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:

$$\psi(\mathbf{r}, t) = Ae^{i k \left( x-\frac{h}{2m\lambda} t \right) }$$

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

2. to which kind of particles is this wave function applicable ?

3. 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?

1 Answer

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!)

• Thanks for your answer. According to my information, photons are only massless in empty space. Sep 6 '20 at 20:04
• I think they are always massless, but when they propagate in matter you have to account for interaction with the environment, thus you must add a potential term to the (relativistic generalization of the) Schrodinger equation, and in principle you can't expect their wave function to be of the form $Ae^{i(kx-\omega t)}$ Sep 6 '20 at 20:11