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Imagine that in a classical sense, a particle has some total energy $E$ and that its potential energy $U(x)$ varies with $x$ in the shape of a well (see the top-left image). Of all the possible positions of the particle, it occupies the centerpoint of the well for the least time, because at the centerpoint, the particle's kinetic energy is highest and, hence, the particle moves fastest at that position. If it moves fastest at that position, compared to at all other positions, then the likelikhood of finding the particle at that position is comparatively low.

In a quantum sense, if a particle once again has some total energy $E$ and a potential energy $U(x)$ that varies with position, the particle is, instead, most likely to be found at its well's center!

Why is that?

My attempt to answer the question: Once outside its quantum well (i.e. in the classically forbidden zone), a particle must work its way to $\psi(x) = 0$. This can be accomplished without compelling the particle to have a very curvy wavelength (a high curvature and, hence, high energy) if the particle has a "bell" at its center and minimums at its well's "walls". Just consider what would happen if the particle were in the shape of the classic probability distribution. From its maximums, occurring at its well's walls, it would have to dip down to 0. That requirement would oblige the particle to carry a greater energy.


Ultimately, my explanation amounts to using the idiom: "nature tends to occupy the lowest energy states." Is there some explanation more satisfactory than that and one that reconciles the probability distribution of the quantum well with its classical counterpart?

enter image description here

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    $\begingroup$ That picture is an 'educational war crime' and total nonsense, for the reasons explained by Timaeus. What's more, for the Quantum Harmonic Oscillator, $\Psi^2$ is only highest in the middle for the Ground State. See the plots for $\Psi_n^2$ for all states: hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc5.html#c1 $\endgroup$
    – Gert
    Sep 20, 2015 at 20:40
  • $\begingroup$ note that for the odd-numbered excite states there is node right at the middle. $\endgroup$ Sep 20, 2015 at 21:51

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A particle that is rolling up and down the walls of a well is nowhere near the ground state potential: the ground state would correspond to the particle being at the bottom of the well, and not moving.

In quantum mechanics, you can't both be "at the bottom of the well" (well defined position) and "not moving" (well defined momentum). So the ground state is spread out a little - peaked in the center, but with some "wings".

Now if you look at a particle that is "in the well, but with excess energy" - such a particle would not have a wave function as given in your diagram. The hyperphysics link that Gert provided in the comments above states the following:

The most probable value of position for the lower states is very different from the classical harmonic oscillator where it spends more time near the end of its motion. But as the quantum number increases, the probability distribution becomes more like that of the classical oscillator [...]

and in fact it shows the probability density function (square of the wave function) for several quantum states. Even $\Psi_3^2$ already shows a tendency for the positions away from the center to become more probable:

enter image description here

This tendency continues as the quantum number increases - until, at very high energy, the quantum distribution and the classical distribution are indistinguishable.

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    $\begingroup$ This. I give my modern physics students the problem of finding the spacial-probability distributions for the classical and quantum SHO at energies corresponding to various states, and make them learn this for themselves. Sometimes I can see the light-bulb come on right through the page while I'm grading. $\endgroup$ Sep 20, 2015 at 21:52
  • $\begingroup$ @Floris I now understand that a particle occupies its centerpoint, for the greatest period of time, in only a few cases. In those cases, you wrote that the rationale for the particle to NOT take the shape of a well at its centerpoint is: "you can't both be 'at the bottom of the well' (well defined position) and "not moving" (well defined momentum)" If the particle's probability did hypothetically take the shape of a well, however, why would being at the bottom of the well imply the particle isn't moving? As I round a maxima or minima, while I don't move up or down, I still move right/left. $\endgroup$
    – Muno
    Sep 27, 2015 at 15:18
  • $\begingroup$ @Muno because of the uncertainty principle you cannot know both the position and momentum precisely. So there can be no state in which the particle is completely stationary at the bottom of the well. I was talking strictly about the lowest energy state - the one described by the bell shaped curve in your question. At higher energies the particle will move near the bottom and its wave function starts to look more and more classical. $\endgroup$
    – Floris
    Sep 27, 2015 at 23:22
  • $\begingroup$ @Floris Ah -- understood. Only with reference to the lowest energy state, does being at the bottom of the well imply that the particle wouldn't move? $\endgroup$
    – Muno
    Sep 28, 2015 at 4:43
  • $\begingroup$ If you "steady state" is "at the bottom of the well", then yes. In the classical sense. The quantum particle can't "sit still", it has something we sometimes call "zero point energy". It is the lowest energy state but it is not (quite) zero. $\endgroup$
    – Floris
    Sep 28, 2015 at 11:59
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In classical physics there are classical forces and also you observe a particle somewhere because it is located there.

In quantum physics there is no reason to think that a particle is located some particular place, and even if it were then there is still no reason to think an observation reveals that position rather than creating a new position. And even if it did you would need additional, state dependent, nonclassical forces to get it to agree with observations. So your potential is just the classical potential and doesn't include quantum forces.

And even if you decided to pretend it had a position before you looked and the position observation just revealed it (and so had extra forces to make that happen) then you would then be unable to do the same for momentum because alternating between a momentum interaction and a position interaction definitely objectively changes the object.

So most people will say it didn't have a position, and a position result was created when you did your observation. A very small minority would model the particle as having a position, but then they have additional forces. Forces that depend on the state, and for which the particle could very well be at rest in a ground state. And if it is at rest then the distribution now is just based on the distribution in the past.

Which is the whole problem with assuming it had a position, if it does then the position now just depends on the position in the past. Which you didn't know. So you've just kicked the question into the past.

There is some work that indicates it could become distributed a certain way based on a series of random interactions in the past, but those theories are aiming to get a particular result and making no new predictions, so I wouldn't know how excited to get about that.

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  • $\begingroup$ This doesn't really answer the question. $\endgroup$
    – DanielSank
    Sep 20, 2015 at 20:42
  • $\begingroup$ @DanielSank Many of my answers address misconceptions or possible misconceptions. In this case there seemed to be a confusion about there being a position and the the measurement process merely revealing it. After that is addressed other questions could be asked about probability current or about the high n classical limit, or about sloshing combinations of different energy eigenstates, but the misconceptions have to be addressed first. $\endgroup$
    – Timaeus
    Sep 20, 2015 at 21:02
  • $\begingroup$ @DanielSank For instance the particle doesn't work to get the wave down to zero, it either doesn't have a position or else it could be sitting at rest in the. Classically forbidden zone or sitting at rest in the middle or many other options. The question seemed more driven by misconceptions than anything else, so I answered those, the cause rather than the symptom $\endgroup$
    – Timaeus
    Sep 20, 2015 at 21:04

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