Timeline for Is there a physical interpretation of Neumann boundary conditions for the free Schrodinger equation on a domain?
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23 events
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| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| Jun 11, 2016 at 20:48 | vote | accept | Neal | ||
| Aug 17, 2015 at 7:52 | history | edited | udrv | CC BY-SA 3.0 |
Added related content based on comments re: boundary conditions vs. confinement.
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| Aug 16, 2015 at 15:24 | comment | added | Neal | @udrv Also, yes, those more general conditions are called Robin boundary conditions. | |
| Aug 16, 2015 at 4:03 | comment | added | udrv | @arivero I agree with what you say. What I tried to point out is that if we impose "total current through boundary"=0, as you suggest, we conserve probability in D while the system can "come and go" across the boundary (in terms of amplitudes). If we impose n . j =0 on the boundary, we make sure the system cannot "cross the boundary" at all (amplitude of crossing boundary at any point is 0). Will follow your suggestion to edit this into the answer. | |
| Aug 16, 2015 at 3:45 | comment | added | udrv | @Neal The idea is correct, only I think you integrate Ψ∆Ψ* - Ψ*∆Ψ. ⟨Ψ|Ψ⟩ is already integrated over the entire support of Ψ. | |
| Aug 16, 2015 at 1:11 | comment | added | Neal | So $n\cdot j = 0$ is equivalent to the condition that the boundary term vanishes when you integrate $\langle \Psi|\Psi^*\rangle$ by parts and that's exactly what you need to have a self-adjoint extension of the Hamiltonian. | |
| Aug 16, 2015 at 1:06 | history | bounty ended | arivero | ||
| Aug 16, 2015 at 0:47 | comment | added | arivero | @udrv I think that your remark on total probability current could be edited into the answer. | |
| Aug 16, 2015 at 0:36 | comment | added | arivero | @udrv Consider the simplest example, the 1-D segment. You do not need n.j=0 in each extreme; it is enough it has opposite value, even if not zero, in both extremes. Because of this, the self-adjoint extensions to the free laplacian in the segment are a four-parameter set. On other hand, the self-adjoint extensions in the half-line, similar to the case you are thinking if you do not consider sum/integrating across al the boundary, are only a one parametric set, $\psi(0)=\alpha \psi'(0)$, where Neumann and Dirichlet and Neumann as the limits zero or infinite of the parameter $\alpha$. | |
| Aug 15, 2015 at 20:09 | comment | added | udrv | @Neal: Technically, what it takes to confine the system in domain D is n. j (t) = 0 at all points along D's boundary for all t. This ensures that there are no "cross-over events" anywhere along the boundary. I think a path-integral approach provides a way to write this nicely. Both the Dirichlet and Neumann conditions provide the simplest examples of how to impose n. j (t) = 0, but one can easily come up with more general (Robin?) conditions. | |
| Aug 15, 2015 at 20:01 | comment | added | udrv | @arivero: Confining ≠ "conserving probability" and technically, given local probability conservation, the total probability to locate the system in a given domain D is not time-independent unless the total probability current through D's boundary is null. In the latter case, if the system starts in D (\Psi_initial = 0 outside D), it must always remain in D. However if \Psi_initial ≠ 0 outside D, the probability to be in D is conserved, but the system is not confined. Cross-over events across the boundary simply balance out. | |
| Aug 15, 2015 at 19:31 | comment | added | arivero | @Neal Yes. The deep thing about fixing boundary conditions is that in order for the hamiltonian to be self-adjoint, no probability must be lost during time evolution. Perhaps it is a bit more generic than identification of points. | |
| Aug 15, 2015 at 19:27 | comment | added | Neal | @arivero What do you mean by "teleportation"? Identification of different sides of the domain, a la gluing a square into a torus? | |
| Aug 15, 2015 at 19:22 | comment | added | arivero | @Neal but really you don't need fencing, you only need probability conservation. So you can also allow for a phase in the bouncing against the well, and you can allow for probability "teleportation" from one side of the boundary to another | |
| Aug 15, 2015 at 19:17 | comment | added | Neal | @udrv So is this the idea? For a domain $D$, there are essentially two ways of fencing in a particle: either impose an infinite potential outside $D$, or impose the condition that the probability current never cross the boundary. Either way, the particle stays confined to the domain. | |
| Aug 13, 2015 at 19:05 | comment | added | udrv | @jac A plane of symmetry does cause n. j = 0 for each eigenfunction individually, because either the eigenfunction or its (normal) derivative vanish on the plane. But it does not mean n. j = 0 for an arbitrary wave function. Write any wave function as \Psi = \Psi_even + \Psi_odd, with \Psi_even(odd) a superposition of even(odd) eigenfunctions, and obtain n. j = i( \Psi*_even d\Psi_odd/dn - \Psi_even d\Psi*_odd/dn)≠0. The Neumann condition, on the other hand, applies to every eigenfunction and means that n. j = 0 for any wave function. | |
| Aug 13, 2015 at 15:43 | comment | added | Neal | Sorry about the sign issues. I added a note to the OP about this. I'm digesting your answer and will be back shortly with more questions, I'm sure. | |
| Aug 13, 2015 at 11:02 | comment | added | jac | Perfect reflection is not the right term, as it could be just a plane of symmetry, which also leads to $\vec n . \vec j=0$, while nothing is being reflected (bounced back). | |
| Aug 12, 2015 at 21:11 | comment | added | udrv | Sorry, meant the Neumann. Dirichlet gives the infinite potential well. This is simply because the wave function vanishes everywhere outside D, including on the boundary, so the probability of the system ever being there is zero. For the Neumann, the wave function need not necessarily vanish, but the probability of the system ever getting out of D is still zero. Hence Neumann = perfect reflection. | |
| Aug 12, 2015 at 20:57 | comment | added | arivero | To clarify... which one is the case of perfect reflection? Dirichlet or Neumann? | |
| Aug 12, 2015 at 20:29 | history | edited | Emilio Pisanty | CC BY-SA 3.0 |
Cleaned up bibliography. Good answer!
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| Aug 12, 2015 at 20:02 | history | answered | udrv | CC BY-SA 3.0 |