[Click here for a PDF of this post with nicer formatting]

## 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.

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

\begin{equation}\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}.

\end{equation}

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

\begin{equation}\label{eqn:sg:40}

P

= \Abs{ \braket{-}{\BS \cdot \ncap ; +} \braket{\BS \cdot \ncap ; +}{+} }^2,

\end{equation}

(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

\begin{equation}\label{eqn:sg:60}

\ket{\BS \cdot \ncap ; +} =

\begin{bmatrix}

\cos\frac{\beta}{2} \\

\sin\frac{\beta}{2} \\

\end{bmatrix},

\end{equation}

so

\begin{equation}\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.

\end{equation}

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

\begin{equation}\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{-},

\end{equation}

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.