What is the explanation for the following result of sequential three Stern Gerlach? Consider the following three Stern Gerlach configurations from Spin and Quantum Measurement


*

*I understand why a and b act so, but can you please explain why c behaves the way it does?

*Will the particle in c still end up in the upper detector if only one particle had been shot?

 A: *

*2)Yes

*1)These cases are why the Stern-Gerlach experiment is the fundamental experiment explaining spin.

The initial beam is unpolarized. It is a mixed state that cannot be described by a wave-function; rather, it requires a density matrix:
$$ \rho_1 = \frac 1 2\left(\begin{array}{cc}1&0\\0&1\end{array}\right)=
\frac 1 2 |\uparrow\rangle\langle \uparrow|
+\frac 1 2 |\downarrow\rangle\langle \downarrow|$$
Which is a "classical" mixture of half up and half down. Note that if you rotate the $z$-axis, $\rho_1$ remains unchanged, as it must.
The first device projects out spin up via:
$$ p_1 = \left(\begin{array}{cc}1&0\\0&0\end{array}\right)$$
so that:
$$ \rho_2 =p_1\rho_1p_1^{\dagger} = \left(\begin{array}{cc}\frac 1 2&0\\0&0\end{array}\right) = \frac 1 2 |\uparrow\rangle\langle \uparrow|=\psi_2\bar{\psi}_2$$
which is half time nothing, and half the time a pure state:
(It's a little unclear why you declare '100' here, and not 50, but OK. It seems spin down is blocked).
The pure state can be written in the $x$-basis:
$$ \psi_2 =|\rightarrow\rangle = \frac 1{\sqrt 2}[
|\leftarrow\rangle + |\rightarrow\rangle]$$
So in (a) and (b) you project out the pure $\pm x$ states so that:
$$ \psi_3^a=|\rightarrow\rangle=\frac 1  2[|\uparrow\rangle + |\downarrow\rangle]$$
$$ \psi_3^b=|\leftarrow\rangle=\frac 1  2[|\uparrow\rangle - |\downarrow\rangle]$$
and the final device just selects the 1/2 that is spin-up ($z$-coordinate).
In device c, your adding both wave functions coherently:
$$ \psi_3^c = \psi_3^a + \psi_3^b = \frac 1 2\big(
[|\uparrow\rangle + |\downarrow\rangle]+
[|\uparrow\rangle - |\downarrow\rangle]
\big) = |\uparrow\rangle $$
(Major) Edit: The OP asked to formulate this answer with density matrices, and I think it is a good idea, although there are some problems.
First: Given a pure state $|\psi\rangle$, the density matrix is:
$$\rho = |\psi\rangle\langle\psi| = p_{\psi}$$
and it's equal (in form) to the projection operator onto $p_\psi$.
Second: With
$$|\uparrow\rangle=\left(\begin{array}{c}1\\0\end{array}\right)$$
$$|\downarrow\rangle=\left(\begin{array}{c}0\\1\end{array}\right)$$
The density matrices of the relevant pure state eigenstates are:
$$\rho_{+z}=\left(\begin{array}{cc}1&0\\0&0\end{array}\right) = p_{+z}$$
$$\rho_{-z}=\left(\begin{array}{cc}0&0\\0&1\end{array}\right)=p_{-z}$$
$$\rho_{\pm x}=\frac 1 2\left(\begin{array}{cc}1&\pm 1\\\pm 1&1\end{array}\right)=p_{\pm x}$$
where it is explicitly stated that they are numerically equal to projection operators for those spin states.
Third: When a state described by $\rho$ passes through a device with projector $p$, the transmitted state is:
$$\rho' = (p|\psi\rangle)(\langle\psi|p^{\dagger})=p\rho p^{\dagger}$$
Fourth: The key to the Stern-Gerlach experiment is that it entangles position and spin. We write the total "state" [note: this is tricky, $\rho$ is not a wave-function] as a column matrix representing the up and lower beams:
$$ \rho = \left[\begin{array}{c}\rho_{\rm top}\\\rho_{\rm bot}\end{array}\right]$$
Here I am using $[$ and $]$ to indicate this array is in position space, not spin-space.
We now have the tools to master the problem.
The beam starts unpolarized:
$$ \rho_0 = \left(\begin{array}{cc}\frac 1 2&0\\0&\frac 1 2\end{array}\right)=\frac 1 2 I_2$$
It passes through a $Z$-device, creating an spin/space entangled state:
$$\rho_1 = \left[\begin{array}{c}p_{+z}\rho_0p_{+z}^{\dagger} \\p_{-z}\rho_0p_{-z}^{\dagger}\end{array}\right]$$
$$\rho_1 = \left[\begin{array}{c}\left(\begin{array}{cc}\frac 1 2 &0\\0&0\end{array}\right)\\\left(\begin{array}{cc}0&0\\0&\frac 1 2\end{array}\right)\end{array}\right]$$
At this point, we block the lower path and renormalize:
$$\rho'_1 = \left(\begin{array}{cc}1&0\\0&0\end{array}\right)$$
Now run this through the $X$-device:
$$\rho_2 = \left[\begin{array}{c}p_{+x}\rho_1'p_{+x}^{\dagger} \\p_{-x}\rho_1'p_{-x}^{\dagger}\end{array}\right]$$
$$\rho_2 = \left[\begin{array}{c}\frac 1 4\left(\begin{array}{cc}1&1\\1&1\end{array}\right)\\ \frac 1 4\left(\begin{array}{cc}1&-1\\-1&1 \end{array}\right)\end{array}\right]$$
Case A:
We take $\rho_{2,{\rm top}}$ and run it through a $Z$-device:
$$\rho_A =  \left[\begin{array}{c}p_{+z}\rho_{2,{\rm top}}p_{+z}^{\dagger} \\p_{-z}\rho_{2,{\rm top}}p_{-z}^{\dagger}\end{array}\right]$$
$$\rho_A = \left[\begin{array}{c}\frac 1 4\left(\begin{array}{cc}1&0\\0&0\end{array}\right)\\ \frac 1 4\left(\begin{array}{cc}0&0\\0&1 \end{array}\right)\end{array}\right]\rightarrow
\left[\begin{array}{c}\frac 1 4 |\uparrow\rangle\\\frac 1 4 |\downarrow\rangle\end{array}\right]
$$
Where the last term is formulated as pure states.
Case B:
We take $\rho_{2,{\rm bot}}$ and run it through a $Z$-device:
$$\rho_B = \left[\begin{array}{c}p_{+z}\rho_{2,{\rm bot}}p_{+z}^{\dagger} \\p_{-z}\rho_{2,{\rm bot}}p_{-z}^{\dagger}\end{array}\right]$$
$$\rho_B = \left[\begin{array}{c}\frac 1 4\left(\begin{array}{cc}1&0\\0&0\end{array}\right)\\ \frac 1 4\left(\begin{array}{cc}0&0\\0&1 \end{array}\right)\end{array}\right]\rightarrow
\left[\begin{array}{c}\frac 1 4 |\uparrow\rangle\\\frac 1 4 |\downarrow\rangle\end{array}\right]$$
All is well.
Case C:
Here we coherently recombine the beams emerging from the $X$-device. Recall that:
$$ \rho_{2,{\rm top}}=\frac 1 2[|\rightarrow\rangle\langle\rightarrow|] $$
$$ \rho_{2,{\rm bot}}=\frac 1 2[|\leftarrow\rangle\langle\leftarrow|] $$
where:
$$ |\rightarrow\rangle=\frac 1{\sqrt 2}\left(\begin{array}{cc}1\\1\end{array}\right)$$
$$ |\leftarrow\rangle=\frac 1{\sqrt 2}\left(\begin{array}{cc}1\\-1\end{array}\right)$$
in the Z-basis column vectors. If we add the density matrices, we are by definition creating a mixed state...thereby destroying coherency. We can't do that. We need to consider the entangled state coming out of the $X$-device:
$$\Psi_2 = \left[\begin{array}{c}
\left(\begin{array}{cc}\frac 1 2\\\frac 1 2\end{array}\right)\\
\left(\begin{array}{cc}\frac 1 2\\-\frac 1 2\end{array}\right)
\end{array}\right]=
\left[\begin{array}{c}\psi_{2,{\rm top}}\\\psi_{2,{\rm bot}}
\end{array}\right]
$$
The final Z-device is a linear projection, so it yields:
$$\Psi_C=\left[\begin{array}{c}
p_{z+}\psi_{2,\rm top}\\
p_{z-}\psi_{2,\rm top}
\end{array}\right]
+
\left[\begin{array}{c}
p_{z+}\psi_{2,\rm bot}\\
p_{z-}\psi_{2,\rm bot}
\end{array}\right] =
\left[\begin{array}{c}
p_{z+}(\psi_{2,\rm top}+\psi_{2,\rm bot})\\
p_{z-}(\psi_{2,\rm top}+\psi_{2,\rm bot})
\end{array}\right]
$$
$$
\Psi_C=\left[\begin{array}{c}
\left(\begin{array}{c}1\\0\end{array}\right)\\
\left(\begin{array}{c}0\\0\end{array}\right)
\end{array}\right]
$$
A: The situation in all three experiments is same upto the second analyzer. Once the beam leaves the upper port of the second analyzer, we can write its state as
$$|\uparrow\rangle_x=\frac{1}{\sqrt{2}}(|\uparrow\rangle_z+e^{i\alpha}|\downarrow\rangle_z)$$
and for the beam leaving from the lower port
$$|\downarrow\rangle_x=\frac{1}{\sqrt{2}}(|\uparrow\rangle_z-e^{i\alpha}|\downarrow\rangle_z)$$
where you can indeed check that $\langle \downarrow| \uparrow\rangle_x=0$. Now if one puts up a z analyzer in front of any of these beams, one would get half of up and half of down polarized beams in the z direction. This is the premise of the experiment a and b. For the case of c, there happens to be a "constructive" interference before we meet the last z analyzer, so the state of the beam going into the last z analyzer would be
$$|\downarrow\rangle_x+|\uparrow\rangle_x\sim|\uparrow\rangle_z$$
where I have used approximation sign as I have not normalized the interference superposed state here. Now for the case of just one particle, I think that for a single particle as there is no other beam for it to interfere with, it would simply not replicate the beam results.  As such a single particle would always collapse into one of the eigenstate. But if you send a stream of particles one at a time, thus eliminating the possibility of interference from the other beam, we would get results similar to the case a and b. At least that is what I think.
