Supermirror, flipper matrices, and workflow with configurable polarizer/analyzer

[SimonHeybrock]

This adds:

• Flipper matrices (seems to be working, according to tests).
• Supermirror skeleton. The actual efficiency function is not implemented.
• PolarizationAnalysisWorkflow, which composes workflows for the polarization and analyzer (each can be He3 or supermirror) into a single workflow.

Fixes #59.

[jokasimr]

It's hard to review this without a description of the procedure. I did not find anything out flippers in the polarization analysis methodology page. What source did you use to implement this @SimonHeybrock?

[SimonHeybrock]

It's hard to review this without a description of the procedure. I did not find anything out flippers in the polarization analysis methodology page. What source did you use to implement this @SimonHeybrock?

@astellhorn was adding information in #56, but I think a file was missing and the PR needs updating. Otherwise, have a look at the attachments in #2 (comment) (probably not most recent version).

[astellhorn]

It's hard to review this without a description of the procedure. I did not find anything out flippers in the polarization analysis methodology page. What source did you use to implement this @SimonHeybrock?

@astellhorn was adding information in #56, but I think a file was missing and the PR needs updating. Otherwise, have a look at the attachments in #2 (comment) (probably not most recent version).

Yes, I forgot to add the new methodology file into the branch, and will update it (best today, but I am on a conference since sunday up to tomorrow, but will have some time today lunch break).

[astellhorn]

now it is under esspolarization/docs

/user-guide/

--> see file General-sans-polarization-analysis-methodology.ipynb
--> there will probably be some small changes needed again

[jokasimr]

Seems fine to me! I just have one question about the flipper implementation.

[jokasimr]

I'm a little bit confused about what's going on here.

What is the flipper matrix? Is it correct that when swap=False then the flipper matrix is

$$F = \begin{bmatrix}1 & 0 \\ 1 - f & f \end{bmatrix}?$$

with inverse

$$F^{-1} = \begin{bmatrix}1 & 0 \\ 1 - 1/f & 1/f \end{bmatrix}$$

Does from_left compute

$$\begin{bmatrix}1 & 0 \\ 1 - 1/f & 1/f \end{bmatrix} \begin{bmatrix}u \\ d \end{bmatrix}=\begin{bmatrix}u \\ (1 - 1/f) u + d/f \end{bmatrix}?$$

In this case that seems to be the same as in the code.

Does from_right compute

$$\begin{bmatrix}u & d \end{bmatrix} \begin{bmatrix}1 & 0 \\ 1 - 1/f & 1/f \end{bmatrix} =\begin{bmatrix} u + (1 - 1/f)d & d/f \end{bmatrix}?$$

In this case the code is different from what I expected, what's wrong?

[SimonHeybrock]

Actually we dot have a vector [u, d] but a matrix [[u, d], [d, u]]. The function returns one of the resulting columns. Does it work out then?

[jokasimr]

Yes then it works out 👍