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

Are there quantum jumps?

  1. Mar 19, 2012 #1

    Jano L.

    User Avatar
    Gold Member

    Imagine thermal radiation interacting with atoms in a hot gas, in interstellar space, or in sodium lamp. Before Schroedinger proposed his wave equation, the basic dogma of quantum theory was that the atoms perform " quantum jumps " between " stationary states ".

    But Schroedinger's equation does not imply that such jumps even exist. Instead, Schroedinger showed that the dipole moment of the atom can oscillate harmonically at the frequencies that are given by differences of the proper values of the Hamilton operator. He did not use nor accept quantum jumps.

    Do you think the picture of atoms being present in preferred states and only jumping between them is still correct today?

    If yes,

    - what is the evidence for the jumps?

    - how long does it take to make such a jump?

    If no, do you think spectroscopic measurements can be explained in the framework of continuous oscillation of atomic dipole moments, along the way Schroedinger proposed?
     
  2. jcsd
  3. Mar 19, 2012 #2
    You are right that quantum jumps do not exist in standard quantum theory; to study them, you must move to a relativistic quantum theory, such as the dynamics of Dirac's equation or a quantum field theory like Quantum Electrodynamics. This is clear from the fact that photons clearly can't be described by the Schrodinger equation--the kinetic term has p2/(2m)... what is the mass of a photon? Without invoking relativity, the Schrodinger equation cannot describe photons, and quantum jumps are not possible without the emission of a photon.

    Quantum jumps certainly do exist in QFT, which is the basis for the Standard Model and models beyond the standard model (like string theory). How long a quantum jump takes is ambiguous: do you mean the expectation value of an excited state's lifetime or the actual jump? Certainly the former depends on the particular problem. (I believe the transition itself may be taken to occur instantaneously--but I'm not sure.) Evidence for quantum jumps couldn't be more abundant: without quantum jumps, there would be no photons and no light for us to see!
     
    Last edited: Mar 19, 2012
  4. Mar 20, 2012 #3

    Jano L.

    User Avatar
    Gold Member

    Hello Jolb. You say

    Where do you see them in the theory? The equations of motion do not provide them: they are partial differential equations for quantum fields, which is a theory very close to Schroedinger's line of thought.

    Surely light can be in the theory without photons - it is described by electromagnetic field, in non-relativistic quantum theory. In QFT, the light is described by an operator of quantum field A defined on spacetime. In a general situation, this field cannot be written as an eigenstate of the atomic Hamiltonian, and even if it was at some moment, the evolution would take it into some superposed state.
     
  5. Mar 20, 2012 #4
    the atoms jumps from one quantum state to another but it does so gradually.
    there is a finite time during which it is in a superposition of both states.
     
    Last edited: Mar 20, 2012
  6. Mar 20, 2012 #5

    Bill_K

    User Avatar
    Science Advisor

    The probability of finding the electron in the final state increases continuously, but the transition itself is instantaneous.
     
  7. Mar 20, 2012 #6

    Jano L.

    User Avatar
    Gold Member

    Bill,
    do you mean that although the wave function is in superposition of two eigenfunctions, the atom itself is either in one or the other corresponding state?
     
  8. Mar 20, 2012 #7
    Haha, nobody knows the answer to how the wavefunction collapses. Quantum mechanics predicts some continuous change of the state as the atom interacts with the field (a Rabi cycle) without jumps. But of course we do get jumps.
     
  9. Mar 20, 2012 #8

    Bill_K

    User Avatar
    Science Advisor

    Yes, that's it exactly. The wave function is a superposition in which the amplitudes of the two states evolve. There is no intermediate state - you can't catch the electron "partially emitted"!
     
  10. Mar 20, 2012 #9

    Demystifier

    User Avatar
    Science Advisor

    There are no true jumps in quantum mechanics (nor in quantum field theory), but there are some continuous very fast processes that look like jumps for most practical purposes. The essential physical mechanism lying behind these fast processes is interaction with the environment, and is better known under the name - decoherence.

    More details on decoherence can be found in the literature listed in
    https://www.physicsforums.com/showpost.php?p=3823258&postcount=2
     
  11. Mar 20, 2012 #10

    Demystifier

    User Avatar
    Science Advisor

    Sometimes it is possible to slow down the effect of decoherence so much that you CAN catch the system in a "partially decayed" state.
     
  12. Mar 20, 2012 #11

    Bill_K

    User Avatar
    Science Advisor

    Demystifier, Thanks, but I don't need a reading list on decoherence - I simply disagree with you!

    Most of physics is done in terms of models. You can do thermodynamics by talking about heat but not atoms. You can do nuclear physics by talking about nucleons but not quarks. Likewise you can do quantum mechanics by talking about quantum systems without talking about their environment.

    Quantum mechanics has been developed as a complete (or nearly complete) self-consistent model. Philosophically, yes, you need to append to it the details of a measurement process, and in that sense an atom about to decay is philosophically no different from a cat about to die. But practically, you want to be able to focus exclusively on the properties of the system itself. To say that the time interval during which an electron changes state is very short but environment dependent - well, it means adding extraneous complexity to the description with substantially nothing to show as payback.

    Unless - can you quantify what you mean by the "very fast processes that look like jumps" solely in terms of the measurable properties of Hydrogen?
     
  13. Mar 20, 2012 #12

    Jano L.

    User Avatar
    Gold Member

    Bill, if I understand your position, you say that the description of the behaviour of the atoms in terms of the wave function and the above assumption about their statistics is all we can hope to get and any more attempt to find out about the processes atoms undergo when they change their states is useless.

    But I think the potential of a more elaborate theory for payback is great. I can give one example, which I think is very practical.

    There is no particular length scale known where Schroedinger's equation should become invalid, so if it applies to atoms, it should apply to molecules too. The equation is just immensely more complicated.

    Consider, for example, light ray passing through a liquid solution of organic molecules, in stationary regime. The intensity of the ray will decrease along the way and the light will be resolved by the medium into frequency components that will propagate with different velocities.

    In order to explain this dispersion of light in terms of behaviour of the molecules, on the classical theory, one assumes that the molecules possess dipole moments oscillating at the frequency of the passing light. This theory fits very well in classical electromagnetism and wave optics.

    It is natural to attempt to use Schroedinger's equation to find the magnitude of these dipole moments. However, if these molecules were only in states corresponding to the eigenfunctions of the Hamiltonian, it is hard to see how the dipole moments could oscillate at foreign frequency.

    One way to get oscillating dipole in wave mechanics is to solve Schroedinger's equation for the wave function of the molecule under action of the electromagnetic field of light.

    I think it can be shown that in the course of time, the wave function evolves in a complicated manner which gives non-zero expectation value of dipole moment. It oscillates with the frequency of the passing wave.

    So it seems that the wave function does not just give probability that atoms are in some discrete states, but can also give some other properties of the atoms and molecules that are foreign to these stationary states. If true, this would be an advantage.

    This result suggests that the superpositions correspond to real states of the molecules.
     
  14. Mar 20, 2012 #13

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Schrodinger's equation becomes invalid as you approach the Compton wavelength of whatever you're trying to describe. At that point you definitely need relativistic QFT.

    I'm not sure what you mean by "foreign frequency." When dealing with molecules, you don't just have atomic electron eigenstates, you also have rotational and vibrational energy levels. The level spacing is on an entirely different scale from the atomic physics. In particular, the vibrational spectrum of molecular bonds is what causes the visible color of materials.

    No one is saying anything different. A system is always in a superposition of physical states. That includes translational, rotational, vibrational, etc. degrees of freedom. What seems to be confusing you is that it is often the case that the Hilbert space of the system is separable, so that we can consider different parts of the problem one at a time and put things together afterwards to describe the system completely. None of these properties are "foreign" when you consider the complete Hilbert space of states.
     
  15. Mar 20, 2012 #14

    Jano L.

    User Avatar
    Gold Member

    fzero,

    By foreign frequency, I mean the frequency of the light that originated elsewhere(daylight, laser). I did not think of excluding other degrees of freedom; let them all enter Schroedinger's equation.

    If we could find eigenfunctions of the full Hamiltonian then, the states corresponding to eigenfunctions would not imply presence of oscillating dipole at the frequency of the external light wave.

    There is no clear agreement on this. Bill and others think that the atoms exist only in certain discrete states and only jump between them.
     
  16. Mar 20, 2012 #15

    Bill_K

    User Avatar
    Science Advisor

    No, I agree with fzero on this.
    And at the opposite end, in the classical limit, Schrodinger's Equation yields to a classical description. While not literally invalid there, it is inappropriate. In fact we frequently get comments here along those lines, in which questioners seem to believe that electromagnetic waves must be discarded and replaced everywhere by photons. The example you gave is in the classical domain, dispersion of a light beam passing through a solution of organic molecules.

    How is this different from the previous case, an atomic transition? Intensity of the light beam. Many many photons are involved, not just one. And at the same time many many atoms. The effect produced is collective. No longer usefully described as a photon striking an atom, rather an atom immersed in an oscillating E field and being polarized by it. And the frequency ω of the beam is arbitrary, not one corresponding to a transition.
     
  17. Mar 20, 2012 #16

    Jano L.

    User Avatar
    Gold Member

    So you mean that the atoms exist in discrete states, jump and emit/absorb photons when the intensity of radiation is low, but exist in states described by superposed wave functions and interact continuously with classical electromagnetic field when the intensity of radiation is high?

    But what is the reason for using these two distinct pictures? Is there some fundamental difference in absorption lines of hydrogen for low and high intensity light, or something else?

    In theory, if we can use semi-classical theory for molecules, we should be able to use it for the atoms as well, no matter intensity of the electromagnetic field.
     
  18. Mar 20, 2012 #17

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, there is, in the following sense. When the density of hydrogen and the light intensity is low, the main features of the absorption spectrum are the dark lines at the characteristic wavelengths associated to the discrete transitions between atomic levels. However, there are higher order effects in perturbation theory where, for example, some of the photon energy is converted to translation of the atomic system, as well as an atomic level transition. Since translations have a continuous spectrum, this is an avenue for the atom to absorb light of any frequency.

    When the density and intensity are low, these events are suppressed by factors of the fine structure constant and other factors, so they have no strong effect on the spectrum. However, if we increase the density and/or intensity, we increase the probability that they occur. This is one of the ways that a macroscopic system differs from a single H-atom. The enhancement of small effects by sheer numbers results in various collective effects that might not be apparent from studying a single part of the system.

    I don't know how familiar you are with higher-order perturbation theory in QM, but it might be a place to start to make better sense of these things.
     
  19. Mar 21, 2012 #18

    Demystifier

    User Avatar
    Science Advisor

    Bill_K, I actually agree with your way of reasoning here. But the most interesting fact you seem not to be aware of is that there are experimentally measurable predictions of decoherence which cannot be obtained by a simple collapse. So yes, in some cases this extraneous complexity pays back quantitatively. (Perhaps not for hydrogen, but still. Quantum mechanics is not only a theory of isolated atoms.)

    For some experiments confirming the existence of decoherence in the real world see e.g. the review
    http://xxx.lanl.gov/pdf/quant-ph/0105127.pdf
    Sec. VIII A

    Also, the experiment discussed here
    https://www.physicsforums.com/showthread.php?t=503861&highlight=implications
    cannot be understood without understanding the concept weak measurement, which, in turn, cannot be understood without styding the response of an environment (serving as a weak-measurement apparatus) and cannot be even approximated by a collapse-model of a measuring apparatus.

    Moreover, your analogies above confirm my points. Yes, you can talk about thermodynamics without atoms, but you cannot predict the value of heat capacity without atoms. Yes, you can do nuclear physics without quarks, but you cannot predict the mass of proton without quarks. So more fundamental description does bring a new value, not only philosophically, but also quantitatively.
     
    Last edited: Mar 21, 2012
  20. Mar 21, 2012 #19

    Jano L.

    User Avatar
    Gold Member

    fzero, I do not see how higher order effects have any bearing on the original question. Perturbation theory is just an approximate way to find the wave function, no matter which order we keep. We can do high - order perturbation theory with time dependent perturbation Hamiltonian and get the wave function, without any jumps in it.

    Sharp lines in resolved light have always some non-zero line width, which is compatible with continuous harmonic oscillation of the atomic charges. No jumps are implied by the wave function - they are an additional assumption about the behaviour of the atoms.

    Forgive me for repeating myself, but I would like to try to ask again: if you think there are jumps,

    - can you describe some evidence for them? Or do you think they are just a useful mode of expression?

    - how long such a jump takes - is it smaller than the period of the radiation involved, or much longer?

    So far I do not understand well your standpoint on these two questions.
     
  21. Mar 21, 2012 #20

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I was addressing your misunderstanding in referring to "foreign frequencies." The point is that absorption spectra are well understood.

    Why would jumps be implied by the wavefunction? In order to have jumps, we must allow for particles to be created and destroyed, so we must turn to quantum field theory. There new states are created by acting with operators at a definite time (QFT obeys locality), so the jumps are instantaneous.

    All of our theoretical understanding suggests that quantum transitions occur at a definite point in time. There is no experimental evidence to the contrary.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook