The solution of wave equation in a box gives some nodes where the wave is $0$ and there wouldn't be a particle present at any time. Did anyone confirmed that experimentally? By example one can send other particles in the nodes and check for their scattering. As far as I am acquainted I haven't seen such experiment. Have anyone know something? Or maybe any other idea for checking this.
$\begingroup$
$\endgroup$
8
-
$\begingroup$ The probability of finding the particle in a single point is in fact always $0$. That follows from the particle's probability distribution $|\psi(x)|^2$ $\endgroup$– GertCommented Sep 12, 2021 at 19:17
-
$\begingroup$ Worth reading: physics.stackexchange.com/q/127334 $\endgroup$– GertCommented Sep 12, 2021 at 19:27
-
4$\begingroup$ Do you think that the box has some special role here except for providing such boundary conditions that produce nodes in its stationary states? If not, then your question reduces to whether it has been verified that a particle is never detected at a node of a wavefuction. This can be verified in a simple double slit experiment where you see that no particles are detected at the final screen at locations where the wavefuction has nodes. $\endgroup$– user87745Commented Sep 12, 2021 at 19:52
-
$\begingroup$ @Gert what you want to say? Modul Psi(x) squared is exactly the probability to find the particle in point x. When it is bigger than 0 you surely will register a particle. E.g. when in the double slit psi (x) is not zero you get particles detected. $\endgroup$– MercuryCommented Sep 13, 2021 at 15:02
-
1$\begingroup$ @Mercury Modul Psi(x) squared is exactly the probability to find the particle in point x. No, it's not. It's the probability density. Please read the link I provided. $\endgroup$– GertCommented Sep 13, 2021 at 15:28
|
Show 3 more comments
1 Answer
$\begingroup$
$\endgroup$
8
A straight forward confirmation comes from electron capture decay. There, the atomic electrons are particles in a box, where the potential comes from their binding to the nucleus. Electrons capture und protons in the nucleus to form neutrons and neutrinos. Electron capture decays happen mostly throught K shell electrons, since these have a non-zero probability distribution at the origin. In contrast, L shell electrons rarely get captured due to their orbitals having zero probability distribution at the origin (which is where the nucleus sits).
-
$\begingroup$ Well that is a nice example. But there is a path for the electron to go around the nuclei which is not broken. In the case of a box there is a whole plane where the particle can not pass as a particle but only as wave. If an experiment is carried out for the box that would probably mean that there are not particles in reality at all but just that particles are emergent phenomena from interaction of fields. That's why I think that such experiment is very important and wonder why it is not carried out. Maybe it is not possible for technical reasons? $\endgroup$– MercuryCommented Sep 15, 2021 at 7:23
-
$\begingroup$ The typical intro example "particle in a box" is one-dimensional $\endgroup$– rflCommented Sep 15, 2021 at 9:44
-
$\begingroup$ Maybe. But if the box is 3 dimensional there are unbroken planes of 0 probability. If z=0 is the plane of the wall every x,y from it which starts a 1d wave forms a standing wave and each wave has the same form as the others. So they all have nodes in 1 plane. $\endgroup$– MercuryCommented Sep 15, 2021 at 15:38
-
$\begingroup$ No. The electron orbitals are exactly the solutions to the particle in a box in 3D. The intersection of three plane nodes is just a point. $\endgroup$– rflCommented Sep 15, 2021 at 18:35
-
$\begingroup$ That's very surprising for me. I think the electron orbitals should be solutions to a particle in a spherical box not rectangular box. To mimic a rectangular box there must be a plane of nucleis in the central plane but this also will be an approximation. $\endgroup$– MercuryCommented Sep 16, 2021 at 5:57