4
$\begingroup$

Consider the following three Stern Gerlach configurations from Spin and Quantum Measurement

enter image description here

  1. I understand why a and b act so, but can you please explain why c behaves the way it does?
  2. Will the particle in c still end up in the upper detector if only one particle had been shot?
$\endgroup$
6
  • $\begingroup$ Hi @alpal . Can you please quote the source of the figure that you have used here for your query? $\endgroup$ Commented Jul 1, 2021 at 4:10
  • $\begingroup$ @quirkyquark Hi. Sure I added it in the main post right before the image. $\endgroup$
    – al pal
    Commented Jul 1, 2021 at 4:13
  • 1
    $\begingroup$ Thanks for the reference. As of what I can see, in the setup c) the two beams that exit the X analyzer are made to interfere with each other and hence the answer given by @JEB makes sense. But this is a bit non-trivial. It is the first time I have seen such interference being used in a SG experiment. $\endgroup$ Commented Jul 1, 2021 at 4:23
  • $\begingroup$ Do consider accepting an answer if it has solved your query. $\endgroup$ Commented Jul 1, 2021 at 12:57
  • $\begingroup$ @quirkyquark Thank you for the reply. Sure I will, but I'm very new to QM, it takes time for me to go over your answers and figure them out. After that I will accept it for sure :) $\endgroup$
    – al pal
    Commented Jul 1, 2021 at 15:47

2 Answers 2

2
$\begingroup$
  1. 2)Yes
  2. 1)These cases are why the Stern-Gerlach experiment is the fundamental experiment explaining spin.

The initial beam is unpolarized. It is a mixed state that cannot be described by a wave-function; rather, it requires a density matrix:

$$ \rho_1 = \frac 1 2\left(\begin{array}{cc}1&0\\0&1\end{array}\right)= \frac 1 2 |\uparrow\rangle\langle \uparrow| +\frac 1 2 |\downarrow\rangle\langle \downarrow|$$

Which is a "classical" mixture of half up and half down. Note that if you rotate the $z$-axis, $\rho_1$ remains unchanged, as it must.

The first device projects out spin up via:

$$ p_1 = \left(\begin{array}{cc}1&0\\0&0\end{array}\right)$$

so that:

$$ \rho_2 =p_1\rho_1p_1^{\dagger} = \left(\begin{array}{cc}\frac 1 2&0\\0&0\end{array}\right) = \frac 1 2 |\uparrow\rangle\langle \uparrow|=\psi_2\bar{\psi}_2$$

which is half time nothing, and half the time a pure state:

(It's a little unclear why you declare '100' here, and not 50, but OK. It seems spin down is blocked).

The pure state can be written in the $x$-basis:

$$ \psi_2 =|\rightarrow\rangle = \frac 1{\sqrt 2}[ |\leftarrow\rangle + |\rightarrow\rangle]$$

So in (a) and (b) you project out the pure $\pm x$ states so that:

$$ \psi_3^a=|\rightarrow\rangle=\frac 1 2[|\uparrow\rangle + |\downarrow\rangle]$$

$$ \psi_3^b=|\leftarrow\rangle=\frac 1 2[|\uparrow\rangle - |\downarrow\rangle]$$

and the final device just selects the 1/2 that is spin-up ($z$-coordinate).

In device c, your adding both wave functions coherently:

$$ \psi_3^c = \psi_3^a + \psi_3^b = \frac 1 2\big( [|\uparrow\rangle + |\downarrow\rangle]+ [|\uparrow\rangle - |\downarrow\rangle] \big) = |\uparrow\rangle $$

(Major) Edit: The OP asked to formulate this answer with density matrices, and I think it is a good idea, although there are some problems.

First: Given a pure state $|\psi\rangle$, the density matrix is:

$$\rho = |\psi\rangle\langle\psi| = p_{\psi}$$

and it's equal (in form) to the projection operator onto $p_\psi$.

Second: With

$$|\uparrow\rangle=\left(\begin{array}{c}1\\0\end{array}\right)$$

$$|\downarrow\rangle=\left(\begin{array}{c}0\\1\end{array}\right)$$

The density matrices of the relevant pure state eigenstates are:

$$\rho_{+z}=\left(\begin{array}{cc}1&0\\0&0\end{array}\right) = p_{+z}$$

$$\rho_{-z}=\left(\begin{array}{cc}0&0\\0&1\end{array}\right)=p_{-z}$$

$$\rho_{\pm x}=\frac 1 2\left(\begin{array}{cc}1&\pm 1\\\pm 1&1\end{array}\right)=p_{\pm x}$$

where it is explicitly stated that they are numerically equal to projection operators for those spin states.

Third: When a state described by $\rho$ passes through a device with projector $p$, the transmitted state is:

$$\rho' = (p|\psi\rangle)(\langle\psi|p^{\dagger})=p\rho p^{\dagger}$$

Fourth: The key to the Stern-Gerlach experiment is that it entangles position and spin. We write the total "state" [note: this is tricky, $\rho$ is not a wave-function] as a column matrix representing the up and lower beams:

$$ \rho = \left[\begin{array}{c}\rho_{\rm top}\\\rho_{\rm bot}\end{array}\right]$$

Here I am using $[$ and $]$ to indicate this array is in position space, not spin-space.

We now have the tools to master the problem.

The beam starts unpolarized:

$$ \rho_0 = \left(\begin{array}{cc}\frac 1 2&0\\0&\frac 1 2\end{array}\right)=\frac 1 2 I_2$$

It passes through a $Z$-device, creating an spin/space entangled state:

$$\rho_1 = \left[\begin{array}{c}p_{+z}\rho_0p_{+z}^{\dagger} \\p_{-z}\rho_0p_{-z}^{\dagger}\end{array}\right]$$

$$\rho_1 = \left[\begin{array}{c}\left(\begin{array}{cc}\frac 1 2 &0\\0&0\end{array}\right)\\\left(\begin{array}{cc}0&0\\0&\frac 1 2\end{array}\right)\end{array}\right]$$

At this point, we block the lower path and renormalize:

$$\rho'_1 = \left(\begin{array}{cc}1&0\\0&0\end{array}\right)$$

Now run this through the $X$-device:

$$\rho_2 = \left[\begin{array}{c}p_{+x}\rho_1'p_{+x}^{\dagger} \\p_{-x}\rho_1'p_{-x}^{\dagger}\end{array}\right]$$

$$\rho_2 = \left[\begin{array}{c}\frac 1 4\left(\begin{array}{cc}1&1\\1&1\end{array}\right)\\ \frac 1 4\left(\begin{array}{cc}1&-1\\-1&1 \end{array}\right)\end{array}\right]$$

Case A:

We take $\rho_{2,{\rm top}}$ and run it through a $Z$-device:

$$\rho_A = \left[\begin{array}{c}p_{+z}\rho_{2,{\rm top}}p_{+z}^{\dagger} \\p_{-z}\rho_{2,{\rm top}}p_{-z}^{\dagger}\end{array}\right]$$

$$\rho_A = \left[\begin{array}{c}\frac 1 4\left(\begin{array}{cc}1&0\\0&0\end{array}\right)\\ \frac 1 4\left(\begin{array}{cc}0&0\\0&1 \end{array}\right)\end{array}\right]\rightarrow \left[\begin{array}{c}\frac 1 4 |\uparrow\rangle\\\frac 1 4 |\downarrow\rangle\end{array}\right] $$ Where the last term is formulated as pure states.

Case B:

We take $\rho_{2,{\rm bot}}$ and run it through a $Z$-device:

$$\rho_B = \left[\begin{array}{c}p_{+z}\rho_{2,{\rm bot}}p_{+z}^{\dagger} \\p_{-z}\rho_{2,{\rm bot}}p_{-z}^{\dagger}\end{array}\right]$$

$$\rho_B = \left[\begin{array}{c}\frac 1 4\left(\begin{array}{cc}1&0\\0&0\end{array}\right)\\ \frac 1 4\left(\begin{array}{cc}0&0\\0&1 \end{array}\right)\end{array}\right]\rightarrow \left[\begin{array}{c}\frac 1 4 |\uparrow\rangle\\\frac 1 4 |\downarrow\rangle\end{array}\right]$$

All is well.

Case C:

Here we coherently recombine the beams emerging from the $X$-device. Recall that:

$$ \rho_{2,{\rm top}}=\frac 1 2[|\rightarrow\rangle\langle\rightarrow|] $$

$$ \rho_{2,{\rm bot}}=\frac 1 2[|\leftarrow\rangle\langle\leftarrow|] $$

where:

$$ |\rightarrow\rangle=\frac 1{\sqrt 2}\left(\begin{array}{cc}1\\1\end{array}\right)$$

$$ |\leftarrow\rangle=\frac 1{\sqrt 2}\left(\begin{array}{cc}1\\-1\end{array}\right)$$

in the Z-basis column vectors. If we add the density matrices, we are by definition creating a mixed state...thereby destroying coherency. We can't do that. We need to consider the entangled state coming out of the $X$-device:

$$\Psi_2 = \left[\begin{array}{c} \left(\begin{array}{cc}\frac 1 2\\\frac 1 2\end{array}\right)\\ \left(\begin{array}{cc}\frac 1 2\\-\frac 1 2\end{array}\right) \end{array}\right]= \left[\begin{array}{c}\psi_{2,{\rm top}}\\\psi_{2,{\rm bot}} \end{array}\right] $$

The final Z-device is a linear projection, so it yields:

$$\Psi_C=\left[\begin{array}{c} p_{z+}\psi_{2,\rm top}\\ p_{z-}\psi_{2,\rm top} \end{array}\right] + \left[\begin{array}{c} p_{z+}\psi_{2,\rm bot}\\ p_{z-}\psi_{2,\rm bot} \end{array}\right] = \left[\begin{array}{c} p_{z+}(\psi_{2,\rm top}+\psi_{2,\rm bot})\\ p_{z-}(\psi_{2,\rm top}+\psi_{2,\rm bot}) \end{array}\right] $$ $$ \Psi_C=\left[\begin{array}{c} \left(\begin{array}{c}1\\0\end{array}\right)\\ \left(\begin{array}{c}0\\0\end{array}\right) \end{array}\right] $$

$\endgroup$
5
  • $\begingroup$ Thanks for the answer. But would you please elaborate further on why we would see the same effect if a single particle is shoot? Because like quirkyquark answer I'm under impression that with single particle there is nothing for it to interfere with! $\endgroup$
    – al pal
    Commented Jul 14, 2021 at 18:37
  • $\begingroup$ I just studied density matrix since you had used it in your answer. Would you please also tell me how density matrix calculation would work on the second and third apparatus the same way you did for $\rho_1$? $\endgroup$
    – al pal
    Commented Jul 14, 2021 at 18:52
  • 1
    $\begingroup$ @alpal (1) That it works with one particle is the core of QM (a la the double slit). If you don't know the path, it takes all paths coherently. (2) I think you should work it out with density matrices, and you should, but I am going to do it b/c I found a mistake in my analysis when I tried. (It's $p_1\rho p_1^{\dagger}$, not $p_1\rho$...though they are equal where I used it. $\endgroup$
    – JEB
    Commented Jul 14, 2021 at 19:23
  • $\begingroup$ @JEB from what I could understand from the rationale, this analysis for (c) works because the beams are taking the two paths coherently. As you've said, "if you don't know the path, it takes all paths coherently". This seems to imply we are not measuring the paths from the second to the third devices, thus we are not making a measurement with the X-device. Is this correct? If so I think it's important to make it very clear, as I was very puzzled before I realized this. $\endgroup$ Commented Jul 14, 2021 at 22:22
  • $\begingroup$ @JEB I could finally understand every part you described except the very last step of your density matrix calculation. Are you switching from density matrix to state vectors before applying very last Z device?! Would you please elaborate further on that? $\endgroup$
    – al pal
    Commented Sep 9, 2021 at 5:05
2
$\begingroup$

The situation in all three experiments is same upto the second analyzer. Once the beam leaves the upper port of the second analyzer, we can write its state as $$|\uparrow\rangle_x=\frac{1}{\sqrt{2}}(|\uparrow\rangle_z+e^{i\alpha}|\downarrow\rangle_z)$$ and for the beam leaving from the lower port $$|\downarrow\rangle_x=\frac{1}{\sqrt{2}}(|\uparrow\rangle_z-e^{i\alpha}|\downarrow\rangle_z)$$ where you can indeed check that $\langle \downarrow| \uparrow\rangle_x=0$. Now if one puts up a z analyzer in front of any of these beams, one would get half of up and half of down polarized beams in the z direction. This is the premise of the experiment a and b. For the case of c, there happens to be a "constructive" interference before we meet the last z analyzer, so the state of the beam going into the last z analyzer would be $$|\downarrow\rangle_x+|\uparrow\rangle_x\sim|\uparrow\rangle_z$$ where I have used approximation sign as I have not normalized the interference superposed state here. Now for the case of just one particle, I think that for a single particle as there is no other beam for it to interfere with, it would simply not replicate the beam results. As such a single particle would always collapse into one of the eigenstate. But if you send a stream of particles one at a time, thus eliminating the possibility of interference from the other beam, we would get results similar to the case a and b. At least that is what I think.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.