Imagine I have a scattering region (denoted as sample). enter image description here

Scattering matrix and transfer matrix gives the same information about scattering. The scattering matrix tells us how incoming modes are scattered into outgoing modes, whereas the transfer matrix tells us how the modes on the left are related to the right (or vice versa).

$$ \begin{pmatrix} \Psi_R^{(+)} \\ \Psi_L^{(-)} \end{pmatrix} = S \begin{pmatrix} \Psi_L^{(+)} \\ \Psi_R^{(-)} \end{pmatrix} $$ Here, $S = \begin{pmatrix} S_{11} & S_{12}\\ S_{21} & S_{22} \end{pmatrix}$

The transfer matrix is written in the following way $$ \begin{pmatrix} \Psi_L^{(+)} \\ \Psi_L^{(-)} \end{pmatrix}= M \begin{pmatrix} \Psi_R^{(+)} \\ \Psi_R^{(-)} \end{pmatrix} $$ Where, $M = \begin{pmatrix} M_{11} & M_{12}\\ M_{21} & M_{22} \end{pmatrix}$. The elements of $M$ and $S$ are related (see http://assets.press.princeton.edu/chapters/s8695.pdf for example).

My question is how to compose transfer matrices and scattering matrices when I have multiple scattering regions. Imagine I have the following series of scattering regions each defined by the scattering matrix $S$ and transfer matrix $M$.

enter image description here

In the transfer matrix approach I just need to multiply a string of transfer matrices to compose them, i.e., the transfer matrix of the whole system is $M_{\text{eff}} = M_1 M_2 M_3 M_4$.

How to compose scattering matrices to get a matrix ($S_{\text{eff}}$) that has two incoming and two outgoing modes?

  • $\begingroup$ It is also the same. It is a matter only of interpretation. S is actually easier to think about as useful. $\endgroup$ Nov 29, 2023 at 2:42
  • $\begingroup$ @naturallyInconsistent the information contained in a scattering matrix is same as that of a transfer matrix. The composition rule of transfer matrix is just taking ordered products, whereas, what is the composition rule for scattering matrices is not clear. $\endgroup$
    – Galilean
    Nov 29, 2023 at 4:57
  • 1
    $\begingroup$ Why don't you convert the S matrix to an M matrix, do your composition, and then convert back to an S matrix at the end? $\endgroup$ Nov 29, 2023 at 4:59
  • $\begingroup$ @naturallyInconsistent yes that would work. But that's bypassing the problem. $\endgroup$
    – Galilean
    Nov 29, 2023 at 5:08
  • $\begingroup$ @naturallyInconsistent yes that would definitely work. Scattering matrices are useful for calculating conductance of a quantum system. If one can reduce the network into a small bunch of scatterers then the job becomes easy and one can calculate local and non local conductances. In the simple example provided in the question, converting from M to S is easy. For for some complicated non-collinear network (aka Chalker Coddington networks) this might be not very easy. Through this simple example I am trying to get a better intuition on how to handle those networks. $\endgroup$
    – Galilean
    Nov 29, 2023 at 5:14

1 Answer 1


Let's set some conventions to begin with. Scattering matrix is given by $$ S = \begin{pmatrix} r'& t \\ t'& r \end{pmatrix} $$ Where, the reflection and transmission is given as below: enter image description here Now introduce two scattering matrices $S_1$ and $S_2$ defined as $$ S_i = \begin{pmatrix} r_i'& t_i \\ t'_i& r_i \end{pmatrix} $$ The composed scattering matrix is $S_{12} = S_1 \circ S_2$ enter image description here One needs to consider infinite number of reflection and transmissions in between the scattering matrices as below from both left and right directions as below (here I have made the assumption that the modes don't get a phase while propagation, otherwise one needs to carry a phase factor also): enter image description here The effective transmission and reflection coefficients of $S_{12}$ are given by: enter image description here

One can then take a series of scattering matrix and perform pairwise compositions. Note that the composition is associative.


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