Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Why Not Mess with the Potential?

  1. Sep 8, 2016 #1
    I was sitting in a lecture the other day and the professor was sort of reviewing the basics of QM for the beginning of a new semester. He was discussing the difference between classical and quantum mechanics and he mentioned that quantum mechanics leaves the potential energy of classical mechanics untouched, and adjusts only the dynamics to account for physical behavior. This is something I've thought about before and have never quite understood. Why is it that the potential energy remains unchanged in the quantum picture? As a naive example, why not try to account for the quantized energy levels of hydrogen by inventing a potential energy function with a succession of "wells" of varying depths at different radial distances, each corresponding to an energy value? I'm quite certain that this wouldn't work, but I mean to say, why not something like it? In short, why is it that the creators of quantum mechanics formulated the theory as a rectification of hamiltonian dynamics, rather than seeking a solution in, say, the form of the coulomb potential. Is there a deep reason that QM leaves the classical potential untouched and concerns itself only with the equations of motion? Is there a historical reason? Have there been attempts to account for the behavior of quantum systems by rethinking the potential energy?
  2. jcsd
  3. Sep 8, 2016 #2


    User Avatar
    Science Advisor
    Homework Helper

    From a purely quantum point of view, the interaction between a proton and an electron must be of electromagnetic origin. The simplest interaction of electromagnetic origin is given by the Coulomb electrostatic repulsion. You can put whatever potential you want between two particles, but you'd be describing something else, not a hydrogen atom. The only novelty that Qm brings in terms of possible interactions between particles is the concept of spin and all interactions of it. The rest of Qm is inherited from classical Hamiltonian mechanics and passes through a so-called quantization, a set of rules to replace functions with operators.
  4. Sep 9, 2016 #3


    User Avatar
    Science Advisor

    You can't reproduce discrete spectra with continuous observables on a classical (connected) phase space. The reason for this is a simple topological fact: The image of a connected space under a continuous map is connected. However, a discrete set is not connected. Hence, the discrete energy levels can't be explained by a continuous energy observable on a classical phase space.
  5. Sep 9, 2016 #4


    User Avatar
    Science Advisor
    2016 Award

    Well, the similarities between quantum theory and classical theory come from the fundamental symmetries of spacetime. To a large extent the observables and the dynamics is determined by the symmetries of spacetime.
  6. Sep 9, 2016 #5
    I get that, but I was asking why. Presumably when the theory was first being created it could have seemed plausible to account for quantum phenomena by making adjustments to classical potentials instead of making adjustments to classical dynamics. Is there a reason that the latter procedure succeeds over the former?

    I don't entirely follow this argument. What mathematics do I need? Can you point me towards some reading?
  7. Sep 9, 2016 #6


    User Avatar
    Science Advisor

    Do you know the intermediate value theorem? It says that if you have a continuous function ##f:\mathbb R\rightarrow\mathbb R##, then the image of an interval is again an interval. For example for ##f(x) = x^2##, we have ##f([-2,2])=[0,4]##. In particular, it is not a disjoint union like for example ##[0,1]\cup[3,4]##. There is a generalization of this theorem in topology that says that if the domain ##X## of a continuous function ##f:X\rightarrow Y## is connected, then ##f(X)## is also connected. Connectedness is a generalization of the idea of an interval. It means that the space can't be written as a disjoint union of two closed sets (like ##[0,1]\cup[3,4]##). The phase space of a classical system is usually connected. For example the phase space of a classical particle is ##\mathbb R^6##. The spectrum of the hydrogen atom is disconnected, because it is the disjoint union of the sets ##[-\frac{1}{n^2},-\frac{1}{n^2}]## (and ##\mathbb R_+##). Hence, our theorem says that there can be no continuous function from the phase space of a particle into the spectrum of the hydrogen atom. In particular, whatever continuous potential you choose, the function ##H(\vec x,\vec p)=\frac{\vec p^2}{2m}+V(\vec x)## can never reproduce the spectrum of a hydrogen atom, so classical mechanics cannot account for the physics of the hydrogen atom.

    The theorem I'm using can be found here. It should be somewhere within the first 50 pages of every standard book on topology.
  8. Sep 9, 2016 #7
    I think I sort of understand. If the Hamiltonian is continuous, then it can't be mapped onto the hydrogen spectrum since any possible mapping would have to also be continuous. Therefore we cannot reproduce discrete spectra from continuous potentials without quantizing the equations of motion, which is why we quantize the equations of motion. I guess two things occur to me in response.

    1. It seems like the argument, if I understand it, assumes that we are using a classical (continuous) potential. Part of my question was why this assumption is justified. Why don't we mess with the potential, like maybe quantize it, rather than quantizing the dynamics.

    2. Isn't it possible to construct a continuous phase space that gives rise to discrete phenomena? Suppose you have a central attractive force surrounded by two concentric "troughs" with rough dissipative walls. Drop a marble into the first trough and let the central force not be strong enough to yank it out. The marble might oscillate to begin, but soon it will lose its energy to friction and settle into equilibrium (probably slightly offset from the true bottom of the trough due to the central force). This equilibrium constitutes a definite energy state. If you perturb it slightly, it will return. Unless, that is, you reach some critical value and give the marble enough kinetic energy to hop into the second trough. When this happens the marble will settle into an equilibrium in the second trough, which will also be a state of stable energy, higher than the first since it is farther from the center of force. This seems like an example of a discrete transition upon passing some critical value, yet the phase space (I think) is continuous. Or what about the transition from laminar to turbulent flow? Aren't there discrete phenomena that can be classically described?
  9. Sep 9, 2016 #8

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    The reason was given - it doesn't work.

    A different (and I think better) argument is that you can explore the same potential with bound states and with scattering. Scattering gives you continuous energies. How can the potential possibly know to change itself from continuous to discrete depending on what kind of experiment is being done?
  10. Sep 10, 2016 #9
    ...it doesn't work because it doesn't work? I don't think that's particularly compelling. Unless you mean that it doesn't work because it's never been found to work. Which means that in principle it could work, but it never has... On the other hand, it seems like something so central that there must be a positive reason, rather than just an observation that it hasn't ever worked.

    Fair enough. But it's generally true that a bound state is confined to a particular region of the potential whereas a scattering state has access to a broader range of the potential. Why not have part of the potential discrete and part of the potential continuous? In the example of my post above, if the marble is given enough kinetic energy to overcome the region of bumps and fly free beyond them, then it could be made to enter a continuous region of potential, couldn't it?

    I'm not saying this makes sense, just that I don't see why it doesn't. Usually we're told stories about failed attempts of past scientists to account for phenomena, like Maxwell's aether, or Bohr's ad-hoc quantization of angular momentum. Didn't anybody ever try to account for quantum phenomena by constructing some kind of funky potential energy function? Why is it so obvious that what had to be changed was not the coulomb potential, but the equation of motion?
  11. Sep 10, 2016 #10


    User Avatar
    Science Advisor

    I don't know what you mean by "quantize the potential". If you mean using a non-continuous potential, then the answer is that in classical mechanics, a potential must always be continuous. It must even be differentiable. Otherwise, the classical equations of motion ##m\ddot x = -\frac{\mathrm d V}{\mathrm dx}## would be ill-defined.

    The proof I have given shows that you can never produce discrete observables from a classical connected phase space. The only option you have is to use a disconnected phase space. That is exactly what Bohr's atomic model does. It restricts the phase space to a disjoint union of sets satisfying the Bohr quantization condition.
  12. Sep 10, 2016 #11
    That's what I meant. Now I see why that's a silly idea... should have been obvious.

    I see mathematically why that's so, but I'm still trying to get it to jive with my intuition. I think maybe my line of thinking is getting at what we mean when we say a discrete spectrum. It's true, from your proof, that we can't produce a strictly discrete spectrum from a continuous phase space. But what if our system visits various locations in a continuous phase space in such a way that it tends to be attracted to certain equilibrium points with a vanishingly small relaxation time. So if the marble sinks to the bottom of the well in like 10^{-20} seconds or something, and always returns there just as rapidly if it's perturbed, unless we perturb it just enough to knock it into the next well. Wouldn't that be a continuous phase space with effectively discrete behavior?
  13. Sep 10, 2016 #12


    User Avatar

    Staff: Mentor

    This is simply not true. Look at the wave functions for a bound electron - there's an exponential tail with increasing ##r##, but there is no region where the bound solution does not apply. Conversely, there is no region in which the potential does not contribute to the scattering solution.
  14. Sep 11, 2016 #13


    User Avatar
    Science Advisor

    No, because still all energies are accessible to the particle and hence we would observe their differences in the emission spectrum of the hydrogen atom. If an energy level isn't kinematically forbidden, you can always excite it by pumping the right amount of energy into the system. Sure, the particle might still be confined to a certain region of space, but it then it would just be oscillating around the local potential minimum. Hence, you can get transitions from any energy to any energy and would see a continuous emission spectrum instead of a discrete one. If you want to get a discrete emission spectrum, you must make all energies except from a discrete set inaccessible, which amounts to using a disconnected phase space. And that's precisely what Bohr and Sommerfeld already did by postulating a quantization condition.
  15. Sep 11, 2016 #14
    Do you mean that the mathematical validity of a solution for a system in some region implies that the characteristics of that region are physically relevant to the behavior of the system even when the probability of the system ever actually visiting that region are vanishingly small? Isn't that like saying the chemical composition of a very distant star is physically relevant to the value of g on earth, since it's true on the star that the value of g on earth is g? The solution may apply in all of space, but aren't there some regions where the electron just doesn't go?

    Once you get it confined to discrete regions of space, you just assign characteristic curvatures (hence characteristic energies) to the local minima so that particles oscillating there are associated with discrete energy values. That's easy enough.
  16. Sep 11, 2016 #15


    User Avatar
    Science Advisor

    By the theorem, this will make the energy non-continuous and hence force you to modify the equations of motion as well, since they require differentiability. We're going in circles. Your idea doesn't work. Any attempt to reproduce discreteness from a classical connected phase space must fail and the use of disconnected phase spaces has already been explored by Bohr and Sommerfeld.
  17. Sep 11, 2016 #16
    I might be confused. The energy, as in the potential energy that you draw on a graph with hills and valleys, is a continuous function. But there are minima on this function with different curvatures. The values of those curvatures form a discrete set, so there is a discrete set of frequencies at which the system can oscillate. I guess you than have to make the leap and say that oscillations of a particular frequency correspond to a certain energy, which is the Einstein-Planck idea. I don't know why that's true, but apparently it is. Then you have a system with a continuous potential energy function, that is capable of displaying oscillations at a discrete set of frequencies. The energy is continuous, but the vibrational behavior is discrete.

    For the record, I don't expect the idea to work, it just isn't entirely obvious to me why something like it couldn't work, and surprising that I can't find past attempts along these lines.
  18. Sep 11, 2016 #17


    User Avatar
    Science Advisor

    The set of minima of the potential might be discrete, but a particle in a potential isn't forced to stay in the minimum forever. There is also the possibility of a particle that oscillates back and forth and it will have a different energy. For example look at the harmonic oscillator. It has a potential ##V(x)=\frac{m\omega^2x^2}{2}## and the energy is given by ##E = \frac{p^2}{2m} + \frac{m\omega^2x^2}{2}##. You can easily see that even though ##x## might be ##0##, still all energies are allowed, since ##p## can be anything. ##p\neq 0## corresponds to a particle that oscillates. The theorem proves that this is not only the case for that particular potential I chose for this example. It's true for all continuous potentials.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted