Is the wavefunction of particles inside a gas spread or localized?
Preliminaries: How do we define 'localized?'
For a single particle, or for multiple non-entangled particles, it is easy to tell from the expressions for the wavefunctions whether they are localized or delocalized. For example, you might say that if the wavefunction is falling off exponentially or faster for large $x$, that is with a form like $\psi(x)\sim e^{-x/\xi}$ with some characteristic length scale $\xi$, then it is localized, while something like a plane wave (which can be considered to be in the $\xi \rightarrow \infty$ limit) is delocalized.
For interacting particles, the many-body quantum state will generically evolve to something that is entangled between the particle. Then there is no longer a wavefunction for an individual particle, and the question of localization is no longer so straightforward. For example, is the two-particle state $\psi(x_1,x_2)\sim e^{-(x_1-x_2)^2}$ localized or delocalized?
A standard way to generalize this idea of localized/extended to many-body systems is by using the concept of entanglement entropy (1), and asking if a particular region is entangled with another distant region of the system. For a one-dimensional system, the entanglement entropy is:
$S(\rho_A)=-\text{Tr}[\rho_A \log \rho_A]$, with $\rho_A$ the reduced density matrix for that system:
$$\rho_A(x_1,x_2,\ldots x_N,x'_1,x'_2,\ldots x'_N)=\int_{|x_i|>x_0} \int_{|x'_i|>x_0} ~\mathrm dx_1 \ldots \mathrm dx_N~\mathrm dx'_1 \ldots \mathrm dx'_N ~\psi(x_1,\ldots, x_N) \psi^*(x'_1,\ldots, x'_N)$$
Here we are looking at a region from $-x_0$ to $x_0$. If $S$ is exponentially small, then the system is localized, and if it is not it is extended. Notice that we've moved from talking about particles to talking about regions. This is more natural when thinking about localization, but for a uniform density of particles at a particular moment in time localization of one implies the other.
The sense of "localized" that we now have is that for a localized system, a measurement at one point does not perturb the quantum state at a faraway point. Using this standard, if you carry out the above calculations on a state like the above two-particle state, or a single particle plane wave state, you will find that they have non-zero entanglement entropy and are extended. However, a state like $\psi(x_1,x_2)\sim e^{-(x_1/\xi)^2}e^{-((x_2-2x_0)/\xi)^2}$ would be localized, as long as $x_0 \gg \xi$.
Eigenstate thermalization
Okay, with these ideas in place I can now state the answer simply: for a quantum system of particles in a box that interact with a hard-shell repulsion, in a highly excited state and dilute limit, and at equilibrium, the entanglement entropy is proportional to the volume of the system.
What this means, roughly speaking, is that every point in the box is equally entangled with every other point. In this sense, the system is extended. Measuring the quantum state at one point will also affect the quantum state at every other point.
The proof of this is basically due to Srednecki, in a foundational paper of quantum thermalization which I encourage you to take a look at (2). For the above system, Srednecki shows that the eigenstates of the system give particle behavior that agrees with Maxwell-Boltzmann statistical mechanics, and furthermore that systems that start far from equilibrium (such as the case you mention where everything starts out localized) will also evolve to an equilibrium state that obeys these predictions. Furthermore, subsequent work has emphasized that any system that has this self-equilibrating property, known as eigenstate thermalization, will also necessarily show volume scaling of entanglement entropy (see, for example, (3)).
Decoherence
All of what I've said so far has been about the pure quantum state, but people often talk about this kind of system in terms of decoherence. What's the connection?
Well, decoherence happens when the system of interest is entangled with many other inaccessible systems- and that's clearly what happens here (4, 5). Since any part of the system is entangled with every other one, for a system of even moderate size it would be practically impossible to observe the coherence between different parts. This means that the system will be functionally indistinguishable from a system with no coherence, or just a classical statistical ensemble. This is the miracle of entanglement- if you have enough of it, things get simpler instead of more complicated. That's why measurements, which invariably produce some complicated entangled state between the system and apparatus, can nonetheless result in gaining knowledge.
Conclusion
There are two valid ways one can describe the state of the box of colliding particles after a long time:
- It is a complex many-body entangled state in which each part is equally entangled with every other part, but in such as way as to reproduce standard statistical results (such as the Maxwell-Boltzmann distribution) for a single-particle measurement.
- Because of the high amount of entanglement, for all practical purposes the particles may also be treated as decohered classical particles, in which case they of course have a well-defined position and momentum.
Neither of these claims is incorrect, and each might be useful in the right context.
It really depends on the boundary conditions. For boundary conditions like a 3D box with reflecting walls, the initial quantum state $\Psi$ will stay a quantum state with the unique wave function depending on variables of each particle: $$\Psi({\bf{r}}_1,...,{\bf{r}}_n, t).$$ If the boundary conditions are such that allow exchange with the environment, then the density matrix approach may become more appropriate.
In both cases the positions of particles are statistically predicted, but in the latter case there may not be interference phenomenon (or it will be less pronounced).
Consider your case for a double slit experiment without and with particle position measurements, measurements before the screen.
To solve your problem exactly, you would have to solve the Schrödinger equation
$$i \frac{\partial}{\partial t} \Psi (\vec r_1 \dots \vec r_N,t)= H \ \Psi(\vec r_1 \dots \vec r_N,t)$$
where $\Psi (\vec r_1 \dots \vec r_N,t)$ is the wave function of the $N$ particles and
$$H=\sum_i^N \frac{p_i^2}{2 m} + \sum_{i<j}^N u_{ij}+V_{\text{ext}}$$
where $u_{ij}$ is some pair potential and $V_{\text{ext}}$ is the box potential. Of course, you also need an initial condition
$$\Psi (\vec r_1 \dots \vec r_N,0) = \tilde \Psi (\vec r_1 \dots \vec r_N)$$
The first thing that you must notice is that it is inappropriate to talk about "the wave function of each particle", because you have to consider the total wave function of the $N$ particles ($\Psi$). If the particles are indistinguishable, this function must posses some symmetries, depending on what kind of particles you are considering (bosons or fermions).
It is really difficult to solve such a problem analytically or numerically. A good starting point to get a qualitative idea would be to solve the corresponding equations for two particles and see what happens.
This has been done numerically by the authors of this article using a gaussian and square potentials, with distinguishable and indistinguishable particles. in the latter case, symmetrized (bosonic) and antisymmetrized (fermionic) wave functions were considered:
$$\Psi'(x_1,x_2) = \frac 1 {\sqrt{2}} [\Psi(x_1,x_2)\pm \Psi(x_2,x_1)]$$
The wave function at $t=0$ is assumed to be the product of two gaussian wave packets:
$$\Psi(x_1,x_2, t=0) = g(x_1,x_1^0,k_1,\sigma) \ g(x_2,x_2^0,k_2,\sigma)$$
Where
$$g(x,x^0,k) = e^{i k x} e^{-\frac{(x-x^0)^2}{4 \sigma^2}}$$
and $k_2 = - k_1$.
In the following plot you can for example see the result of the collision of two distinguishable particles with equal mass $m$ interacting via a square potential (the curves are the probability densities obtained by the wave functions):
The numbers in the top left corners indicate the time (in units of $100 \times $ time step) and the edges of the frames correspond to the walls of the box.
You can see that there is indeed a "spreading" of the wave packets (to be more precise, thier modulus squared, that is, the probability densities) after the collision (cfr. 1 and 120).
Quoting from the article:
In Fig. 5 we show nine frames from the movie of a repulsive m–m collision in which the mean kinetic energy equals one quarter of the barrier height. The initial packets are seen to slow down as they approach each other, with their mutual repulsion narrowing and raising the packets up until the time (50) when they begin to bounce back. The wavepackets at still later times are seen to retain their shape, with a progressive broadening until they collide with the walls and break up.