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Is Energy Really Quantized ? (electric Potential enrgy)

  1. Aug 11, 2012 #1
    We all know that the elctrons in the nucleus of an atom cant jump to any enrgy state but rather specific sets of energy levels which we call sub-shells (orbitals), implying that energy is quantized just like electric charge.

    But here is an excerpt from my Phyiscs Book which says otherwise :

    When moving a particle
    between two points A and B that are
    a distance x apart,

    Work done by electric field= FΔx= -ΔPotential Energy
    F = -ΔPotential Energy/Δx

    If the two points are close together, x is very small. The change in
    potential energy must therefore also be very small; otherwise, the
    force would be extremely large. This argument applies as the displacement
    becomes infinitesimal (in the limit Δx → 0) and means that the electric
    potential energy cannot jump discontinuously when a particle moves
    between two points A and B an infinitesimal distance apart.

    Which one is correct, the book or the theory (we all take for granted)
  2. jcsd
  3. Aug 11, 2012 #2


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    Staff: Mentor

    A physical particle cannot be in one single specific location, it always has some distribution in space.

    Eigenstates are quantized, "the energy of a particle" is not, unless you require that it is in an eigenstate (which is usually done for spectroscopy and so on).
  4. Aug 11, 2012 #3
    In reality, the electric field itself (and every other field save maybe gravity) is also quantized. Instead of a particle moving through an electric potential and gaining energy continuously, in reality, the energy discretely reaches the particle by the exchange of a photon (or virtual photon). The study of this and related phenomena is called "quantum field theory." It is an advanced topic that confuses even the top experts.

    Introductory quantum students are often left in the dark about this fact. In intro quantum, we study particles in an external potential--e.g. particle in a box, particle in a delta potential, particle in a 1/r potential ("hydrogen atom"). However, these problems are all called "semiclassical" because realistically the fields that give rise to such potentials are quantized and are mediated by the exchange of particles--to treat this, you need QFT.

    For example, in introductory quantum mechanics, if you study the semiclassical treatment of the hydrogen atom, you'll learn that if a hydrogen atom is prepared in an excited state, then that state is an eigenstate of the hamiltonian and no matter how long you wait, the state will not decay to its ground state. This is not what is experimentally observed, and it has to do with the fact that there are fluctuations in the vacuum due to the electric field being quantized. So the intro quantum treatment, i.e., the semiclassical approximation, has its obvious downfalls.
    Last edited: Aug 11, 2012
  5. Aug 11, 2012 #4


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    The bound state energy levels of a system are discrete, and so the transition energies between them are discrete. But the unbound states (scattering states) above them are continuous.

    In particular, a photon can have any energy whatsoever. Blackbody radiation emitted by a light bulb, or from the sun, is a continuous spectrum. Every time a photon scatters off an atom its energy is changed.
  6. Aug 12, 2012 #5
    hms: I'd say you are comparing two different situations....different models....but that doesn't mean the question isn't a good one!!

    If you consider the field of a point charge as a function of 1/r2 obviously something goes wrong with the models as r approaches zero.....But experimental observations shows it works pretty well outside that range. So what is 'right' is based in large part on what we can verify experimentally.

    Try this Wikipedia article for some related insights:


    and especially checked the SUPERPOSITION section.

    While the electromagnetic field IS quantizied [e.g; F = qE] moving an infinitesimal distance r would not seem to necessarily encounter a quantum jump. Seems like that depends on your model.

    The electric field is dependent on position, but also the configuration of source charges....those ARE [again] quantized, but must there positions and configurations also be....seems like you could get to arbitrarily small increments. Note the description in the Wikipedia article for Gauss's law which involves ρ the charge density and obviously that is [again] quantized...yet we use a continuous mathematical model. Or we can quantize our model as Jolb describes.

    [Not to confuse things further, but based on past discussions in these forums nobody knows whether space and time are quantized or continuous...and some research papers claim that distinction is meaningless anyway. Do we need to worry about increments of Planck length or Planck time....]

    Anyway, in science and engineering, you always have to understand the limitations and constraints of the problems you are solving and the models you utilize. Does the mathematical model fit the physical situation?
  7. Aug 12, 2012 #6
    What means "fluctuations on the vacuum due to electric field been quantized"? The reason that excited states may fall into others with lower energy is explained by Fermi's golden rule and it can be explained with the existence of a "virtual photon" with energy [itex]\hbar[/itex][itex]w_{ab}[/itex] (a is the current state and b the possible new state) whose hamiltonian would cause a perturbation that gives Fermi's transition rate.

    With this approach it's only needed to add a new phenomena to classical QM.
  8. Aug 12, 2012 #7
    Well, if you add a photon as a perturbation, then you're not dealing with a hydrogen atom in a vacuum. You're dealing with a hydrogen atom with a photon flying by. I was referring to an absolutely empty vacuum, with no electric field besides that of the proton and electron.

    You are right that a photon can cause "stimulated emission," but even in the absence of any photon, an excited hydrogen atom can decay to its ground state in a process called "spontaneous emission."

    Let me quote Shankar's Principles of Quantum Mechanics, Second Edition, Chapter 18 "Time Dependent Perturbation Theory." This comes after a discussion of stimulated emission and Fermi's Golden Rule. From pp.506-507:
    The general formalism, illustrated by the preceding example, may be applied to a host of other phenomena involving the interaction of atoms with radiation. The results are always in splendid agreement with experiment as long the [sic] electromagnetic field is of macroscopic strength. The breakdown of the above formalism for weak fields is most dramatically illustrated by the following example. Consider a hydrogen atom in free space (the extreme case of weak field) in the state |2,l,m>. What is the rate of decay to the ground state? Our formalism gives an unambiguous answer of zero, for free space corresponds to A=0 (in the Coulomb gauge), so that H1=0 and the atom should be in the stationary state |2,l,m> forever. But it is found experimentally that the atom decays at a rate R≃109 second-1, or has a mean lifetime τ≃10-9 second. In fact, all excited atoms are found to decay spontaneously in free space to their ground states. This phenomenon cannot be explained within our formalism.

    So are we to conclude that our description of free space (which should be the simplest thing to describe) is inadequate? Yes! The description of free space by A=Ȧ=0 is classical; it is like saying that the ground state of the oscillator is given by x=p=0. Now, we know that if the oscillator is treated quantum mechanically, only the average quantities <0|X|0> and <0|P|0> vanish in the ground state, and that there are nonzero fluctuations (ΔX)2=<0|X2|0> and (ΔP)2=<0|P2|0> about these mean values. In the same way, if the electromagnetic field is treated quantum mechanically, it will be found that free space (which is the ground state of the field) is described by <A>=<Ȧ>=0 (where A and Ȧ are operators) with nonvanishing fluctuations (ΔA)2, (ΔȦ)2. The free space is dormant only in the average sense; there are always quantum fluctuations of the fields about these mean values. It is these fluctuations that trigger spontaneous decay.​

    A very eloquent passage I couldn't hope to improve on (except fixing his typo). He's using mostly the standard notation: H1 is the perturbation Hamiltonian, A is the magnetic vector potential, and |n,l,m> are the eigenstates of the hydrogen atom.

    "Quantizing the field" is just terminology which means: treat A and Ȧ as operators with nonzero commutators, in contrast to the classical versions, which are real valued and have identically zero commutators. It's exactly the same concept as treating X and P as operators with nonzero commutators.
    Last edited: Aug 12, 2012
  9. Aug 13, 2012 #8
    Hi, but I thought when we do EM field quantization, we view the photons that make up an EM field as harmonic oscillators, which have discrete energy levels? (From Shankar)
  10. Aug 14, 2012 #9
    So, QFT considers that the spontaneous decay is caused due to fluctuations on the vacuum state A = |0> and A is no longer a field but and operator... QFT it's a very hard theory, it's easier to think on a virtual photon that may appear causing the spontaneous decay :).
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