# Why is the position space free particle wavefunction a function of momentum?

This is one of those little things that has always confused me. If someone said to you "in quantum mechanics, the eigenfunctions of a free particle are $\exp(ipx/\hbar)$" how would you know that this is the position space wavefunction, and not the momentum space one? The wavefunction itself is an explicit function of both.

There are a great many other simple quantum systems (square well, harmonic oscillator and so on) in which the momentum doesn't enter the real space wavefunction at all, but those particles still have an associated momentum.

So I guess my question is: what does that $p$ actually mean?

No, the (elementary solution for the position representation of the) wavefunction of a free particle, $\psi(x) = \mathrm{e}^{\mathrm{i}px}$ is not an "explicit function of both" position and momentum. It is a function of position - the momentum of the plane wave is fixed, and the momentum space wave function of this is its Fourier transform $\tilde{\psi}(k) = \delta(k-p)$.
By the way, note that neither of these are square-integrable functions, and hence not actual possible wavefunctions. Actual wavefunctions are, for example, Gaussian wavepackets made out of superpositions of $\mathrm{e}^{\mathrm{i}px}$ for different $p$ with a certain width $\Delta p$.