-23}J[/itex]. Let's call a classical energy one joule. Then to get to the classical regime the quantum number must be on the order of [itex]n \approx 10^{23}[/itex]! A long way to go from 500.
     
  5. Jan 17, 2015 #4

    Avodyne

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    I was able to calculate the "tunneling probabilty" analytically in the limit of large ##n##. The result I got was ##c n^{-1/3}## where ##c=\Gamma[1/3]/(2^{5/3}3^{2/3}\pi)=0.129##. This appears to match your curve.
     
  6. Jan 17, 2015 #5
    Thanks, Avodyne. This was also my guess. but only guess. A week ago I posted the numerical fit based on the first 100 points on my blog:

    ogony100a.gif

    My fit was a/n^k, a = 0.1465, k = 0.359871. Your formula matches the asymptotic part better.

    It would be useful (for instance for using in textbooks) to see the details of the arguments used in your theoretical derivation. Would you like to post it?
     
  7. Jan 18, 2015 #6
    Haborix, if we accept the asymptotic formula by Avodyne, then to get the quantum tunneling probability less than 1/1000000 we need n of the order of [itex]10^{15}.[/itex]
    Indeed A long way to go from 500.
     
  8. Jan 18, 2015 #7
    dextercioby: Looking at the first 30-40 probability density plots one gets indeed the impression of a quick descend of the nonclassical tail areas. Textbooks usually stop at that, as for instance in Thaller, Visual Quantum Mechanics, Springer (2000), p. 168:
    thaller.gif

    Further calculations need more number crunching, so that the wrong impression of a fast descend of tunneling probabilities is being formed.
     
  9. Jan 18, 2015 #8

    Avodyne

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    To do the calculation, I used the asymptotic formula for the (unnormalized) quantum wave function found here:
    http://en.wikipedia.org/wiki/Hermite_polynomials#Asymptotic_expansion
    (3rd equation from the end of this section). Normalize the wave function correctly, square it, change integration variables from ##x## to ##\phi##, and integrate over ##\phi>0##. The exponent ##2\phi-\sinh(2\phi)## can be replaced by the first term in its Taylor expansion, ##-4\phi^3/3##, to get the leading behavior of the integral.

    I'm also surprised by the slow fall off with ##n##, because the wave function falls exponentially fast in the forbidden region. However, it's also starting from a height that goes to infinity in the classical limit. Apparently these two effects offset.
     
  10. Jan 19, 2015 #9
    Avodyne, I am trying to understand your derivation. I checked Wiki, then went to the cited monograph by Szego. Szego is somewhat more precise than Wikipedia:

    szego.gif

    Do you mean you were using (8.22.13) , which is supposed to be valid only for [itex]x>=(2n+1)^{1/2}[/itex], and applied it to the whole real axis?
     
  11. Jan 19, 2015 #10
    For completeness: here is the comparison of the calculated probabilities with Avodyne's asymptotic formula (in red), for [itex]100\le n\le 512[/itex] tunelling_fit.png
     
  12. Jan 19, 2015 #11

    Avodyne

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    No, I used it only for ##x\ge(2n+1)^{1/2}##. This is the classically forbidden region. So I computed
    [tex]\int_{\sqrt{2n+1}}^\infty|\psi(x)|^2 dx[/tex] using (8.22.13) to approximate ##\psi(x)##.
     
  13. Jan 20, 2015 #12
    Update: in order to look at it from a different angle I did the animation for the Wigner quasiprobability distribution

    Details are in the description on Youtube. Here the meaning of the clasically forbidden tails is less evident.
     
  14. Jan 20, 2015 #13
    Avodyne, if you expanded only the tails, how did you normalize? Afterwards? Did you normalize [itex]n^{-1/3}[/itex]?
     
  15. Jan 21, 2015 #14

    Avodyne

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    The eigenfunction is ##\psi_n(x)=N H_n(x)e^{-x^2/2}## with ##N=\pi^{-1/4}(2^n n!)^{-1/2}##. Eq.(8.22.13) gives an approximation to ##H_n(x)e^{-x^2/2}## that is valid for ##x>\sqrt{2n+1}##. I multiplied this by ##N## to get an approximation to ##\psi_n(x)## valid for ##x>\sqrt{2n+1}##.
     
  16. Jan 21, 2015 #15
    OK. Thanks. All clear now. Evidently you also used [itex]dx/d\phi = \sqrt{2n+1}\sinh\phi[/itex]
     
    Last edited: Jan 21, 2015
  17. Jan 21, 2015 #16
    Avodyne: I am trying to reconstruct your argument. Yet I am getting a different coefficient, approximately twice your value. Perhaps I am making some mistake. Would you like to look at my calculations in the pdf document here?

    Update: I double checked my calculations. Could not find a mistake. Therefore my hypothesis is as follows: while the shape of the asymptotic may indeed be like [itex]n^{-1/3}[/itex], the constant factor can only be found through a fit to the exact data. One must not expect a proper probability normalization constant from the asymptotic behavior in a particular region. Shall we agree with such a proposal?
     
    Last edited: Jan 21, 2015
  18. Jan 21, 2015 #17

    Avodyne

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    I think you lost a factor of 2 between your (16) and (17). I claim that, in the large ##n## limit, your ##d_n##, (13), is just ##1/(2\pi)##. But the ratio between (17) and (16), which should be ##d_n##, is ##1/\pi## (again in the large ##n## limit).
     
  19. Jan 21, 2015 #18
    Uploaded a new version with several misprints corrected, including missing 2 in (13). Checking again.

    Update: our results indeed differ by the factor [itex]2[/itex]. Are you sure you multiplied by 2 in order to take into account both tails, left and right?
     
    Last edited: Jan 21, 2015
  20. Jan 21, 2015 #19

    Avodyne

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    OK, I've realized several things.

    First, I was not multiplying by 2 to get both forbidden regions.

    Second, after I do this, my result is numerically too large by about a factor of 2. I checked this by comparing it to the actual probability for ##n=100##.

    Third, that asymptotic formula we're both using is actually only valid for values of ##\phi## that are pretty big, ##\phi>0.25## for ##n=100##. For smaller values of ##\phi##, where the bulk of the tunnelling probability comes from, it's actually a terrible approximation. It's only good for getting the extreme tail of the wave function.

    Fourth, the "transition" formula, eq.(8.22.14), works well in the tunnelling region near the boundary of the allowed region, so this is the one to use. I'm not sure what Szego's ##A(t)## is, so I'm using the wikipedia version of the formula, which involves a standard Airy function. Plotting this approximation and the exact wave function for ##n=100##, the curves lie right on top of each other in the forbidden region.

    Finally, using the "transition" formula, I get the probability that ##|x|>\sqrt{2n+1}## is ##cn^{-1/3}##, where ##c=2/(3^{2/3}\Gamma[1/3]^2)=0.133975##. So the same ##n## dependence as before, but now with a coefficient that matches the numerical results well.

    EDIT: here is the approximate function I'm now using for the normalized energy eigenfunction:
    [tex]\psi_n(x)\simeq 2^{1/4}n^{-1/12}\mathop{\rm Ai}\bigl(2^{1/2}n^{1/6}(x-\sqrt{2n+1})\bigr)[/tex]where the Airy function ##{\rm Ai}(x)## is defined here: http://en.wikipedia.org/wiki/Airy_function
    In Mathematica, it is AiryAi[x].
     
    Last edited: Jan 21, 2015
  21. Jan 21, 2015 #20
    Thanks a lot for your thoughts and for your work! I will play now with Airy in order to reproduce your results.
     
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