**Required math: calculus **

**Required physics: **SchrÃ¶dinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Chapter 2, Post 53.

We’ve seen that in the general scattering problem, we can write the particle stream magnitudes on each side of the potential by using a scattering matrix. In general, the wave function in the left hand region where is

where

On the right,

We can relate the coefficients by using the matrix

This expresses the outgoing particle streams on each side in terms of the incoming streams.

We can also express the streams on the right in terms of the streams on the left by using a *transfer matrix*. That is, we can write

By solving the scattering matrix equation for and in terms of and we can express the transfer matrix in terms of the scattering matrix.

The element is just the determinant of so we have

Conversely, we can express the scattering matrix in terms of the transfer matrix:

In the special case where the only incoming particles are from the left, and from the scattering matrix, we have for the reflection coefficient

For the transmission coefficient

If the incoming particles are from the right only, and we get

Now suppose we have a potential which is non-zero in only two isolated regions. For example, we could have two finite square wells separated by a gap, or a double delta function well. To the left of the leftmost non-zero region, the wave function is

In between the two regions, we have

and to the right of the second region we have

We cannot say what the wave function within either region is unless we specify the potential there, of course.

In terms of the transfer matrix, the transition from region 1 to region 2 is given by

Between regions 2 and 3, we have

Thus the overall transfer matrix is the product of the two individual ones:

This result generalizes to any potential that consists of a number of distinct regions where it is non-zero.

As an example, we can consider the delta function well again, except this time we’ll position the well at some arbitrary location , so we have

In this case, we have two regions, so we can write

Following the same analysis as in the original delta function at the origin, we have a couple of boundary conditions at . From continuity of the wave function we have

The first derivative is discontinuous, and we get the condition

Solving these two equations in terms of and we can read off the transfer matrix

where

We can now reconsider the double delta well problem by applying the product of transfer matrixes above. The potential is

We already have the transfer matrix for the part of the potential, which we’ll call since it’s on the right hand side. We can get the other transfer matrix by substituting for :

The transfer matrix for the combined potential is then so we get

As a check, we can work out the transmission coefficient for the case and compare it with our earlier result. From above, we have

The determinant conveniently works out to

so we get

which is the same as the result we got previously.