Can the de Broglie wavelength become infinite?

i am new to the concept of de Broglie wavelength.

We all know that $\lambda=\frac hp$ and $E_\text{kin}=\frac{p^2}{2m}$ implies $p=\sqrt{2E_\text{kin}m}$ and therefore $\lambda=\frac{h}{\sqrt{2E_\text{kin}m}}$

So, say an object like a baseball of mass $m$ is lying at some height and such that it has no kinetic energy, i.e. it is at rest with $E_\text{kin}=0$. Substituting the value of $E_\text{kin}$ in the above equation then it gives us "$\lambda=\frac{h}{\sqrt{2\cdot 0\cdot m}}=\lambda=\frac h0\implies \lambda=\infty$".

What does that mean?

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.

• Hi, sir; just did a little modification by removing the pic. Instead used $\LaTeX$ & made it look as close as the pic:P Just one thing I added is $\ge$ instead of $\gt$ in the Uncertainty Principle. I hope this wouldn't bother you but if it is doing so, then please feel free to rollback the edit; it's your answer, after all. BTW, I would latexify all the equations in your answers as far as I can; just a charity-work:) Good-night! – user36790 Oct 30 '15 at 17:24
• @user36790 Thanks, No problem. I am an old dog and reluctant to learn new tricks like Latex and mathwhatnot. – anna v Oct 30 '15 at 17:25