# Schroedinger equation of confined particle with a dirac sink potential

I'm searching for the solution of the 1D Schroedinger equation for a particle confined in a region $-a<x<a$ and with a potential energy of $$U(x)=-\frac{\hbar^2\Omega}{m}\delta(x)$$ where $\Omega>0$

So basically I have a particle confined in a potential box but with a "Dirac" sink of potential in $x=0$. Since the potential is parity invariant I search for solutions of definite parity and since I have a dirac sink I apply the condition of continuity of the wavefunction and jump discontinuity of its derivative. I get the following results and I'm pretty sure they're correct (as, up to this point, my teacher gave us this result): $$\phi^+(x)=N^+\big[\big(1+\frac{\Omega}{ik}\big)e^{ik|x|}+\big(1-\frac{\Omega}{ik}\bigr)e^{-ik|x|}\big]\\\phi^-(x)=N^-\big(e^{ikx}-e^{-ikx})$$ where $N^-$ and $N^+$ are normalization constants and I used $+$ and $-$ to label even and odd parity. $k$ is my wavevector $\sqrt{2mE/\hbar^2}$, which contains my eigenvalue E.

Now, if I'm not mistaken, I should apply the condition $\phi(a)=0$ for each wavefunction ($\phi(-a)$ is redundant as they are even and odd wavefunctions).

Focusing on the values $E<0$, which means that $k$ is imaginary, so $k=i\tilde{k}$, I find out that my oscillating, complex exponentials become decreasing or increasing exponentials, which cannot cannot satisfy $\phi(a)=0$. Indeed, the condition applied on $\phi^-(x)$ gives me $e^{\tilde{k}a}=e^{-\tilde{k}a}$ which holds only for $\tilde{k}=0$, but this would lead to a vanishing wavefunction. For the even wavefunction the same holds, plus the coefficients can't vanish simultaneously. So I have NO eigenvalues for $E<0.$

For $E>0$ I have $k>0$ and real. For the even wavefunction, applying $\phi(a)=0$ I noticed that $(ik+\Omega)e^{ika}$ is the complex conjugate of $(\Omega-ik)e^{-ika}$. So, for my condition to hold, $(ik+\Omega)e^{ika}$ should be real but this is possible, again, only for $k=0$. For the odd wavefunction I have instead that $ka=n\pi$, so $k_n=n\pi/a$ and I finally have my eigenvalues. $$E_n=\frac{\hbar^2k_n^2}{2m}$$ Does this even make any sense? I find strange that my eigenfunctions can only be odd and that I didn't find eigenvalues below the zero value of potential. If my assumptions are correct and I didn't make any mistake, is there any physical reason for the even eigenfunctions to disappear? I studied the particle in a box model and the "dirac sink" model and there is not such a thing. I don't get why "mixing" the two together leads to this result.

• Problems 20-21, pgs.39-41, in fulviofrisone.com/attachments/article/453/… refer to a Dirac wall, but should give you a good idea what's going on. – udrv Sep 5 '16 at 3:14
• The eigenfunctions will be $\sin$ or $\cos$ functions on $(-a, 0)$ and $(0, a)$ with a kink at $0$. Draw graphs to see the possible solutions. – Keith McClary Sep 5 '16 at 3:34
• Keith, I got complex exponentials But applying the conditions my odd wavefunctions vanish anyways. I could Try to transform my exponentials in sin and cos But I don't think that would change anything – Luthien Sep 5 '16 at 10:16