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## Homework Statement

Given the series of three Stern-Gerlach devices:

Represent the action of the last two SG devices as matrices ##\hat{A}## and ##\hat{B}## in the ##|+z\rangle, |-z\rangle## basis.

## Homework Equations

##|+n\rangle = cos(\frac{\theta}{2})|+z\rangle + e^{i\phi}sin(\frac{\theta}{2})|-z\rangle##

## The Attempt at a Solution

A Stern-Gerlach device with one beam blocked off can be represented as a projection of the incoming state onto the outgoing state. A projection operator can be written as the outer product of the selected basis state with itself, e.g. ##|+z\rangle\langle\,+\,z\,|## would be the projection operator for an SGz device with the minus size blocked.

The projection matrix of the first SGn is then:

##

\hat{A}\rightarrow{|+n\rangle\langle\,+\,n\,|} =

\begin{pmatrix}

cos(\frac{\theta}{2}) \\

e^{i\phi}sin(\frac{\theta}{2})

\end{pmatrix}

\begin{pmatrix}

cos(\frac{\theta}{2}) & e^{-i\phi}sin(\frac{\theta}{2})

\end{pmatrix} =

\begin{pmatrix}

cos^{2}(\frac{\theta}{2}) & e^{-i\phi}sin(\frac{\theta}{2})cos(\frac{\theta}{2}) \\

e^{i\phi}sin(\frac{\theta}{2})cos(\frac{\theta}{2}) & sin^{2}(\frac{\theta}{2})

\end{pmatrix}

##

Therefore, the beam coming out of the positive side of the SGn should be:

##

|+n\rangle = \hat{A}|+z\rangle \rightarrow

\begin{pmatrix}

cos^{2}(\frac{\theta}{2}) & e^{-i\phi}sin(\frac{\theta}{2})cos(\frac{\theta}{2}) \\

e^{i\phi}sin(\frac{\theta}{2})cos(\frac{\theta}{2}) & sin^{2}(\frac{\theta}{2})

\end{pmatrix}

\begin{pmatrix}

1 \\

0

\end{pmatrix} =

\begin{pmatrix}

cos^{2}(\frac{\theta}{2}) \\

e^{i\phi}sin(\frac{\theta}{2})cos(\frac{\theta}{2})

\end{pmatrix}

##

Should this be equal to the given ##|+n\rangle##?

I feel as though I'm missing something completely obvious or misunderstanding the question.

Thank you for any help you can provide.