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How can one say that the number of nodes in the ground state must be nodeless . And how one can ensure that, when one gets up in the energy spectrum, for consecutive States the difference of number of nodes must be one. Any qualitative or mathematical explain will do.

Galilean
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  • If you're talking about a confined particle, the modes are standing waves. The quantization condition just requires that the wave function vanish at the boundaries. The lowest such mode is a half wavelength that looks somewhat like an inverted parabola, and has no nodes. The excited states are just harmonics. See https://en.wikipedia.org/wiki/Harmonic. – Kyle Arean-Raines Sep 01 '16 at 13:29
  • @KyleArean-Raines I guess the OP wants to know, if it is mathematically ensured that every possible potential V and the according Hamiltonian for, in this case probably time independent bound case, has a solution with no nodes except for the boundary. Next step would be to show that going to the next solution with the next higher energy increases the nodes by 1. This is only that simple in 1D though. – mikuszefski Sep 01 '16 at 14:15
  • @mikuszefski ah yes. Missed the "arbitrary" part. Interesting question. – Kyle Arean-Raines Sep 01 '16 at 14:18
  • I have thought about this myself. I suspect (but have not demonstrated) that if you play around with potential functions that have multiple regions of attraction (separated by repulsive regions) that you would find cases where the lowest energy state is not nodeless. – Lewis Miller Sep 01 '16 at 14:30
  • @LewisMiller I guess not, I am quite sure that the structure of the operator always results in a nodeless ground state. Or lets say there is a nodeless solution. The question then is, is this solution the one with the smallest eigenvalue? – mikuszefski Sep 01 '16 at 14:43
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    One usually expects the ground state of a double-well potential to possess one node, so I don't think the conjecture about nodeless ground states is true. It's fairly easy to see why (for a double-well potential): the wave function should vanish in regions where the potential is large to minimise potential energy. Combining this with the constraint of minimising curvature (kinetic energy) leads to a ground state that is negative under parity and therefore possesses a node. – Mark Mitchison Sep 01 '16 at 14:43
  • @MarkMitchison No – mikuszefski Sep 01 '16 at 14:45
  • For 1d see https://en.wikipedia.org/wiki/Ground_state, and also https://arxiv.org/pdf/quant-ph/0702260.pdf. – udrv Sep 02 '16 at 04:41
  • @udrv I like the second link. The wiki article only states that one can decrease the energy by removing a node, but is this necessarily an eigenfunction. That would be required as well. – mikuszefski Sep 02 '16 at 08:12
  • @mikuszefski Yeah, somebody should actually fix the wiki article. – udrv Sep 02 '16 at 15:54
  • Partial duplicate of 337858. – Cosmas Zachos Jun 08 '17 at 21:38
  • @mikuszefski The WP article is fine. It demonstrates the nodeful w.f. is not a variational energy minimum. The nodeless deformation need lower the energy, not minimize it. Further deformation and variation will lead to the actual absolutely minimizing ground state eigenfunction, but the point is it will have no node. The existence of such an absolute minimum is standard Courant stuff. – Cosmas Zachos Jun 08 '17 at 21:44
  • @CosmasZachos I did not say it wasn't. Maybe you had udrv in mind. Anyway, from a mathematical point of view the question would be. If there is a no-node function that lowers the energy compared to the one-node eigenfunction (And this part of the proof is crystal clear) is there a necessarily a no-node function that is an eigenfunction this no-node construction can be deformed into? Due to completeness properties, or similar? I guess one little extra sentence on this would not hurt in the article, especially as it is titled *Ground state*. Cheers – mikuszefski Jun 09 '17 at 06:03
  • This is a reductio ad absurdum article proving that any nodeful state cannot be the ground one, not a method to find the ground state. Rayleigh-Ritz variation methods suffice. You are extending the question into a grand vision of deformability among all wave functions, in the face of Morse theory, or something. Nobody proposed such procedures. The point of the theorem is, if you think you have the ground state and it has a node, find your mistake, because it isn't. The mumbling about taking wiggles out is to help one intuit the role of curvature, not a bypass of variational techniques. – Cosmas Zachos Jun 09 '17 at 13:56

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