Are de Broglie wavelength and wavelength of a wave function the same thing? I know that de Broglie proposed that all matter exhibit wave particle duality. He proposed that the Waves of these matter are called matter waves and he also proposed that they had a wavelength which he called Dr Broglie wavelength. Today we know that all matter have a characteristic wave function which gives us the probability of finding the particle at a given position. So are de Broglie wavelength and wavelength of a wave function the same thing?
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$\begingroup$ Related: physics.stackexchange.com/q/592011 and links therein $\endgroup$– Nihar KarveCommented Jun 17, 2021 at 6:35
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2 Answers
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The de Broglie wavelength was proposed , together with the Heisenberg uncertainty principle, at the time quantum theory was developing. With the establishment of quantum mechanics one can see a relationship, but not a derivation.
The HUP comes out naturally from the commutators of the operators of quantum mechanics. The de Broglie wavelength is vaguely related to the "wave" in wavefunction, because a wavefunction is a mathematical solution of a boundary condition problem, and it is responsible for the interference effects measured in experiments ( as the single electron double slit) but differs according to the boundary conditions, so assuming just the mass has no meaning for elementary particles.
See the answer here to see the correlation of the DeBroglie wavelength to the solutions of the hydrogen atom. In this sense it is still useful to estimate space occupation for complex systems. Note though that if has not much to do with the wavefunction itself. If one wanted to do a double slit experiment using hydrogen, the DB wavelength would be useful to estimate the width of the slits and the distance between them in order to see interference, for example. The real wavefunction of the experiment would depend on those calculated numbers and the interference would be related.
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$\begingroup$ can you explain more on this part of your answer? "so assuming just the mass has no meaning for elementary particles" $\endgroup$– P.A.MCommented Apr 18 at 16:57
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The word wave is used a lot in physics - wave theory, em waves, matter waves, wave functions, waviform, waveguides, wave packets, microwave, retarded wave etc. They all differ in their meaning.
Are de Broglie wavelength and wavelength of a wave function the same thing?
No. Wavefunctions are full fledged solutions of Schrodinger's famous equation that contain all the observable information about a system. They aren't waves - they are imaginary functions that can be interpreted as probability densities.
de Broglie's enlightening extension of wave nature to all matter uses the simple relation that a wavenumber $k=p/\hbar$ can be associated to all momenta $p$. This wave nature is real and is the extension of the wave-particle duality to matter as photons were to light. The associated waves of say, definite energy electrons, do correspond to the predicted $k$.
However, there is a subtlety that de Broglie's formula sidesteps. First one must note that de Broglie's thesis came in, if not nascent, then the adolescent years of quantum mechanics and preceded the formulation of Schrodinger's eqn. As such, its limited in its correct applicability.
The subtle issue is that a particle like electron when sharply localized in space - its mass may occupy only so and so region - has no single associated momentum value that can be attributed to it. This is a consequence of the quantum uncertainty theorem. As a result it isn't clear how its matter wavelength can be calculated and if it would have any relationship to its momentum like the deBroglie relation.
If instead, one tries to extract information about the localized electron by solving the Schrodinger eqn., the full mechanics, one realizes that its wave function, aphysical as it may be, comes out to be as if de Broglie waves corresponding to all those spread out momentum values had been added up.
So while a free particle may follow de Broglie's eqn as a purely monochromatic matter wave, as per quantum mechanics it would be infinitely spread out, and being unnormalizable, not form just by itself, a physically realizable solution to Schrodinger eqn. For that we construct several such waves to construct wavepackets that physically and better represent reality though then we aren't talking about the original de Broglie wave, are we.
