0
$\begingroup$

Reading the wikipedia article about the particle in the box, there is this image:

https://en.wikipedia.org/wiki/Particle_in_a_box

Animations from B to F show wave function of a particle in a box starting from ground state up to excited states. The animation C shows wave function behavior in the first excited state and at the middle point both real part (blue) and imaginary part (red) of the wavefunctions are zero all the time. Does it mean that particle will never ever be found at that point? Similarly, for second excited state in the picture D we have two points where this happens, so does that mean that there are two points in space where particle will never ever be found?

I can't really tell from the picture if this continues for higher excited states. Is there some law that says for the n-th excited state there will be n points in space where probability of finding particle is zero?

$\endgroup$
8
$\begingroup$

The probability of finding a particle at a point is always zero.

Recall that $\rho(x) = \lvert\psi(x)\rvert^2$ is a probability density, not a probability, and so the probability to find the particle somewhere inside the interval $[a,b]$ is given by $$ P([a,b]) = \int_a^b \rho(x)\mathrm{d}x.$$ Since points have zero measure ($\int_a^a \rho(x)\mathrm{d}x = 0$ regardless of $a$), this is always zero for single points. So it is not evident that there is any meaning to saying "the particle will never be found at $x_0$" because quantum mechanics only allows us to talk meaningfully about a region (however small) in which the particle can be found.

This is supported by the fact that the "eigenstates" $\lvert x\rangle$ of the position operator are not actual states since they are non-normalizable ($\langle x \vert x\rangle$ cannot be made finite/well-defined), so there is no actual measurement whose result could be $\lvert x\rangle$, a fully localized particle.

However, you are asking about the nodes of the wavefunction of a particle trapped in a box. Indeed, even though you should not think of them are "points where the particle can never be found", the $n$-th excited state has $n$ of these nodes in its wavefunction.


garyp suggests an alternative interpretation of "the particle can never be found at $x_0$" in the comments that actually then is correct for the nodes:

For the nodes $n_i$, we have $$\lim_{a\to 0}\frac{P([n_i-a,n_i+a])}{P([x_0-a,x_0+a])} = 0$$ for any $x_0$ that isn't a node itself. This means, in words, that if we take regions of equal size centered around the points $n_i$ and $x_0$ and shrink them, it becomes more and more likely to find the particle around $x_0$ compared to finding it around $n_i$, until in the limit the ratio becomes zero suggesting it is infinitely more likely to find the particle "at" any other $x_0$ then it is to find it "at" $n_i$. Note that the latter part of this sentence should only be understood heuristically due to the actual probability of finding a particle at a point being zero as discussed at the beginning.

$\endgroup$
  • 2
    $\begingroup$ I think it's fair to say that the ratio of the probability of finding the particle in a region symmetrically placed around a node to the probability of finding the particle in a region of the same size anywhere else approaches zero as the size of the region goes to zero. That's a wonky way of saying: roughly speaking, the answer to the OPs question is yes, provided a loose interpretation of the question is taken. But I take your point about the details, and I agree that the clarification you present is an important one that I hope will help straighten out misconceptions. $\endgroup$ – garyp Sep 21 '16 at 13:47
  • $\begingroup$ @garyp Nice, that's correct, and now I wonder why I've never heard this before! $\endgroup$ – ACuriousMind Sep 21 '16 at 14:06
  • $\begingroup$ Why are our measurements of particles not considered points? $\endgroup$ – Yogi DMT Sep 21 '16 at 15:26
  • $\begingroup$ @ACuriousMind thx a lot for this answer. I hope you don't mind if I ask two potentially naive questions :( (i): as you explain, mathematically we are seeing as you nicely discuss here that because position is a continuous observable, its corresponding state probability calculations deal with volume measures, so it s clear the answer to "what is the chance that the particle is fully localised at $x_0$" is always 0. But physically, is this to be understood in terms of the Heisenberg uncertainty relation? (ii) regarding HUP: I never understand what people mean when they ... $\endgroup$ – user929304 Mar 5 '17 at 23:10
  • $\begingroup$ (...) say "if we want to know a particle s position exactly its momentum becomes infinitely large...". But if something s momentum uncertainty is very large, doesn't that mean it must be moving all over the place, so how can we even claim to know its position at all? $\endgroup$ – user929304 Mar 5 '17 at 23:11

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.