5
$\begingroup$

In Physics (both in Statistical & Quantum Mechanics) when we describe the probability function of finding a particle between $x$ and $x+dx$, we write $\int_{x}^{x+dx} p(x) \,dx = P(x)$.

Here in some books the interval is chosen as $(x,x+dx)$ and in some other books it's chosen as $[x,x+dx]$. Depending on the chosen interval, the writing will either say: “between $x$ and $x+dx$“ or “from $x$ to $x+dx$“, respectively.

Though the probability in finding a particle at the endpoints $x$, $x+dx$ is zero, so it's not necessary to include or exclude the boundary points, as the integral will give the same result, but for the Physics and Physical arguments, which one is correct and why?

Edit:

My understanding (I may be wrong): if we consider a spherical shell at a distance $x$ from the origin with width $dx$, then the particle would be in a small shell of $dx$ from the distance $x$. My question is if we define the open interval and closed interval in the probability definition, then we are respectively excluding and including the fact that the particle may not or may access the position $x$ and $x+dx$ respectively. Both can't be correct according to my understanding. So, which one should be correct and why?

$\endgroup$
3
  • 6
    $\begingroup$ Why do you think there should be a physical difference? $\endgroup$ – BioPhysicist May 25 at 12:51
  • 3
    $\begingroup$ If both integrals evaluate exactly the same why would you expect one to be more correct than the other from a physics standpoint? $\endgroup$ – Charlie May 25 at 12:51
  • 1
    $\begingroup$ Unless there is a singularity at the end points choosing one or the other will not make a difference $\endgroup$ – nwolijin May 25 at 14:32
12
$\begingroup$

There are already answers from the "physical" viewpoint. Let me throw in some mathematics, too, which is honestly the juice of a physical theory:

In Quantum Mechanics, the integral defining a probability function is made with respect to the Lebesgue Measure. A neat thing about the Lebesgue Measure $\mu_L$ on $\mathbb{R}$ is that:

$$\mu_L (\left(a,b\right)) = \mu_{L} (\left[a,b\right]) = b-a, ~ \forall~ -\infty<a<b<+\infty$$ $$ \mu_L (\{a\}) = \mu_L (\{b\}) = 0$$

As such, a probability measure on $\mathbb{R}$ is defined according to whichever interval you want, the result will be the same. This is only because the points on the real axis (in particular the two endpoints of the interval) are Lebesgue-measure $0$.

Why is Lebesgue integrability relevant?

Wavefunctions or probability densities are required to be Lebesgue integrable and not required to be continuous or Riemann integrable. This is of utmost importance since at an abstract level it allows both the completion of the pre-Hilbert space $\mathscr{L}^2(\Omega\subseteq\mathbb{R}^3,d\mu)$ with respect to the norm induced by the scalar product, and the von Neumann's theory of self-adjoint extensions of symmetric operators which is the basis of all the mathematical (standard, i.e. Hilbert space) framework of states and observables (to be more precise, the domains of self-adjointness of operators contain functions which are not continuous on subsets of $0$-Lebesgue measure. This forces us to define the Schrödinger equation not in the strong sense, but in a weak sense).

$\endgroup$
2
  • $\begingroup$ Thus, $\int_{[a,b]}f(x)\,\mathrm{d}x=\int_{]a,b[}f(x)\,\mathrm{d}x$. $\endgroup$ – Filippo May 25 at 14:21
  • $\begingroup$ Absolutely. In physical applications, we tend to transform the Lebegue integration to (improper) Riemann integration, however. $\endgroup$ – DanielC May 25 at 14:25
11
$\begingroup$

The only physical difference would be if a particle were exactly localized at either one or both endpoints. Mathematicians consider these exotic situations important, but physicists do not. The simple reason is that even if you measure the location of a particle to be $x$, to 1000 decimal places, generically in the 1001st decimal place you will find that it is either before or after $x$. And obviously you can never measure forever. (And that's besides the fact that quantum mechanics and relativity make it impossible to exactly localize a particle.) So in any physical situation you can safely assume that there are exactly 0 particles localized at the point $x$.

$\endgroup$
5
$\begingroup$

for the Physics and Physical arguments, Which one is correct and why?

Physics is, at its core, an experimental science. In the end, all that matters is consistency with experiment. Both approaches produce the same predictions so both approaches are equally valid physically.

This may feel a bit disturbing. You may feel that they are different so only one should be correct. However, this occurs often in science. When you have two different concepts that produce the same experimental predictions, what that tells you is that nature simply doesn’t care about that distinction. That distinction is only philosophical. That is the case here. It is a distinction that nature does not distinguish.

$\endgroup$
4
$\begingroup$

For "well-behaved" functions, even mathematically there is no difference as the sets {x} and {x+dx} are so called zero sets, in the sense that they have a zero measure with respect to the standard integration measure.

$\endgroup$
2
  • $\begingroup$ Is there any book, which I can read to clear my concept and misunderstandings regarding this matter? @photon $\endgroup$ – User_New2021 May 25 at 13:09
  • $\begingroup$ @User_New2021 There is no real physical difference between finding the particle at $x$ or infinitesimally close to $x$. This is consistent with the math, as a single point does not bring any contribution to the integral. For your doubt, I would suggest any calculus textbook really. $\endgroup$ – Davide Dal Bosco May 25 at 13:22
4
$\begingroup$

Since we are talking about a probability density, the probability of hitting a specific point $x$ is zero - only the probability of getting into an interval is finite, so it doesn't make difference whether we take open or closed interval. The exception are singular distributions, e.g., when using delta-functions.

Strictly speaking, this is a question about probabilistic calculus, measure theory, etc. - not about physics really.

$\endgroup$
2
  • 3
    $\begingroup$ Strictly speaking, this is a question about probabilistic calculus, measure theory, etc. - not about physics really. Why give an answer then? Do you think questions not pertaining to physics are still allowed on PSE? $\endgroup$ – BioPhysicist May 25 at 13:51
  • $\begingroup$ @BioPhysicist I began writing it as a comment, but than changed my mind, because it was getting a bit long and the people seem to post teh answers that are beyond the point. This is a kind of stuff that a physicist should know... but it is studied in math courses. I would be fine with migrating this question to Math or cross-validated. $\endgroup$ – Roger Vadim May 25 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.