I'm trying to rework this old homework problem, and I am having problems arriving at the same solution on the answer sheet:

Let $$V(x)=\begin{cases}\infty &\text{ if } x < 0\\ \alpha\delta(x-a) &\text{ if } x \geq 0 \end{cases}$$ where $a$ and $\alpha$ are of the appropriate units.

A particle starts out in the well $0 < x < a$, but because of tunneling its wave function gradually leaks out through the delta-function barrier.

(a) Solve the time independent Schrödinger equation for this potential; impose appropriate boundary conditions, and determine the energy $E$. An implicit equation will do.

Here is what I know:

$$\psi(x)=\begin{cases}Ae^{ikx}+Be^{-ikx} &\text{ if } 0 \leq x \leq a\\ Fe^{ikx} &\text{ if } x > a \end{cases}.$$

$\psi(0)=A+B$ and this implies $-A=B$. I also understand that $\psi$ is continuous at $a$. What I am stuck on is why $\psi'(a)$ is discontinuous. Specifically, why is it that:


  • 2
    $\begingroup$ Hint: Consider integrating the Schrödinger equation over an infinitesimal region about $x=a$ ($x\in [a-\epsilon,a+\epsilon]$). $\endgroup$
    – Will
    Commented Jul 1, 2013 at 4:00
  • $\begingroup$ It becomes something like: $\Delta\left(\frac{d\psi}{dx}\right)=\frac{-2ma}{\hbar^2}\psi(a)$. I'm wondering why that negative sign disappears. $\endgroup$
    – emka
    Commented Jul 1, 2013 at 4:09
  • $\begingroup$ What is $\Delta$? Try again. Integrate $\int^{a+\epsilon}_{a-\epsilon} dx \left[-\frac{\hbar^2}{2m}\frac{\partial^2\psi(x)}{\partial x^2}+V(x)\psi(x)\right] = \int^{a+\epsilon}_{a-\epsilon} dx E\psi(x)$, in the limit as $\epsilon\rightarrow 0$, what do you get? Do not input an explicit $\psi(x)$. $\endgroup$
    – Will
    Commented Jul 1, 2013 at 4:18
  • $\begingroup$ Right. So $\int_{a-\epsilon}^{a+\epsilon}\frac{-\hbar^2}{2m}\frac{d\psi^2}{dx^2}+\int_{a-\epsilon}^{a+\epsilon} V(x)\psi(x)=0$. Then $\int_{a-\epsilon}^{a+\epsilon}\frac{-\hbar^2}{2m}\frac{d\psi^2}{dx^2}=-\int_{a-\epsilon}^{a+\epsilon} V(x)\psi(x)$. The right hand side becomes $\frac{2ma}{\hbar^2}\psi(a)$. The left hand side becomes $\frac{d\psi}{dx}\bigg|_{a+\epsilon}-\frac{d\psi}{dx}\bigg|_{a-\epsilon}$. I'm a bit rusty with all this stuff. $\endgroup$
    – emka
    Commented Jul 1, 2013 at 4:26
  • 3
    $\begingroup$ But you have just shown that the Schrödinger equation requires the derivative of the wavefunction to not be continuous at $x=a$. $\endgroup$
    – Will
    Commented Jul 1, 2013 at 4:41

1 Answer 1


Note: I have tried to make my answer a little more general, with detail, so that it will be useful for more people.

The question is what boundary conditions do we apply to our wavefunction either side of a Dirac delta function?

In your example we have the potential $$V(x)=\begin{cases}\infty &\text{ if } x < 0\\ \alpha~\delta(x-a) &\text{ if } x \geq 0 \end{cases}$$ We are interested in the boundary conditions either side of $x=a$. What information do we have? Well, due to the probabilistic interpretation of the wavefunction we require continuity of the wavefunction. That is, our first boundary condition is $$(1)~~~~~~~~~~~~\boxed{\psi_{-}(a) = \psi_{+}(a)}$$ where the $\pm$ subscripts represent the right and left sides of $x=a$ respectively. What other conditions can we set? Well, usually we would ask that the first derivative is also matched either side of $x=a$ (you should be asking yourself why do we do this?), but in this case this is not the right condition to impose. Let's see why.

Where does the boundary condition on $\frac{\partial \psi}{\partial x}$ come from?

Our wavefunction is a solution to the 1-D time-independent Schrödinger Equation: $$H~\psi(x) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)+V(x)\psi(x) = E~\psi(x)$$ Looking at this equation, we see that we can get a boundary condition on $\frac{\partial \psi}{\partial x}$ at any point $a$ by integrating it w.r.t $x$ over the region $[a-\epsilon,a+\epsilon]$, taking $\epsilon\rightarrow 0$: $$\lim_{\epsilon\rightarrow 0}\left[-\int^{a+\epsilon}_{a-\epsilon}dx\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)+\int^{a+\epsilon}_{a-\epsilon}dx~V(x)\psi(x)\right] = E~\lim_{\epsilon\rightarrow 0}~\int^{a+\epsilon}_{a-\epsilon}dx~\psi(x)\\ \implies -\frac{\hbar^2}{2m}\left[\frac{\partial \psi_{+}(a)}{\partial x}-\frac{\partial \psi_{-}(a)}{\partial x}\right] + \lim_{\epsilon\rightarrow 0}\int^{a+\epsilon}_{a-\epsilon}dx~V(x)\psi(x) = 0$$ where we have used the continuity of $\psi$ to evaluate the RHS as zero. Note that for any $V(x)$ the second term doesn't necessarily vanish. But, if $V(x)$ is continuous, this term will vanish for the same reason the RHS did, and in those cases we yield the usual boundary condition $$\boxed{\frac{\partial \psi_{+}(a)}{\partial x}=\frac{\partial \psi_{-}(a)}{\partial x}}~~~~~~~(\mbox{when $V(a)$ is continuous at $x=a$})$$ We note that the for example given here, $V(x)$ is definitely not continuous at $x=a$, where it diverges to infinity. So this second term doesn't vanish, in fact $$\lim_{\epsilon\rightarrow 0}\int^{a+\epsilon}_{a-\epsilon}dx~V(x)\psi(x) = \alpha\lim_{\epsilon\rightarrow 0}\int^{a+\epsilon}_{a-\epsilon}dx~\delta(x-a)\psi(x) = \alpha~\lim_{\epsilon\rightarrow 0}~\psi(a) = \alpha~\psi(a)$$ so rearranging our results, the second boundary condition for the problem is $$(2)~~~~~~~~~~~~\boxed{\frac{\partial \psi_{+}(a)}{\partial x}-\frac{\partial \psi_{-}(a)}{\partial x} = \frac{2m\alpha}{\hbar^2}\psi(a)}$$

Explicitly using the form of your wavefunctions (using $B=-A$ to eliminate $B$) $$\psi_{+}(x) = Fe^{ikx},~~~~~~\psi_{-}(x) = A\left(e^{ikx}-e^{-ikx}\right)$$ this boundary condition becomes $$\boxed{ikFe^{ika}-ik\left(Ae^{ika}+e^{-ika}\right) = \frac{2m\alpha}{\hbar^2}\psi(a)}$$ as required.

  • 4
    $\begingroup$ This answer (v1) is beautifully written and formatted, so I think we should eventually make an exemption this one time around and allow this full homework answer. But in accordance with our homework policy, I'm temporarily deleting this. $\endgroup$
    – Qmechanic
    Commented Jul 2, 2013 at 7:18

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