Killtech said:
I probably either don't understand the calculus of optical interference or measurement well enough.
You seem to be using an incorrect representation of both the degrees of freedom in the state space and the operation of the beam splitter.
Take just a single beam splitter. It has two input arms and two output arms, and the operator describing it has to tell us how the amplitude in each input arm gets split between the two output arms. However, the amplitude in each input arm is not just a single amplitude, because the particles we are testing have spin, which we are considering to be spin 1/2 (photon polarization can be treated this way at least for this discussion), so each particle has two spin degrees of freedom. That means the total state space that the beam splitter acts on has four degrees of freedom (spin up/down for each input arm, mapped to spin up/down for each output arm), so the operator describing the beam splitter has to be a 4 x 4 matrix, not a 2 x 2 matrix.
If we want to look at the state in a particular arm of a particular beam splitter, we then have to take the appropriate two degrees of freedom from the four that describe the total state coming into or going out of that beam splitter. In a multiple beam splitter experiement, we can then take those two degrees of freedom and combine them with two others to get the full input state for the next beam splitter.
Now let's apply the above to your setup. At the first beam splitter, we only have nonzero amplitude coming in on one arm. So the 4-component input vector will look something like this:
$$
\begin{pmatrix} a \\ b \\ 0 \\ 0 \end{pmatrix}
$$
It's worth taking a moment to unpack what this vector means. It means we have amplitude ##a## for spin up in the first input arm, amplitude ##b## for spin down in the first input arm, and amplitude ##0## for spin up and spin down in the second input arm.
The beam splitter operator matrix will look like this:
$$
\frac{1}{\sqrt{2}} \begin{pmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 1 & 0 \\
0 & 1 & 1 & 0 \\
1 & 0 & 0 & 1
\end{pmatrix}
$$
As you can easily verify, the result of applying this operator to the input vector is:
$$
\frac{1}{\sqrt{2}} \begin{pmatrix} a \\ b \\ b \\ a \end{pmatrix}
$$
Can you see how this describes what you already intuitively know that the beam splitter does?
Now, your setup has a second beam splitter downstream of the "A" output arm of the first. The full input state of that beam splitter will be the first two components of the output vector above (can you see why?), combined with two new amplitudes coming in on the other arm, so it will look like this:
$$
\begin{pmatrix} \frac{1}{\sqrt{2}} a \\ \frac{1}{\sqrt{2}} b \\ c \\ d \end{pmatrix}
$$
Now you can just apply the same beam splitter operator as above to this input state to get the output state, and the two pairs of components of that output state will tell you what to predict for each detector.
