I'm working through some problems for a QM exam and I've realised I don't really understand the concept of parity of solutions. I'm looking at a simple finite potential well problem: $$V(x)=0, \quad |x|>d$$ $$V(x) = V_0 \quad |x|<d$$ where $V_0$ is positive. I'm looking at bound states, so that $0<E<V_0$. I divide the x-axis into three regions, from left to right, in the standard way, so that my wavefunction is as follows:

$$ \psi(x) = \begin{cases} Ce^{ax}, & x\in I\\ A \sin kx + B\cos kx, & x\in II \\ De^{-ax}, &x \in III \end{cases}$$

Where $a^2\equiv \frac{2m(V_0-E)}{\hbar^2}$ and $k^2\equiv \frac{2mE}{\hbar^2}$. Now this is the general wavefunction, but to make any progress I know I need to consider even and odd solutions (because the potential is symmetric). These I can find by imposing symmetry/antisymmetry on the wavefunction, so I get

$$ \psi_S(x) = \begin{cases} Ce^{ax}, & x\in I\\ B\cos kx, & x\in II \\ Ce^{-ax}, &x \in III \end{cases}$$

$$ \psi_{AS}(x) = \begin{cases} Ce^{ax}, & x\in I\\ A\sin kx, & x\in II \\ -Ce^{-ax}, &x \in III \end{cases}$$

But what ARE those functions? Am I right in thinking that "half" the eigenvalues of energy correspond to eigenfunctions which are symmetric (and the other half - antisymmetric)? So that the entire energy spectrum looks like this: $$\{ E^S_n , E^{AS}_n \}_{n \in \mathbb{N}}$$

Is my understanding correct?

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    $\begingroup$ Your question is excellently answered in Shankar (p. 157-159 + ex. 5.2.6, you can find the solution to the exercise at physics.uc.edu/~argyres/710/index.html) $\endgroup$
    – Rexcirus
    Commented Jun 21, 2014 at 17:02
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    $\begingroup$ Yes, you have it right. What are you uncertain about? $\endgroup$
    – rob
    Commented Jun 21, 2014 at 21:09
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    $\begingroup$ Because the same sort of argument, if applied to a simpler case (regular infinite potential well), gives me a spectrum of $c(n+1/2)^2$ for the symmetric case, and $cn^2$ for the antisymmetric case - this can't be right? ($c=\frac{\hbar^2 \pi^2}{2m L^2}$) $\endgroup$ Commented Jun 21, 2014 at 22:17
  • $\begingroup$ Never mind. It actually works after multiplying and dividing by $4$. The day is saved. Thanks to everyone for clearing this up for me. $\endgroup$ Commented Jun 22, 2014 at 10:02
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    $\begingroup$ Related: physics.stackexchange.com/q/44003/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Aug 8, 2015 at 0:19

1 Answer 1


It would be better to specify that $V_0 < 0$ as otherwise your problem is that of a potential barrier (no bound states) and not that of a finite potential well.

The case of a finite potential well is fully developed here (Wikipedia entry), the case of a rectangular potential barrier you can find here (Wiki).


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