[Eternal Student (OP)]

Hi.

You wrote contradicting statements in the same paragraph.

   I don't see that anything was contradictory.   However, some sentences did have a lot of negations in them:    Don't, doesn't, not   etc.   and it may not be clear what they meant.

Precession or wobbling is a widely observed phenomenon in macroscopic spinning objects.

   By your own words, that is a macroscopic phenomenon.

Assuming its complete absence in microscopic objects is an extraordinary assumption, IMO.

   1.   Something which happens in a macroscopic scale doesn't always happen at the microsocopic level.
   2.    I didn't suggest that precession can't happen.    However, as it turns out, precession of spin in a magnetic field is handled slightly differently.    The state of an electron in a magnetic field is a superposition of the up and down spin states ( taking the magnetic field to be in the z-axis).   This state evolves over time in the magnetic field and we can determine the expected component of spin in the x, y, and z directions.   The x-component  varies as  Sin ωt   ,   the y-component  ~  Cos ωt   and the z-component  remains fixed.   So that, superficially it looks as if the spin is precessing around the z-axis.    However, that is only the expected value for the spin  - what you would obtain if you had a large number of electrons all spinnning like this, stop time at t, measured the components of spin for several of them at this time t - and then averaged appropriately.      Whenever you measure the component, say the x-component of spin for an individual atom you won't get the value ~ Sin ωt,  you will only ever get the value  + or - 1/2.   We can't say what the actual spin of an individual atom was doing, only that the expected value for a measurement of the spin is a function of time and corresponds to precesssion of the spin about the z-axis.   
     Since most of classical physics and macroscopic phenomena only involve large numbers of particles,  the expected values for spin are most naturally used and will be what you want to know for any classical model.   So the apparent precession of a particles spin in a magnetic field does naturally emerge.

Best Wishes.

[varsigma]

In a single Stern-Gerlach apparatus, a beam of electrons is divided in two, you don't recombine the beams but you can show that sending either part of the split beam--each is spin polarized but oppositely--through a second doesn't conserve spin states.

If you have a spin up polarized half-beam, and it goes through a second apparatus which is rotated by some angle, maybe perpendicular, the spin up state doesn't distribute like in Boolean logic. If you build up a sequence and try to determine spin states for each beam you can't use the "previous state", you can't build what's known as a distributive lattice with the logic.

[hamdani yusuf]

   I don't see that anything was contradictory.

I do. You said it doesn't affect anything, right after saying it may affect the next SG apparatus.

[hamdani yusuf]

I didn't suggest that precession can't happen.    However, as it turns out, precession of spin in a magnetic field is handled slightly differently.   

Who handled what?

[hamdani yusuf]

Whenever you measure the component, say the x-component of spin for an individual atom you won't get the value ~ Sin ωt,  you will only ever get the value  + or - 1/2. 

The SG apparatus can be turned to +/- 45 degree. How do you think the distribution will be, if the source is still unaligned?
What if the source is first aligned vertically using a vertical homogenous magnetic field?
What if the source is first passed through a vertical SG apparatus?

[hamdani yusuf]

In a single Stern-Gerlach apparatus, a beam of electrons is divided in two, you don't recombine the beams but you can show that sending either part of the split beam--each is spin polarized but oppositely--through a second doesn't conserve spin states.

Afaik, Stern and Gerlach used silver atoms instead of electron.

[hamdani yusuf]

If you have a spin up polarized half-beam, and it goes through a second apparatus which is rotated by some angle, maybe perpendicular, the spin up state doesn't distribute like in Boolean logic. If you build up a sequence and try to determine spin states for each beam you can't use the "previous state", you can't build what's known as a distributive lattice with the logic.

What if it's not perpendicular? In the first configuration of the picture below, the spin from previous stage is maintained.

[hamdani yusuf]

spin : Stern and Gerlach experiment

The animation shows an unrealistic scenario labelled "classical magnets" where the moving magnets are expected to maintain their original orientation after passing the diverging magnetic field.

I suspect that some "failures of classical physics" to explain physical experiments were caused by missing the relevant factors in those experiments or misunderstanding how the experimental setup/apparatus works. Similar situation can be seen in 3 polarizers paradox.

Here's another video with some narration.

Classroom Aid - Electron Spin and the Stern-Gerlach experiment

[hamdani yusuf]

MIT 8.05 Quantum Physics II, Fall 2013
View the complete course: http://ocw.mit.edu/8-05F13
Instructor: Barton Zwiebach

In this lecture, the professor talked about position and momentum in quantum mechanics, Stern-Gerlach Experiment, etc.

Stern-Gerlach starts at 40:00

[Eternal Student (OP)]

Hi.

I do. You said it doesn't affect anything, right after saying it may affect the next SG apparatus.

    I think this is a language problem.  The fault is mine and I should have tried to use clearer phrases.  I don't think you really want me to go over it all again, it would take at least as much space as last time.

Afaik, Stern and Gerlach used silver atoms instead of electron.

   Yes but the Original Post didn't.  It was talking about electrons.

The SG apparatus can be turned to +/- 45 degree. How do you think the distribution will be, if the source is still unaligned?

   For any individual electron, you can only ever get a value of  +(1/2) ħ     or  - (1/2) ħ  when you measure the component of spin along any direction.
   For many electrons, or the same experiment run many times over,  you get those values in a certain proportion.
   The proportion follows from the square of the coefficients for the wave function you had (on entry to the SG), when that wave function is expressed in a basis of spin states for the SG they are about to enter.   You'd need to know what the electron wave function was on entry,  for example if the electrons have just come out from some other SG and we know exactly what state they are in.   If we know nothing about the state of the electrons on entry, then we can say very little about how many would be deflected either way by the SG.   Every electron is going up or down (well, diagonally up or diagonally down) none will go straight or only half of a unit diagonally etc.  but we can say nothing about the intensity or numbers of electrons that pile up at each of the two possible exit points.

Best Wishes.

[hamdani yusuf]

Yes but the Original Post didn't.  It was talking about electrons.

The experiment using electron or ion add complexity to the resulting pattern caused by Lorentz force to the moving charged particle.

https://arxiv.org/pdf/1504.07963.pdf
Observing the spin of free electrons in action
(Stern-Gerlach experiment by free electrons)
Patent:139350140003006698 ,Tuesday, September16, 2014 1
Hosein Majlesi 2
Independent Researcher

Abstract
Stern-Gerlach experiment by free electron is very important experiment because it answered some questions that
remain unanswered for almost a century. Bohr and Pauli considered its objective observation
as impossible while some other scientists considered such observation as possible. The experiment on free
electrons has not been conducted so far because the high magnetic field gradient predicted there was thought
as impossible to generate. This paper proves that it is not only possible but also observable using a high
vacuum lamp which is deionized well. To obtain a high magnetic field gradient, it is not necessary to have
a very strong magnetic field and it is possible to observe the phenomenon using a very sharp pointed magnet
and adjusting the voltage in a certain distance from free electron beams. that objective observation
requires your consideration of some technical points simultaneously.In this experiment, no electric field and
no magnetic field does not change with time.

Figure 7: Figure 7-a represents the spiral path of
electrons caused by interaction. Figure b shows the
spiral path of electrons when confronted with inhomogeneous magnetic field

[hamdani yusuf]

 If we know nothing about the state of the electrons on entry, then we can say very little about how many would be deflected either way by the SG.   Every electron is going up or down (well, diagonally up or diagonally down) none will go straight or only half of a unit diagonally etc.  but we can say nothing about the intensity or numbers of electrons that pile up at each of the two possible exit points.

In my question, it's unaligned, i.e. fresh from the oven.
The answer is 50-50, aligned with the axis of SG apparatus.

[varsigma]

Afaik, Stern and Gerlach used silver atoms instead of electron.

Yes, sorry. I was generalising without thinking about it. It's a beam of fermionic states of silver atoms.

[varsigma]

What if it's not perpendicular? In the first configuration of the picture below, the spin from previous stage is maintained.

No it isn't. I think you misunderstand what the diagrams are about.

Any quantum physicist will tell you the spin state of fermionic bits of matter is only conserved until you measure it.

That is what an SG apparatus does though. As soon as you have a spin polarized beam go through a second magnetic field the first state is erased because the apparatus "prepares or measures" the state--each particle can only have one state.

The state is entangled with whatever prepared it, until you entangle it differently, in which sense a measurement you don't know about is what entanglement "really is". In that case, since we don't know about particle interactions until we measure a state, we are the ones who need to preserve it, not the experiment.

[hamdani yusuf]

Any quantum physicist will tell you the spin state of fermionic bits of matter is only conserved until you measure it.

Passing them through an SG apparatus is counted as a measurement.

[varsigma]

Passing them through an SG apparatus is counted as a measurement.

But you also say that the spin state is "maintained" from a previous stage. What does that mean?

[hamdani yusuf]

Passing them through an SG apparatus is counted as a measurement.

But you also say that the spin state is "maintained" from a previous stage. What does that mean?

Do you mean this?

What if it's not perpendicular? In the first configuration of the picture below, the spin from previous stage is maintained.

It's my commentary on the first diagram, where z+ stream from first stage is maintained in second stage. Nothing changes into z-.

[alancalverd]

Starting from ES's initial experiment, my interpretation is that

1. We begin with an unpolarised stream of neutral atoms (never mind charged particles at this stage)  with their spins randomly and isotropically distributed

2. An inhomogeneous field polarises and separates the stream into two streams, with spin up U and spin down D.

3. Now consider only the U stream. Again, an inhomogeneous field perpendicular to the first, separates this into UL and UR streams.

4. Suppose that the second field was at 45 deg rather than 90. Same principle would surely split U into U(45) and U(135)

5. So any polarised stream S passing through a field at angle X to the plane of polarisation would be split into S(X) and S(-X)

6. And after N distinct iterations of X we would have 2^N distinct streams each separated by X degrees.

[alancalverd]

Apropos practicality, I guess that nowadays we could use a collimated stream of neutrons rather than silver atoms.

[Eternal Student (OP)]

Hi.

Everyone's worried about using charged particles.   The experiment has been done and the Lorentz force wasn't an insurmountable problem.   For example, you can just add an E field to counter the effect.

If the experiment is conducted using charged particles like electrons, there will be a Lorentz force that tends to bend the trajectory in a circle. This force can be cancelled by an electric field of appropriate magnitude oriented transverse to the charged particle's path.
     https://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment#Experiment_using_particles_with_+1%E2%81%842_or_%E2%88%921%E2%81%842_spin

Apropos practicality, I guess that nowadays we could use a collimated stream of neutrons rather than silver atoms.

   I'm sure you could.  Just try to use a particle of spin 1/2 so that  the quantum number m_z takes only two values.   A Neutron would satisfy that and Silver atoms also did in the original experiment (as you may know, Silver has an electronic structure where most orbitals are full and contribute 0 net spin,  only the unpaired electron in the 5s orbital contributes spin).

Best Wishes.