# Particle as wave, stable?

1. Schrodinger equation solutions for a free particle is a sum of terms of the form:

$$\psi(\mathbf{r}, t) = Ae^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$$

however, a single element of this form can not normalize, thus, can not exists alone. That is, a particle must be an addition of several terms, a wave packet. (TODO: verify a Gaussian wave packet normalizes :-).

1. The restriction:

$$\omega = \frac{\hbar k^2}{2m}$$

applies to previous wave function. That means that each component of the wave packet has a different propagation speed. As consequence, the particle spreads.

The question: particles tends to disperse (dissolve) ? If so, how to explain the stable existence of protons, fermions, ... ?

• Free wave packets disperse. Do learn about Schroedinger wavepackets in oscillator potentials, which do not! – Cosmas Zachos Sep 8 at 19:32
• – Cosmas Zachos Sep 8 at 22:18

Free particules tend to disperse, this is true. However this is not a dispersion like the dispersion of some energy into space. This is a probabilistic dispersion.

Think of Brownian motion for example. A particle starting from a known position undergoes Brownian motion. If you look at the probability density, it tends to disperse exactly like heat spreads (it is actually the same differential equation). However, after some time, your particule is still at a single place. This place is known with growing uncertainty but it is still a single place.

QM is not as simple as a probabitity dispersion: something that would be at a single place with growing uncertainty. This would be a pure particle model with stochastic motion, not the wave/particle duality. But understanding the (squared modulus of the) wave function as a probability density helps. This is not really like a spreading of energy (a dissolution).

The question: particles tends to disperse (dissolve) ? If so, how to explain the stable existence of protons, fermions, ... ?

It is important to have the correct frame of reference,in physics the mathematical models we are using are not "explaining" the data, i.e. the data is not created by the mathematics, but is modeled by the mathematics . Successful physics models are the predictive ones, not just the descriptive ones.

Schrodinger's equation has plane wave solutions and it is true that a plane wave cannot be normalized, It is evident a single plane wave cannot be used to model free particles. Schrodinger's equation is very successful (predictive) in modeling particles in potential wells, and the theory of quantum mechanics that developed has in its postulates that the wavefunction describing particles are probability distributions.

The modeling of single particles using plane waves (probability waves) uses the wavepacket solutions, to model single particles . These can be made narrow enough to restrict the particle to the dimensions experiments have measured, and thus model stable particles like protons and electrons if necessary for visualization.

Fortunately, physics models of quantum mechanics have progressed to quantum field theories, and it is not necessary to describe mathematically single free particles in order to predict the behavior of experiments. Experimental predictions are done with QFT calculations using Feynman diagrams as you will learn if you continue your studies.

Maybe this answer of mine will help.

It's right, you cannot normalize a plane wave.

But... remember that plane waves cannot exist in nature, because nothing is actually infinite. Plane waves do not exist, no matter what waves we're talking about.

But! They are a nice approximation for many situations. You can approximate many usual waves to be plane, and it works well, because it is a very good approximation.

In the same way, tehre are never 100% free particles, but if they are far enough from other objects, the free particle approximation is a good one. Also is its solution.

So yes, if we were extremely strict, we shouldn't admit plane waves, but we do because tehy work well under many situations.

Note, altough many states are not normalizable, mean values can be normalized, because the divergence vanishes. Some sort of...

$$\langle A\rangle = \dfrac{\langle\psi|A|\psi\rangle}{\langle\psi|\psi\rangle}$$

The terms that would cause the divergence just cancel out. SO, numerator and denominator do not work well separately, but put together, they give a finite value.

So that's why we can make an exception, admittin plane waves

• The question is not about plane waves: it is about dispersion and its significance. – Cosmas Zachos Sep 9 at 0:30

The solutions of the Schroedinger equation (SE) are not "particles." They are wave functions. (More precisely, a wave function can be written as a normalizable superposition of the solutions of the SE.) The modulus square of a wave function gives a probability distribution to observe a particle at a specific point. This is the essence of the wave-particle duality.