[Eternal Student (OP)]

Hi again.

Thank you for your time and consideration of the problem so far.

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)

   But if that angle, X, is 0 degrees then none at all are split and sent to your S(-X) collection point.   We also  know that when X = 90 degrees,  S(X) and S(-X) have equal intensity.
    Assuming continuity, we have that for X → 0 degrees,  the intensity at the S(-X) collection point → 0.    The question is, how fast does does S(-X) → 0  compared to the rate at which X→0 ?

    Something for every reader to try that is quite fun:
    Try using this interactive simulation produced by St. Andrews University,  U.K.

https://www.st-andrews.ac.uk/physics/quvis/simulations_html5/sims/MeasurementUncertainty/measurement-outcome-uncertainty.html

It may take a minute or two of fiddling to work out what the controls are and what the numbers are.  For example, the results ignore the losses from the first SG (or first two SG if you've selected three SG) - you'll work it out if you fiddle with it.

We're also going to be ignoring losses in the first SG,  assuming you sent a random mixed state of electrons into the first SG, you would have already lost half of them.  We'll just be intersted in those electrons that do actually take the correct deflection path from the first SG.  All we need to start from is a stream of electrons entering the second SG for which we know they are in a state with m_z = +(1/2)ħ.
    You should see that if you rotate the second at 10 degrees,  you obtain 0.992   or  99.2% of the electrons hitting the screen at a place corresponding to having spin component +(1/2)ħ in that direction (10 degrees off the z-axis).
    If you rotate the SG at 20 degrees then you obtain 0.970.

Obviously there's nothing special about what we called the z-axis.   Whenever we have the electrons enter an SG aligned at +10 degrees to a previous SG axis,   then we will get  0.992  of the electrons being deflected to the +(1/2)ħ collection point for the current SG axis.

Get your own calculator and check that  0.992²  =  0.984   >   0.970.     That is to say that sending the electrons through two SG, each at 10 degrees to the former would get more of your electrons to a certain collection point than sending the electrons through just one turned at 20 degrees.

Similarly if we had 9 (nine) SG,  each at 10 degrees to the former, then we obtain approximately  0.992⁹  ≈ 0.93    =  almost all of them arriving at one collection point.      That's clearly a lot more then sending them through a single SG at 90 degrees  (where only  0.5  would have arrved at that collection point).

   Sadly the simulation won't let you make smaller rotations than 10 degrees - but hopefully you've already spotted the pattern.    If you had 90  SG apparatus,  each only at 1 degree to the former,  then you get an even bigger proportion (> 0.93) of the electrons arriving at the final collection point  (where particles would have gone if you used just one SG at 90 degrees and corresponds to having a final spin component of +(1/2)ħ along that last axis measured).

    Anyway, returning to the ideas in the original post:   I don't think you will end up with a uniform intensity smudge of electrons all over the screen when you use N (large N) SG apparatus each slightly rotated relative to the former.   I think there is going to be one bright (or high intensity) spot on the screen and only some very dim spots elsewhere.   The evidence is as presented above.
   The more SG you use and the smaller the rotation of each one relative to the former one,  the brighter that single spot will be and the dimmer the rest becomes.   The rotating SG idea as suggested in the original post may be equivalent to passing the electrons through an infinite set of SG, each only infinitesimally rotated relative to the former one.    So quite possibly you do end up with just one bright spot on the screen and 0 intensity for all the other spots (i.e. no other spots).   As I said,  I don't know.... just seems like a reasonable possibility.

Best Wishes.

[alancalverd]

Quote from: alancalverd on Today at 09:41:26
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)
   But if that angle, X, is 0 degrees then none at all are split and sent to your S(-X) collection point.   We also  know that when X = 90 degrees,  S(X) and S(-X) have equal intensity.

Mea culpa - I should have written S_x and S_-x to denote "two diverging streams with spins oriented in the directions X and -X" which is exactly what the original SG experiment generated.

There's obviously a first-order correction to be applied either between each stage of the apparatus to render the streams parallel or to allow for the divergence by constructing ever-larger magnets with fields diverging in two directions, but I can't imagine why anyone would want to blow neutrons through a magnetic trombone anyway.

[varsigma]

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

No that isn't what "happens".
Nothing about the spin state is "maintained in the second stage". Where is it maintained or "stored". You should realize, there isn't any room for this in the beams. You seem to be making a mistake about classical information; it isn't what most people think it is, although it is what (some) people say it is, which makes the stuff interesting.

[varsigma]

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.

Again, the diagram at the top doesn't say what you say.
It says the z_ beam is dumped, which is why it says "No z_" at the output.

The z+ part is maintained because the two SG "gates" are oriented the same. But the combination dumps 1/2 the beam.
This might seem like a detail, but one thing it empasises is that half the beam can still be split in two, but not by the top pair of SG's

[Eternal Student (OP)]

Hi.

I should have written Sx and S-x

   I don't think the notation is a serious problem, I knew what you meant and my reply just continued with that notation.

I can't imagine why anyone would want to blow neutrons through a magnetic trombone anyway.

    I don't expect I will ever do the experiment or have a practical application for it.   However, you've got to ask "what if" questions at least some of the time when you're studying any science.
    As I think you've said yourself a few times:  We know what we know.  However, we aren't even aware of what we don't know and there's almost certainly a whole lot more of that.

    I thought I knew a lot about SG apparatus but it seems I may not.  Moreover, the more carefully I look, the more chasms in my knowledge I'm noticing.  Things I may have just stepped around seem to be right dead centre and in the way.

   For example, there is a connection with a wider issue that is called "the measurement problem" and I'm sure you've heard of that before.   There are several separate things involved with "the measurement problem" but the one I'll be interested here is the following:
    What constitutes a measurement and when does it happen?

How short do you want the next section to be?   Short --> don't pull down any spoilers.   It can be dull and my problems do not need to be yours.

Spoiler: show

     There are various ways in which a measurement could have happened, or specfic places where it may have happened.  Some texts will declare that when the electron enters the next SG, that's where a measurement is made.   Other texts suggest the measurement happens only on exit from the SG.   Many texts just say that an SG is a measurement device and don't trouble themselves over exactly where or how that happens.   Some texts are very precise and would directly dispute that an SG should be treated as a measurement device.   These would suggest that there is only smooth evolution of the wave function according to the TDSE (time dependent Schrodinger equation) while inside the SG, there is no abrupt or discontinuous change in the wave function associated with a wave function collapse what-so-ever when a particle passes through an SG.   Measurement happens only when the particle exits the SG and a definite displacement for the particle (up or down the axis of the SG) is observed.   So it's always something else, like a screen that is actually making the measurement.
     A screen forces the location of the particle to be determined because only one piece of the screen will glow.   However, there are no simple rules about what a measurement device can be or when a measurement has been made.  A second SG apparatus placed beyond the first SG also forces a determination of the particles displacement by a previous SG because the particle can only get into the channel for the second SG if it's been deflected by the first SG the correct way.   So the aperture of the second SG and its placement can also be considered as a measurement device.

 

   When you have a big series of (static, non-rotating) SG, then it doesn't matter if the measurement was made on entry, on exit or half-way in the middle of the SG,  in any of those situations there will have been only one wave function collapse (one per SG) that could lead to a new current wave function.  There's more I could say but this is enough for the point I'm trying to make. 
    In the case of my very long and rotating SG, the particle doesn't enter or exit an SG between each change of alignment in the magentic field of the SG.  So now we do have to pause and consider where a measurement might occur.   We can't be sure that measurement and wave function collapse is happening frequently while inside the SG.
     
   

Spoiler: show

   If we can't assume measurements are frequently made while particles are inside the SG then we can't analyse the situation in the way suggested in earlier posts.
      I've been re-reading an old textbook  (Quantum Mechanics, Franz Schwabl) where a treatment of the evoluton of the wave function according to a given Hamiltonian and determined by the TDSE (Time dependent Schrodinger Equation) is favoured when considering what happens inside an SG.   Reduction of the wave function (wave function collapse) is not performed until the particle emerges and its displacement along the axis of the SG is observed.   To say that more clearly - the SG is NOT considered to cause any wave function collapse, there is only smooth evolution of the wave function by the TDSE while the particle is inside the SG.  The sad thing (possibly good thing) is that, for a static SG, that still produces exactly the same results.   For a static SG you couldn't be sure if a measurement was performed inside the SG or only when the displacement after exit from the SG was measured because the screen would show the same thing in either case. 
    Anyway, you could run the TDSE for a slightly more complicated Hamiltonian (for the rotating SG, the magnetic field is changing alignment).   Then see what you'll get when the only measurement assumed to occur is the one that happens at the screen.   Like the static SG, it may give the same results as assuming measurements were made inside the SG.   

   Anyway, I don't know  - but that's OK and probably a good thing.

Late Addittion:  @varsigma has posted while I was writing this.   I don't think I would disagree with much that was said.  Stern-Gerlach stuff still gets mentioned and examined in many articles and papers.  I suspect there's a lot we don't know and only a few things that do seem to work.   It's easy to assume we (human beings) know a lot about it because all the usual problems we might get in, say a university course on Quantum Mechanics,  are just using the stuff we do know about and skirting around all the bits we don't.  For the static sets of SG the explanations are all available online in various places,  I think @hamdani yusuf  's diagram came from Wikipedia which is one place where a fair sized discussion for that diagram already appears.

Best Wishes.

[alancalverd]

What constitutes a measurement and when does it happen?

I think it is shorthand for the point at which a particle or photon interacts with something else.

The phrase causes a lot of confusion because a standard textbook "explanation" of Heisenberg involves bouncing a photon off an electron  in the hope of measuring the latter's speed and position, added to folk talking about "wave function collapse" as though it was a real phenomenon and not just a mathematical description of one.
 
I always revert to the simple mantra: we describe propagation with wave equations, interaction with particle equations.

[hamdani yusuf]

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

AFAIK, not all metal atoms produce the same effect. Gemini says that neutron can't be used in Stern Gerlach experiment to produce the same results as silver atoms.

[Eternal Student (OP)]

Hi.

Gemini says that neutron can't be used in Stern Gerlach experiment

    I don't know where Gemini was getting its information.  It seems that neutrons have been put through a SG.
https://journals.aps.org/pr/abstract/10.1103/PhysRev.96.1546
Sadly that article is pay-walled.          This is someone else's interpretation of it:
https://physics.stackexchange.com/questions/197252/how-was-the-neutrons-spin-measured

Actual results obtained will depend on:
   (i) the spin of the particle because that influences the magnetic moment which influences the force that will act,     
   (ii) any other angular momentum of the particle, for example electrons in orbit around a nucleus will have orbital angular momentum in addition to spin angular momentum.
   (iii) the gyromagnetic ratios (or g-factors) for the particle (constants relating magnetic moment to spin or other forms of angular momentum) because that is also governing the force that will act.
   (iv) the mass of the particle because that provides inertia to resist the force that acts.
   (v) other things that might be taken for granted.  E.g.  a fast particle doesn't spend long in the SG and won't be deflected much etc.

    So, you aren't going to get exactly the same deflection for a neutron and silver atom but the general behaviour should be the same.

Best Wishes.

[alancalverd]

Another nail in the coffin of chatbots. Or are they nailing down the coffin of truth?

[varsigma]

It's about what classical information is. What does it mean to say "half the beam is dumped", or "half the beam is spin up in the z direction"?

If you think you know what those mean in the context of spin-polarized half beams, then you understand what QM says about the nature of "information".

Have at it

ed. any discussion will necessarily include that patterns, in nature, are the things that convey meaning. All patterns we observe have meaning, we just have to apply the right interpretation to squeeze it out of the data, or the pattern we think we can see.
There is a big, big difference between classical statistical data, in which we expect or hope to see a pattern, and the statistics of tiny, tiny bits of matter. That's one reason SG has beams of lots and lots of them, it's then a statistical measure of the, you know, the interaction, the thing that happens to a dumped beam.

[varsigma]

The diagram of SG gates as I call them,  is really about how you can compose these quantum measurements. That is, first accepting that a beam of ionized silver atoms interacts with the apparatus such that two half beams emerge (a classical observation!) and they are polarized in an up-down sense. We assume that differentiating between up-down or left-right, or at some angle, is an important detail, which it is for classical measurements, all of them are distance measurements in the end and without angles and directions that isn't meaningful.

So the diagrams

[varsigma]

 


ok. the beams going in initially are in an unknown polarization, and until they go through a first gate, that won't be an observable. If half the beam is dumped. the remainder can still be spin-polarized in any chosen direction, but the effect is the same as in the initial case, if two half (or quarter) beams emerge in a known spin state, nothing can be known about what that was before they went in.
 Or in the above case, nothing that is known makes the least difference and appears to be redundant.

I'll try to unpack that a little. Nothing can be known about the x spin state for a half beam which is known to be z spin up, in the diagrams.

apologies for the poor editing.

[hamdani yusuf]

Physics Misunderstood This Experiment For Years

The Stern-Gerlach experiment was the first detection of quantum spin, except we didn't know it for 4 years. At the time, we didn't even know quantum spin was a thing.

TIME CODES

00:00 Intro
01:03 Angular Momentum
02:09 Stern-Gerlach Experiment
03:38 Experiment Results
04:55 Schrodinger Breaks Everything
07:04 Quantum Spin
08:03 Closing Thoughts
08:30 Supporter Shoutout
08:53 Featured Comment

And some interesting comments.

s Isaac Asimov said ?The most exciting phrase to hear in science, the one that heralds new discoveries, is not 'Eureka!' but 'That's funny?'?

And this is another great example of that.

Some other fun historical facts about this experiment that really should have failed. They were out of funds when conducting the experiment, so they wrote to an American banker and got money from him (they were not expecting that!).

And when they tried to see the experimental results, they initially saw nothing. But Stern was smoking very cheap cigarettes, and sulfur in the smoke from these cigarettes made the faint traces of silver visible.

"The Stern-Gerlach Experiment (ESI College Physics Film Program 1967)" is a video on youtube that also shows the behavior in a homogenous/non-homogenous field... the magnet doesn't 'align with the poles' but will sit stabilly with either n-s or s-n alignment.... part of that though, is if you spin the magnet around it's magnetic axis, then it will oscillate back and forth....
But then, the curl of an electron beam through a magnetic field is around the magnetic lines of flux - so the atoms will interact with the field, and if they are slightly up or down, interact in such a way as to fully force an alignment through the length of the magnet, and then come out....
That's part of the 'measurement problem' that the quantum thing interacts with the measurement device - and that's not even part of the equations.  polarized light is the same - as it interacts with a polarization filter, the resulting photon is probably no longer in the same polarization/alignment as it was; that's why quantum cryptography works - if you put  a filter in the way - even if it's aligned the same as the resulting detectors - the signal from the origin vs the signal through it's own filtering is detectable, so a man-in-the-middle attack is identifiable.

[varsigma]

A question about Stern-Gerlach or any experiment with beams that halves the intensity.

As in polarization of light. We assume that any kind of beam-splitting apparatus outputs a pair of beams with half the intensity of the input. Or that say, if we order some equipment it will be specified for a certain frequency of light.

So if you repeat this with enough beam-splitters in a chain, when do you stop seeing any outputs? This is measurement and statistics in action. All experiments should gather a representative sample, by repeating some measurement process often enough.

[alancalverd]

when do you stop seeing any outputs?

Never.

If the incident flux is N photons (or whatever) per second and you have n lossless splitters, the average exit flux from the nth splitter is N/2ⁿ photons per second, so you just have to wait a bit longer for the next photon.

[varsigma]

If the incident flux is N photons (or whatever) per second and . . .

An if that gets bigger as you increase the number of splitters. You assume that it's possible to count the photons as they leave the source, or at least to a good approximation you can, by using known laws.

Also, I can see that if halving a beam of particles by halving the intensity is the statistical sampling, then I don't need to split a beam into two, I just need to halve it. That's what a polarizing optical filter does, roughly, and there is a dependence on the frequency. So if I dump half the beam of light by absorbing it with a filter, I should get a result--a representative sample. I should be able to derive a Haar measure from these results so that I have a formal "statistical" theory.

[alancalverd]

No need to count the photons leaving the source. If you know n, or any other attenuation factor, you just count the arriving photons and multiply by the attenuation factor to get N.

To avoid a frequency or energy dependence, use a grid attenuator for uncharged particles or photons. For charged particles, use a magnetic or electrostatic deflector. 

[varsigma]

No need to count the photons leaving the source. If you know n, or any other attenuation factor, you just count the arriving photons and multiply by the attenuation factor to get N.

To avoid a frequency or energy dependence, use a grid attenuator for uncharged particles or photons. For charged particles, use a magnetic or electrostatic deflector. 

For a beam of light which isn't monochromatic, you have a mixture of frequencies. It would be better to know that you have a beam which is at or close to a given frequency. Using an attenuator will mean having information about the frequency, because it can be controlled at the classical level.

The information I'm after is the number of filters required to attenuate a given intensity to zero. You claim this will never happen, but if a beam of light is actually a lot of individual particles which the experiment won't multiply, then the number of filters required must be finite. Algebra, huh?

[alancalverd]

Not if the beam is continuous - eventually one photon will get through. It's all down to Heisenberg: there are no nonzero solutions to the wave equation!

Important to remember that light isn't a wave or a particle, but a phenomenon for which we need two mathematical models.

[varsigma]

Not if the beam is continuous - eventually one photon will get through. It's all down to Heisenberg: there are no nonzero solutions to the wave equation!

Indeed. And I will need to admit the observation, to my theory, that polarizing filters (in the optical domain) can appear completely dark, but light will "get through". In fact, this can be demonstrated with just three filters.

Now, instead of thinking about attenuation of a continuous input, I have to consider rotation angles for the filters.