Is it accurate to say “a wavefunction is a function of particle positions or momenta”?

Something has been bothering me for a while. I encounter this kind of statement everywhere:

While a single particle is described by a wave function $\Psi({\vec r};t)$, a system of two particles, call them 1 and 2, is described by a wave function $\Psi({\vec r}_1,{\vec r}_2;t)$ depending on both particle positions. Leon van Dommelon

and

A free quantum particle is described by the Schrödinger equation:

$\frac{\hbar^2}{2m} \nabla^2 \ \psi(\mathbf{r}, t) = i\hbar\frac{\partial}{\partial t} \psi (\mathbf{r}, t)$ where $\psi$ is the wavefunction of the particle at position r and time t. Wikipedia

Many other examples of this kind of phrasing are on Physics Stack Exchange as well.

The problem is that it seems to me that the wavefunction is not a function of particle positions at all, for positions are not defined until the system is collapsed with a measurement. It would be weird if the wavefunction was a function of random eigenvalues. Instead, it looks like the wavefunction is a function of coordinates in Euclidean position space $\mathbb{R}^{3n}$ that maps a list of $x_1, y_1, z_1, x_2, y_2, z_2\ldots$ to a complex number.

Am I wrong? Is there a good reason to say that the wavefunction is a function of particle positions (or momenta)?

• I think they're just using the term "position" as a loose synonym for "group of coordinates". It's also convenient and intuitive to think of them as particle positions, since they correspond to all the potential positions the particle may have. It seems like a natural extension of the term "position" into the realm of quantum mechanics. – Brionius Jan 5 '15 at 23:26
• The modulus squared of a wavefunction gives a probability density of finding a particle at a certain position $\mathbf x$. Hence the probability of finding the particle in a region $D$ at an instant $t$ is just $\int_D|\psi(\mathbf x,t)|^2\ \text d^3\mathbf x$ – Phoenix87 Jan 5 '15 at 23:31
• Re "... where ψ(r,t) is the wavefunction of the particle at position r and time t." There's nothing wrong with that. It is not saying that r is the particle's position. It is saying that ψ is a function of position and time -- which is exactly what ψ(r,t) means. – David Hammen Jan 5 '15 at 23:36
• It may help to remember that the position of a mass in Newtonian mechanics isn't defined as the position of a point mass, either. It's defined as the coordinates of the center of mass for extended objects for which internal degrees of freedom don't matter for the description of their movement. So even though the Earth is 6500km in radius, for the purposes of applying Kepler's law to it we can reduce that to $(x, y, z)$ or $(r, \phi, \theta$), neglecting all the internal structure of its mass distribution. – CuriousOne Jan 6 '15 at 0:08
• While this does not answer your question, I highly recommend reading this question and answer. The use of wave functions to represent quantum states of multiple particles is much easier to understand if you learn about the so-called "second quantization". Don't be afraid of the name, it's actually a stupid overly scary name for something which is quite simple. – DanielSank Jan 6 '15 at 1:53

However, a "position" in this sense, a.k.a. spatial coordinate, can be considered a potential position of the particle. You can then think of a wavefunction as the amplitude associated with a universe in which the particle does exist at that position. (This is basically the foundation of the many-worlds interpretation of quantum mechanics.) It's often useful for conceptual understanding to think of wavefunctions in this sense, and so this is a way in which it can be considered valid to talk about $\vec{r}$ as particle position.