Lets say we have a finite square well symetric around $y$ axis (picture below).

enter image description here

I know how and why general solutions to the second order ODE (stationary Schrödinger equation) are as follows for the regions I, II and III.

\begin{align} \text{I:}& & \psi_{\text{I}}&= Ae^{\kappa x} \\ \text{III:}& & \psi_{\text{III}}&= Be^{-\kappa x} \\ \text{II:}& & \psi_{\text{II}}&= C \cos(k x) + D\sin(kx) \end{align}

But now i got to a point where i have to start applying a boundary conditions to get a speciffic solution. So i start with the 1st boundary condition which is $\psi_{\text{I}}\left(-\frac{d}{2}\right)=\psi_{\text{II}}\left(-\frac{d}{2}\right)$ for the left potential shift and $\psi_{\text{II}}\left(\frac{d}{2}\right)=\psi_{\text{III}}\left(\frac{d}{2}\right)$ for the right potential shift. These leave me with a system of 2 equations (one for left and one for right potential shift):

\begin{align} {\scriptsize\text{left potential shift:}}& & Ae^{-\kappa \frac{d}{2}} &= C \cos\left(k\tfrac{d}{2}\right) - D\sin\left(k \tfrac{d}{2}\right)\\ {\scriptsize \text{right potential shift:}}& & Be^{-\kappa \frac{d}{2}} &= C \cos\left(k\tfrac{d}{2}\right) + D\sin\left(k \tfrac{d}{2}\right)\\ \end{align}

Question 1:

From here on authors of most books don't seem to explain much. Most of them only say that we must use $\boxed{C\!=\!0}$ to solve for odd solutions and $\boxed{D\!=\!0}$ to solve for even solutions. What is this argument based on?

  • $\begingroup$ I remember calculating this problem (darkly). I am not sure if I understand your question though. However, you also need to make sure that the derivatives of your wave functions are smooth at the boundaries, i.e. $\frac{\partial \psi_I}{\partial x} = \frac{\partial \psi_{II}}{\partial x}$ at $-d/2$ and the same on the other side. That will give your more equations and with that you can setup a system of equations and solve it for the three regions. I am not sure why your books try to split this up into even and odd solutions though. $\endgroup$
    – seb
    Commented Mar 30, 2013 at 16:00
  • $\begingroup$ actually, shouldn't it be the other way around, i.e. $C=0$ for odd solutions and $D=0$ for even solutions, since the sine is odd and the cosine is even? $\endgroup$
    – seb
    Commented Mar 30, 2013 at 16:02
  • $\begingroup$ I know that i have to also equate the derivatives but how do i solve this? And yes i did edit the post accordingly =) $\endgroup$
    – 71GA
    Commented Mar 30, 2013 at 16:03
  • $\begingroup$ I would just like to add to DaniH's answer that this gets A LOT easier to calculate if you put the left wall at $x=0$ and the right wall at $x = L$ where L is the length of your box. I don't know if you read German, if not just use google translate and take a look at the equations at this Wikipedia post $\endgroup$
    – seb
    Commented Mar 30, 2013 at 16:10
  • $\begingroup$ oh, and you solve this by "looking at it" ;-). at some point you get to $\psi(x=0) = C \cdot 1 + D \cdot 0$ that is if you put the left side of the box at the origin. Then you assume that this equation needs to be equal to zero, because you want your wave function to be zero at the wall (from physics), which in turn gives you $C = 0$. This is the same as dividing your solutions into even and odd when you don't put the left wall at the origin. $\endgroup$
    – seb
    Commented Mar 30, 2013 at 16:14

1 Answer 1


The short answer: your Hamiltonian commutes with the parity operator. Therefore, the eigenfunctions which diagonalize the Hamiltonian can be searched within the eigenfunctions diagonalizing the parity operator, which are the sets of even and odd functions. Hence, you can use the condition $C=0$ to look for the odd solutions since $\sin(x)$ is odd, and $D=0$ to look for the even solutions since $\cos(x)$ is even.

This result is natural since the square well potential is even $V(x)=V(-x)$ and therefore we have $$H\psi(x)=E\psi(x)$$ $$H\psi(-x)=E\psi(-x)$$

Using linearity, symmetric and antisymmetric combinations of eigenfunctions will also be solutions. These correspond to even and odd parity solutions respectively.

The long answer: the first thing I want you to note is that, as the Schrödinger equation is a second order differential equation, there are two boundary conditions. In this case, the usual choice is to solve the equation for Cauchy boundary conditions in which the value of the wavefunction and the derivative at boundary are specified. Hence you need to match $$ \psi_I \left(-\dfrac{d}{2}\right)=\psi_{II} \left(-\dfrac{d}{2}\right) $$ $$ \psi'_I \left(-\dfrac{d}{2}\right)=\psi'_{II} \left(-\dfrac{d}{2}\right) $$ and similar at the right side $x=d/2$.

Considering the left side of the well, you get

$$ A \exp[-\kappa d/2]= C \cos(kd/2)-D\sin(kd/2)$$

$$ \kappa A \exp[-\kappa d/2]= k C\sin(kd/2)+kD\cos(kd/2)$$

Dividing both expressions, one gets $$ \dfrac{\kappa}{k}=\dfrac{C \cos(kd/2)-D\sin(kd/2)}{C\sin(kd/2)+D\cos(kd/2)}$$

Performing the same calculation for the other side, we get $$ \dfrac{\kappa}{k}=\dfrac{C \cos(kd/2)+D\sin(kd/2)}{C\sin(kd/2)-D\cos(kd/2)}$$

Equating both expressions gives, after some algebra, the condition


which assures you that you can look for odd /even solutions separately since either $C$ or $D$ but not both must be zero.

  • $\begingroup$ Oh my god this is soooo awesome anwser! Please people vote this up! $\endgroup$
    – 71GA
    Commented Mar 30, 2013 at 16:35

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