The wave function described by Schrodinger's equation is interpreted as describing the probability of a particle in at any point in space, i.e. a probability distribution. Since this distribution cannot be a repeating wave (probability peaks at a particular place), how can we have a wavelength of these matter waves (de Broglie wavelength) ?
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2$\begingroup$ "probability peaks at a particular place" you are thinking of classical statistical probability for random motions. Probability is a more general term. en.wikipedia.org/wiki/Probability $\endgroup$– anna vCommented Feb 9, 2016 at 19:30
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The wavefunction associated to a free particle of momentum $p$ is $$ \psi_p(x) = \mathrm{e}^{\mathrm{i}px/\hbar}$$ which is obviously just a plane wave with wavelength $h/p$, fully compatible with the deBroglie relation.
Strictly speaking, this function itself is not an admissible wavefunction because it is not square-integrable and hence $\lvert\psi_p\rvert^2$ not a normalized probability distribution, but superpositions of these plane waves can be square-integrable, and they are the actual wavefunctions.
The particle doesn't have a definite momentum in those states, so its de Broglie wavelength is also not definite - it is smeared out. If the superposition is sharply peaked around a certain momentum, the resultant wave-packet behaves very much like a wave with the associated de Broglie wavelength.
answered Feb 9, 2016 at 18:48