# Question on Finding Transmission Coefficient

1. Feb 27, 2011

### arenaninja

1. The problem statement, all variables and given/known data
Find the transmission coefficient for the potential $$V(x)=-\alpha\left[\delta\left(x+a\right)+\delta\left(x-a\right)\right]$$, where alpha and a are positive constants.

2. Relevant equations
$$T \equiv \frac{\left|F\right|^{2}}{\left|A\right|^{2}}$$

3. The attempt at a solution
I'm technically not sure on where to begin this problem. After reading the section (twice now), I noticed that it pretty much explicitly omits considering potentials which are not zero. And also, there is another formula for the transmission coefficient
$$T=\frac{1}{1+ \beta^{2}}$$
but, again, I'm not too sure on whether I should be using this one. My first instinct is to solve the Schrodinger wave equation, but it looks to be fairly mmessy with those dirac deltas.

Am I missing something really obvious? To recap, it looks like the book's section wanted to purposely omit non-zero potentials, and yet this problem appears to have a non-zero potential, leaving me clueless as to how to deal with it short of starting from scratch.

2. Feb 27, 2011

### fzero

The potential is zero in the 3 regions: $$x< -a, -a<x<a, a<x$$, so you can solve for the wavefunctions in those regions. You can obtain relations between the coefficients by demanding that certain physical conditions be satisfied. For example, the wavefunction must be continuous at $$x=\pm a$$. However, because the potential is not continuous, we cannot demand that the derivative of the wavefunction is continuous. However, Schrodinger's equation must still be valid, so we can obtain relationships across the singular points by integrating the Schrodinger equation around $$x=\pm a$$:

$$\int_{x=\pm a-\epsilon}^{x=\pm a+\epsilon} \left( - \frac{\hbar^2}{2m} \psi''(x) + (V(x)-E) \psi(x) \right) =0.$$

3. Feb 28, 2011

### arenaninja

ahhh, that's right. Thanks a lot, it makes a lot more sense from your first sentence and I can make some more progress on this one now.