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I'm having difficulty understanding de Broglie matter wave hypothesis. It is a mass or a particle hypothesis? According to de Broglie a particle with mass $m$ moving at a constant speed has an associated matter wave with a frequency
\begin{equation*} \nu\:=\:E/h \end{equation*} where $E$ is particle energy. Suppose this is just a mass relationship. Then, we can conceptually imagine the particle composed of two halves traveling at the same speed. Since each part has now half of the total energy they have an associated frequency that is half of the original \begin{equation*} \nu_{1/2}\:=\:\frac{1}{2}\:\nu \end{equation*} and so in general by splitting the particle in fractions of any proportions we can get all sorts of matter frequencies associated with the particle parts. In a sense this is the situation with a molecule where each atom that composite it has an associated frequency different from the whole (without considering the waves associated with the individual particles that compose the atoms themselves). So, is this interpretation correct or I'm missing something?

@Andrew: I read about bi-photons a while ago and was searching for a physical interpretation in the same lines. If I understood correctly, each photon has its own frequency but when they get entangled they behave very much as a single object with a frequency proportional to the total energy. I guess there are other requirements for a combination of two particles to be treated as a composite beside that both particles travel at the same speed. In any case I guess we can write a wave function for the composite traveling at a constant speed as $\Psi=\Psi_1(x_1,t)\Psi_2(x_2,t)$ where $\Psi_1=e^{i(k_1 x_1-\omega_1 t)}$ and $\Psi_2=e^{i(k_2x_2-\omega_2t)}$. Then assuming that $x_1= x_2\equiv x$ and $v_1=v_2\equiv v$ we get $\Psi(x_,t)=e^{i((k_1+k_2)x-(\omega_1+\omega_2)t)}$ which has a frequency that is the sum of the individual frequencies. I suppose this is equivalent to the center of mass approach that you suggest. Nevertheless, I just found out a similar question posed in this forum (Validity of naively computing the de Broglie wavelength of a macroscopic object) that treats the subject in some detail.

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In it's simplest form, de Broglie's hypothesis is meant to be applied to fundamental, indivisible particles, like an electron (an electron is fundamental and indivisible to within our current experimental precision at least). In that case it doesn't make semse to talk about half an electron, or to divide the mass of the electron among its parts. There is a single well defined frequency/energy for an electron at rest (but not a position :) ).

For composite particles like molecules things are more complicated. De Broglie's hypothesis only applies to free particles, so you shouldn't apply it naively to the bound degrees of freedom within a composite particle. You can however think of a de Broglie wavelength for the center of mass degree of freedom, which is particularly useful when you do experiments that don't probe the internal structure of the object.

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  • $\begingroup$ Thanks for your answer. I know that de Broglie model for his hypothesis was the behavior of a single electron. However, I've seen that atoms and molecules are treated as single particles in matter wave diffraction experiments. In that case we are dealing with a composite object for which each of its parts in some sense have the same average velocity (the atoms as a whole in the molecule or the nucleus components in the atom). How is that the composite object gets a single overall wave frequency in those cases when the frequency of their parts are apparently different? $\endgroup$ – Bruce C Apr 2 '15 at 4:32
  • $\begingroup$ You can't assign de Broglie wavelengths to the individual particles in a molecule, their wave function is not a plane wave. But if you write a wavefunction for the center of mass of the molecule, that will obey a free Schrodinger equation (assuming you can ignore intermolecular forces). One thing to google to learn more about the wave function of the center of mass is positronium. $\endgroup$ – Andrew Apr 2 '15 at 13:00
  • $\begingroup$ I modified my original question above (tried to write it as a comment but it was too long and the system wouldn't let me do it). $\endgroup$ – Bruce C Apr 2 '15 at 16:22
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If it helps, you can think of the de Broglie frequency of a composite particle (like an atom) as a beat frequency. This is not a formal result. The de Broglie wave was a conceptual "step" in the development of quantum mechanics, and was considered to be superseded when the full theory of Schrodinger, Heisenberg, Dirac, etc. was developed. It would be interesting to see an analysis using the full theory of the frequency and wavelength of an atom from its bound constituents. I have not seen it and do not have a reference I am sure contains it, however in a 2007 paper by Wignall (open access here: http://iopscience.iop.org/article/10.1088/0026-1394/44/3/N01/meta ) there is mention of "...the possibility of writing the solution of the N-particle Schrodinger equation for a bound system as the product of a free de Broglie wave with angular frequency equal to the stated additive absolute mass, representing the behaviour of the composite’s centre of mass, times the solution of an (N-1)-particle Schrodinger equation in terms of reduced masses," and some references are given.

Wignall was proposing the use of de Broglie frequency as a replacement for the international standard of mass, so obviously his answer to your question is that it is a mass concept, or can be. He has written numerous papers on the subject you can easily find.

It is oversimplified to view a hydrogen atom as just an electron and a proton, because there are binding energies. The binding energies are what is referred to as reductions to the additive absolute mass in Wignall's comment above. Even a proton is not a fundamental particle, but composed of quarks bound by the strong force, and in addition to binding energy (negative) much of the proton's mass is kinetic energy of the component quarks (positive). Interference experiments have been done with neutrons, which also are composite particles, not fundamental.

The justification for the beat frequency is that for a composite particle to be detected, there has to be some probability of detecting all its constituents. If we multiply the two probability waves together, which would be the standard technique for determining coincident probability (of detecting both of them), we get sum and difference frequencies. See https://en.wikipedia.org/wiki/Envelope_(waves) . The sum of the frequencies, obviously, corresponds to the frequency of the sum of the masses. The trouble with the beat frequency heuristic, which I have not been able to solve, is what to do with the difference frequencies.

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  • $\begingroup$ Thanks for that. I wonder if the differences cancel out when one adds the two (or n) parts together with a Brownian or Zitterbewegung motion between them. en.wikipedia.org/wiki/Zitterbewegung $\endgroup$ – Tom Andersen Jun 26 '16 at 5:51
  • $\begingroup$ @Tom, yes and no. It does not add quite the same way as EM waves, which can result in zero energy. If the waves cancel in one place they will increase in another so that the total probability across all space of finding the energy (particles) is still 100%. This is handled more precisely by the complex number and squared probability of the Schrodinger et. al. theories. $\endgroup$ – Robert Shuler Jun 27 '16 at 16:46

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