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