I read a paper about boundary conditions and imaginary potentials. In this paper the author considers a single, non-relativistic quantum particle of mass $m > 0$ in $1$ dimension with a soft detector in finite interval $[0,L]$, $L>0$, with Schrödinger equation \begin{equation} i\hbar \frac{\partial\psi}{\partial t} = - \frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}- iv \Theta(x)\psi(x) \end{equation} where $v>0$ is a constant and $\Theta(x)=1$ for $x\geq 0$ and $0$ otherwise. At $L>0$ he considers the Neumann boundary condition $\frac{\partial \psi_t}{\partial x}(L)=0$.

The author writes, that for $x<0$, being an eigenfunction of $-(\hbar^2/2m)\partial^2/\partial x^2$ means to solve an ODE in $x$, whose general solution has the form \begin{equation} f(x)=d_k e^{ikx}+c_ke^{-ikx}\,. \end{equation}

My question is: How did he find this eigenfunction and how do you compute an eigenfunction of a wave function with boundary condition?

  • 1
    $\begingroup$ how do you compute an eigenfunction of a wave function with boundary condition? It's nothing unusual. To get bound states (eigenstates) you need BCs. $\endgroup$
    – Gert
    Sep 10, 2020 at 14:21

1 Answer 1


A solution in the form of plane wave superposition $d_k e^{i k x} + c_k e^{-i k x}$ is a standard solution always used to solve ODEs. To see this, you should refer to the first paper where it was ever used. I don' t know what it is. The point is that exponential solutions work.

If you want some interpretation, my personal one is that using plane waves means going into the frequencyu (Fourier) space, where derivatives become products. But this is just some interpretation, practically when you solve differential equations, the first soultion you look for always is in the form of $e^{i k x}$.

  • $\begingroup$ thank you very much! $\endgroup$
    – uzizi_1
    Sep 10, 2020 at 13:09

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