0
$\begingroup$

Lets suppose we have spin 1/2 particle and some sort of Stern-Gerlach experiment setup (SG from now on).

So one SGz setup means one that is oriented for spin projection in z direction, similarly for other directions. So, if we prepare a beam of spin 1/2 particles and pass it through SGz and then just put some detector, we will find two distinct points on our detector. One for spin up in z and other in spin down in z (from now on Uz and Dz).

But, there is a Feynman sequential setup and analysis of spin one particle in his book, you can check it out here: https://www.feynmanlectures.caltech.edu/III_05.html. In a nutshell, there is conclusion to be drawn and it is this:

  • If you pass a beam through SGz and then block Uz and pass Dz through SGx but you block Ux for example, you are left with just Dx.

  • If you now pass this Dx through another SGz it again splits in in Uz and Dz. I have nothing against that.

Feynman points out now that, if you just pass the Dz beam through SGx without blocking anything and THEN let it through SGz again you get Dz like SGx wasnt even there.

So this was a confusing moment for me because I thought that act of passing the beam through magnetic field x is enough to make the state collapse and split the beam into Ux and Dx and that stoping or blocking the beam wasnt important part.

So now it looks to me that there is connection between spin and positinonal measurements. Because what it means is, even though Dz went through SGx it wasnt changed. I thought that simply by entering magnetic field of SGx aparatus a particle is forced to choose, so to speak, but it seems that it is not. It is forced to choose only when we take positional measurement!

So if we let particle through just SGz and never take some sort of measurement of its trajectory, it is still not in a definite state, until positional measurement by which we actually see what is the deflection and conclude what is the state. Till that measurement particle is deflected in both directions at the same time. Because surely, if we pass a particle only through one SGz it is not the same as leting it through SGz and then through SGx.

I mean, it is the same if we are concidering the outcome of third - SGz again - aparatus, without measuring SGx outcomes. But if we decide to measure SGx, particle acts as it went through SGx and is detected in Ux or Dx state.

So in light of this, my question is: If I just let spin 1/2 particle through SGz, but never detect its deflection with some kind of a plate on which I would be able to see where the particles landed, will the particles enter the definite up and down states or not? My thought was that it would, because of the magnetic field, but from Feynmann explanation it seems to me that still it wont.It seems to me that what you need to do to force a particle into definite up or definite down state is the plate and position measurement.

$\endgroup$
  • $\begingroup$ Really? I did not see that. $\endgroup$ – Žarko Tomičić May 14 at 6:06
  • $\begingroup$ I edited a bit but I dont see the roblem..if there is something unclear I will be happy to answer. $\endgroup$ – Žarko Tomičić May 14 at 6:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.