When solving the Schrodinger equation in case of finite potential well, we get the following equations after separation: $$(1)-\frac{\hbar^2}{2 m} \frac{d^2 \psi_1}{d x^2} = ( E_1 - V_o) \psi_1$$ $$(2)-\frac{\hbar^2}{2 m} \frac{d^2 \psi_2}{d x^2} = E_2 \psi_2 $$ $$(3)-\frac{\hbar^2}{2 m} \frac{d^2 \psi_3}{d x^2} = ( E_3 - V_o) \psi_3$$

I was wondering why all of $E_1$, $E_2$, $E_3$ should be equal. Using the argument for energy, I get that they should. But mathematically they pose no problem whatsoever as solutions to the equation. Moreover, if I use the fact that the second order derivative of $\psi$ should exist (which I haven't seen anyone else doing), I get the additional relation that $E_1 = E_3$ and $V = E_1 - E_2$. I know this looks absurd when viewed in terms of energy, but why isn't this actually valid?


This is what I'm saying should be done:

$\psi = \begin{cases} \psi_1, & \mbox{if }x<0\mbox{ (the region outside the box)} \\ \psi_2, & \mbox{if }0<x<L\mbox{ (the region inside the box)} \\ \psi_3 & \mbox{if }x>L\mbox{ (the region outside the box)} \end{cases}$

where $\psi_1 = Ae^{\alpha x}$, $\psi_2 = Csin(kx) + Acos(kx)$, $\psi_3 = Fe^{- \alpha x}$, along with some relation between $A$, $C$, $D$, and $F$ after continuity of $\psi$ and $\frac{\partial \psi}{\partial x}$.

Now if we impose existence of $\frac{\partial^2 \psi}{\partial x^2}$, for the first and second regions we have $LHD( \frac{\partial \psi}{\partial x} ) = RHD( \frac{\partial \psi}{\partial x} )$ (left and right hand derivative). Now substituting values from the Schrodinger equation, we have

$( E_1 - V_o) \psi_1 \mid_{x=0} = E_2 \psi_2 \mid_{x=0} \implies E_1 - V_o = E_2$ (since $\psi_1 \mid_{x=0} = \psi_2 \mid_{x=0}$). So we get $E_1 - E_2 = V_o$. Similarly from second and third region we get $E_3 - E_2 = V_o$ and hence $E_1=E_2$.

  • $\begingroup$ But you know those three equations work in different areas rigth? $\phi_1$ is equal to $\phi_2$ only at the boundary, and even there their second derivative is not the same. $\endgroup$
    – Victor
    Commented Sep 22, 2016 at 16:43
  • $\begingroup$ What are you getting at? If you mean that both $\psi_1$ and its derivative should be equal to the corresponding values for $\psi_2$, then I know that. I'm asking why isn't the condition for existence of second derivative imposed on the wavefunction here. $\endgroup$
    – Akshit
    Commented Sep 22, 2016 at 16:47
  • $\begingroup$ But the second derivative exist, if you look af the solution it is clearly there, even at boundaries, where it decreases as $e^{-x}$ outside the box. Are you asking why can not we say that the second derivative is not the same in the boundary, same as we do with first the first derivative? I am sorry I don't understand what you mean by "condition for existence of second derivative" could you post how to derive $V=E_1+E_3$ $\endgroup$
    – Victor
    Commented Sep 22, 2016 at 17:08
  • $\begingroup$ Sorry, I had another typo. It was supposed to be $E_1-E_2$ instead of $E_1+E_2$ $\endgroup$
    – Akshit
    Commented Sep 22, 2016 at 18:12
  • $\begingroup$ You can find the full solution for the finite walled box here: hyperphysics.phy-astr.gsu.edu/hbase/quantum/pfbox.html#c1 (note: this site is a bit temperamental, if it doesn't load immediately try again a little later). $\endgroup$
    – Gert
    Commented Sep 22, 2016 at 18:45

1 Answer 1


You have to solve the Schroedinger equation for the whole system, which includes the 3 regions. The Schroedinger equation has wave function solutions for the whole system for each eigenvalue E that are composed of the wave solutions in the 3 regions. So you have for one solution only one energy which holds for the whole system.


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