Wave function of free particle According to wiki, the wave function for a free particle is:
$$ \psi(\mathbf{r}, t) = Ae^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)} $$
that with the necessary restrictions and uni-dimensional became:
$$ \psi(\mathbf{r}, t) = Ae^{i k \left( x-\frac{h}{2m\lambda} t \right) } $$

*

*this wave function must fulfill normalization, thus
$$|A|^2 x \rvert_{-\infty}^\infty = 1 $$
something impossible for any $A $?


*to which kind of particles is this wave function applicable ?


*In case it is applicable to photons, we could compare with the electromagnetic waves:
$$ Ae^{i k \left( x-vt \right) } $$
(with $v=c$ in empty space)
we can see that in newtonian $v$ is a term related to the medium, being a universal constant in the case of empty space, while in quantum it is $\frac{h}{2m\lambda}$, something that depends only in particle characteristics $m$ and $\lambda$. How to match both facts?
 A: About your first question, the wave function of a free particle does not belong to the vector space $L_2(\mathbb{R})$ of the functions having the square modulus integrable in $\mathbb{R}$. Formally it lives in a generalized vector space, whose elements are known as"distributions" in mathematical literature. I don't remember the mathematical details.
However, physically this reflects the Heisemberg's principle. In this representation you know exactly the momentum of the particle, which is $\hbar k$ (simply apply the momentum operator $-i\hbar \partial/\partial x$ to the function and check what you get); so Heisemberg's principle states that you have total uncertainty on the position, and indeed the probability distribution in space is a constant $|A|^2$. The non integrability is a pathology in the formalism due to the fact that you don't have any uncertainty in the momentum value.
Coming to the second question: this can be applied to any free particle if you take the general form $Ae^{i(kx-\omega_k t)}$, where $\omega_k$ is the dispersion relation, which depends on the specific problem. For example this is a solution of Schrodinger equation with no external potential if $\omega_k = \hbar k^2/2m$.
The problem with photons is that, being fully relativistic ($m=0$) objects they are not correctly described by the Schrodinger equation. In fact the correct dispersion relation for phonons is NOT $\omega_k=\hbar k^2/2m$ clearly, because this would not make sense given $m=0$. Instead it is $\omega_k = c k $, as we already know for electromagnetic waves; but this would be consistent with a fully relativistic generalization of Schrodinger equation. (And that was question 3 :) Hope this helped!)
