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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.

Charlie
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    "The problem is that probability tells me that the fact that the probability is zero doesn't mean that the event is impossible."...why does probability tell you that? – ACuriousMind Nov 06 '14 at 13:40
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    @ACuriousMind It seems like Charles is talking about the distinction between almost never and never, e.g. throwing a dart at a unit square and trying to hit a diagonal. – JohnnyMo1 Nov 06 '14 at 13:49
  • @JohnnyMo1: Ah, that makes sense. – ACuriousMind Nov 06 '14 at 13:52
  • How about respecting the uncertainty principle, if you localize the particle at any particular point, that is with zero uncertainty, then its momentum tends to infinity and thus will not be in this potential well –  Nov 06 '14 at 20:27

3 Answers3

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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.

ACuriousMind
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  • Thanks a lot, you clarified a lot of things. I've already noticed that in any point the probability is $0$ since we're in the continuum. Then I thought that the difference between the points would be the possibility/impossibility of the events. But now it makes more sense. – Charlie Nov 06 '14 at 14:37
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    Actually, this made me think that finding the particle at $x=L/2$ is indeed impossible because $\rho(L/2)=0$, which is different from other values of $x$, although all sets of the form ${x}$ have zero Lebesgue measure. – Mateus Sampaio Nov 06 '14 at 15:31
  • @MateusSampaio: Hm, but does the true probability $P$ care about whether $\rho$ is zero there or not? I mean, can you justify that events with $\rho = 0$ are somehow "special", i.e. different from one where just $P$ vanishes? I can't see how one would go about that. Do you have an idea, or is it just intuition? – ACuriousMind Nov 06 '14 at 15:44
  • I made the same question to myself, but couldn't find a good answer for it. What is bothering me is the fact that I imagine there's a difference between $x=L/4$ and $L/2$ for example, but I cannot see how this could be made precisely. It's true that if $\rho_1(x)-\rho_2(x)\neq0$ only for a set of zero measure, all probabilities will be the same for both distributions. I thought about something like the limit$$\lim_{\varepsilon\to0}\frac{P(B(x,\varepsilon))}{2\varepsilon}=\rho(x)$$that holds for continuous $\rho$, but not sure if the limit exists in the general case and I guess it won't exist. – Mateus Sampaio Nov 06 '14 at 16:04
  • An idea would be a concept like some kind of "essential pre-image", that would be related to the essential range, but I couldn't find anything about this. I guess this discussion is too deep for the question posted and is more about probability theory and philosophy. – Mateus Sampaio Nov 06 '14 at 16:04
  • Let's suppose that the energy level is $E_2$ so the wavefunction is given by $$\psi_2(x)=(\frac {2}{L})^{\frac {1}{2}} \sin(\frac {2 \pi x}{L})=\psi_{2}^*(x)$$.

    Now calculate the "expectation value" of position operator $\hat{x}$:

    $$2=<E_2|\hat{x}|E_2>=\frac{2}{L}\int_0^L\psi{2}^*(x)x\psi_2(x)dx$$

    So $$_2=\frac{2}{L}\int_0^L x\sin^2(\frac {2 \pi x}{L})dx=\frac {L}{2}$$

    Now you say that you "will observe the particle" at $x=\frac {L}{2}$

     –  Nov 06 '14 at 16:17
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    @VictorVahidiMotti: That's not the issue here. We are not talking about expectation values, but about the probability of specific events occuring, while the expectation value is the event occuring on average. And saying that an expectation value $a$ of an operator $A$ means "You will measure $a$ as the value of $A$" is simply false - there's a reason you can also calculate a standard deviation from that $a$. – ACuriousMind Nov 06 '14 at 16:21
  • I see that I have made another mistake here as well. Since the position and Hamiltonian operators do not commute we cannot use the eigenfunction of energy to calculate the expected value of position, am I right? –  Nov 06 '14 at 17:08
  • @VictorVahidiMotto: It is not required that a state be an eigenstate of an operator for it to have an expectation value w.r.t. to that operator. In fact, the expectation values of eigenstates are boring - they're the eigenvalue, and the standard deviation is really zero - if you know you've got an eigenstate, that is the one time you really know what you're gonna measure in advance. (Note that the position eigenstates are $\delta(x - x_0)$, which aren't really allowed wavefunctions.) – ACuriousMind Nov 06 '14 at 17:17
  • Do you mean that my calculation above is correct for the expectation value of the position? I think I made a mistake. Because I am using the state of an operator, here the Hamiltonian, for the position operator. While this is correct for the momentum operator, because position and energy operators do not commute I cannot use the same state basis for both. –  Nov 06 '14 at 17:52
  • @VictorVahidiMotti: Of course you can use operators in bases that are not their eigenbasis! What else would most matrices be? And I believe you calculation is correct, simply because there's no preferred direction in the setup, so the probabilities should be distributed symmetrically about the middle of the box, and so the expectation value will lie in the middle. – ACuriousMind Nov 06 '14 at 18:00
  • We are therefore Expecting the possible event of zero probability –  Nov 06 '14 at 20:32
  • @VictorVahidiMotti: The expectation value is not the most likely value. It is the expected average value after many repeated measurements. Though unfortunate, the mathematical terms do not always mean what we intuitively take them to mean. – ACuriousMind Nov 06 '14 at 20:55
  • I was referring to your point above about almost never or zero probability: ...."It means you do not expect them to ever happen." Back to the question, How about respecting the uncertainty principle, if you localize the particle at any particular point, that is with zero uncertainty, then its momentum tends to infinity and thus will not be in this potential well – –  Nov 06 '14 at 21:04
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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.

leftaroundabout
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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.

Yakk
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