Wavefunction, probability and impossible events A friend of mine asked me a question, which I considered trivial at first, but after a while gave rise to some doubts.
For instance, we have a potential well in 1 dimension defined by
$$
V(x)=
\begin{cases}
+\infty &\text{if}& x<0 \text{ and } x>L\\
0 &\text{if} &0\leq x\leq L
\end{cases}
$$
We know the wave function that describes the particle in the potential at a given energy level
$$
E_n=\frac{\hbar^2\pi^2n^2}{2mL^2}
$$
Now if we take the state at the energy level $E_2$ we have a wavefunction that behaves like $\psi_2\sim\sin(\frac{2\pi x}{L})$.
We are interested in the probability density, so we take the square modulus, which would be $0$ at $L/2$.
According to this fact I would say that it's impossible to find the particle in the position $L/2$, which can be said as: the event: "find the particle at $L/2$" is impossible.
The problem is that probability tells me that the fact that the probability is zero doesn't mean that the event is impossible. Of course to get the probability I should integrate over a length, but how can I say that the event IS impossible?
Isn't it?
Maybe it's a stupid question and I'm missing something, but I just can't fulfill my purpose.
 A: If you look closely, you will find that the event "find the particle at $x$" always has zero probability of occuring:
Since $\rho(x) = \lvert \psi(x) \rvert ^2$ is a probability density, it must be integrated over some subset of $\mathbb{R}$ to actually gain a probability. The probability to find the particles at one of the positions in a subset $S \subset \mathbb{R}$ is given by
$$ P(S) := \int_S \rho(x)\mathrm{d}x $$
Now, for "Find the particle at $x$", we would have to take $S = \{x\}$. But that is a null set w.r.t. to the usual integration measure on $\mathbb{R}$, so $P(\{x\}) = 0 \;\forall\; x \in \mathbb{R}$.
So the probability to find the particle at any particular point is zero. Huh. What gives?
The distinction to be made here is that between the formal almost never and the impossible. 
Every event that belongs to the probability space our probability density lives on is possible. Even if it is assigned probability zero, it is an imaginable outcome of whatever is happening. If events are impossible, you cannot even feed them to $P$ above as arguments.
Events that are assigned probability $0$ happen almost never. That does not mean they are impossible, but that the probability of them occuring is smaller than any $\epsilon > 0$, implying it is zero. It means you do not expect them to ever happen, not even in an infinitely long time, but you cannot rule them out. 
Lastly, note that, due to the nature of measurements, $S$ will never be a null set in any practical application, since we can only detect particles in some interval $[x,x+\delta x]$, so no events you would ordinarily think should occur are assigned zero probability.
A: Like ACuriousMind said, the crucial thing is that you always need to integrate over some interval to get an actual probability. In that light, the probability is neither zero for


*

*$P_{x_0}$, a particle in the interval $[x_0 \pm \Delta x]$ (for $x_0 \neq L/2$), nor for

*$P_{L/2}$, a particle in the interval $[L/2 \pm \Delta x]$.


The crucial difference between those cases is however that $P_{x_0}$ scales simply with $\Delta x$: on a small interval, the integral over the sine-squared looks like the integral over a constant, so as you make the interval smaller the probability shrinks yet always stays significant. OTOH, around $L/2$, the sine-squared looks like a parabola touching $0$, so $P_{L/2}$ scales $\propto (\Delta x)^3$ – i.e. the probability becomes small-cubed, which as we know is zero.
A: We do basic calculus and much of physics using $\mathbb{R}$.  This implies that there are uncountably infinite many "points" in even the smallest region of space or smallest interval in the space where are considering.
There is little reason to be certain this is true when talking about the actual universe.  If position is granular on any scale -- say, the Planck length -- then the math on $\mathbb{R}$ becomes a mere approximation of what is going on.  It can be a very good approximation, good enough that you cannot tell the difference, especially with sufficiently well behaved functions.
If, however, position is granular, then the probability returned at any actual granular location where it "could" occur becomes strictly non-zero.  You can express this in the language of measure theory (countable disjoint unions behave differently than uncountable disjoint unions) if you want.
Now, the degree of granularity does not matter (Planck length?  Much smaller?  Much larger?) to this result: any finite granularity works.  So we can return to our nice, well behaved, familiar $\mathbb{R}$ by treating such questions (what is the probability that something happens at x?) as maps from regions to probabilities, not from points to probabilities.  In short, you don't talk about points in $\mathbb{R}^n$, but instead very small regions (arbitrarily small even) in $\mathbb{R}^n$.
The density of the probability at a given point can be finite and reflect this information relatively nicely.  So long as our functions are well behaved, the probability density at a given point becomes quite meaningful, as it reflects what the probability of a small region around that given point would have at any reasonably small level of granularity.
This does not mean you can talk about the probability at a given point with any concrete meaning.  You can talk about the probability in a region, not at a point.  The density at a point says "for a small region around that point, the product of the n-dimensional volume of that region and this value is the probability of the region, with increasing precision as the region gets smaller", which is physically meaningful.
This also aligns us better with experiment, as our ability to determine where something is is not infinitely precise -- we can only talk about regions.
