To elucidate the question I have in mind, let us consider a very simple QM system that has exact solution, say Particle in a 1D Infinite Well.

Since this is a well studied problem I will simply state the result for origin at one of the boundaries, $$\Psi_n(x,t)=\sqrt{\frac{2}{L}}sin\left(\frac{n\pi{x}}{L}\right)e^{-i\omega{t}}$$

For those who are unfamiliar with Quantum Mechanical Treatment of Particle in a Box, you may refer here.

Furthur, a Node is any point where the probability of finding a particle is zero.

By Bohr-Copenhagen interpretation, the probability of finding a particle in a radius of $dr$ centered at $r$ is given by $(\Psi)^*\Psi{dr}$. Node is then given by the solutions for $r$ where $(\Psi)^*\Psi{dr}=0$.

Coming to my question, lets consider $n=2$ without loss of generality which has a node at $x=L/2$.

Argument 1: Physically this means that the $x=L/2$ eigen state of Position Operator is a Forbidden State, since the Probability of a particle collapsing into this state is absolute exact zero.

Argument 2: In wave mechanics treatment, these solutions are what we call as Standing Waves, where the nodes especially are points that doesn't permit transfer of energy (I am assuming this holds equally well for transfer of Information, in a more general sense (correct me if I am wrong!)).

By the two independent arguments above I am forced to conclude that, there is no way particles can traverse between nodes other than for some very weird phenomena where the particle vanishes with absolutely no trace at $x\to{L/2}^-$ and appear back at $x\to{L/2}^+$! But the fact that this argument is proved wrong by experiments (repeated observations have found particles observed in 1st half to be later observed in the second half) makes it all the more interesting to me. How exactly does this happen then?

To clarify my question a bit more, I am not interested in questions like 'Why Schrodinger Equation?' or such, rather I am interested in 'How to make physical sense of nodes?'

Remark :

  1. Particle in a box is purely used as an illustration, this question is very general and is by no means restricted to Particle in a Box problems.

  2. As a heads-up, for those of you who try to use Uncertainty Principle to explain this, please give a very vigorous mathematical way of looking at it rather than a handwavy qualitative argument, because I would like to remind you that Uncertainty principle has already been accounted for in Schrodinger Eq. and nodes comes out as a solution to the same.

  3. I am not interested in disproving the existence of nodes in QM systems. Its a theoretical and experimentally verified fact. Rather I am interested in the "implications" of the fact that nodes exist.

  4. You might also want to think of Feynman Path Integrals that seems to uphold the classical notion of continuity of paths. (motivative by a comment make by $@{anna}$ - Thanks for that) Nevertheless please feel free to Hypothesis what you think is happening.

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    $\begingroup$ Note there are no position eigenstates(or if you prefer, the position eigenstates are non-normalizable and therefore do not represent physical states), as the spectrum of the position operator is continuous. You can only measure whether the particle is in some interval, so you need to be more careful with what the vanishing of the wave function at a point actually means. $\endgroup$
    – J. Murray
    Mar 7, 2021 at 18:00
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    $\begingroup$ I think you are overthinking this. Particles don't have a trajectory in quantum mechanics, so there is no need for a particle to travel from the left side to the right side of the box. The wavefunction gives you the probability distribution to find the particle at any location, which is all we can know. The node means that the probability to find the particle in the center of the box is very small. $\endgroup$
    – Andrew
    Mar 7, 2021 at 18:21

2 Answers 2


As Anna points out, the physical position of the particle is described by a probability distribution, say $$\rho= \psi^* \psi$$ and sticking to your example, we can define $$dP=\rho dx = \frac{2}{L} sin^2 (\frac{n\pi x}{L}) dx$$

The probability of finding the particle at an exact location is not defined due to the fact that the particle will be localised (not allowed by the uncertainty principle), so it makes sense to calculate the probability over a small but finite interval, so that we will get over this interval a probability $$P = \int_x^{x+\delta x} \rho dx = \int_x^{x + \delta x} \frac{2}{L} sin^2 (\frac{n\pi x}{L}) dx$$

Now if all the math is correct, and we consider your case with $n=2$ and for $x=L/2$, we get $$P=\frac{8\pi^2}{3L^3} (\delta x)^3$$

This is a non-zero probability.

  • $\begingroup$ I understand your point, but I would like you to explain how you would apply uncertainty principle here. The principle simply states that one cannot simultaneously measure both ${p}$ and ${x}$ with arbitrary precision. Here I have no intention of measuring $p$ or Energy. Hence I should be able to measure $x$ with arbitrary precision, of course at the expense of $p$, of which I dont care. Essentially I want to ask what stops $\delta{x}$ in your calculation from being zero? $\endgroup$
    – YouFoundMe
    Mar 7, 2021 at 8:01
  • $\begingroup$ In fact, as I already mentioned in the remark in my original question post, nothing stops $\delta{x}$ from being zero because of the simple reason that, nodes are a solution of Schrodinger Eq. and Schrodinger Eq already accommodates Heisenberg Uncertainty principle, so you dont have to account for it separately unless one has plans to measure conjugate observables simultaneously. Please do correct me if I am wrong. Thank you for your time! $\endgroup$
    – YouFoundMe
    Mar 7, 2021 at 8:04
  • $\begingroup$ Hi. While you cannot measure position and momentum simultaneously with arbitrary position, assuming that you can have an exact value for x implies an infinite uncertainty for momentum. This is not defined. And yes, you are right in that the uncertainty principle is implicit in solution, but why does this mean that we should not obey it in this analysis? $\endgroup$
    – joseph h
    Mar 7, 2021 at 8:11
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    $\begingroup$ Hey and sorry I should have mentioned this earlier. I think we are deviating from my original question. I am not interested in "existence of nodes in QM". Its a very well documented result both theoretically and experimentally (Thanks to the wonderful work of @Anna we do have an excellent experimental result cited in this very thread). I am rather interested in the "implication" of the existence of nodes. I hope I have clarified my original question. $\endgroup$
    – YouFoundMe
    Mar 7, 2021 at 8:44
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    $\begingroup$ Thanks so much man! I really appreciate it. I am hoping that some hardcore quantum physicist might notice the question and may be able to provide insights. The answer by @YoungAndria seems to be very reasonable, although I dont like that idea I am still unable to come up with a contradiction. $\endgroup$
    – YouFoundMe
    Mar 7, 2021 at 11:00

Are you aware that the quantum mechanical wave function is a mathematical function that cannot be measured? It is the probability distribution, $Ψ^*Ψ$ that is measurable , which means the accumulation of many measurements of particles with the exact same boundary conditions.

In quantum mechanics individual particles do not travel in space, do not form orbits, they form probability loci, as the orbitals of the hydrogen atom show.


It means that there is a probability for a particle to be found at (r,theta,phi) when measured, and it is not a sequential in space path that describes the orbital that each individual particle in the accumulation of probabilities will find itself.Its location is random weighted by the QM probability distribution, $Ψ^*Ψ$ .

The zero nodes of the wavefunction for this complex system, show up as zero probability loci.Look at the experimental verification


Fig. 3. Experimental observation of the transverse nodal structure of four atomic hydrogen Stark states. The images in the middle show experimental measurements for (n1,n2,m) = (0,29,0), (1,28,0), (2,27,0) and (3,26,0). Interference patterns are clearly observed where the number of nodes corresponds to the value of n1. The results may be compared to TDSE calculations shown to the left (for details see text), revealing that the experimentally observed nodal structures originate from the transverse nodal structure of the initial state that is formed upon laser excitation. A comparison of the experimentally measured (solid lines) and calculated radial (dashed lines) probability distributions P(R) is shown to the right of the experimental results. In order to make this comparison, the computational results were scaled to the macroscopic dimensions of the experiment. Please note that, since P(R) = ∫ P(R,α) R dα, the radial probability distributions P(R) have a zero at R=0, even if the two-dimensional images P(R,α) do not.

  • $\begingroup$ Yes I absolutely agree with you. And I am actually interpreting the wavefunction exactly as you have mentioned. My question is how does one explain the fact that particles are seen on either side of the nodes although the node points are "Forbidden", how does one interpret this experimental finding? The question boils down to how does particles move between nodes when they are not allowed "at the nodes"? $\endgroup$
    – YouFoundMe
    Mar 7, 2021 at 6:57
  • $\begingroup$ Thank you very much for your response! I would like to clarify that I am not questioning the fundamentals of Quantum Mechanics, instead I am merely wondering about the implication of these fundamental postulates. I am hoping that I conveyed by question properly. $\endgroup$
    – YouFoundMe
    Mar 7, 2021 at 7:10
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    $\begingroup$ "that particles are seen on either side of the nodes although the node points are "Forbidden" " Probability locus means that each individual particle does not go through the nodes, as it has zero probability, it will be found when measured where the probability is larger than zero. Its path is not continuous, there is no path, just a proabbility locus. $\endgroup$
    – anna v
    Mar 7, 2021 at 7:14
  • $\begingroup$ okay, but how did you conclude the "non-existence of path in QM". QM have everything to do with probability of finding a particle but it has little to do with defining the path of a particle. I agree that QM says nothing about the path. But just because it says nothing about paths, it does not conclusively prove the non existence of path in QM. In fact the entire QM perspective in "Feynman Path Integrals" seems to uphold the classical notion of continuity of paths. I didn't include this in my original question because I thought it might lead to unnecessary complications. $\endgroup$
    – YouFoundMe
    Mar 7, 2021 at 7:39
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    $\begingroup$ Perhaps you might want to include a mathematical/physical explanation for "non-existence of paths in QM" in your original post. Thanks a lot! $\endgroup$
    – YouFoundMe
    Mar 7, 2021 at 7:41

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