# Why doesn't a marble rolling on a table ever reflect back at the edge?

In light of the fact that when encountering a potential step there is a non-zero probability of a particle reflecting back, I was just wondering why it is that a marble doesn't "reflect" back from the edge of a table, but rather rolls off.

Is it because the potential difference is extremely insignificant when compared to the total energy of the marble (so not just kinetic, but also internal energy), so that the reflection coefficient essentially evaluates to 0?

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It's not clear to me that the edge of a table actually qualifies as a potential step for a marble. A better question might be: why doesn't a marble ever "quantum tunnel" down through the table and fall on the floor? In which case, the standard answer is: it might, but the probability is infinitesimally small, since it falls off exponentially with the thickness of the table in terms of the marble's de Broglie wavelength. – Nathan Reed Mar 13 '13 at 2:44
@NathanReed Why would it not be a potential step? It's a negative potential step, but the coefficient for the transmission is the same as if the particle encounters a positive one. – Ryker Mar 13 '13 at 3:39
Well, the marble doesn't suddenly change its energy because it rolled off the edge of the table. Don't confuse the notional "height" of a potential well with height in space; they're unrelated. A potential step would be something that would resist or enhance the marble's horizontal motion, such as a horizontal electric field if the marble is charged. If it rolls against the field it slows down, gaining potential and losing kinetic energy, and if it rolls the other way it speeds up, losing potential and gaining kinetic energy. – Nathan Reed Mar 13 '13 at 4:19
Ryker has just made the puzzle more picturesque by stating it in terms of the marble and the table. Criticizing the imagery doesn't address the interesting underlying issue, which is that the probability of reflection from a step-function potential doesn't superficially seem to obey the correspondence principle. – Ben Crowell Mar 30 '13 at 15:31
@BenCrowell Well, in all honesty, I cannot take the credit for making the problem picturesque. That is due to my QM professor :) – Ryker Mar 30 '13 at 20:33

John Rennie has pointed out that decoherence is a practical reason why we would not actually see such an effect. However, this is not an objection in principle, only in practice. In principle, we could isolate the marble from its environment. What is interesting about this example is that on the surface, it doesn't seem to obey the correspondence principle, and the decoherence argument fails to address that.

If you look at various treatments of the step potential, you'll see that people usually identify the classical limit with the limit $V_0 \ll E$, where $V_0$ is the height of the step and $E$ is the particle's energy. However, for a fixed value of $V_0/E$, the expression for the probability of reflection is simply a fixed value. This is because the probability of reflection is equal to the square of the familiar expression from wave kinematics for the amplitude of reflection at a boundary between media, $(v_2-v_1)/(v_2+v_1)$ (ignoring phase).

Based on the correspondence principle, we would have expected that the probability would depend on the particle's mass $m$ and on Planck's constant $h$. This is what happens, for example, with tunneling; the WKB probability of tunneling through a rectangular barrier depends on the unitless quantity $(w/h)\sqrt{m(V_0-E)}$, where $w$ is the width of the barrier.

Comparing the result for tunneling with the step function, we can see that there is no way to construct any such unitless quantity. The step-function potential "ramps up" discontinuously, so there is nothing with units of length that would play the role of $w$. Using only the variables $m$, $h$, $V_0$, and $E$, there is only one unitless quantity that can be constructed, which is $V_0/E$. But in reality, the step function can't really be discontinuous. It has to ramp up over some finite distance $w$. In the case of the marble rolling off the edge of the table,$w$ is roughly the diameter of the marble. This dimensional argument makes it plausible that we really do have a criterion for the classical limit: it should obtain when the distance over which we ramp up is large compared to the wavelength.

To see that we really do obtain the classical limit in this way, consider that for a particle whose wavelength is short compared to $w$, we can break the ramp up into a series of rectangles (as in the WKB approximation). A standard computation shows that the probability of reflection from a thin rectangular barrier, with $E>V_0$, is of the form $1-(\ldots)k^2\delta^2$, where $k$ is the wavenumber inside the barrier, $\delta$ is the width of the barrier, $(\ldots)$ depends only on $E/V_0$, and the barrier is "thin" in the sense that $k\delta\ll 1$. Stringing together a series of such barriers and multiplying the probabilities, the dependence on $\delta$ to the second power ensures that we get a product that approaches 1 in the classical limit $kw\ll 1$. (The exact solution to the Schrodinger equation for this ramp potential is given in Vern.)

There is a published paper by Branson on this topic, but unfortunately only the first page is available without going through a paywall.

D. Branson. 'The correspondence principle and scattering from potential steps', American Journal of Physics, Vol.47, 1101-1102, 1979. First page available at http://www.deepdyve.com/lp/american-association-of-physics-teachers/correspondence-principle-and-scattering-from-potential-steps-tKM85ATfDZ/1

Vern, "Airy wave packets as quantum solutions for recovering classical trajectories," BYU senior thesis, http://www.physics.byu.edu/faculty/vanhuele/Research/VernThesis.pdf

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This does not address Nathan Reed's point in the comments. How is this at all a "step" potential? – Pricklebush Tickletush Mar 31 '13 at 0:34
@AlecS: I addressed Nathan Reed's comment in a comment of my own. – Ben Crowell Mar 31 '13 at 14:56

I can think of two reasons (I'm sure there are others!):

Firstly, as the comments to your question have said, the moving marble is a 2D system and the $x$ and $y$ motions are separable i.e. (ignoring friction and air resistance) the acceleration in the $y$ direction when the marble rolls off the end of the table does not affect its horizontal, $x$, motion. In the $x$ direction the marble would behave as a free particle and there would be no reflection at any point.

If you replaced the end of the table by a ramp down to the floor then there could potentially be reflection because now you have a horizontal force acting. But ...

The marble is too big to behave as a single coherent object. The different bits of the marble will rapidly decohere with each other due to interactions with its environment, and this makes it impossible for the marble as a whole to exhibit coherent behaviour like reflection at a potential energy change. It also explains why the marble won't be diffracted by a pair of slits, or indeed show any other quantum mechanics behaviour of the type seen in smaller objects.

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in other words the trajectory of the marble is not the solution of a quantum mechanical potential, but an incoherent superposition of 10^23 or so solutions. – anna v Mar 13 '13 at 9:47
@annav: yes, well put :-) – John Rennie Mar 13 '13 at 10:40
@JohnRennie Given that the question is hinting on a QM analogy, the last paragraph of your answer is the most relevant part, and thus makes it a very good answer. – JKL Mar 13 '13 at 18:50
IMO this answer fails to address the point, as explained in my answer. – Ben Crowell Mar 30 '13 at 15:29