Can Spin State Recombination Restore the Original Quantum State?

[Mentz114]

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 Z_{+} 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 between 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.

 

[PeterDonis]

I tried relating it to the Bloch sphere but so far have failed.

That would be expected, since your representation uses three numbers to specify a state, but only two are needed (because the Hilbert space of a qubit, which is what you are working with here, is two-dimensional).

 

[Mentz114]

That would be expected, since your representation uses three numbers to specify a state, but only two are needed (because the Hilbert space of a qubit, which is what you are working with here, is two-dimensional).

This is true. I think my state is a bag of measurement results and latent results so I have been trying SU(2) and SU(3) spin projectors but so far no luck there.

It is a question of constructive and destructive phase differences so I'm going to tinker some more with projectors before I drop it.

 

[Mentz114]

It is a question of constructive and destructive phase differences so I'm going to tinker some more with projectors before I drop it.

The action of the second SG apparatus on $|Z_+\rangle$ is to project into the $|X_+\rangle$ and $|X_-\rangle$ states which gives
<br /> \begin{align*}<br /> C + D &amp;= (\hat{X}_+ + \hat{X}_-)|Z_+\rangle = |X_+\rangle\langle X_+|Z_+\rangle + |X_-\rangle\langle X_-|Z_+\rangle \\<br /> &amp;= \alpha e^{-i\pi/4} |X_+\rangle + \alpha e^{i\pi/4} |X_-\rangle<br /> \end{align*}<br />
where I have adopted the convention that the scalar bracket $\langle S_1| S_2\rangle$ introduces a phase change of half the signed angle between the states ^{note A}. The normalization constant $|\alpha|^2=1/4$ follows from $| (\hat{X}_+ + \hat{X}_-)|Z_+\rangle|^2=1$.

If this superposition is acted on by the projectors $\hat{Z}_+=|Z_+\rangle\langle Z_+|$ and $\hat{Z}_-=|Z_-\rangle\langle Z_-|$ the result is
\begin{align*}
\hat{Z}_+ (\hat{X}_+ + \hat{X}_-)|Z_+\rangle &= \alpha e^{-i\pi/4}|Z_+\rangle\langle Z_+| X_+\rangle + \alpha e^{i\pi/4} |Z_+\rangle\langle Z_+| X_-\rangle\\
&=|Z_+\rangle\\
\hat{Z}_- (\hat{X}_+ + \hat{X}_-)|Z_+\rangle & = \alpha e^{-i\pi/4}|Z_-\rangle\langle Z_-| X_+\rangle + \alpha e^{i\pi/4} |Z_-\rangle\langle Z_-| X_-\rangle\\
&= \alpha e^{-i\pi/2}|Z_-\rangle + \alpha e^{i\pi/2} |Z_-\rangle = \alpha (i + 1/i)|Z_-\rangle = 0
\end{align*}
Thus $|\langle Z_+ | (\hat{X}_+ + \hat{X}_-)|Z_+\rangle|^2=1$ and $|\langle Z_- | (\hat{X}_+ + \hat{X}_-)|Z_-\rangle|^2=0$

The interference terms are clear now and the phase difference can be seen to cancel all the $|Z_-\rangle$ probability.

note A : I cannot justify this except as a kind of hidden variable that applies to super-posed states.

 

[Mentz114]

The action of the second SG apparatus on $|Z_+\rangle$ is to project into the $|X_+\rangle$ and $|X_-\rangle$ states which gives
<br /> \begin{align*}<br /> \psi_{C + D} &amp;= (\hat{X}_+ + \hat{X}_-)|Z_+\rangle = |X_+\rangle\langle X_+|Z_+\rangle + |X_-\rangle\langle X_-|Z_+\rangle \\<br /> &amp;= \tfrac{1}{\sqrt{2}} |X_+\rangle + \tfrac{1}{\sqrt{2}} |X_-\rangle<br /> \end{align*}<br />
This result depends on the inner products of the basis vectors $\langle X_{\pm}|Z_{\pm}\rangle$. These are all $1/\sqrt{2}$ except $\langle X_{-}|Z_{-}\rangle = -1/\sqrt{2}$ which provides the negative amplitude required to create interference.

If this superposition is acted on by the projectors $\hat{Z}_+=|Z_+\rangle\langle Z_+|$ and $\hat{Z}_-=|Z_-\rangle\langle Z_-|$ the result is
\begin{align*}<br /> \hat{Z}_+ |\psi_{C + D}\rangle &amp;= \tfrac{1}{\sqrt{2}}|Z_+\rangle\langle Z_+| X_+\rangle + \tfrac{1}{\sqrt{2}} |Z_+\rangle\langle Z_+| X_-\rangle\\<br /> &amp;=(\tfrac{1}{ {2}} +\tfrac{1}{ {2}} )|Z_+\rangle\\<br /> \hat{Z}_- |\psi_{C + D}\rangle &amp; = \tfrac{1}{\sqrt{2}}|Z_-\rangle\langle Z_-| X_+\rangle + \tfrac{1}{\sqrt{2}} |Z_-\rangle\langle Z_-| X_-\rangle\\<br /> &amp;= (\tfrac{1}{ {2}} - \tfrac{1}{ {2}} ) |Z_-\rangle = 0<br /> \end{align*}

Also $\langle X_{\pm}| X_{\mp}\rangle = \langle Z_{\pm}| Z_{\mp}\rangle = 0$ ensures that $\hat{X}_+|\psi_1\rangle =\hat{X}_-|\psi_1\rangle = 1/2$

Finally, mission accomplished in a few lines. The trick was to use the correct basis vectors and pay attention to the inner products.

[The basis vectors are $|X_+\rangle = (1/\sqrt{2}, 1/\sqrt{2}),\ \ |X_-\rangle = (1/\sqrt{2}, -1/\sqrt{2})$ and $|Z_+\rangle = (1, 0),\ \ |Z_-\rangle = (0, 1)$]