How to properly relate quantum mechanical energy to the classical world

Question

The Equation of Interest

The energy equation derived from the "Particle in a 1-D box" problem is as follows:

$E=\dfrac{n^2h^2}{8mL^2}$

where $n$ is the principle quantum number, $h$ is Planck's constant, $m$ is the mass of the particle (e.g. electron), and $L$ is the length of region of space (i.e. box).

The Goal

I would like to illustrate the harmonious nature of this energy equation derived from the quantum mechanical nature of the system and classical mechanics. In theory we should be able to define a mass and region of space for a clear comparison.

The Example: Electron vs. Car

I want to compute the lowest possible energy of an electron using the equation above. We define $m=9.11\times10^{-31} \textrm{kg}$ and the region of space as a reasonable value for the diameter of an atom, $L=0.2 \textrm{nm}$.

Plugging this into our equation along with Planck's constant ($h=6.626\times10^{-34} \textrm{ m}^2 \textrm{kg s}^{-1}$) we obtain

$E_{\textrm{electron}} = \dfrac{(6.626\times10^{-34} \textrm{ m}^2 \textrm{kg s}^{-1})^2}{8 *9.11\times10^{-31} \textrm{ kg} * (2\times 10^{-10} \textrm{m})^2} = 1.51 \times 10^{-18} \textrm{ J}$.

Now let us imagine a car with $m=2000 \textrm{ kg}$ and a region of space as $L=1 \textrm{ m}$. The resulting energy is

$E_{\textrm{car}}=2.74\times10^{-71} \textrm{ J}$.

Now in both cases, the energy is so small that we might as well say its zero (from a classical standpoint anyway). Of course $E_{\textrm{car}} << E_{\textrm{elecron}}$.

*So we cannot stop here as quantum theory and classical mechanics are virtually indistinguishable (with respect to raw energies) at this point correct?

We can, however, convert these energies (which are purely kinetic energies) into something which is clearly distinguishable such as velocity by using the equation

$E = \dfrac{1}{2}mv^2$

Alternatively,

$v = \sqrt{\dfrac{2*E}{m}}$

where $m$ is the mass of the object and $v$ is the velocity. If we plug in our previously determined energies we get the following results...

$v_{\textrm{electron}} = 1.8\times 10^6 \textrm{ m s}^{-1}$

and

$v_{\textrm{car}} = 1.7\times 10^{-37} \textrm{ m s}^{-1}$

Here we see that the velocity of the car is essentially zero (or technically speaking, so slow that it wouldn't ever be observed as moving in billions of lifetimes). This jives with classical theory. However, we see that the velocity of the electron is staggering, even though the energy of the electron is 'essentially zero'

The Conclusions

So, in order to properly relate the energy equation derived from the quantum mechanical system of a particle in a 1-D box, one cannot simply compare the raw energies of the particle (i.e. electron) to that of an ordinary object such as a car because both energies result in a number that is essentially zero which defeats the point. As such, one must relate the energy to something appropriate such as the velocity of the object to properly illustrate how the energy equation still relates to everyday life.

The Question

Are my conclusions correct or can the raw energetics simply be used to make a proper comparison between quantum/classical theory without resorting to recasing the results into a physical property such as velocity?

#not-homework

Answer 1

*So we cannot stop here as quantum theory and classical mechanics are virtually indistinguishable (with respect to raw energies) at this point correct?

This is wrong. Just because you can't immediately, instinctively comprehend the difference between the $10^{-18}\ \mathrm{J}$ and $10^{-71}\ \mathrm{J}$ and therefore think they are both "indistinguishably small" doesn't mean it's true. The difference is massive. Our brains just haven't evolved to deal with numbers whose logarithms are very far from 0, because for almost all of our existence as a species, it never had to. Perhaps this site may give some very qualitative though fun context on how brains perceive the magnitude of a number.

In order to cope with this limitation of ours, we have developed some strategies. One can be understood as a form of chunking, where we simply modify numbers in a way so they become closer to something we understand. You probably won't be able to immediately give an example of an object that weighs 200,000 ounces or $3\times 10^{-27}$ solar masses, though you may find it quite a bit easier if I say 6 tonnes instead. This is because even if we inherently know the concept of an ounce or a solar mass, we can't really comprehend the magnitudes of 200,000 or $3\times 10^{-27}$ (the latter of course being much worse), and in order to get our heads around it more easily, we instead invent a new unit to measure weights in which we can express the same information using numbers closer to 1. The whole idea of scientific notation for numbers and SI prefixes for units can be understood as a consequence of this.

Another way is the one you use; instead of reframing the units and numbers, reframe the situation. You may feel there is no difference between $10^{-18}\ \mathrm{J}$ and $10^{-71}\ \mathrm{J}$, but the 53 orders of magnitude in difference is about the same as the number of atoms in a human body and the number of atoms in the entire observable Universe! Notice how I simply took the ratio into a slightly more familiar context (you know the Earth is just a small part of the observable Universe, and a human being is puny in comparison to the Earth).

So all of your calculations are correct, but if Homo sapiens were smarter, they would be entirely unnecessary!

Answer 2

You are treating the car as an elemental particle without inner structure, where the only degree of freedom is the overall translation. Within this model, it is true that QM seems to produce very little effect for cars.

However, cars that have only overall movement but no other structure is not at all useful. You would not call something a "car" if it does not have internal combustion. You need materials with different properties to construct different mechanical and electronic components of the car to make it useful, safe and comfortable. None of these are possible without quantum mechanical effects.

Even though combustion and ignition seem quite violent and energetic reactions, on a molecular level it is quite rare to achieve energies E-18J. In fact, using Bolzmann distribution, one find that such energy level actually correspond to extremely high temperature of almost 60000K, that is 10 times the temperature of the center of the sun. Of course, we rarely actually deal with electron in boxes, and in chemical reactions one is dealing with the CHANGE of energy, not the energies themselves, so normally one come across energy figures magnitudes smaller than that, but nevertheless quantum mechanical effect is enormous on a chemical level.

In fact, without quantum mechanical effect, there will be no stable atoms, no stable molecules, no reaction and no chemistry. It will just be a bunch of chaotic movement of particles that no one can understand, and there will be no computers to help either. In fact, there would be no life as well.

You may be baffled at the thought of this, because rarely does such small number seem to matter in life. In life, when we think about energy, we think about events that only happen a small number of times, like monthly electricity bills or a high MPG car, or say the electricity supply of a billion household. But the number of molecules is much greater, usually ~10^23, and each carrying a piece of quantum mechanical effect. Put together, a small energy on a microscopic level can be magnified to reach quite amazing results.

Answer 3

without resorting to recasing the results into a physical property such as velocity?

The energy is a perfectly physical property. In fact, for many quantum mechanical systems, the energy is the easiest physical property to measure.

can the raw energetics simply be used to make a proper comparison between quantum/classical theory

Yes, the scales of the energies are sufficient to make a comparison between quantum and classical systems. As you and Nicolau Saker Neto suggest in teh comments to his answer, convert the energy of the electron into eV: $1.5\times 10^{-18}\mathrm{J} \times \frac{1\mathrm{eV}}{1.6\times 10^{-19}\mathrm{J}} \approx 10 \mathrm{eV}$. The ground state of the Hydrogen atom is $-13.6 \mathrm{eV}$, comparable to the electron in the $2.0\mathrm{nm}$ box. So the energy you calculated for the electron is similar to that of most of chemistry, which is dominated by quantum mechanics. So that's a reasonably-sized quantum mechanical energy. But it's not noticeable compared to the joules of macroscopically-sized objects moving at macroscopic speeds.

The $1.7\times 10^{-37}\mathrm{J} \approx 1.0 \times 10^{-18}\mathrm{eV}$, on the other hand, is negligible compared to both macroscopic physics and to ordinary chemistry.