# Can Spin State Recombination Restore the Original Quantum State?

• I
• Mentz114
In summary, the conversation discusses an experimental setup involving silver atoms being sent through Stern-Gerlag filters and detectors to study the recombination of particles through different ports. The speaker shares their difficulties with a geometrical representation of the data and the need to consider hidden variables in superposition states. They also propose using basis vectors and inner products to simplify the calculations and successfully demonstrate the interference effects.
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 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.

Douglas Sunday
Mentz114 said:
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).

PeterDonis said:
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 said:
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
\begin{align*} C + D &= (\hat{X}_+ + \hat{X}_-)|Z_+\rangle = |X_+\rangle\langle X_+|Z_+\rangle + |X_-\rangle\langle X_-|Z_+\rangle \\ &= \alpha e^{-i\pi/4} |X_+\rangle + \alpha e^{i\pi/4} |X_-\rangle \end{align*}
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.

The action of the second SG apparatus on ##|Z_+\rangle## is to project into the ##|X_+\rangle## and ##|X_-\rangle## states which gives
\begin{align*} \psi_{C + D} &= (\hat{X}_+ + \hat{X}_-)|Z_+\rangle = |X_+\rangle\langle X_+|Z_+\rangle + |X_-\rangle\langle X_-|Z_+\rangle \\ &= \tfrac{1}{\sqrt{2}} |X_+\rangle + \tfrac{1}{\sqrt{2}} |X_-\rangle \end{align*}
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*} \hat{Z}_+ |\psi_{C + D}\rangle &= \tfrac{1}{\sqrt{2}}|Z_+\rangle\langle Z_+| X_+\rangle + \tfrac{1}{\sqrt{2}} |Z_+\rangle\langle Z_+| X_-\rangle\\ &=(\tfrac{1}{ {2}} +\tfrac{1}{ {2}} )|Z_+\rangle\\ \hat{Z}_- |\psi_{C + D}\rangle & = \tfrac{1}{\sqrt{2}}|Z_-\rangle\langle Z_-| X_+\rangle + \tfrac{1}{\sqrt{2}} |Z_-\rangle\langle Z_-| X_-\rangle\\ &= (\tfrac{1}{ {2}} - \tfrac{1}{ {2}} ) |Z_-\rangle = 0 \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)##]

Last edited:

## 1. What is spin state recombination?

Spin state recombination is a process in which the spin states of two particles combine to form a new spin state. This can occur in materials such as semiconductors and in chemical reactions.

## 2. How does spin state recombination affect materials?

Spin state recombination can have a significant impact on the properties of materials. For example, it can lead to changes in conductivity and magnetic behavior.

## 3. What factors influence spin state recombination?

There are several factors that can affect spin state recombination, including temperature, pressure, and the presence of impurities or defects in the material.

## 4. Can spin state recombination be controlled?

Yes, spin state recombination can be controlled by adjusting the conditions in which it occurs. For example, by changing the temperature or pressure, the rate of recombination can be altered.

## 5. What applications does spin state recombination have?

Spin state recombination has a wide range of applications, including in the development of new materials for electronic and magnetic devices, and in chemical reactions for the production of specific compounds.

• Quantum Physics
Replies
9
Views
1K
• Quantum Physics
Replies
1
Views
496
• Quantum Physics
Replies
5
Views
1K
• Quantum Physics
Replies
3
Views
856
• Quantum Physics
Replies
2
Views
252
• Quantum Physics
Replies
4
Views
1K
• Quantum Physics
Replies
16
Views
1K
• Quantum Physics
Replies
1
Views
773
• Quantum Physics
Replies
3
Views
1K
• Quantum Physics
Replies
2
Views
785