A different angle on this that I DON'T believe is in conflict with Terry Bollinger's answer: whether you express a wavefunction in position co-ordinates, or, as its Fourier transform, i.e. in momentum co-ordindates, the two models are precisely the same. So neither the expression of a position co-ordinate wavefunction (such as you find from the solution of the chemist's wonted form of the Schrödinger or Dirac equation for the hydrogen atom) nor its Fourier transform are any less "real" or "physical" than the other.
More technically, most often in quantum mechanics we take the space of functions to be the separable (i.e. having countable basis) Hilbert space $\mathcal{L}^2(\mathbb{R}^N)$ of square Lebesgue integrable functions (well almost: see footnote). Such functions are a proper subset of the so-called tempered distributions, which are mathematical objects (broadened notions generalising functions) for which:
The distribution itself and its Fourier transform constitute precisely the same information
One can be reversibly transformed into the other by a one-to-one transformation (i.e. the Fourier transform) which is also unitary when we restrict to $\mathcal{L}^2(\mathbb{R}^N)$. If you need more information on tempered distributions, I give more discussion here in my answer to the question "What restrictions on time boundary conditions does it have to use Fourier transform to solve wave equation?" as well as here, in answer to the question "Are all scattering states un-normalizable?" and here.
What Terry's answer is about is that plane waves and delta functions, the "eigenfunctions" of the momentum and position observables, respectively, do not belong to the quantum state space $\mathcal{L}^2(\mathbb{R}^N)$ of physical wavefunctions: they are not Lebesgue integrable. So these observables actually have no eigenfunctions in this quantum space. It helps our intuition to extend the Hilbert space to the more advanced concept of a Rigged Hilbert space so that we can talk about these rigorously as vectors of the broadened notion of the observables in the advanced framework. But physical wavefunctions have finite extent and thus always have a nonzero spread in their Fourier transforms.
Footnote: More precisely: we take our quantum state space to be the Hilbert space of equivalence classes of Lebsesgue integrable function modulo the relationship equality almost everywhere (see Wiki page "Almost Everywhere"). So we're talking about $\mathcal{L}^2(\mathbb{R}^N)\,/\,\sim$, where $\sim$ is equality almost everywhere. Another way of saying the same thing, we can take it to be the span of a complex countable set of basis functions, such as quantum harmonic oscillator eigenfunctions)
