Doppler shift in De Broglie's wavelength

Question

If I were to take a particle, moving at a non relativistic velocity, it will have a wavelength (according to De Broglie). However, If I were to change the referance frame of the motion, I know that the wavelength will change because the momentum changes. But will the change be synonymous with the shift of wavelength, if I did the same problem, just considering it wave and applying doppler's shift.

If so, why? And if not so, then also, why?

What would be the changes if I were to take a relativistic frame instead of a non relativistic one?

PS: Due be noted that I am a high school students, who just has a surface level understanding of lorentz transformations and relativity.

Answer 1

The basic answer is that in Newtonian physics, the wavelength of a wave doesn't change when you switch reference frames. This is simply because in Newtonian physics, time is universal, and the distance between two events at the same time are the same in all reference frames. So if the crests of the wave are 1 µm apart in one frame, they're 1µm apart in all reference frames. The frequency of the wave does change — that's the Doppler shift you know & love — but so does the speed of the wave involved, in such a way that the wavelength stays the same.

Under Lorentz transformations, on the other hand, both the frequency and the wavelength shift because measurements of both space and time change. You can use these transformation laws to find out what the new frequency and wavenumber of the wave are, and you find that they obey the sort of Doppler-shift law you would expect for light waves and they change as you would expect for Newtonian velocity transformations at low speeds.

Answer 2

If $\psi(x,t)$ is a solution of the time-dependent Schroedinger equation in a frame at rest, then, using the chain rule for partial derivatives, you will find that from a frame frame moving at speed $-U$ the solution becomes $$ \tilde \psi(x,t)= e^{imUx/\hbar -i\frac 12 mU^2t/\hbar}\psi(x-Ut,t). $$ In particular, wavefunctions do not transform as scalars. A plane wave $e^{imvx/\hbar}$ will be seen as $e^{im(v+U) x/\hbar}$, but I would not call this a Doppler shift.

Thw weird transformation occurs because the Schroedinger wavefunction is related to the Klein-Gordon equation solution (which is a Lorentz scalar) by $\psi_{\rm KG}(x,t) = e^{-imc^2 t} \psi_{\rm Sch}(x,t)$ and the $c^2$ in the exponential cancels the factors of $1/c^2$ in $$ t'= \gamma (t+ U x/c^2), $$ when you charge to the moving frame.