What is the interpretation of zero probability in physics?

Question

Impossibility of an event implies vanishing of it's probability. But the reverse is not true. This post in math stack exchange posts says why zero probability doesn't necessarily mean impossible events. Then why do we act like it is, in physics ,i.e., how is vanishing probability both necessary and sufficient for the impossibility of an event in physics?

As an example, the probability of choosing a specific real number from the set of all real numbers is zero but yet if someone really picks up that very number it turns out that the event was not truly impossible afterall...

Similarly, can a particle be found where wave-function vanishes identically? I mean whenever we integrate square modulus of a wavefunction in some interval and the result is found to be exactly zero, we interpret it as an impossibility of the particle to be in the region of integration. Is this interpretation correct? If yes, why so? If not, how should we correctly interpret zero probability generally in physics?

Answer 1

The square of the wavefunction $\vert\psi(x)\vert^2$ is a probability density, not a probability. The probability of finding the system in a small bin of width $dx$ centred at $x_0$ is very nearly $\vert\psi(x_0)\vert^2 dx$ and thus very nearly $0$ if $\vert\psi(x_0)\vert^2= 0$, but the exact calculation yields $$ P=\int_{x_0-dx/2}^{x_0+dx/2} dx \vert\psi(x)\vert^2 $$ which will be vanishingly small but nevertheless non-zero even if $\vert\psi(x_0)\vert^2=0$ since there will presumably be nearby point in the interval $[x_0-dx/2,x_0+dx/2]$ where $\vert\psi(x_0)\vert^2\ne 0$ exactly.

Note that this is a feature of continuous probability distributions where the distribution is $0$ at isolated points. If the $\vert\psi(x)\vert^2$ is exactly $0$ on the interval, the probability of finding the system in that interval is exactly $0$.

If instead you are dealing with discrete outcomes, and - say - you prepare a system in the $\vert \uparrow \rangle$ state, there is $0$ and exactly $0$ probability of finding it in the $\vert \downarrow \rangle$ state.

Answer 2

In probability theory, an event is possible if it is non-empty. In the context of random variables, we can say that it is possible for a random variable $\xi$ to take on the value $x$ if $\xi(\omega)=x$ for some $\omega\in\Omega$, where $\Omega$ is the space of elementary outcomes in the probability space on which $\xi$ is defined.

In physics, we don't have access to probability spaces; we only have probability distributions. In other words, if we have some random variable $X$ representing the outcome of a position measurement of a particle in some state $\lvert\alpha\rangle$, we can find the probability density of $X$ by $p_X(x)=\lvert\langle x\vert\alpha\rangle\rvert^2$, but this density does not uniquely define a random variable on a probability space, so we can consider $X$ to be any random variable with this density. Therefore, we do not actually have sufficient information to say that it is impossible to find the particle in a node (point where the wavefunction vanishes). However, it is also important to remember that any measurement you make is going to have some non-zero uncertainty, so there isn't really any need to worry about the fact that individual points have zero probability, since in practise, you can really only measure the particle to be in an interval, rather than at a particular point.