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."

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    $\begingroup$ related physics.stackexchange.com/q/199394/1335 $\endgroup$
    – arivero
    Aug 9, 2015 at 11:16
  • $\begingroup$ 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" $\endgroup$
    – arivero
    Aug 13, 2015 at 16:16
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    $\begingroup$ @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. $\endgroup$
    – Neal
    Aug 13, 2015 at 17:25
  • $\begingroup$ Related: physics.stackexchange.com/q/30374/975 $\endgroup$
    – Joe
    Aug 16, 2015 at 10:36

2 Answers 2


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 ${\bf n} \cdot {\bf \hat j}$.

  • $\begingroup$ To clarify... which one is the case of perfect reflection? Dirichlet or Neumann? $\endgroup$
    – arivero
    Aug 12, 2015 at 20:57
  • $\begingroup$ 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. $\endgroup$
    – udrv
    Aug 12, 2015 at 21:11
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    $\begingroup$ 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. $\endgroup$
    – Neal
    Aug 16, 2015 at 1:11
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    $\begingroup$ @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. $\endgroup$
    – udrv
    Aug 16, 2015 at 3:45
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    $\begingroup$ @udrv Also, yes, those more general conditions are called Robin boundary conditions. $\endgroup$
    – Neal
    Aug 16, 2015 at 15:24

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

  • $\begingroup$ This is good, but what about Neumann conditions on two boundaries in one-dimensional space? $\endgroup$
    – Ruslan
    Aug 10, 2015 at 14:25
  • $\begingroup$ 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 $\endgroup$
    – arivero
    Aug 10, 2015 at 14:41
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    $\begingroup$ @arivero There's not room in the post to add a third column, so I added a postscript instead :) $\endgroup$
    – Neal
    Aug 13, 2015 at 15:38
  • $\begingroup$ @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? $\endgroup$
    – Neal
    Aug 13, 2015 at 15:42

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