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Are de Broglie's waves and wave functions practically the same thing?

  1. Jul 8, 2010 #1
    I've just finished my first year on a Physics degree course at uni, and I'm a bit confused about some of the concepts we've covered. I was hoping someone could straighten things out for me.

    We were told about wave-particle duality and de Broglie and how he came up with the idea about particles behaving like waves (in the same way that light waves are like 'particles' called photons). We were also taught about wave functions, as in the Schroedinger equation, which, it seems, tell you the probability of finding a particle in any given location.

    As I understand it, the wave function itself isn't actually 'real', in that it's not a wave that is oscillating through space - all it tells you is the probability of where a particle will be found if you look for it. Is this the only sense in which particles behave like waves?

    If so, surely that mean that light waves are also just a wave function to tell you where a photon may be? In which case how can it also be an oscillation of magnetic and electric fields?

    Are de Broglie's waves and wave functions the same thing? Is the 'wave' of 'wave-particle duality' a wave function?
  2. jcsd
  3. Jul 8, 2010 #2
    Hello Mr Turtle,

    In the de Broglie-Bohm interpretation of quantum mechanics the wave function and the de Broglie wave are the same thing, and both refer to something which objectively exists. Particles are supposed to exist in addition to the wave (hence wave particle duality) and one can show that over the course of time the particles naturally become distributed as the square of the wave (hence the additional probability interpretation of the wave function). The latter is therefore a kind of 'equilibrium distribution' analagous to thermal equilibrium.

    In most other interpretations of quantum mechanics (where people are often not clear or avoid the question of what is supposed to objectively exist) then the answer is that either they are not the same thing or people make no definitive statement either way.

    So since no-one knows which - if any - interpretation of QM is 'correct' then the truth is that nobody knows the answer to your question. Sad, isn't it..

    Oh well..

  4. Jul 8, 2010 #3
    It's Miss Turtle, actually :)
    Thanks for the response, although I'm not sure I quite followed all of it... You're saying that de Broglie's waves and the wave function are the same thing, then? Or at least, probably?

    What you said about particles becoming distributed as the square of the wave - what if you only have one particle? We did a lot about the 'particle in a box' situation, and that has a wave as described by the Schroedinger equation, and yet doesn't seem to have much to do with averaging out over time. I realise that what they've taught us in our first year is probably a gross over-simplification of the real nature of things, but I was also reading a book called In Search Of Schroedinger's Cat, which admittedly was written in the early 80s, but it said (as far as I can recall, and I may be wrong) that all too often physics students go away with the idea that the wave function is a real wave oscillating in space when in fact it's nothing of the sort.

    And again, if the wave of wave/particle duality is the same as the wave function then what about light and its oscillating fields?
  5. Jul 8, 2010 #4
    It's Miss Zenith too.. All girls together eh? We should swap make-up tips later.

    Look I've only got about two minutes before the heavy brigade delete all my posts for not making it clear to you that the de Broglie-Bohm interpretation of QM is not the officially-approved interpretation of QM. I'm not in fact supposed to mention it to newbies, even though its greatest benefit is that it's the only one that is clear and actually makes sense. All the others just sound massively mysterious and make it easier for the boys to attract women.
    Repeat the experiment measuring the position of your one particle a million times. Plot all the million positions on a graph. It'll look like the square of the wave function.

    In orthodox QM that's true always. In the deBB approach, it's possible for the particle not to be distributed as the square of the wave (since they are logically separate entities), but the dynamics - with the wave pushing the particle - are such that it will become so distributed in the course of time. Once so distributed, it will stay that way. Hence the probability = psi squared rule.
    It's a 'stationary state' i.e. the shape of the square of the wave doesn't change over time. Think of the probability aspect as referring to an 'ensemble' i.e. a large numbers of copies of the system. Repeat the experiment lots of times on a system repeatedly set up in an identical way.
    No-one knows what it is. Believing that it is an objectively existing real wave is a perfectly consistent point of view (one held by, at least, the de Broglie-Bohm interpretation, the many-worlds interpretation, and the GRW interpretation of QM).
    To do light properly you have to do relativistic QM (quantum field theory and all that). I wouldn't worry about this yet..
  6. Jul 8, 2010 #5
    In terms of most undergraduate physics courses the de Broglie "wave" is just the hypothesised wave associated with matter of wavelength = h / momentum.

    Once a wave was suggested you needed an equation to describe it, which in the non-relativistic case is the Schrodinger equation, and the solutions of this (with suitable boundary conditions) give you a wave-function.

    The wave-function is defined in configuration space which, for N particles, has 3N dimensions, so doesn't directly relate to a wave in normal 3D space (eg for two particles, the wave function amplitude (squared) gives the probability of finding each particle at a specific location, and each particle's location requires 3 coords, so you have a function of 6 coords in total)

    For relativistic and/or massless particles like the photon you can derive a wave-function from a relativistic variant of the Schrodinger equation (Dirac, Klein-Gordon) or QFT. For a photon, it turns out to be the classical Maxwell EM wave, but it can't be interpreted as a position probability, eg see The Maxwell wave function of a photon, and in this case QED/QFT techniques are far more useful.

    Schrodinger originally thought the wave could have some real physical interpretation rather than a probability interpretation, but he couldn't find a workable physical model.

    I don't know how the de Broglie-Bohm people think of the physical basis of the "pilot wave", is it some magic force-field or similar? :wink: :smile:
  7. Jul 8, 2010 #6
    Indeed - de Broglie conceived of the idea of the wave function. Using de Broglie's 'optical analogy' (and with a bit of oral stimulation from the 'Dark Lady of Arosa') Schroedinger was able to derive an equation for the time evolution of the wave.
    This can be taken to be due to the presence of non-local forces between the particles.
    But de Broglie did (his model was presented in full at the 1927 Solvay conference - this is essentially the modern de Broglie-Bohm interpretation). Unfortunately he was beaten-up afterwards until he agreed to change his mind and support the Copenhagenists.
    That's exactly what it is! How do you think the Harry Potter stuff works?
  8. Jul 8, 2010 #7
    The "real nature of things" is that nobody has a clue what the "real nature of things" is. The math they are teaching gives the right answers, the rest is open for research and opinion. Mostly opinion.. :wink:

    Again, you can choose to simply accept the math, or some ontological interpretation, and work out specific problems in your career, or you can try and shed light on these issues in your career. It's up to you. I like the foundational stuff, but that's just a personal preference.
  9. Jul 8, 2010 #8

    You spilled the beans - Debb can no longer be a 'hidden' variable theory. People will now know that the pilot wave is an advanced technology indistinguishable from magic.
  10. Jul 8, 2010 #9
    Drat and triple drat... How could I have given the game away so easily? It's all that wine I drank earlier. I'll be expelled from the Magic Council.
  11. Jul 8, 2010 #10
    zenith8, seriously though, how do you interpret the pilot-wave physically?

    I appreciate it has to be a speculative "guess" or similar, but this seems to be "the elephant in the room" wrt pilot-wave theories and is rarely addressed.

    Apart from this aspect I find the de Broglie-Bohm model pretty convincing, and the material you have posted in the past has been very enlightening and informative (so thanks for those efforts) :smile:
  12. Jul 8, 2010 #11
    Thank you. Glad to be of service.

    Sigh, this will hardly help Miss Turtle, but here goes (stolen from post #15 of the last https://www.physicsforums.com/showthread.php?t=340864" where somebody asked this question):


    In non-relativistic quantum mechanics the pilot wave, or wave field, is a real field objectively existing in 3d space that is represented mathematically by the 3N-dimensional wave function of Schroedinger theory. It is (or appears to be) a time-dependent distribution of energy (or more strictly energy-momentum) in space.

    So now you tell me that you don't know what energy is. Well, it doesn't seem to bother most physicists, but let's just say that energy is something which:

    (1) is conserved.

    (2) exists in different forms

    (3) can be stored

    (4) can be transferred through space or from one material body to another

    Now in the de Broglie-Bohm interpretation of QM or pilot-wave theory - which is what you're referring to - electrons (say) exist as particles in addition to the pilot wave. Because the wave field is a repository of energy it can exert a force on the particles (the so-called 'quantum force'). Like all such a fields it has a potential energy function (Bohm's 'quantum potential' Q) and the force is given simply by [tex]-\nabla Q[/tex] .

    Remember that in general, potential energy is a property of fields, and the potential function Q represents the potential energy available to the particle at a specific position in the wave field.

    Depending on the prevailing circumstances, some (or all) of a particle's energy-momentum can be transferred and temporarily stored in its wave field. Once stored in the field, energy-momentum can be returned to the particle if circumstances change, and its kinetic energy will then increase (it will accelerate). This has the interesting consequence that the motion of a quantum particle need not be in a straight line even if there is no external field present.

    For example, if the pilot-wave passes an obstacle (such as a couple of slits) then its form will change (it will develop an interference pattern in this case) and energy will be transferred to and from the particle travelling through it according to the usual equations; the electron trajectory will then deviate from its classical (Newtonian path). It will end up getting guided into places where there is constructive interference in the pilot wave, and so after multiple experiments we see an 'interference pattern' developing in the positions of particle detections on a screen placed on the other side of the slits.

    Even though Feynmann (and God knows how many textbook writers) said no-one knew how to do this in terms of electrons following trajectories. He was just wrong.

    So to summarize the properties of the pilot wave and the quantum potential:

    (1) the pilot wave exhibits the usual wave properties (e.g. reflection, transmission, diffraction, interference etc.) and obeys the principle of linear superposition. The whole experimental field of 'matter wave optics' depends on this being the case, thus indicating pretty unequivocally that the wave field objectively exists (in order for it to act in such a manner, and be acted upon).

    (2) Since the Schroedinger equation is homogeneous, the pilot wave is not a radiated field and there is no source term for the field.

    (3) The environment surrounding a quantum particle (in part) determines the shape of the pilot wave..

    (4) The pilot wave is the repository of potential energy in a quantum system.

    (5) The pilot wave acts on the quantum particle similar to an external field and receives or imparts energy and momentum to the particle.

    (6) The quantum potential represents a portion of the energy contained in the pilot wave and is the amount of potential energy available to the particle at its specific position in the pilot wave field.

    (7) The magnitude of the quantum potential is independent of the intensity of the pilot wave.

    (8) Non-local connections between particles in a many-particle quantum system are facilitated through the operation of the quantum potential.

    If you want to know what it is at a deeper level than that, or what gauge boson is exchanged(!), then the answer is that nobody knows. But that doesn't stop you from asking or trying to find out.
    Last edited by a moderator: Apr 25, 2017
  13. Jul 8, 2010 #12
    Sorry. One point :smile:.

    According to the Davisson-Germer experiment,
    If an electron is accelerated through a potential V, its momentum becomes [tex]p =\sqrt{2meV}[/tex].

    So the de Broglie's wavelength becomes,
    [tex]\lambda = h/p[/tex]
    This wavelength is confirmed by this experiment.

    Also in the bound state such as hydrogen atom, and hydrogen molecule, the "classical" Coulomb force(F) influences the kinetic energy (T).
    Because also in the quantum mechanics, the Virial theorem is satisfied.

    [tex]2\langle T\rangle = - \sum \langle F_k \cdot r_k \rangle[/tex]

    When the force F is the Coulomb force, the Virial theorem becomes,

    [tex]\langle K \rangle = - \frac{1}{2} \langle V \rangle[/tex]

    (I wonder why the name "de Broglie's wave" is not used so much in QM, though this concept is used both in the Schrodinger and Dirac equations.)
  14. Jul 8, 2010 #13
    Thanks for the comprehensive reply, but yes I want to know what guage boson is exchanged :), or how the heck the potential acts on the entire universe instantaneously, which is basically what you mean by (8) right?

    I think the pilot-wave would be more appealing if you drop the requirement for a wavefunction of the universe, and make it a more "local" phenomenon (but not einstein local obviously). So you could postulate communication via spacetime wormholes or compactified kaluza-klein dimensions or "off-brane" signalling or tachyons which would give you a non-local/ftl (though not instantaneous) mechanism for the potential.

    This could be experimentally tested (in theory) if the ftl mechanism is finite and has an upper bound then assuming it's not too great it ought to be detectable eg by switching deflectors in an Aspect experiment really quickly and determining a point where the correlations no longer violate a Bell inequality.

    I'm sure this is exactly the discussion Miss Turtle was hoping for :smile:
    Last edited by a moderator: Apr 25, 2017
  15. Jul 8, 2010 #14
    Actually, having thought about this a little more, there are some subtleties involved.. At 1st year undergraduate level one normally encounters 'de Broglie waves' in the situation where you are given a series of things with different mass (an electron, a proton, a soccer ball, a person etc..) and one is invited to 'write down their de Broglie wavelengths.'

    Then a criterion is stated that 'quantum effects' start to become important when the object's size is comparable to it's de Broglie wavelength. One notes that a soccer ball's wavelength is very tiny and that it's size is quite big. Hence a soccer wall doesn't tunnel through walls (exit lecturer under a hail of tomatoes).

    Now in doing the sums we use the de Broglie relations [tex]p=\hbar k[/tex] and [tex]E=\hbar \omega[/tex]. Here [tex]p[/tex] and [tex]E[/tex] are the momentum and energy of a corpuscle associated with a vibration of wave vector k and frequency [tex]\omega = \hbar k^2 / 2m[/tex]. Note here that we are assuming that the wave in question is a plane wave with a fixed wavelength (whose phase presumably satisfies the classical Hamilton-Jacobi equation). How do we generalize this?

    OK - here's the science bit. [Miss Turtle - I apologize for this..]

    First, let's assume we have a variable external potential (and hence a variable particle momentum). Then the de Broglie wavelength is given by [tex]\lambda = h / p = h / [2m(E-V)]^{1/2}[/tex]. Here p is the momentum of a classical particle in the external force, and hence [tex]\lambda[/tex] is associated with a classical Hamilton-Jacobi wave. This is not what we want (remember Hamilton-Jacobi is just a way of writing ordinary classical mechanics as a wave equation).

    The correct quantum generalization is difficult for believers of orthodox quantum theory to do, because it requires a notion of particle momentum that is valid for all quantum states. Note that the above expression for the wavelength employs a variable momentum which already goes beyond what is valid in the usual interpretation. But in the case of the general Schroedinger wave function this definition has no particular significance and we can invoke the de Broglie-Bohm theory where one does have a definition of the continuously variable momentum.

    The classical definition of wavelength is the distance over which the phase increases by [tex]2\pi[/tex] i.e. [tex]S(x+\lambda n, t) - S(x,t) = 2\pi \hbar[/tex], where n is a unit vector. Taylor expanding, and assuming [tex]\lambda[/tex] is small or all derivatives of S higher than the first are zero, then

    [tex]\lambda(x,t) = h / |\nabla S(x,t)|[/tex]

    This is now a definition of 'local wavelength' where the wave is not periodic. Given that [tex]\nabla S[/tex] is just the particle momentum in de Broglie-Bohm theory, then the general definition of the de Broglie wavelength (using the quantum Hamilton-Jacobi equation) is:

    [tex]\lambda(x,t) = h / [2m(-\partial S / \partial t - V - Q)]^{1/2}[/tex]

    You can see this is not only variable in spacetime but it is state-dependent. Therefore it is not something that is associated with a particle independent of the context, unlike we implicitly assumed at the start with the soccer balls and stuff. It also now takes into account the existent of the quantum potential Q (i.e. due to the wave 'pushing' on the particle, as it must if the predictions are to agree with QM..).

    Well, Miss Turtle, you don't know what you've started, do you? Can we do girl stuff now?
    Last edited: Jul 8, 2010
  16. Jul 8, 2010 #15
    OK - now you see the danger of too much philosophy. But it's' the opposite problem to what you might think.. If people hadn't been banned from thinking about alternative interpretations to QM during most of the 20th century, then maybe some smart young thing like Miss Turtle would have sat down and figured out the answers to your questions (which are supposed to be meaningless, remember?).

    The pilot-wave isn't in the standard model - hence no gauge bosons - and yet.. it provides a wonderful causal explanation for everything that one normally insists cannot be given an explanation. Why is this? Is it a coincidence? If even 0.0001% of the effort that is expanded on string theory and other stuff were devoted to this, then maybe we might have a better idea.

    OK - the potential affecting the entire universe simultaneously.. well, not really. Only the relative small numbers of things (on a universe-wide scale) that are entangled because they interacted in the past.

    Everyone apart from a few (mostly harmless) lunatics now accepts that Bell experiments very strongly imply that non-local interactions exist (or that the universe is not real, causal influences act backwards in time, or there are an infinite number of universes - which for the sake of argument let's just discount). So the fact that deBB is nonlocal can't be held against it. And yet it is the case that we really know very little about non-locality and it is only the advent of quantum mechanics that has demanded its explicit recognition in quantum physics.

    Remember nonlocality also occurs in relativistic QM (though people often deny it). It's just that it is not possible to send a controllable signal between events with a spacelike separation. The causal connection remains.

    What is the means by which the non-local connection is actualized? No-one knows. The possibilities are:

    (1) the connection is done through ordinary (simply-connected) three dimensional space and is mediated through particles or fields that travel at superluminal speeds.

    (2) physical space has more than three dimensions (despite the complete lack of evidence for this)

    (3) the causal connection travels at sub-light speed but physical space is not simply connected.

    Non-local connections have at least the features of

    - not decreasing with distance

    - they cannot be shielded against

    - they are highly selective in what they effect

    Maybe option (3) seems the best, but who knows?

    Whaddya think, Miss Turtle?
    Last edited: Jul 8, 2010
  17. Jul 9, 2010 #16
    What appears to be non-local effects may not necessarily be so. If you imagine de Broglie waves to be sub-quantum fluctuations of the same EM medium that sponsors radiation then as a particle is accelerated the fields can be modified in a manner where the fluctuations precede the particle in its trajectory. This works even for photons if you consider that in the process of formation there is a moment where the energy packet must be accelerated to the speed of light.
  18. Jul 9, 2010 #17
    I'm betting on 2. You mean lack of experimental evidence, there is plenty of theoretical/mathematical evidence. Given the impressive track record in Physics of mathematical models preceding experimental confirmation that's pretty persuasive.

    I'm not sure the connection can't decrease with distance, or why it couldn't be shielded against. Also, since (I think?) there is no experimental evidence for quantum entanglement on cosmological scales you don't require that for the pilot wave.

    But I've dragged the discussion into a clumsy speculative area, so I'd better stop here. :smile:
  19. Jul 9, 2010 #18
    OK this has all gone way over my head and I have to admit to skim-reading some of it, but it's nice to think that if I work hard for the rest of my degree then some day, maybe, just maybe, I might have a clue what you're all talking about. Sorry to have started such a complex debate - there was me thinking it was a relatively simple question that would at least have a yes or no answer. I suppose I should have known better, this is quantum mechanics after all.
  20. Jul 9, 2010 #19


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    Somehow everything tangentially related turns into a de Broglie-Bohm thread around here..

    To give an alternative (historical rather than interpretational) answer to the original question, the de Broglie wavelength, which originated with de Broglie's 1924 thesis is not the wave function, because Schrödinger developed his famous equation after that, influenced by de Broglie's ideas. By 1927 de Broglie had largely united the two with a proper 'pilot-wave theory' (and the nucleus of of it is already there in the thesis), which is now de Broglie-Bohm.

    So in short, if you're talking de Broglie's waves (1924) they're not really the same, if you're talking 1927 they are.
    Last edited: Jul 9, 2010
  21. Jul 9, 2010 #20
    Indeed - see my post #14 for the mathematical details.
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