# Why does the wave description say that probability oscillates, while the phase interpretation says constant amplitude?

The wave description of a particle illustrates an oscillating probability of the particle being found in any point in space.

When a particle travels, it carries along with it a phase that oscillates in the complex plane. When the particle travels, the amplitude never changes although the phase does, which doesn't matter. The only property that matters is the amplitude.

Why does the wave description say that probability oscillates and the phase interpretation says constant amplitude? What am I missing?

Let me see if I can put this in simple terms. I wrote a simulation of this once.

You start with a simple wave like $exp(i \omega_1 xt)$ that sits there oscillating on the real x line. That represents a particle whose energy you know, but whose position you don't, because the probability of it's being anywhere in x is everywhere equal. Now you introduce some uncertainty in energy, adding in a second wave with frequency $\omega_2$. When you do that, you get a beat frequency between the two waves. Where the waves reinforce, the wave is strong and the probability is high. Where they cancel, the wave is weak and the probability is low. If you add in another wave at, say, $\omega_3$, you still get a beat pattern, but the peaks are further apart. Every additional wave you add causes the peaks to appear further apart. So that's how more uncertainty about energy causes less uncertainty about position (because there are more places where the probability is low). This is five waves added together, with slightly different $\omega$: Take this to its extreme:

Add in an infinite number of waves, all of slightly differing $\omega$, and the peaks become infinitely far apart, so the one peak you see has all its probability in one place.

I assume you're referring to the wave equation of a free particle in quantum mechanics. The free Hamiltonian is just the kinetic energy $p^2/2m$ (for a massive particle). So, the eigenstates of the Hamiltonian are simply the momentum eigenstates, i.e. particles with constant momentum.

Mathematically, you can write the position-space wavefunction as $\psi(x) = e^{-i \omega x}$. As you point out, the phase here cancels in $|\psi|^2 = 1$. Considering the uncertainty principle, you can see that a particle with definite momentum must have completely indefinite position, i.e. the probability of finding at a given position is the same everywhere. And indeed, this wavefunction is not normalizable, since a momentum eigenstate is a singularity of the uncertainty principle: $\Delta x = \hbar/2 \Delta p$

• the picture i had in my head was a beam of particles, like a laser or something – Ben Steen Apr 7 '13 at 17:32
• Right, so a laser emits photons at a (very nearly) fixed energy/momentum. In empty 1-d space, how can you say "where" the laser beam is? It goes on indefinitely. – chase Apr 7 '13 at 23:51
• and the beam can be represented by a sin wave, illustrating its oscillating phase with time. if you choose to pick a point at a time where a 'node' is, you'll get zero intensity. – Ben Steen Apr 8 '13 at 14:21
• but shouldnt the amplitude oscillate in time but never change magnitude? – Ben Steen Apr 8 '13 at 14:21
• im confused why people say the phase oscillates in time and the amplitude stays the same (the reason for having complex numbers). but the intensity of a traveling beam does oscillate with time – Ben Steen Apr 8 '13 at 14:23