I hope this question is not too straightforward for this Q&A site.

I have been reading a set of notes in which the transmission and reflection amplitudes for the delta potential Schrodinger equation are calculated. Unfortunately, my calculations don't match up with the notes' calculations, and I'm not sure why. I'm either misunderstanding how to do the calculation or the book is wrong, and I would appreciate if you guys could help me understand which.

We are given the following Schrodinger equation

$$-\frac{1}{2} \psi "(x) + c_0 \delta(x) \psi(x) = E \psi(x)$$

The solution to the equation is (if x<0)

$$\psi(x) = Ae^{ipx} + Be^{-ipx},$$ $$\psi'(x) = ip \left( Ae^{ipx} - Be^{-ipx} \right),$$

and if $x>0$

$$\psi(x) = Ce^{ipx},$$ $$\psi'(x) = ipCe^{ipx}.$$

Integrating the Schrodinger equation over a small interval $(-\epsilon, \epsilon)$ gives us

$$-\frac{1}{2} \left[ \psi'(\epsilon) - \psi'(-\epsilon) \right] + c_0 \psi(0) = E \int_{-\epsilon}^\epsilon \psi(x) dx.$$

Letting $\epsilon \rightarrow 0$ gives us the following condition according to my calculation:

$$\frac{ip}{2} \left[-C+A-B \right] +c_0 C = 0.$$

However, the notes say that this condition should be

$$\frac{ip}{2} \left[C-A+B \right] c_0 C = 0.$$

Who is right? This leads to different transmission and reflection amplitudes (using continuity of the wavefunction as well: $A+B=C$). According to the notes they are

$$T=\frac{C}{A} = \frac{p}{p+ic_0}$$ $$R = \frac{B}{A} = -\frac{ic_0}{p+ic_0}$$

whereas according to my calculations they are


because using continuity our boundary condition becomes

$$\frac{ip}{2}\left[-(A+B)+A-B \right] + (A+B)c_0=0$$

$$-ipB+(A+B)c_0 = 0$$





and I didn't really calculate $R$ because it didn't seem worth it until I knew whether or not I was correct or not. Anyway, your assistance would be appreciated in clarifying this matter.

  • $\begingroup$ Your boundary condition is correct but your expression for $T$ is not. $\endgroup$
    – Praan
    Commented Dec 8, 2015 at 20:16
  • $\begingroup$ Is the book's expression for $T$ correct? $\endgroup$ Commented Dec 8, 2015 at 20:20
  • $\begingroup$ The expressions from your notes for $R$ and $T$ are correct. $\endgroup$
    – Praan
    Commented Dec 8, 2015 at 20:21
  • $\begingroup$ Let me add a detailed computation for T in my original post. $\endgroup$ Commented Dec 8, 2015 at 20:22
  • $\begingroup$ Added detailed calculation for T. $\endgroup$ Commented Dec 8, 2015 at 20:31

2 Answers 2


Your calculation for $T$ is correct through the step

$$ - i p B + (A + B) c_0 = 0. $$

After this, you wrote

$$ (c_0 - i p) B = c_0 A. $$

You've made an algebra error here and lost a sign---the correct statement is

$$ (c_0 - i p) B = - c_0 A. $$

Therefore we have

$$ B = - \frac{c_0}{c_0 - i p} A. $$

$$ C = A + B = \left( 1 - \frac{c_0}{c_0 - i p} \right) A = \frac{- i p}{c_0 - i p} A. $$

This yields

$$ T = \frac{p}{i(c_0 - i p)} = \frac{p}{p + i c_0}, $$

as desired.

  • 1
    $\begingroup$ I'm sorry. I feel so silly for not seeing this sooner. Thank you for your help. $\endgroup$ Commented Dec 8, 2015 at 20:41
  • $\begingroup$ No problem, it happens :) $\endgroup$
    – user_35
    Commented Dec 8, 2015 at 20:57

The boundary condition on the derivative you derived is correct. Also you can drop the integral over the wave function on the right side of your condition since it vanishes in the limit $\epsilon \rightarrow 0$ because the wave function is continuous. However, the expression for $T$ from the notes/book is correct.

From the continuity of the wave function and the boundary condition on the derivative: \begin{align} 1 + R & = T \\ 1 - R & = T \left( 1 + \frac{2ic_0}{p} \right), \end{align} we find from adding both equations \begin{equation} 2 = 2T \left( 1 + \frac{ic_0}{p} \right) \Rightarrow T = \frac{p}{p+ic_0}, \end{equation} and \begin{equation} R = T - 1 = -\frac{ic_0}{p+ic_0}. \end{equation}

  • $\begingroup$ I'm not sure I'm understanding how you get your first two equations from the boundary conditions. I've added a detailed calculation of how I calculated T as well from the boundary condition + continuity, however I don't understand where I've gone wrong. $\endgroup$ Commented Dec 8, 2015 at 20:33
  • $\begingroup$ @Dargscisyhp The first equation is the continuity of the wave function where I divided both sides by $A$ and the second line is the condition on the derivative after dividing by $A$. In the third line of your calculation there should be $-c_0A$ on the right side. Minus sign errors are the worst :-) $\endgroup$
    – Praan
    Commented Dec 8, 2015 at 20:38
  • $\begingroup$ I'm sorry. I feel so silly for not seeing this sooner. Thank you for your help. $\endgroup$ Commented Dec 8, 2015 at 20:41

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