In the text, it reads that the momentum of a particle will change if it is moving at speed close to light speed. In the general case, the wavelength is given as
$$
\lambda = \frac{h}{p}
$$
and
$$p = \frac{mv}{\sqrt{1-v^2/c^2}}$$
when $v \to c$, $p\to\infty$, so is it say that the wavelength is ZERO? I don't understand why the wavelength will change to zero if it is moving at speed very close to light speed?
[Physics] How does the wavelength change in relativistic limit
special-relativityspeed-of-lightwave-particle-dualitywavelength
Related Solutions
The basic confusion comes with identifying Photons with Light, i.e. the quantum mechanical entity that a photon is, with the classical electromagnetic wave. The classical electromagnetic wave emerges from a confluence of photons, the quantum mechanical entities, in an analogous way that a crystal symmetry emerges from the quantum mechanical entities of molecules.
Photons, as quantum mechanical entities, have wave functions which have real and imaginary parts which will contribute in interactions to the probability distributions. They build up the classical electric and magnetic fields of the beam in synergy of wave functions, not interacting , but by their complex wave functions being in phase with each other. In vacuum a wave front is built up having the same frequency h*nu, photon energy and light beam frequency.
When the light beam hits a medium, if it is opaque, the photons scatter and are absorbed and turn into infrared eventually. In a transparent medium the organization of the wavefront changes, as the photons scatter elastically, the path on each photon changes, each individual photon is not following the shortest ray path. The collective wave function built up from the scattered photon changes the velocity of the wavefront that they build up accordingly. Thus the wavelength changes because of the changed path length of the constituent photons of the beam, with respect to the ray direction.
Note that the scattering has to be elastic for phases to be retained and a coherent beam to come out. It is not a matter of absorbing and re-emission as is wrongly stated sometimes. If the photon excites an energy level the deexcitation will have arbitrary phases and direction with the original beam.
A more general principle in the quantum framework is the Heisenberg uncertainty relation
$$\large{\color{red}{\Delta \mathrm{x}\Delta \mathrm{p}\ge \frac{\hbar}{2} \\ \Delta \mathrm{E}\Delta \mathrm{t}\ge \frac{\hbar}{2} }}$$
It tells us that when the momentum is zero the position is indeterminate, actually it could go from zero to infinity to obey the principle.
This is consistent with the wavelength being infinite in the de Broglie relation, and it means that the localization of the particle is indeterminate.
What do these generalized relations mean? The Heisenberg uncertainty principle is in one to one correspondence with the quantum mechanical commutator of the two observables. It tells us that if the wavefunction for the problem under consideration is acted upon by the momentum operator and a momentum eigenvalue is obtained the position is indeterminate. It is probable to find the particle anywhere within the boundary conditions, according to the probability distribution given by the square of the wave function.
As we know that quantum mechanics applies to the dimensions of molecules and atoms, and we do observe them in fixed within a width locations in a crystal, and in scattering experiments with fixed momentum beams the particle again are localized within nanometer widths, it is obvious that the simple sinusoidal waves are not adequate to model matter. One uses wave packets to do that, which give localization and can model a particle .
A wave packet solution to the wave equation, like a pulse on a string, must contain a range of frequencies. The shorter the pulse in time, the greater the range of frequency components required for the fast transient behavior.
Quantum mechanics ascribes a special significance] to the wave packet; it is interpreted as a probability amplitude, its norm squared describing the probability density that a particle or particles in a particular state will be measured to have a given position or momentum. The wave equation is in this case the Schrödinger equation. It is possible to deduce the time evolution of a quantum mechanical system, similar to the process of the Hamiltonian formalism in classical mechanics. The dispersive character of solutions of the Schrödinger equation has played an important role in rejecting Schrödinger's original interpretation, and accepting the Born rule.
In other words a single frequency plane wave, which is what the de Broglie relation implies is not a good model for physical matter particles.
Best Answer
Lorentz contraction! The measured de Broglie wavelength in the direction of propagation vanishes because that's what special relativity says happens. The wavelength has to go as $h/p$ as you wrote, so why does it surprise you that when $p$ gets large the wavelength gets small?