[NOTES/QM-13010] Dirac Delta Function Potential -Direct integration of the Schr\"{o}dinger equation

The Initial set up 

We integrate the Schr\"{o}dinger equation to derive boundary condition on the derivative of the eigenfunction at $x=0$. We rewrite the Schr\"{o}dinger equation for the delta function potential \begin{equation} -\frac{\hbar^2}{2m} \DD[u(x)]{x} - g \delta(x) u(x) = E u(x) \end{equation} in the form \begin{equation} \DD[u(x)]{x} + \frac{2mg}{\hbar^2} \delta(x) u(x) = \frac{2mE}{\hbar^2} u(x). \end{equation}

General  solution of the eigenvalue equation

We want to solve for the bound state energy, hence $E$ is negative and we set $E=-|E|$ and rewrite the Schr\"{o}dinger equation as \begin{equation} \DD[u(x)]{x} +\frac{2mg}{\hbar^2} u(x) -\alpha^2 u(x) =0, \label{eq01} \end{equation} where $\alpha^2 = \frac{2m|E|}{\hbar^2}$. Since $\delta(x)$ is zero for $x\ne 0$, the Schr\"{o}dinger equation for $x<0$ and $x>0$, both, takes the form \begin{equation} \DD[u(x)]{x} -\alpha^2u(x) =0 \end{equation} which has two independent solutions $e^{\alpha x}$ and $e^{-\alpha x}$ and we write the most general solution as \begin{eqnarray} u(x) = \begin{cases} u_1(x) = A \exp(\alpha x) + B \exp(-\alpha x)& x < 0 \\ u_2(x) = C \exp(\alpha x) + D \exp(-\alpha x) & x> 0 \end{cases} \end{eqnarray}

Apply boundary condition at infinity

Using the boundary condition $u(x) \to 0 $ as $x\to\pm \infty$ we get $B=C=0$ and the solution for the eigenfunction becomes \begin{eqnarray} u(x) = \begin{cases} u_1(x) = A \exp(\alpha x) & x < 0 \\ u_2(x) = D \exp(-\alpha x) & x> 0 \end{cases} \end{eqnarray} Demanding that the wave function be continuous at $x=0$ ( $u_1(0)=u_2(0)$) gives $D=A$ and we get \begin{equation} u(x) = \begin{cases} u_1(x) = A \exp(\alpha x) & x < 0 \\ u_2(x) = A \exp(-\alpha x) & x> 0 \end{cases} \label{eq02} \end{equation}

Matching condition at \(x=0\)

We integrate the Schr\"{o}dinger equation, \eqref{eq01}, from $-\epsilon$ to $\epsilon$ and take the limit $\epsilon \to 0$. \begin{eqnarray} \int_{-\epsilon}^{\epsilon} \DD[u]{x}\, dx + \frac{mg}{\hbar^2} \int_{-\epsilon}^{\epsilon} \delta(x) u(x) \,dx - \alpha^2 \int_{-\epsilon}^{\epsilon} u(x) \, dx &=& 0.\\ \text{or} \qquad \dd[u]{x}\Big|^\epsilon_{-\epsilon} + \frac{2mg}{\hbar^2} u(0) - \alpha^2 \int_{-\epsilon}^{\epsilon} u(x) \, dx &=& 0. \end{eqnarray} The solution $u(x)$ is continuous at $x=0$, hence in the limit $\epsilon \to 0$ the region of integration shrinks to zero and the last terms vanishes and we get the boundary condition on the first derivative at $x=0$ as \begin{equation} \boxed{\dd[u_2]{x}\Big|_{x=\epsilon} - \dd[u_1]{x}\Big|_{x=-\epsilon} + \frac{2mg}{\hbar^2} u(0) = 0.} \label{eq03} \end{equation} Now using the explicit solution, \eqref{eq02}, we get

\begin{equation} u(0)=u_1(0)=u_2(0)=A \end{equation} and \begin{eqnarray} \dd[u]{x}\Big|_{x=\epsilon} &=& \dd[u_2]{x}\Big|_{x=\epsilon}=-A \alpha e^{-\alpha\epsilon},\\ \dd[u]{x}\Big|_{x=-\epsilon} &=& \dd[u_1]{x}\Big|_{x=-\epsilon}=A \alpha e^{-\alpha\epsilon}. \end{eqnarray} The boundary condition, \eqref{eq03}, in the limit $\epsilon \to 0$ becomes \begin{eqnarray} -A\alpha e^{-\alpha \epsilon} -A \alpha e^{\alpha \epsilon} + \frac{2mg}{\hbar^2} A =0 \\ \frac{2mg}{\hbar^2}= 2\alpha \Rightarrow \frac{m^2g^2}{\hbar^4} = \alpha^2 = \frac{2m|E|}{\hbar^2}. \end{eqnarray}

The final answer

Thus we get the bound state energy as \begin{equation} |E|= \frac{mg^2}{2\hbar^2} \Longrightarrow E=-|E|=-\frac{mg^2}{2\hbar^2}.\label{eq04} \end{equation} The final form of the energy eigenfunction \begin{equation} u(x) = A \exp(-\alpha |x|), \end{equation} after normalization \begin{equation} \int_{-\infty}^\infty |u(x)|^2 \, dx = 1 \Rightarrow 2 \int_0^\infty |A|^2 \exp(-2\alpha x)\, dx =1, \end{equation} is given by \begin{equation} u(x) = \alpha^{-1/2} e^{-\alpha |x|},\qquad \alpha = \frac{2mg}{\hbar^2}. \end{equation} and the corresponding energy is given by \eqref{eq04}.