When does the wavefunction wavelength equal the De Broglie wavelength?

[etotheipi]

For instance, in the case of the infinite square well, the wavelength of the wavefunction is $\frac{2L}{n}$. This also turns out to be the De Broglie wavelength, and and we can find the possible energies directly from the Schrodinger equation, or by using the De Broglie relations.

However, if the wavefunction is a measure of probability amplitude, and the De Broglie wavelength is a measure of the particle's momentum, why are they equal? And when are the not equal - am I missing something more fundamental here? Thank you!

 

[Dr_Nate]

The de Broglie wavelength is a nice way to introduce QM to students, but as far as I'm aware it doesn't have a lot of higher order applicablilty. The wave function is the more complete concept.

From the wave function, you can calculate many properties, including the momentum of the particle. But, the momentum is not always a particular number (look up uncertainty relations). The de Broglie wavelength suggests that momentum is always one number. What you've found is a case where they both agree.

 

[jtbell]

But, the momentum is not always a particular number (look up uncertainty relations).

And neither is the de Broglie wavelength. In general a wave function can be considered as the superposition (sum) of a set of wave functions with definite momentum and de Broglie wavelength. The mathematics (Fourier analysis) was well-known long before QM existed.