I'm working through Griffith's Intro to Quantum Mechanics, attempting to solve problem 2.27.

Consider the double delta-function potential $$ V(x)= -\alpha [\delta(x+a)+\delta(x-a)] $$ where $\alpha$ and $a$ are constants.

b) How many bound states does it possess? Find the allowed energies, for $\alpha=\hbar ^2 /ma$ and for $\alpha=\hbar ^2 /4ma$, and sketch the wave function

I'm having trouble right off the bat with this problem. A cursory google search has informed me that I need to split the wave function up into cases of even and odd parities in order to reduce the amount of constants in the problem. I'm not sure what this means, or how to approach it mathematically.

Before searching for a hint, I split up the potential into three regimes: $$ x<-a, \, -a<x<a, \, x>a $$ and found the wave function for each case. I ended up with the following result: $$ \psi (x)= \begin{cases} Be^{kx} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (x<-a) \\ Ce^{-kx} + De^{kx} \ \ (-a<x<a) \\ Ee^{-kx} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (x>a) \end{cases} $$

From there, I used the continuity conditions for $\psi$ and $\psi'$ to solve for each constant. My general approach seems to be correct, barring my mistake with the parity.

To reiterate, I understand that the even and odd cases will reduce the amount of constant from 4 to 2, allowing us to solve for them using the continuity conditions for $\psi$ and $\psi'$. My confusion is regarding how to apply the parity of the wave function when starting the problem from scratch.


Hi, all. So I've figured out my original question. With the hints given, I was able to to make some significant progress. Now I'm faced with solving the discontinuous derivative at $\pm a$.

I've found that: $$ \psi (x) = \begin{cases} Be^{kx} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (x<-a) \\ C(e^{-kx} + e^{kx}) \ \ (-a<x<a) \\ Be^{-kx} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (x>a) \end{cases} $$

When applying the continuity conditions, I found that $$ \psi_{-}(a) = \psi_{+}(a) = Be^{-ka}=C(e^{-ka} +e^{ka}) \\ \Rightarrow B=C[1+e^{2ka}] $$

I've been trying to figure out the derivative, but I'm unsure of how to approach this. If I could ask for the following hint, I think I can work the rest out. Why does the discontinuity of $d\psi/dx$ at $x=a$ imply the following? $$ -kBe^{-ka} - C(ke^{ka}-ke^{-ka}) = -\frac{2m\alpha}{\hbar ^2}Be^{-ka} $$

  • 2
    $\begingroup$ The hint is the following. Let $P : x \to -x$ be the parity operator. Then $[P,H] = 0$. This means that you can work in a basis of even and odd wavefunctions. So you need to treat two separate cases: either $\psi(x) = \psi(-x)$ or $\psi(x) = -\psi(-x)$. $\endgroup$
    – user159249
    Commented Sep 26, 2017 at 21:47
  • $\begingroup$ @Marty, Is it as simple as solving the Schrodinger equation for both $\psi (x)$ and $\psi (-x)$? $\endgroup$
    – NoVa
    Commented Sep 26, 2017 at 21:51
  • $\begingroup$ Related: physics.stackexchange.com/q/44003/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Sep 28, 2017 at 7:59
  • $\begingroup$ The edit part (v4) seems to belong in a separate answer, not in the question. $\endgroup$
    – Qmechanic
    Commented Oct 7, 2017 at 11:45

1 Answer 1


Your approach is correct, and you can use the parity argument once you have all the pieces of your wavefunctions, as you do.

Because the potential is symmetric (in the sense that $V(x)=V(-x)$), the solutions will be symmetric, i.e. $\psi(x)=+\psi(-x)$ or antisymmetric, i.e. $\psi(x)=-\psi(-x)$. Simply apply this to your solution. In the specific case of the symmetric solution, this will lead to $$ B=E\, ,\qquad C=D\, . \tag{1} $$ Note that $C=D$ means your wave function in the middle region is a $C\cosh(kx)$, which is symmetric. You can guess at the result for the antisymmetric case.

You don't really save work overall by using the symmetry. From Eq.(1) you reduce the number of unknowns from $4$ to $2$ as you correctly understand, but you also loose some boundary conditions, in the sense that the boundary conditions on $\psi$ and its derivative at $x=-a$ will work out to the same as the boundary condition at $x=a$. However, you need to do this twice: once for the symmetric, and once for the antisymmetric case.

The real advantage of setting things up this way is that the algebra in each case is simpler (1/2 as hard) and also the solution is more insightful.

If you have a bit of experience you can apply this from scratch by noticing right away that the wavefunction in the middle part must be, in the symmetric case $C\cosh(kx)$, since this is the only symmetric function containing exponentials of the type $e^{kx}$ and $e^{-kx}$.

  • $\begingroup$ With your hint, I was able to get to the derivative when using the continuity conditions. If it's not too much trouble, could you give me one last hint on how to approach this part? (see edit) Thank you $\endgroup$
    – NoVa
    Commented Sep 27, 2017 at 0:01
  • 1
    $\begingroup$ @Kosta The discontinuity of $\psi'$ is governed by a rule given here: en.m.wikipedia.org/wiki/Delta_potential. Simply apply this at $x=a$. BTW you need to set $x=a$ in your current continuity condition for $\psi$. $\endgroup$ Commented Sep 27, 2017 at 0:29
  • $\begingroup$ I'm sorry, I'm still confused. The wiki article did give me the idea of setting $\psi(x) = \psi(a)$ into SE, giving me: $$ \psi ''(a) = -\frac{2m\alpha}{ \hbar ^2} Be^{-ka} $$ for $E=\alpha$. This is the RHS of my question, which leads me to believe that $\psi ''(a)$ is the LHS somehow, which I have noted as a linear combination of the first derivative of the $(-a <x<a)$ and $(x>a)$ cases. I feel like I'm missing some piece of information here. $\endgroup$
    – NoVa
    Commented Sep 27, 2017 at 1:17
  • $\begingroup$ Also, I fixed the typo. $x=a$ in my expression for $B$ now. $\endgroup$
    – NoVa
    Commented Sep 27, 2017 at 1:18
  • 1
    $\begingroup$ Because of the potential has an infinite step-discontinuity, the derivative at that point will have a finite discontinuity. Integrating your $\psi''$ gives $$ \int_{a-\epsilon}^{a+\epsilon}dx\psi''(x)=\psi'(a^+)-\psi(a^-)=-\frac{2m}{\hbar^2}\int_{a-\epsilon}^{a+\epsilon}dx (V(x)-E)\psi(x)\, . $$ Your last line is exactly as above, with $\psi'(a^+)=-Bk\psi(a)$ etc. $\endgroup$ Commented Sep 27, 2017 at 7:19

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