1. What does a de Broglie wave look like?

  2. Are de Broglie waves transverse or longitudinal?

  3. Can they be polarized?

  4. What about the de Broglie wave of a ground state neutral spin-zero Helium 4 atom?

  5. What experimental evidence do we have that supports the detailed nature of a de Broglie wave?

I have always assumed that de Broglie waves were mathematically identical to electromagnetic waves, but I just realized there is no basis for this assumption, and in fact it must be false, unless there is an analogue to both the magnetic and the electric components of the electromagnetic wave. So what does a de Broglie wave "look like"?

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    $\begingroup$ A Bulk of questions ? Users would suffer. Please consider revision..! Or atleast, a way of improvement.. $\endgroup$ Commented Oct 28, 2012 at 12:17
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    $\begingroup$ Like Crazy Buddy says, it would be much better if you can focus the post on one specific question. $\endgroup$
    – David Z
    Commented Oct 28, 2012 at 12:30

4 Answers 4


A deBroglie wave has two interpretations, which are generalizations in different domains, and which are conflated for a single particle. This leads to a lot of confusion.

  • A classical field describing the motion of a single particle or many coherent bosons in a Bose-Einstein condensed state.
  • A wavefunction, a probability wave over configurations of particles.

historically, Schrodinger interpreted the deBroglie wave as the first thing initially, as a physical scalar wave. This is the wrong interpretation, as it is not equivalent to matrix mechanics, and it is experimentally untenable since a physical wave doesn't allow for entanglement. The battle over this was settled by Schrodinger (and Einstein and deBroglie too, who understood the deBroglie wave was like the solution to the Hamilton Jacobi equation, something that lives in configuration space), who demonstrated that the wave was in configuration space in 1926, and proved that with this interpretation, the Heisenberg formalism was a consequence of the wave formalism.

To quickly answer the questions

  1. It looks like a solution to the Schrodinger equation--- a wave peaked at where the particle is most likely to be (or where there are the most particles, in the field interpretation, see below), whose complex phase is twisting in the direction of motion at a rate which is proprotional to the local velocity of the particle (or the local velocity of the superfluid flow in the BEC in the field interpretation).
  2. Neither--- they aren't deformations in a material, so the idea is nonsensical. If you have a sound wave in a solid, you can ask if it is transverse or longitudinal, because it is motion of atoms. The deBroglie waves are wave of possibilities (except, you can ask this question in the field intepretation, see below).
  3. If the deBroglie wave is for a spinless particle, it has no analog of polarization. There is only one component. If you have a deBroglie wave for a particle with spin, it has several components. For the spinning electron, there are two components, for the two different spins, so that there are two deBroglie waves. The polarization for electron waves is spin-1/2, so it isn't like a photon polarization which is spin 1.
  4. The ground state of a He atom is highly entangled--- the configurations where one electron is on one side of the atom, the other electron tends to be on the other side, due to the repulsion. The entanglement is highest in the case of He (actually, highest of all in the case of the H negative ion, but this ion is marginally unstable, since it is only the entanglement which keeps it bound at all), because as the nucleus becomes more highly charged, the innermost electrons mutual repulsions are relatively weaker compared to their attraction to the nucleus. The precise description was worked out in the 1930s using the variational approximation, and it is essentially as exact as you like, because the variational ansatz, after you take into account rotational invariance and the spin being all locked up between the two electrons, is parametrized by one function of 3 dimensions which you can approximate very very precisely as an exponential of a polynomial with appropriate asymptotics.
  5. The experimental evidence for the new quantum mechanics, with its entanglement, in the 1920s-1930s consisted of the following: The precise spectrum of the H-ion and He atom, which could be worked out variationally. The approximate spectrum and specific heat of metals, where the electrons form a quantum Fermi gas, the spectroscopic entanglement of radiation with atoms which followed from the Heisenberg Jordan Dirac treatement of electrodynamics, and which resolved the paradoxes of photon absorption and emission in the older, entanglement free, Kramers Bohr Slater theory. In the 1940s, you get more precise evidence in the Lamb shift and countless condensed matter systems, and by the 1960s, you have Bells theorem and superconductivity. Essentially the only thing we haven't verified experimentally is quantum computation.

The points above require a little more discussion, regarding the field and particle intepretation.

When deBroglie understood the matter waves, it wasn't clear if these are physical waves in space, like an electromagnetic wave, or if they are something more abstract, like the solution to the Hamilton Jacobi equation. The Hamilton Jacobi solution is over all classical configurations, and it tells you what the integrable motion frequencies are. Einstein established the character of the deBroglie waves in 1924, by showing that the semiclassical limit description, they are the solution to the Hamilton Jacobi equation. When Schrodinger found the right equation, Einstein and Schrodinger discussed the interpretation, and it became clear that the Schrodinger equation too was to be thought of as a wave over configurations.

What this means is that the wave for 2 electrons is in 6 dimensions, for 3 electrons in 9 dimensions, describing all possible mutual positions of these. This led Einstein to ask how physical these waves are, considering that if you have a powderkeg in quantum mechanics, you can set up a situation where its wave is superposed between exploded and unexploded. This observation of Einstein's is the origin of Schrodinger's cat, and it is the reason Einstein could never be convinced to take the quantum formalism seriously as a description of physical reality--- it was just too enormous to be physical. It looked like a statistical description of something else. This has not been a common interpretation, because if it is a statistical description of something else underneath, we don't know exactly what that other thing could be.

But before chatting with Einstein, Schrodinger believed his equation described ordinary scalar waves in space. This interpretation made the amplitude $|\psi|^2$ a charge density, and the Schrodinger current an actual electromagnetic current.

While this interpretation is incorrect for the fundamental quantum deBroglie wave, it is correct for a Bose Einstein condensate. If you have many Bosons in a superposition state where they all share the same quantum state, their wavefunction becomes a classical field which obeys the Schrodinger equation, a Schrodinger field. The Schrodinger field description does not require linearity, it is just a scalar field (or a vector/tensor field for bosons with spin) which describes the density and matter-current in a Bose Einstein condensate. In this context, it is called a Gross-Pitaevsky equation, or in other contexts, a Bogoliubov-deGennes equation, or something else, but this field interpretation is very important, because it is the only limit in which Schrodinger waves turn into waves in space.

In this context, the deBroglie wave shared by the Bose particles turns into a classical scalar field, and it has an intepretation which is identical to the one proposed by Schrodinger. But such a description cannot describe entanglements in nature, and the simplest case where entanglement is seen to be necessary is in the ground state of the Helium atom.


Like @anna v noted, a DeBroglie wave is just a mathematical formulation to describe a probabilistic event.

One way to think about it is that a the de broglie wavelength of a particle is the scale at which classical mechanics fails completely, and you can see quantum behaviour. At much larger scales, the system can be well approximated by classical physics.

If you read about the electron double slit experiment, you can kind of infer that it isn't a 'wave' in the everyday, classical sense of the word. The wave just describes the probability of a particle being in a particular place. So the higher the amplitude of the wave at that point, the higher the probability that the particle will be there. Conversely, if the amplitude is zero anywhere, the particle will never be in that place, no matter how long you observe the particle.

So to answer your questions - Only the last two are valid, since they aren't waves like we experience them everyday. I'm unsure about your question regarding the Helium atom... but the fact that theoretically, using the probabilistic interpretation of quantum mechanics we can completely and accurately describe the behaviour of the Hydrogen atom to ridiculous precision, is among other things, proof that this works. The larger atoms are harder, because it is more than two bodies interacting, and the math gets very complicated very quickly.

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    $\begingroup$ You can describe He to ridiculous precision too, it's not 2 body, but it has a very very good variational solution. You can also describe Li to precision, although maybe not ridiculous. The higher atoms are needed to establish entanglement is physically real, that the DeBroglie waves are in configuration space. $\endgroup$
    – Ron Maimon
    Commented Oct 29, 2012 at 13:33
  • $\begingroup$ @RonMaimon - That's true. But I meant as far as I know (I could be wrong) Hydrogen is the only exactly solvable system, without variational/perturbation methods. But the overall success of QM across basically everything (except gravity, ofcourse. :P) is proof enough for me! $\endgroup$
    – Kitchi
    Commented Oct 29, 2012 at 13:42
  • $\begingroup$ The H-atom is only exactly solvable in the nonrelativistic limit, and I am not sure why any convergent variational series for He is not considered an exact solution--- it converges to the right answer. You shouldn't be too certain before verification of quantum computation, we don't know how entangled things can get before they crap out, although all evidence now suggests "as entangled as you make them". $\endgroup$
    – Ron Maimon
    Commented Oct 29, 2012 at 14:09

1) You can not see it. It is a scalar wave: that is, it is a wave of a scalar quantity, as opposed to a wave of a vector quantity like the electric or magnetic fields. It is therefore more similar to pressure waves in air rather than to electromagnetic waves.

2) None of them. They are scalar. Transverse or longitudinal are an attribute of vector waves, associated with the vector lying in the transverse plane or in a parallel line to the wave vector. If the wave is made of scalars, none of these make sense.

3) No. Polarization only makes sense for vector waves.

4-5) I do not understand the other two questions.

I believe that many of your doubts could be claryfied by studying a textbook on quantum mechanics. I studied on my professor's notes (in Italian), so If you do not speak italian you probably are not interested in them.

EDIT: substituted the word "field" with the word "quantity". Since I was speaking of mathematical fields in a quantum context, not of quantum fields, my words were open to misunderstandings. But not anymore (at least on that front!). Thanks to @Ron Maimon to make me notice it.

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    $\begingroup$ I don't know what you mean by a scalar wave. A wave of existence and non-existence, not a wave of motion at all? $\endgroup$
    – Jim Graber
    Commented Oct 28, 2012 at 19:40
  • $\begingroup$ No, I mean "scalar" as opposed to "vector". The distintion is purely mathematical. You'll recall that by wave we mean any function that solves the D'Alambert equation . If this function is a scalar function, that is, it maps any point of space to a number (which could represent the pressure, the temperature, sea level or something else), then the wave will be scalar. If the function is a vector (like electromagnetic waves, stress waves in solids, or something else) then the wave will be a vector wave. $\endgroup$ Commented Oct 29, 2012 at 10:08
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    $\begingroup$ It is not good to mix up deBroglie waves, which are waves on configurations, with scalar waves, which are waves in physical space. This confusion started with Schrodinger, and it isn't a complete confusion, but if you mention it, you need to say at some point that fields and wavefunctions are distinct things. $\endgroup$
    – Ron Maimon
    Commented Oct 29, 2012 at 13:34
  • $\begingroup$ @RonMaimon The scalar or vectorial adjectives are purely mathematical. Therefore, I believe they can be applied to physical observables (like the electric field) or to mathematical quantities alike, without concern of their reality. Though in my previous comment I forgot to mention that a wavefunction doesn't have to be mathematically a wave, I was just clarifying the different between a scalar and a vector. A wavefunction COULD be seen as a vector wave in the Heisenberg rapresentation,but it would not be a vector in physical space and therefore I would not link it to physical vector fields $\endgroup$ Commented Oct 29, 2012 at 13:52
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    $\begingroup$ @FerdinandoRandisi: It is also not a scalar field in the Wikipedia sense--- it doesn't associate a scalar value to every point in space. That's only for a single particle. For two particles, it associates a scalar value to every pair of points in space, and for 3 particles, it associates a value to every triplet of points in space. The dimension goes up like crazy, it's not a field at all. $\endgroup$
    – Ron Maimon
    Commented Oct 29, 2012 at 14:51

A de Broglie wave was a hypothesis that fitted the two slit experiments for electrons.It is not a wave in a field or a medium, so it is neither transverse nor longitudinal.

It is a quantum mechanical phenomenon that is well described by solutions of the quantum mechanical equations which give a wave nature, i.e. sine/cosine dependence for the probability of finding a particle in a particular location at a particular time. Thus the wave nature of particles is displayed in the QM probability of finding them at (x,y,z) at time t. The experimental evidence is the two slit experiment and the multitude of elementary particle experiments which agree completely with quantum mechanical solutions.

The photon is the dual particle of the electromagnetic radiation. It is fortuitous that the wave nature given by the solutions of Maxwell's equations coincides in frequency with that in the quantum mechanical solutions. When the photon is considered as a particle its wave nature also describes the probability of finding the photon at that particular (x,y.z) at time t.

That said we have to keep in mind that due to the Heisenberg Uncertainty principle particles are wave packets of probability, i.e. there is a width to their momentum and position distributions, and thus not a single wavelength or frequency describes them but a packet..

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    $\begingroup$ Historically, it was used to fit the Bohr conditions from the idea that the deBroglie waves are standing. The closest historical thing to a 2-slit experiment was the Davisson Germer (sp?) experiment on diffracting electrons through solids, which was complicated by the issue of effective electron mass changing, something which was explained only by Bethe in the early 1930s. $\endgroup$
    – Ron Maimon
    Commented Oct 29, 2012 at 13:36
  • $\begingroup$ This might be silly, but I would like to ask that the photon is comprised of oscillations in the mutually perpendicular electric and magnetic field which has a physical existence, what field can we attribute to the matter waves - a probability field? $\endgroup$ Commented Nov 2, 2019 at 13:03
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    $\begingroup$ @AjinkyaNaik The photon is a quantum mechanical particle in the standard model, a point particle. It does not have electrc and magnetic fields except in its wave function , $Ψ$ which $Ψ^*Ψ$ is the probability of finding the photon at (x,y,z,t). see cds.cern.ch/record/944002/files/0604169.pdf $\endgroup$
    – anna v
    Commented Nov 2, 2019 at 17:14
  • $\begingroup$ @annav Thank you so much! $\endgroup$ Commented Nov 3, 2019 at 7:51

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