Help understanding De Broglie relation

[bob900]

I'm having trouble understanding some things about the De Broglie relation, λ=h/p

1. A massive system, such as an atom, is a composite of several particles (protons/neutrons, electrons), each of which is a matter wave with its own wavelength, according to De Broglie relation. But the atom considered as a single particle, has its own (lower) wavelength. How do these individual λ_proton, λ_electron, etc. waves combine into the single λ_atom wave? In other words, what does it mean that the atom is a matter wave with a single wavelength - isn't it a composite of its many individual component waves?

2. Actually, even a single particle does not have a single well defined wavelength - the Fourier decomposition of its wave function produces many component wavelengths - so what does the λ in λ=h/p stand for then?

 

[Darwin123]

I'm having trouble understanding some things about the De Broglie relation, λ=h/p

1. A massive system, such as an atom, is a composite of several particles (protons/neutrons, electrons), each of which is a matter wave with its own wavelength, according to De Broglie relation. But the atom considered as a single particle, has its own (lower) wavelength. How do these individual λ_proton, λ_electron, etc. waves combine into the single λ_atom wave? In other words, what does it mean that the atom is a matter wave with a single wavelength - isn't it a composite of its many individual component waves?

A composite body can be treated as one wave, in the de Broglie sense, if and only if all its composite particles are fully entangled. The wave has to be in a state where the individual particles are fully indistinguishable. This means that the probability distribution of the particle is invariant to particle exchange.
The hidden assumption here is that the ground state of the atom or molecule is an entangled state of the individual components. In other words, probability distribution of the nucleus is invariant to an exchange of protons and invariant to an exchange of neutrons. Therefore, the ground state of the nucleus can be treated as the wave from a single object (i.e., the nucleus).
Excited states do not have to be in an entangled state. Therefore, the de Broglie relations may not work so well for an excited nucleus.
The same goes for molecules. The ground state of the molecule is assumed to be an entangled stated of the individual protons, neutrons and electrons. Therefore, the entire nucleus acts as a composite body.
This gets into the physics of entangled states. Many body quantum mechanics is a little different from single body quantum mechanics, which you are probably more used to.
A multiparticle wave function has to be decomposed into individual entangled states. This is because the Hamiltonian (i.e., energy) of a multibody system is invariant to exchange of particles which are the same type. Multibody wave functions have to be decomposed into eigenfunctions of the exchange operator.
The ground state of bound fermions is unique since two fermions can't exist in the same state (Pauli's exclusion principle). Therefore, one can be sure that the ground state of a collection of fermions will be entangled. Thus, the ground state can always be treated as a single composite particle by the de Broglie relations.
The issue really is what sort of systems are entangled and which are not entangled. Although de Broglie didn't understand entanglement, entanglement is fundamental in understanding the de Broglie relations.

2. Actually, even a single particle does not have a single well defined wavelength - the Fourier decomposition of its wave function produces many component wavelengths - so what does the λ in λ=h/p stand for then?

If the wavelength of the wave is uncertain, then the momentum of the particle is uncertain.
The Fourier decomposition of it wave function produces a distribution of momentum states. The square of the amplitudes of this Fourier decomposition determine the probability of the detected particle have the specific momentum associated with the amplitude.