Free Particle Wave Function

Why aren't sines and cosines good enough to represent matter waves in quantum mechanics? One reason for this is the need to distinguish between waves with positive and negative energies, and hence freqencies. If we replace $k$ and $\omega$ by $-k$ and $-\omega$ in the cosine form for a plane wave, we get $\cos (-kx + \omega t) = \cos [-(kx - \omega t)] = \cos (kx - \omega t)$. In other words, changing the signs of $k$ and $\omega$ results in no change in a plane wave expressed as a cosine function. The two quantum mechanical states, one with wavenumber and frequency $k$ and $\omega$ and the other with $-k$ and $-\omega$, yield indistinguishable wave functions and therefore would represent physically indistinguishable states in this case. However, since wavenumber and frequency are related to momentum and energy, these states are in reality quite different. The cosine form is thus insufficiently flexible to represent quantum mechanical waves. On the other hand, if we replace $k$ and $\omega$ by their negatives in the complex exponential form of a plane wave we get $\psi = \exp [-i(kx - \omega t)]$, which is different from $\exp [i(kx - \omega t)]$. These two wave functions are distinguishable and thus correspond to distinct physical states.

Another answer has to do with the probability interpretation of the quantum mechanical wave function. A plane wave

$$\psi (x,t) = \sin (kx - \omega t)$$ (10.8)

represents a particle with a definite momentum $\Pi = \hbar k$ and energy $E = \hbar \omega$ in the old view. According to the uncertainty principle, a particle with definite momentum, i. e., with $\Delta \Pi = 0$, has completely indefinite position $\Delta x = \infty$. However, the relative probability of finding the particle is proportional to the square of the wave function, or

$$P = \vert \psi \vert^2 = \sin^2 (kx - \omega t) .$$ (10.9)

This probability distribution oscillates between zero and one, depending on the position $x$ (and time $t$). Thus, the position of the isn't completely indefinite, in that we know that the particle is not located where $kx - \omega t = m \pi$, $m$ being any integer.

On the other hand, if we represent a particle with definite momentum and energy by the wave function

$$\psi (x,t) = \exp [i(kx - \omega t)] ,$$ (10.10)

then the probability distribution is given by

$$P = \vert \psi \vert^2 = \exp [i(kx - \omega t)] \exp [-i(kx - \omega t)] = 1 ,$$ (10.11)

i. e., the probability of finding the particle anywhere in space and time is uniform. This is more in line with our view of the uncertainty principle.