Consider a charged particle with charge $q$ trapped in a box of length $L$ with finite constant potential $ V_0 $ on both ends. A constant (static) electric field of magnitude $F$ is applied from $- \infty $ to $+ \infty$.

I have divided the whole domain in three regions

  • from $-\infty $ to $0$ as region I
  • from $0 $ to $L$ as region II
  • from $L$ to $+\infty$ as region III


  • The Schrodinger equation for region II is $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + qFx =E\psi \, .$$
  • The Schrodinger equation for regions I and III is $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + qFx +V_0=E\psi \, .$$


  • Region I $$ \newcommand{\Ai}{\operatorname{Ai}} \newcommand{\Bi}{\operatorname{Bi}} \psi(x) = c_4 \Ai[\alpha(V_0 + Fqx - E_n)] + c_5 \Bi[\alpha(V_0 + Fqx - E_n)]$$

  • Region II $$\psi(x) = c_1 \Ai[\alpha( Fqx - E_n)] + c_2 \Bi[\alpha(Fqx - E_n)]$$

  • Region III $$\psi(x) = c_3 \Ai[\alpha( V_0 + Fqx - E_n)]$$ (The $\Bi$ part is excluded because it blows up on $+\infty$.)

where $c_1 , c_2 , c_3 , c_4 , c_5$ are constants, $\alpha = \left( 2^{1/3}m /\hbar^2 \right) \left(Fmq / \hbar^2 \right)^{2/3}$ , $\Ai$ and $\Bi$ are Airy functions of first and second kind respectively, and $E_n$ are energy eigenvalues.

Applying the boundary conditions gives the following four equations:

$$ \begin{align} c_4 \Ai[\alpha(V_0 - E_n)] + c_5 \Bi[\alpha(V_0 - E_n)] &= c_1 \Ai[\alpha( - E_n)] + c_2 \Bi[\alpha(- E_n)]\\ c_4 \Ai'[\alpha(V_0 - E_n)] + c_5 \Bi'[\alpha(V_0 - E_n)] &= c_1 \Ai'[\alpha( - E_n)] + c_2 \Bi'[\alpha(- E_n)]\\ c_1 \Ai[\alpha( FqL - E_n)] + c_2 \Bi[\alpha(FqL - E_n)] &= c_3 \Ai[\alpha( V_0 + FqL - E_n)]\\ c_1 \Ai'[\alpha( FqL - E_n)] + c_2 \Bi'[\alpha(FqL - E_n)] &= c_3 \Ai'[\alpha( V_0 + FqL - E_n)] \end{align} $$

How do I calculate the bound states $E_n$ from these equations? Also, I am bogged down by the fact that on both ends of the box $\psi$ behaves differently from what is seen in trivial problems? Also, I can use computational software like MATLAB, so if someone can help me with the computational technique to find $E_n$, that is perfectly fine.

  • 1
    $\begingroup$ I fixed the first infinity symbol and hbar symbol. Please look at how I did it and fix the rest of them. It's not hard to Google search to find out how to type symbols in TeX. $\endgroup$
    – DanielSank
    Oct 25, 2014 at 17:41
  • $\begingroup$ I gather from the equations that you are considering a one-dimensional problem. That does not seem realistic to me. $\endgroup$
    – Urgje
    Oct 25, 2014 at 18:57
  • $\begingroup$ @Urgje My motive is to probe the stark effect in Quantum dots. Though it has been probed but there is a condition - BDD condition which accounts for mass discontinuity across barriers (which here I have not included in the equations to simplify them as I am looking on HOW to solve the questions) which has not been worked upon in research domain. So I have simplified the model to 1 dimensional problem and I aim to see how BDD affects the results. So yes the problem is not realistic but useful to study. And since, I am an undergrad student , it is solvable (without an advisor sadly). $\endgroup$
    – user57568
    Oct 25, 2014 at 19:05
  • $\begingroup$ I see. But be aware that the spectral properties can be quite different. And what is BDD? The phrase turned out to be ungooglelable. $\endgroup$
    – Urgje
    Oct 25, 2014 at 19:11
  • $\begingroup$ BDD = Ben Daniel Duke condition $\endgroup$
    – user57568
    Oct 25, 2014 at 20:11

3 Answers 3


First, since the potential is unbounded from below at the left side, the particle has continuous spectrum. This means that what you just have to do is compute the coefficients $c_i$, taking $E$ (real, never complex!) as input parameter.

One of these coefficients, $c_3$, is in fact arbitrary, because it only influences normalization, not smoothness of wavefunction. Thus, to compute the connection coefficients you have to find $c_1,c_2,c_4,c_5$.

And finally, your equation is a simple system of $4$ algebraic equations with $4$ unknowns. If I take $c_3=1$ and use arbitrary letters to denote all those Airy functions and their derivatives, I get $$\left\{\begin{align} c_4A+c_5B&=c_1C+c_2D,\\ c_4G+c_5H&=c_1I+c_2J,\\ c_1K+c_2L&=M,\\ c_1N+c_2O&=P. \end{align}\right.$$

Solve it for $c_i$ and you've solved almost the whole of your problem. Now if you need normalized wavefunctions, you have to use some scheme for normalization of unbounded states, e.g. so called "normalization by Dirac delta", discussed e.g. in [1].


  1. L.D. Landau & E.M. Lifshitz, Quantum Mechanics. Non-Relativistic Theory, $\S5$.

It is a transcendental system of equations where $E_n$ is unknown complex number(s). It should be solved numerically. If your potential well is rather deep, you can find real $E_n$ in the infinite well and use them as the initial approximations for numerical iterations in the real case.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Jan 6, 2015 at 18:06
  • $\begingroup$ @DavidZ too bad MathJax doesn't seem to work there... $\endgroup$
    – Ruslan
    Jan 6, 2015 at 18:07
  • 1
    $\begingroup$ @Ruslan use this $\endgroup$
    – David Z
    Jan 6, 2015 at 18:08
  • $\begingroup$ Why is this downvoted? This is exactly right. The potential is unbounded on one end, so there are no bound states (corresponding to real energies). However if the potential $V_0$ is deep enough, there may exist energies $E_n-i\Gamma_n/2$, albeit complex, but describing quasi-stationary "states" that are slowly "evaporating" away from inside the well. $\Gamma$ would correspond to a mean life-time $\tau$ of such a state as $\tau=\hbar/\Gamma$ $\endgroup$
    – xaxa
    Apr 6, 2016 at 20:05
  • $\begingroup$ @xaxa for all I know about Hermitian operators, complex energies can't be eigenvalues of a Hamiltonian (unless it's non-Hermitian). Do you mean these energies exist in some special, non-eigen-solution, sense? It was me who downvoted this anwer, maybe wrongfully due to my ignorance. But I still don't quite see why it would be right. $\endgroup$
    – Ruslan
    Apr 14, 2016 at 16:21

both ends of the box ψ behaves differently from what is seen in trivial problems, this is true because potential at one end of the box is not the same as potential on the other end, because there is constant electric fields from -infinity to +infinity , and electric fields flows from higher to lower potential.so it is fine.