From de Broglie hypothesis we know that for a particle with Momentum p, Wavelength(L)
$$L=h/p$$
Now suppose we look at the same particle from another reference frame, so the momentum P gets changed and if this law is to be conserved then the wavelength also has to be changed but this is absurd as wavelength is not dependent not reference frame. So,where lies the fallacy?
Using $\lambda=\frac{h}{p}$ and $\omega=\frac{E}{\hbar}$ we can write the following relations: $$\tag{1}\vec{p}=\hbar\vec{k}\label{1}$$ $$\tag{2}E=\hbar\omega\label{2}$$ In general, the four-momentum is: $$P=\left(\frac{E}{c},\vec{p}\right)$$ Using (\ref{1}) and (\ref{2}): $$P=\left(\frac{\hbar\omega}{c},\hbar\vec{k}\right)$$ We can define the four wave-vector $K$ as: $$K=\left(\frac{\omega}{c},\vec{k}\right)$$ So the four-momentum becomes: $$P=\hbar K$$ All the four-vectors are frame-independent.
The answer to this question depends on whether we are dealing with relativistic systems or not. The answer by Arthur refers to the relativistic case where the wavelength changes because of Lorentz contraction. In my opinion the non-relativistic case of Schroedinger waves is rather more interesting because what happens is quite subtle. Suppose we have a solution $\psi(x,t)$ to the Schroedinger equation $$ i\hbar\frac{\partial \psi}{\partial t} = -\frac {\hbar^2}{2m} \frac {\partial^2\tilde \psi }{\partial x^2} + V(x)\psi, $$ and look at it from a moving frame where the equation becomes $$ i\hbar\frac{\partial \tilde\psi}{\partial t} = -\frac {\hbar^2}{2m} \frac {\partial^2\tilde \psi }{\partial x^2} + V(x-Ut)\tilde\psi. $$ Then some manipuations with the chain-rule for partial derivatives shows that we must have $$ \tilde \psi(x,t)= e^{imUx/\hbar -i\frac 12 mU^2t/\hbar}\psi(x-Ut,t). $$ The extra phases mean that quantum wavefunctions do not transform as scalars under Galilean transformations. For example the wavefunction for an electron in the ground state of a hydrogen atom is a real function, but seen by someone running at speed $U$ towards the atom, the electron has a net momentum $p=mU$ towards the runner and the wavefunction has an $e^{-imU/\hbar}$ factor.
How this arises is quiate interesting. To get the non-relativistic wavefunction $\psi(x,t)$ for a particle of mass $m$ from a relativistic wavefunction $\phi(x,t)$ that does transform as a scalar, we write $$ \phi(x,t)= e^{-imc^2 t/\hbar}\psi(x,t). $$ In other words the non-relativistic $\psi$ does not include the phase coming from the $mc^2$ rest-mass energy. Suppose now we make a Lorentz transform $$ x\to x'= (x-Ut)/\sqrt{1-U^2/c^2}, $$ $$ t\to t' = (t+Ux/c^2)/\sqrt{1-U^2/c^2}. $$ Even when $U$ is very small compared to $c$ --- so that we are effectively making a Galilean transform --- the $c^2$ in the $mc^2$ cancels with the $c^2$ in the $x/c^2$ term and the result is the extra factor of $e^{imUx/\hbar}$ in $\tilde \psi$.
I first learned all this from a problem at the end of Chapter 22 in Gordon Baym's "Lectures on Quantum Mechanics."
