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Let $\Omega$ be a domain in $\mathbb{R}^n$. Consider the time-independent free Schrodinger equation $\Delta \psi = E\psi$.[*] Solutions subject to Dirichlet boundary conditions can be physically interpreted as the stationary states of a particle trapped in an infinite well around $\Omega$. Are there any good physical interpretations of solutions subject to Neumann boundary conditions?

For the other two of the "big three" elliptic equations, I understand the physical interpretation of both boundary conditions. So I guess I'm asking how to best fill in the missing entry of this table:

$$ \begin{array}[c|ccc] \mbox{} & \mbox{Dirichlet condition} & \mbox{Neumann condition} \\ \hline \\ \Delta \psi = \psi_t & \mbox{zero temperature boundary} & \mbox{perfect insulating boundary}\\ \Delta \psi = i\psi_t & \mbox{infinite potential well} & \mbox{???} \\ \Delta \psi = \psi_{tt} & \mbox{fixed-boundary membrane} & \mbox{free-boundary membrane} \end{array} $$

PS- If one imposes mixed conditions, i.e. Dirichlet on some boundary components and Neumann on others, then one can interpret solutions as symmetric solutions to the equation on the (larger) domain defined by reflecting across the Neumann boundary components.


[*] My convention is $\Delta = -\nabla\cdot\nabla = -\sum \partial_i^2$, so the operator, after fixing boundary conditions, is self-adjoint and positive-(semi)definite on its domain of definition. I believe this is called the "geometer's laplacian."

Neal
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    related https://physics.stackexchange.com/q/199394/1335 – arivero Aug 09 '15 at 11:16
  • In 1D the boundary is colapsed to a point and so generic BC are "generic point interactions in the circle". I wonder if the same trick can be done in any dimension AND if in classifies all the possible "point interactions" – arivero Aug 13 '15 at 16:16
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    @arivero I'm not sure if that intuition extends to higher dimensions. For instance, in a multiply connected domain in the plane, like a thickened figure-8, collapsing the boundary to a point produces a space that's not even a manifold. In 1d, any function on an interval can be extended to a periodic function, which then descends to a function on the appropriate circle, so the analogue in higher dimensions would be functions on the appropriate torus. That's my mathematical intuition, at least; I can't really speak to the physical intuition. – Neal Aug 13 '15 at 17:25
  • Related: https://physics.stackexchange.com/q/30374/975 – Joe Aug 16 '15 at 10:36

2 Answers2

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Let wave function $\Psi$ be defined on domain $D \in \mathbb{R}^n$. The Neumann condition $\frac{\partial \Psi} {\partial {\bf n}} = 0$ on the boundary $\partial D$ has a simple interpretation in terms of the probability current of $\Psi$. For $\Delta \Psi = i \partial\Psi/\partial t$ (although it's usually taken as $i \partial\Psi/\partial t = - \Delta \Psi$), the probability current at an arbitrary point ${\bf x} \in \mathbb{R}^n$ is $$ {\bf j}({\bf x}) = i [ \Psi^*({\bf x}){\bf \nabla}\Psi({\bf x}) - \Psi({\bf x}){\bf \nabla}\Psi^*({\bf x}) ] $$ and the normal current on $\partial D$ reads $$ {\bf n} \cdot {\bf j} = i\; [ \Psi^* \frac{\partial \Psi}{\partial {\bf n}} - \Psi \frac{\partial \Psi^*}{\partial {\bf n}} ] $$ (has the wrong sign, I know, but I accounted for OP's form of the Sch.eq. as $\Delta \Psi = i \partial\Psi/\partial t$).

Setting $\frac{\partial \Psi} {\partial {\bf n}} = 0$ amounts to ${\bf n} \cdot {\bf j} = 0$ everywhere on $\partial D$, thus confining the corresponding system within $D$ without an infinite potential well, as under Dirichlet conditions ($\Psi = 0$ on $\partial D$). This is the case of perfect reflection on $\partial D$.

There is a mention of this in Section 5.2 of Visual Quantum Mechanics: Selected Topics with Computer-Generated Animations of Quantum-Mechanical Phenomena, by Bernd Thaller (Springer, 2000); Google Books link.

As for applications, one answer to another post, Can we impose a boundary condition on the derivative of the wavefunction through the physical assumptions?, pointed to the use of Neumann conditions in R-matrix scattering theory.

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Clarification following @arivero's observation on conditions necessary to trap the system within domain D:

We can say that the system described by $\Psi$ is trapped in domain D if the total probability to locate it in D, $P_D$, is conserved in time: $dP_D/dt = 0$. In this case, if the system is initially located within D, such that $\Psi({\bf x},t=0) = 0$ for all ${\bf x} \notin D$ and $P_D(t=0) = 1$, then it will remain in D at all $t > 0$, since $P_D(t) = P_D(t=0) = 1$. If initially $P_D(t=0) < 1$ (the system has a nonzero probability to be located outside D), then we still have $P_D(t) = P_D(t=0) < 1$.

Conservation of $P_D$ is equivalent to a condition of null total probability current through the boundary $\partial D$. Note that it is not necessary to require null probability current at every point of $\partial D$, but only null total probability current through $\partial D$.

The difference can be understood in terms of path amplitudes (path-integral representation). In the former case, the amplitude $\Psi({\bf x}_1 t_1, {\bf x}_2 t_2; {\bf x} t)$ that the system "goes" from point ${\bf x}_1 \in D$ at time $t_1$ to point ${\bf x}_2 \notin D$ at time $t_2 > t_1$ while passing through point ${\bf x} \in \partial D$ at some time $t$, $t_1 < t < t_2$, is nonzero $\forall {\bf x} \in \partial D$: $\Psi({\bf x_1} t_1, {\bf x_2} t_2; {\bf x} t)≠0$. If however we demand null probability current at every point of $\partial D$, then $\Psi({\bf x}_1 t_1, {\bf x}_2 t_2; {\bf x} t)=0$, $\forall {\bf x} \in \partial D$.

In other words, null total probability current on $\partial D$ enforces weak trapping in the sense that overall $P_D(t) = $ const. and "cross-over events" across the boundary balance out. Null local probability current at every point of $\partial D$, ${\bf n} \cdot {\bf j} = 0$, corresponds to strong trapping in the sense that the system "does not cross" at any point ${\bf x} \in \partial D$. Imposing the strong trapping condition is equivalent to requiring that the weak trapping condition be satisfied by any wave function $\Psi$, as opposed to one selected $\Psi$. In this case the system is essentially confined within D at all times. Incidentally, the strong trapping condition follows from the requirement that the restriction of the system Hamiltonian on domain D remain self-adjoint.

Derivation of probability current conditions:

The free Schroedinger equation for $\Psi$, $i\partial\Psi/\partial t = \Delta\Psi$ as above (OP's choice of sign), implies local conservation of the probability density $\rho({\bf x,t}) = \Psi({\bf x},t)\Psi^*({\bf x},t)$: $$ \frac{\partial \rho({\bf x},t)}{\partial t} + {\bf \nabla}\cdot {\bf j}({\bf x},t) = 0 $$ Integrating this over domain D yields $$ \int_D dV\;\frac{\partial {\rho({\bf x},t)}}{\partial t} + \oint_{\partial D}dS\;{{\bf n}\cdot{\bf j}} = \frac{d}{dt}\int_DdV\;{\rho({\bf x},t)} + \oint_{\partial D}dS\;{{\bf n}\cdot{\bf j}} = 0 $$ which after denoting $P_D = \int_DdV\;{\rho({\bf x},t)}$ becomes $$ \frac{dP_D}{dt} + \oint_{\partial D}dS\;{{\bf n}\cdot{\bf j}} = 0 $$ Imposing $dP_D/dt = 0$ necessarily means $\oint_{\partial D}dS\;{{\bf n}\cdot{\bf j}} = 0$. Note that $\oint_{\partial D} dS\;{{\bf n}\cdot{\bf j}} = 0$ does not require ${{\bf n}\cdot{\bf j}} = 0$ at every point on $\partial D$, whereas ${{\bf n}\cdot{\bf j}} = 0$ does imply $\oint_{\partial D} dS\;{{\bf n}\cdot{\bf j}} = 0$ and $dP_D/dt = 0$.

Self-adjoint restriction of the free Hamiltonian on domain D:

A self-adjoint restriction of $H\Psi = \Delta \Psi$ on D requires that $ \int_D dV\;\Phi^* (\Delta\Psi) = \int_D dV\;(\Delta\Phi^*) \Psi$ or $ \int_D dV\;[\Phi^* (\Delta\Psi) - (\Delta\Phi)^* \Psi] = 0$. Use $\Phi^* (\Delta\Psi) = {\bf \nabla}\cdot(\Phi^* {\bf \nabla}\Psi) - {\bf \nabla}\Phi^* \cdot {\bf \nabla}\Psi$ and Green's theorem to obtain $$ \int_D dV\;[\Phi^* (\Delta\Psi) - (\Delta\Phi)^* \Psi] = \oint_{\partial D} dS\;[\Phi^*\frac{d\Psi}{d{\bf n}} - \Psi\frac{d\Phi^*}{d{\bf n}}] = 0 $$ If the last condition above is to be satisfied by arbitrary $\Phi$, $\Psi$, it must hold locally: $$ \Phi^* \frac{d \Psi}{d{\bf n}} - \Psi \frac{d \Phi^*}{d{\bf n}} = 0 $$

This means $\frac{1}{\Psi}\frac{d\Psi}{d{\bf n}} = \frac{1}{\Phi^*}\frac{d\Phi^*}{d{\bf n}} = a({\bf x}), \forall {\bf x} \in \partial D$. The case $\Phi = \Psi$ shows that $a({\bf x}) = a^*({\bf x})$. Therefore the restriction of $H$ on $D$ is self-adjoint if and only if wave functions in its support satisfy a boundary condition $$ \frac{d\Psi}{d{\bf n}} = a({\bf x})\Psi $$ for some given real-valued function $a({\bf x})$. In particular, this means every wave function $\Psi$ also satisfies the strong trapping condition ${\bf n} \cdot {\bf j} = 0$.

Finally note that the strong trapping condition means $\Psi^*\frac{d\Psi}{d{\bf n}} - \Psi\frac{d\Psi^*}{d{\bf n}} = 0$, but does not imply that $\Phi^*\frac{d\Psi}{d{\bf n}} - \Psi\frac{d\Phi^*}{d{\bf n}} = 0$ for arbitrary $\Psi$, $\Phi$. If we consider the latter expression as the matrix element of a "local current operator" ${\bf n} \cdot {\bf \hat j}$, then the strong trapping condition requires that diagonal elements of ${\bf n} \cdot {\bf \hat j}$ are 0, whereas self-adjointness requires that the entire space of wave functions is in the kernel of each ${\bf n} \cdot {\bf \hat j}$.

udrv
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  • To clarify... which one is the case of perfect reflection? Dirichlet or Neumann? – arivero Aug 12 '15 at 20:57
  • 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. – udrv Aug 12 '15 at 21:11
  • 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). – jac Aug 13 '15 at 11:02
  • 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. – Neal Aug 13 '15 at 15:43
  • @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. – udrv Aug 13 '15 at 19:05
  • @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. – Neal Aug 15 '15 at 19:17
  • @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 – arivero Aug 15 '15 at 19:22
  • @arivero What do you mean by "teleportation"? Identification of different sides of the domain, a la gluing a square into a torus? – Neal Aug 15 '15 at 19:27
  • @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. – arivero Aug 15 '15 at 19:31
  • @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. – udrv Aug 15 '15 at 20:01
  • @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. – udrv Aug 15 '15 at 20:09
  • @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 $\alpha$. – arivero Aug 16 '15 at 00:36
  • @udrv I think that your remark on total probability current could be edited into the answer. – arivero Aug 16 '15 at 00:47
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    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. – Neal Aug 16 '15 at 01:11
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    @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. – udrv Aug 16 '15 at 03:45
  • @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. – udrv Aug 16 '15 at 04:03
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    @udrv Also, yes, those more general conditions are called Robin boundary conditions. – Neal Aug 16 '15 at 15:24
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Suppose you want to analyze the stationary behavior of a particle in a potential well that is symmetric with respect to $x=0$ (picture below). In order to simplify your calculation, you could use as a boundary condition that $\frac{\partial \psi(x)}{\partial x}=0$ and solve the Schroedinger equation for x>0 only. This boundary condition then just reflects the symmetric nature of the problem. On top of that, it is equivalent to requiring that the expectation value for the impulse is $0$ in $x=0$, which is what you expect of a stationary solution for a symmetric potential well. You can also extend this to the non-stationary Schroedinger equations with symmetric potential well and symmetric initial conditions, i.e. $\Psi(x,t=0)$ is symmetric around $x=0$.

So, the 'label' you could use is 'plane of symmetry for potential function'. enter image description here

jac
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  • This is good, but what about Neumann conditions on two boundaries in one-dimensional space? – Ruslan Aug 10 '15 at 14:25
  • it is more an invitation for the OP to add a third column... "Dirichlet - Neumann" for mixed conditions, as the limits of the well could still be Dirichlet – arivero Aug 10 '15 at 14:41
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    @arivero There's not room in the post to add a third column, so I added a postscript instead :) – Neal Aug 13 '15 at 15:38
  • @jac What about domains where the boundary is not totally geodesic? For instance, a disk in $\mathbb{R}^2$ where half the boundary is given Dirichlet and the other half is given Neumann? – Neal Aug 13 '15 at 15:42