- #1

Mentz114

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The picture shows an experimental setup where one or more silver atoms are sent from an oven through 3 Stern-Gerlag (SG) filters with outputs from E and F going to detectors. If C and D remain coherent they can recombine to restore the original state and the particles all go through port E. If either C or D is blocked the detectors at E and F will show equal frequencies of [itex]Z_{+}[/itex] and ##Z_{-}##. This is easy to show by calculating the appropriate amplitudes and so the probabilities of results but is somewhat long winded and I find it unsatisfactory from the point of view of recombination. After using a lot of pencil and paper I came up with this.

These diagrams are interpreted by looking at the intersections of the axes with the circle. In A the vertical axis (z by convention) intersects at ##\theta = 0## which means that a z-basis measurement gives ##Z_+## but an x-basis measurement would give ##X_+## and ##X_-## with equal probabality. Realigning the apparatus to the x-direction is equivalent to rotating A by ##\pi/2## which gives C and a measurement in the -x-direction requires a rotation of ##-\pi/2## which gives D. Writing this algebraically we can represent A by the triplet ##(0, -\pi/2, \pi/2)##. So

\begin{align*}

A &= \left(0, -{\pi/2}, {\pi/2}\right)\\

C &= A + (\pi/2, \pi/2,\pi/2) = (\pi/2, 0, \pi)\\

D &= A - (\pi/2, \pi/2,\pi/2) = (-\pi/2, -\pi, 0)

\end{align*}

and obviously (adding components) ##\tfrac{1}{2}(C+D) = A##. In this representation the projection operator is a rotation and the relationship betwen the C and D is made somewhat clearer, perhaps. It certainly satisfies my aim of brevity and no 'interference'.

The problem is that it is not right. When adding components we have to distinguish beteen ##-\pi## and ##\pi## although they represent the same point on the circle. The algebra of recombination is therefore more than just arithmetic in the reals.

My question is - has this geometrical representation been worked out properly ? I tried relating it to the Bloch sphere but so far have failed.