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## Question: Cascading Stern-Gerlach ([1] pr. 1.13)

Three Stern-Gerlach type measurements are performed, the first that prepares the state in a $$\ket{S_z ; + }$$ state, the next in a $$\ket{ \BS \cdot \ncap ; + }$$ state where $$\ncap = \cos\beta \zcap + \sin\beta \xcap$$, and the last performing a $$S_z$$ $$\Hbar/2$$ state measurement, as illustrated in fig. 1.

fig. 1. Cascaded Stern-Gerlach type measurements.

What is the intensity of the final $$s_z = -\Hbar/2$$ beam? What is the orientation for the second measuring apparatus to maximize the intensity of this beam?

## Answer

The spin operator for the second apparatus is

\label{eqn:sg:20}
\BS \cdot \ncap
= \frac{\Hbar}{2} \lr{ \sin\beta \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} + \cos\beta \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} }
= \frac{\Hbar}{2}
\begin{bmatrix}
\cos\beta & \sin\beta \\
\sin\beta & -\cos\beta
\end{bmatrix}.

The intensity of the final $$\ket{S_z ; -}$$ beam is

\label{eqn:sg:40}
P
= \Abs{ \braket{-}{\BS \cdot \ncap ; +} \braket{\BS \cdot \ncap ; +}{+} }^2,

(i.e. the second apparatus applies a projection operator $$\ket{\BS \cdot \ncap ; +}\bra{\BS \cdot \ncap ; +}$$ to the initial $$\ket{+}$$ state, and then the $$\ket{-}$$ states are selected out of that.

The $$\BS \cdot \ncap$$ eigenket is found to be

\label{eqn:sg:60}
\ket{\BS \cdot \ncap ; +} =
\begin{bmatrix}
\cos\frac{\beta}{2} \\
\sin\frac{\beta}{2} \\
\end{bmatrix},

so

\label{eqn:sg:80}
P
= \Abs{
\begin{bmatrix}
0 & 1
\end{bmatrix}
\begin{bmatrix}
\cos\frac{\beta}{2} \\
\sin\frac{\beta}{2} \\
\end{bmatrix}
\begin{bmatrix}
\cos\frac{\beta}{2} &
\sin\frac{\beta}{2} \\
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix}
}^2
=
\Abs{
\cos\frac{\beta}{2}
\sin\frac{\beta}{2}
}^2
=
\Abs{\inv{2} \sin\beta}^2
=
\inv{4} \sin^2\beta.

This is maximized when $$\beta = \pi/2$$, or $$\ncap = \xcap$$. At this angle the state leaving the second apparatus is

\label{eqn:sg:100}
\begin{bmatrix}
\cos\frac{\beta}{2} \\
\sin\frac{\beta}{2} \\
\end{bmatrix}
\begin{bmatrix}
\cos\frac{\beta}{2} &
\sin\frac{\beta}{2} \\
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix}
=
\inv{2}
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
=
\inv{2}
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
=\inv{2} \ket{+} + \inv{2}\ket{-},

so the state after filtering the $$\ket{-}$$ states is $$\inv{2} \ket{-}$$ with intensity (probability density) of $$1/4$$ relative to a unit normalize input $$\ket{+}$$ state to the $$\BS \cdot \ncap$$ apparatus.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.