The wavelength of a particle of momentum p is calculated using De Broglie relation.
The de Broglie relation was postulated for what is called a matter waves. Now according to the statistical interpretation of QM we know that there is no matter waves. It is an abstract wave called the wave function which describes the particle probabilistically.
1- Is de Broglie relation still valid although matter waves do not exist ? and why?
2- If the answer to 1 is affirmative then what is de Broglie wavelength of a particle that is localized in space? (because now the particle will not have sharp momentum, hence not a unique de Broglie wavelength.)
Best Answer
Actually, according to quantum mechanics, for every particle (matter), there is a complex valued wave function and it is the modulus of that wave function that is the probability that the particle is at a given location.
For example, a wavefunction for a uniform plane wave of matter could be:
$$\psi(\vec x, t) = A e^{i(\vec k \cdot \vec x + \omega t)} = A(cos(\vec k \cdot \vec x + \omega t) + i\cdot sin(\vec k \cdot \vec x + \omega t))$$
This wavefunction is a complex valued wave (notice the sine and cosine functions) that has a wavelength ($1/|\vec k|$ = the de Broglie wavelength) and a frequency ($\omega$). It is this complex wave that could result in quantum interference effects.
However the probability for a particle to be at a position $\vec x, t $ is:
$$P(\vec x,t) = |\psi(\vec x, t)|^2 = |A|^2 $$
So it is constant everywhere in space and time and is not a wave. Of course, an actual wavefunction for a particle would be something like this plane wave multiplied by a gaussian to result in a wave packet that is localized in space and time and that travels the way a classical particle would travel (see wave packet for more information).
But the key point is that there is a complex valued wavefunction with wave-like properties but that the probability does not have to have any wave like properties.
EDIT for edited question:
For a particle localized in space, there is still an average position and an average momentum so the average de Broglie wavelength will be the one that corresponds to that average momentum. According to Heisenberg there will be an uncertainty in position and an uncertainty in momentum and thus there will be an uncertainty in the de Broglie wavelength also.
What this really means is that whenever you have a localized particle, you have a wave packet which means there is a superposition of a number of plane waves with similar momentum, in other words a superposition of a number of different de Broglie wavelengths (see wave packet for more information).
If you are wondering if the concept of de Broglie wavelengths is useful in this situation, then the answer is that it is not very useful - just work with your mathematically defined complex valued quantum mechanical wavefunction instead.