Traversing between nodes (Zero Probability) in Quantum Mechanics

Question

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.

Answer 1

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.

Answer 2

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.