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Until what point can the de Broglie wave be thought as a real wave?

I mean, is it made of something? What amplitude does it have? Is it a sine wave?

How can it be related to the wavefunction of the particle?

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All matter is made of waves—at least it can be represented that way, and it behaves that way. Of course, matter also behaves like particles. This is one of the odd-but-true conclusions of quantum mechanics.

The de Broglie wavelength gives the wavelength of any "matter wave." These waves aren't waves in the classical sense with amplitude and the like; they are wave functions, which express the probable location of a particle as something that looks like a wave. You can think of it in a sense that, looking very closely, the location and motion of a particle becomes blurry and starts to look like a wave instead of just one point.

We usually think of matter as a wave only when making observations on a scale comparable to the de Broglie wavelength, which is very small for most things. Using the de Broglie relation $\lambda=\frac{h}{p}$, you can calculate the wavelength of a particle with momentum $p$. As an example, a electron with a kinetic energy of 10 eV has a wavelength of 0.39 nm. That is the wavelength of the electron's wavefunction. If you perform an experiment that would highlight that wave behavior, such as pass a beam of electrons of that energy through a diffraction grating with a spacing comparable to that wavelength, you would see an interference pattern, just as you do with light, because electrons behave like waves (when we want them to).

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Do you mean that de Broglie's wave is the same as the wavefunction $\Psi$? If I calculate the wavefunction for a free electron it's wavelength would be $\frac{h}{p}$? Does this formula apply if the particle is subject to a potential? –  jinawee May 21 '13 at 15:02
    
As far as I know, yes. The de Broglie wavelength of a particle is the wavelength of its wave function in free space. I think that this is true when the particle is subject to a potential as well, as is the case with electrons in atomic orbits. The difference is that their momentum is quantized. Still... I haven't learned that much about quantum mechanics yet so if someone else would confirm that it would be better. –  krs013 May 21 '13 at 16:52
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De Broglie s' matter wave, is, in some sense, the prehistory of the wavefunction, but its original interpretation is false.

De Broglies relations makes a link between particles characterics (momentum, energy) and wave characteristics (frequency, wavenumber). De Broglie thought (like Einstein and Schrodinger), that it was a real "matter wave".

But, thanks to Quantum Mechanics, we know now that this is false, the "wave" or "wavefunction" is not a real "matter wave". So De Broglie, Einstein, and Schrodinger were wrong.

The correct interpretation - the Quantum mechanics interpretation - is that the "wavefunction" represents a probability amplitude. So it is nothing that complex numbers which allow you to calculate probabilities.

In fact, in Heinsenberg representation, where operators depend on time, and states (wave functions) are constant, we can interpret wavefunctions - a set of complex numbers - as some kind of initial conditions. So it has nothing to do with a real matter wave.

If you want to know the exact behaviour of the wavefunction, you will have to use the Schrodinger equation.

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