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  4. How many spots are produced from Stern-Gerlach apparatus that rotates?
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How many spots are produced from Stern-Gerlach apparatus that rotates?

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Offline Eternal Student (OP)

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How many spots are produced from Stern-Gerlach apparatus that rotates?
« on: 15/03/2024 15:01:23 »
Hi.

      Hopefully, some of you have heard of Stern-Gerlach apparatus.



Image taken from:  https://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment    where some more information is also avaialble.

     Basically, we could send some electrons through a piece of Stern-Gerlach apparatus aligned in the z-axis direction and they would have their path deflected up or down dependng on whether they have a z-component of spin +1/2   or  -1/2.   We only end up with two spots or locations on the screen rather than a smeared out line - suggesting that this spin is quantised.  I expect you've seen or heard the usual explanations.
    Similarly we could turn the appratus sideways, so that the magnetic field is aligned in the y-axis.   Then we have two spots separated left and right instead of up and down.
     More interestingly we can send the electrons through a z-axis aligned apparatus and then send them through a y-axis aligned apparatus.   What we find is that electrons with the z-component of spin +1/2   will still have an equal chance of having the y-component of their spin reported as being + or - 1/2 when it is measured.   Overall, we end up with  four spots or locations for the electrons at the end with an equal intensity on the screen:   1/4  went up and left,    1/4 went up and right,  1/4 down and left,  1/4 down and right.

    If you were able to trace the path of an electron throughout the whole experiment then you expect to observe electrons that were near the North magnetic pole of the first apparatus to be deflected left or right with equal probaility in the second apparatus.   Similarly, those near the South pole of the first apparatus were deflected left or right with equal probability in the second appratus.     At the end of the experiment then,  the electrons near the North pole of the last appartus were not necessarily near the North pole of the first appartus, they could have come from the South pole in the first apparatus.    I hope this makes sense, even if you need to read it twice.    The electrons can move in the two directions independantly in both pieces of apparatus.   Those which are on the Left in the final apparatus could have had a total movement that was  Up then Left   or   Down then Left.

    Now, what happens if you do the experiment slightly differently?   Instead of having two pieces of apparatus, with one just in front of the other,   we will use just one very long Stern-Gerlach apparatus and slowly rotate the apparatus while the electrons are travelling through it.    Initially the apparatus would be aligned in z-axis direction when the electrons enter it,  it would then be rotated so that the magnetic field was aligned in the y-axis direction by the time the electrons exit from it. 
     You would expect the electrons to get deflected according to their spin as usual.   What path will the electrons actually take now?   If an electron had started to move toward the North pole of the appartus when it entered the apparatus, that means it had z-component of spin +1/2.  In principle it can still have a y-component of spin that is + or -1/2 as shown in the original experiment, which could suggest it will move away from the North pole and toward the South pole of the apparatus when the apparatus is rotated and now measuring the y-component of spin.
   On the other hand, perhaps it ALWAYS stays near the North pole of the apparatus even during the rotation.   Will we only get two spots on the screen when the electrons exit the apparatus?     If it's still just the usual four spots on the screen, then you've got to ask yourself some questions:   Will some electrons that were at the North pole initially or early on, move to the south pole later in the experiment,  while an equal number that were at the South pole move to the North pole?  (i.e. so you may have equal total numbers at the North and South poles of the apparatus but some electrons have switched sides during the experiment?)
    I honestly don't know what we will get and would like some opinions.   I suspect we are only going to get two spots on the screen at the end and the situation will be similar to sending polarised light through polarising filters.   (Birefly recall experiments where polarised light is sent into a polarising filter.   If you use one polarising filter at 90 degrees to the incoming light, then you get nothing through.   If you use two polarising filters, the first at 45 degrees orientation to the incoming light, the second at 45 degrees to the first (so now a total of 90 degrees to the original light) then you do get some light through at the end.   The more filters you add and the more gradually you ask the photons questions about the orientation of their polaristaion, then the more light will be passed through at the end).

Best Wishes.
« Last Edit: 15/03/2024 15:06:05 by Eternal Student »
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Offline Halc

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #1 on: 15/03/2024 15:23:39 »
Quote from: Eternal Student on 15/03/2024 15:01:23
Overall, we end up with  four spots or locations for the electrons at the end with an equal intensity on the screen:   1/4  went up and left,    1/4 went up and right,  1/4 down and left,  1/4 down and right.
OK, as expected. Measurements taken along perpendicular axes would register no particular correlation.  I take it that the deflection is small enough that a single second device measures both outputs of the first device.

Quote
Now, what happens if you do the experiment slightly differently?   Instead of having two pieces of apparatus, with one just in front of the other,   we will use just one very long Stern-Gerlach apparatus and slowly rotate the apparatus while the electrons are travelling through it.    Initially the apparatus would be aligned in z-axis direction when the electrons enter it,  it would then be rotated so that the magnetic field was aligned in the y-axis direction by the time the electrons exit from it.
Not sure. A quantum measurement is already taken (collapsed so to speak) by the initial z-alignment, so at no point is the new angle not yet measured. I think it will carry this collapsed state through the rotation, leaving again two dots. Before, you were measuring a previously unmeasured y component. Not the second time.

That's my take anyway. Good question. I'm not enough of an expert to properly justify my response. Yes, it's a lot like sending light through a series of slowly changing angled filters.
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Offline Eternal Student (OP)

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #2 on: 15/03/2024 15:55:53 »
Hi.

Quote from: Halc on 15/03/2024 15:23:39
I take it that the deflection is small enough that a single second device measures both outputs of the first device.
   Yes, that is the situation I'm trying to go for.     Historically, the up / down separation was sufficiently large that they needed two pieces of Stern-Gerlach apparatus for the the second  y-axis alignment,  one for each beam coming out of the first apparatus.    We're going to keep the deflections small enough that we will be able to get all the output from the first apparatus into the central channel of the second apparatus (with some tolerance).   As a thought experiment only, it won't make any difference whether you do need two pieces of apparatus for the second measurement.   All we wanted to demonstrate was just that the y-axis component of spin, when measured, can be  + or - with equal probability independantly of the first measurement.

Quote from: Halc on 15/03/2024 15:23:39
That's my take anyway.
   Thank you.   Also, this isn't anything I'm actually doing or desperately need an answer for.  It's just something that was of interest, so you don't have to spend too long here.   The thread will be here for a long time, if anything does come to mind. 

Best Wishes.
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Offline alancalverd

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #3 on: 15/03/2024 17:43:56 »
Quote from: Eternal Student on 15/03/2024 15:01:23
More interestingly we can send the electrons through a z-axis aligned apparatus and then send them through a y-axis aligned apparatus.
Ahem! SG1 device splits a single beam up/down relative to its incoming vector. You now have two beams U, D with different vectors, each of which is split left-right by SG2.

Problem is that vectors U and D are not parallel so you really need two SG2s to do the second split, but assuming you can do it, you get four exit beams UL, UR, DL, DR so I think
Quote from: Eternal Student on 15/03/2024 15:01:23
At the end of the experiment then,  the electrons near the North pole of the last apparatus were not necessarily near the North pole of the first apparatus, they could have come from the South pole in the first apparatus. 
is potentially misleading.

However the question is what do you mean by "slowly rotating"  the device? I can imagine a magnetic field that rotates with distance along the z axis, or in practice a series of SGs each slightly rotated with respect to the previous one, so if we ignore the essential divergence of the beams at each point, we'd get a circular distribution at the exit.   But I rather think we are ignoring the weight of the elephant!

And then it looks as though you have addressed the question in your second post anyway.

You might be interested in the umpteen ways we manipulate proton spins  in an MRI system!
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Offline Eternal Student (OP)

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #4 on: 15/03/2024 23:12:16 »
Hi.

Quote from: alancalverd on 15/03/2024 17:43:56
so you really need two SG2s to do the second split
    Seems like a splitting of hairs - but OK... as mentioned in post #3 or my second post,  have two SG apparatus, one for each beam emerging from the first.
   We'll create an overall image of the dots by taking both screens and super-imposing them on top of each other.   Then we have the usual 4 dots,  one pile of electrons went:
 (i)   Up and left.
 (ii)  Up and right
 (iv) Down and left
  (v) Down and right
with equal numbers of electrons piled up in each of these final locations -  so  1/4  of the original input number of electrons at each final location.

Quote from: alancalverd on 15/03/2024 17:43:56
I can imagine a magnetic field that rotates with distance along the z axis,
   OK... but I'm having the electrons generally travel along the x-axis all the time.   The SG apparatus can only be aligned with the magnetic field in the y-axis direction or  z-axis direction,  please.

Quote from: alancalverd on 15/03/2024 17:43:56
the question is what do you mean by "slowly rotating"  the device?
    I mean put the device in or on something which you can slowly rotate around the x-axis.   Do that while the electron is travelling through it (along the x-axis).


* SG.jpg (35.72 kB . 978x466 - viewed 1247 times)


Quote from: alancalverd on 15/03/2024 17:43:56
or in practice a series of SGs each slightly rotated with respect to the previous one
    That is exactly how you might try to imagine the situation.

We can write  | ↑ >   for the state where an electron had spin up when the z-component of spin was measured.   Now,  it's true that   
    |↑>   =   (1/√2)   |←>     +  (1/√2)  |→>       (in words, the spin up is an equal proportioned superposition of the spin left and spin right states,  there may be a - sign instead of a plus sign, I can't recall which one.   The spin down state  would be  |↓>  =  (1/√2)   |←>     -     (1/√2)  |→>   , i.e. it combines the left and right with a minus sign if the up state had the + sign).

     In either of these two cases,  if we immediately measure the y-component of spin then we force a collapse to the left or right spin state with equal probability because it was an equal proportion of these two states in the state |↑>  .    That explains the standard  SG  experiment where we measure the z-component and then immediately measure the y-component of spin.   As is usually the case with Quantum Mechanics, once we do measure the component of spin in the y-axis we have forced a collapse.  We have destroyed the original state that was |↑> and have only a state that is  |→>.   If we then went to measure the component of spin the z-axis again, it will have an equal chance of collapsing to the |↑>  or  |↓> state and seemingly forget that it was state  |↑> just a few moments ago.

      However, if the second SG apparatus hadn't been precisely at 90 degrees to the first,   let's say it's only at an angle of θ degrees   ( θ << 90 degrees),   then  the  |↑>   state    is not likely to be an equal combination of the states   
|points at an angle θ off the vertical>      and
|points at an angle 180+θ (degrees) off the vertical>   
For brevity, we'll call the first state   |nearly straight up>    and the second one  |nearly straight down>.
What we have, I would think, is that    |↑>   =   √(0.9)  |nearly straight up>    +   √(0.1)  |nearly straight down>
That is to say,  that most of the electrons will collapse to the state that has spin component +1/2 when you measure the component of spin in the "almost vertical orientation" from the state where we knew they had spin +1/2 when we measured the component of spin in the (perfectly) vertical axis. 
    Now, if you make the rotation of each successive SG apparatus sufficiently small, then the electrons will almost always collapse to the state where they have spin component +1/2  in the new orientation being measured.   I don't know.... that's the main issue being asked.
    With a continuously rotating SG apparatus, you are measuring the spin component many times (an infinite number of times) with only infinitessimal changes in the orientation on each measurement.   On each measurement, only the state that the electron was in just prior to entering the next piece of of the SG apparatus is relevant information and determines the probability for the next collapse to + or - 1/2 spin component in the new orientation of measurement.
    As previously mentioned, we have some interesting results concerning polarised light passing through a series of filters that are only slightly rotated from the previous filter.  If we had an infinite set of such filters with only infinitessimal rotations between each, then we should get the result that any photon which passed the first filter will end up passing all of the filters even if that means it emerged with a polarisation that is 90 degrees to the polarisation it had when we sent it in.    By analogy, instead of being "blocked" or "allowed to pass" each filter we have an electron that is either attracted or repelled from the North pole at each place (or time) along the path taken by the electron.   If it was attracted (had z-component of spin +1/2) on first contact with the SG apparatus then it may continue to be attracted at every place where the component of spin in the new orientation is being measured.    (I don't know).
    There may be many other ways of imagining the problem or breaking it into smaller chunks,  imagining it as a succession of slightly new measurements all along the way is only one possibility.

Quote from: alancalverd on 15/03/2024 17:43:56
we'd get a circular distribution at the exit.
   Why do you think that?   Do you mean two spots at any moment of time  BUT, if the apparatus was rotating then these trace out a circle on the screen over time?   I would still call that only two spots.    Alternatively, did you mean that you get a circular distribution of electrons on the screen in every moment?

Quote from: alancalverd on 15/03/2024 17:43:56
You might be interested in the umpteen ways we manipulate proton spins  in an MRI system!
    Always willing to hear some interesting things and always grateful for any expertise.

Best Wishes.
« Last Edit: 16/03/2024 00:07:17 by Eternal Student »
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Offline Bored chemist

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #5 on: 16/03/2024 13:22:42 »
In the interests of pointless pedantry, Drs Stern and Gerlach did their experiment on Earth- which is rotating.
So we already know what happened.
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Offline paul cotter

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #6 on: 16/03/2024 13:49:27 »
I remember this experiment from my long distant college days but I had forgotten it's name. I can't add anything of value, I do however have a question: how are the silver atoms entrained in a beam? the only way I can think of is electrostatic acceleration, that would lead to charged particles which are ruled out by the terms of the experiment.
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Offline Eternal Student (OP)

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #7 on: 16/03/2024 14:54:29 »
Hi.

Quote from: paul cotter on 16/03/2024 13:49:27
how are the silver atoms entrained in a beam?
    Put silver in a furnace chamber,  it will eventually start to vaporise.   Put a small hole in the side of the chamber and a beam of essentially atomised silver streams out of it.
    Probably not something that passes health and safety standards but it was all done in a vaccum chamber that was well sealed.   I don't know the full history but I'm sure there are some full explanations and accounts of what was done.   There are a few critera they wanted to meet  (e.g. I think they wanted a thin flat beam so they used slits inside instead of round holes,  probably also ensured the beam was reasonable collimated by passing it through two slits etc.)

Quote from: Bored chemist on 16/03/2024 13:22:42
on Earth- which is rotating.
   It is,  although not in the way we (possiby just I) want to rotate it.   The centre of rotation and axis are in the wrong places.

Best Wishes.
« Last Edit: 16/03/2024 23:05:17 by Eternal Student »
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Offline hamdani yusuf

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #8 on: 18/03/2024 15:57:51 »
Quote from: alancalverd on 15/03/2024 17:43:56
However the question is what do you mean by "slowly rotating"  the device? I can imagine a magnetic field that rotates with distance along the z axis, or in practice a series of SGs each slightly rotated with respect to the previous one, so if we ignore the essential divergence of the beams at each point, we'd get a circular distribution at the exit.
IMO, it depends on how strong the magnetic field divergence is, and the rate of change of the SG apparatus axis along the path of the atoms.
If the divergence is strong enough, and the rate of change of the axis is slow enough, then the final result is two dots along the axis at the end of the SG apparatus.
« Last Edit: 18/03/2024 16:05:01 by hamdani yusuf »
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Offline hamdani yusuf

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #9 on: 18/03/2024 16:01:46 »

The results can be explained by assuming that the passing through SG apparatus changes the orientation of the atoms spin. Just like how polarizers change the polarization axis of the passing light.
« Last Edit: 18/03/2024 16:14:42 by hamdani yusuf »
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Offline Eternal Student (OP)

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #10 on: 19/03/2024 03:30:38 »
Hi.

   Thank you for your opinions @hamdani yusuf .

Quote from: hamdani yusuf on 18/03/2024 15:57:51
If the divergence is strong enough, and the rate of change of the axis is slow enough, then the final result is two dots along the axis at the end of the SG apparatus.
   I'm not sure what you meant by "slow enough" change of the axis.
   If the rate of rotation of the SG apparatus was so slow that it didn't change orientation at all, then we just have a standard experiment with static SG apparatus.  We have the results for that and we do get two spots, so you would certainly be correct.   

Quote from: hamdani yusuf on 18/03/2024 16:01:46
The results can be explained by assuming that the passing through SG apparatus changes the orientation of the atoms spin. Just like how polarizers change the polarization axis of the passing light.
    Sorry, no,  it isn't simple to assume the SG actively changes the spin.   Polarising filters and polarised light can also have similar problems.   
    I'm not sure which way to approach that discussion.   I've written something and edited it for the last hour.  I think it may be best to just keep it short.
    An electron doesn't have a complete set of spin information.   The SG apparatus cannot be a device that acts on the spin because the spin information does not exist.   A measurement of the component of spin in the axis aligned with SG apparatus has to be made first.

     If a SG device was a simple machine that just twisted the spin of a particle when it passes through, then it ought to do the same thing every time.   It doesn't.  See your own diagram,  last line and focus on the last or right-most SG apparatus.   Send in electrons one at a time.  They were all the same as far as you can make them  (for example, if we assume the SG is a machine to change spins then they have just come out of the previous machine with spin +1/2 in the x-axis and we would reasonably accept that one electron is the same as another electron).   The last SG apparatus spits out those electrons sometimes with spin +1/2 and sometimes -1/2 in the z-axis,  that is utterly random.   The SG isn't following any deterministic rules that a machine acting on the spin should have.

    It is far easier to recognise that a measurement of spin in the z-axis was being made.   The randomness of the final result or output from the SG is then adequately explained because the wave function collapse was inherently random.
    Once the appropriate component of spin is known, the SG apparatus does nothing at all to change that.   The electron is attracted or repelled to/from the North pole but that is a change in trajectory and not in the spin.   At no point along the SG apparatus is the spin made more, less or twisted in direction.   The only important event for the final spin result was right at the start (where we would assume a measurement was made).

Best Wishes.
« Last Edit: 19/03/2024 03:38:36 by Eternal Student »
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Offline hamdani yusuf

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #11 on: 19/03/2024 04:52:01 »
Quote from: Eternal Student on 19/03/2024 03:30:38
I'm not sure what you meant by "slow enough" change of the axis.
It can be measured by degree of axis change per cm.
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Offline hamdani yusuf

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #12 on: 19/03/2024 04:57:03 »
Quote from: Eternal Student on 19/03/2024 03:30:38
Sorry, no,  it isn't simple to assume the SG actively changes the spin.   Polarising filters and polarised light can also have similar problems.   
We know pretty sure that polarizing filters change the polarization of light. Quarter wave plates can turn a linearly polarized light into a circularly polarized light.
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Offline hamdani yusuf

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #13 on: 19/03/2024 05:53:02 »
Quote from: Eternal Student on 19/03/2024 03:30:38
If a SG device was a simple machine that just twisted the spin of a particle when it passes through, then it ought to do the same thing every time.   It doesn't. 
It does in some cases, such as in the top diagram. If the second SG apparatus is inverted up and down, then all streaming atom will go down, and none goes up.

Quote
See your own diagram,  last line and focus on the last or right-most SG apparatus.   Send in electrons one at a time.  They were all the same as far as you can make them  (for example, if we assume the SG is a machine to change spins then they have just come out of the previous machine with spin +1/2 in the x-axis and we would reasonably accept that one electron is the same as another electron).   The last SG apparatus spits out those electrons sometimes with spin +1/2 and sometimes -1/2 in the z-axis,  that is utterly random.   The SG isn't following any deterministic rules that a machine acting on the spin should have.
It can be explained by assuming that there is some precession or non-zero residual random spin in the other axis. If the second SG apparatus is aligned in the same axis as the first one, then this residual random spin has no perceivable effect. On the other hand, if the second SG apparatus is aligned in the perpendicular axis to the first one, then this residual random spin brings the only perceivable effect. In between, the effect is a combination between the two factors, depending on which one is stronger.
« Last Edit: 19/03/2024 06:53:58 by hamdani yusuf »
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Offline hamdani yusuf

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #14 on: 19/03/2024 06:44:43 »
The video I posted in my other thread shows the effect of homogeneous field compared to diverging field. Homogeneous field only twist dipole objects in it without moving them from their existing position. Whereas diverging field can move them from their existing position.

Quote from: hamdani yusuf on 09/02/2024 13:59:27
I searched for electrogravity on Youtube, and this video shows up in the results.
...
You can skip to the experiment part at around 5:30.

The first SG apparatus (shown on the left side) in the diagram can be replaced by a homogenous  magnetic field, and the end result will be the same, in my hypothesis.
« Last Edit: 19/03/2024 21:27:28 by hamdani yusuf »
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Offline alancalverd

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #15 on: 19/03/2024 10:43:09 »
Worth being a bit pedantic and pointing out that it's more complicated with electrons, or any charged particle, in a magnetic field, because their path will also be be deflected by the Lorentz force.

That said, if we start with an electrically neutral atom we know that application of a homogeneous field results in alignment, and a divergent field produces separation, so sequential application of N divergent fields, each slightly rotated with respect to its predecessor, would be expected to produce a circular distribution consisting of 2N opposing arcs.
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Offline Eternal Student (OP)

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #16 on: 19/03/2024 17:41:06 »
Hi.

1.       That video seems to be presenting an alternative view of what gravity might be.   I only watched a minute, its relevance to the thread seemed limited.

2.   
Quote from: hamdani yusuf on 19/03/2024 05:53:02
It does in some cases, such as in the top diagram.
     Yes, which is also consistent with taking a measurement.    The component of spin in the z-axis was already known from the first apparatus.   No wave function collapse was required in the second apparatus.
      Assuming a measurement was made explains ALL the situations.   Meanwhile, assuming the SG apparatus was just a machine acting on spin doesn't.

3.   
Quote from: hamdani yusuf on 19/03/2024 05:53:02
It can be explained by assuming that there is some .... random spin in the other axis.
   No, not quite.   The phrasing needs to be very carefully constructed and precise here.    This (un-measured) component of spin along a different axis isn't something that affects anything.   I don't mean it doesn't affect the next SG apparatus and what happens in that, I mean it doesn't affect anything, not anything what-so-ever or at all.  A property that the particle has really should be something that matters to the particle.   It has got to affect something about its behaviour, future evolution... or be important to something involving the particle and what it does.  If we bend the rules for declaring what is a property of the particle, then we can make up a new thing, let's say "the shoe size" of the particle and assume that is a property of the particle.  Provided the value of this "shoe size" doesn't appear in any equation governing the behaviour or future evolution or anything about the particle, then it doesn't matter,   we can assume it's a property of the particle.   Similarly we can say that "the colour of my wall",  "your weight in ounces", or "the Zippy-Zappy coefficent of Kryptonite"  are all properties of the particle.   Clearly, we don't want to bend the rules for declaring something to be a property of the particle:   A property of the particle really should be something that matters to the particle somehow.
     Until measured, the component of spin along a new axis doesn't matter to the particle.  It doesn't matter to anything at all.    The value of that component doesn't appear in or influence any equation governing the evolution of the particle or its behaviour.  The time dependant Schrodinger equation just involves an uncollapsed wave function.   After measurement, things are different.   The wave function is updated and the reported value of the property would matter,  the evolution and behaviour of the particle is now going to be different - the behaviour now and all future evolution of the particle certainly does depend on that reported value from the measurement.   If it was one thing, then the wave function would have been updated to ψ1,  if it was the other thing then the wave function would have been updated to ψ2.    It's not just that we have assumed the Schrodinger equation and QM was the right way to determine the behaviour of the particle,   we have some experimental results concerning "Hidden varibles" to back some of this up.   Rather than assuming the (un-measured) component of spin was something but it's random,  we should recognise that this un-measured component (or the "residual spin" as you phrase it) cannot meaningfully be considered as a property of the particle.  It's no more meaningful than giving the particle a "shoe size".   Whatever value you might think this quantity should take - whether that is fixed number like 1 or 2   or whether it is a genuinely random number generated by rolling dice at every milli-second, it doesn't matter because it doesn't affect anything about the behaviour or future evolution of the particle.   It won't affect anything about the behaviour or future evolution of the particle until it has been measured.

I hope that makes some sense.   It's not sufficient to imply the un-measured components of spin were some random value.  They could be fixed valued, randomly generated, fluctuating every second or bright purple  -  it just wasn't anything that matters to the particle and affects it in any way and therefore not something we can even consider to be a property of the particle.   Until it is measured, this un-measured quantity is just "shoe-size" - some thing that isn't sensibly a property of the particle at all  (despite our prejudice from Classical physics).

Best Wishes.
« Last Edit: 19/03/2024 17:46:49 by Eternal Student »
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Offline hamdani yusuf

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #17 on: 19/03/2024 21:13:02 »
Quote from: alancalverd on 19/03/2024 10:43:09
Worth being a bit pedantic and pointing out that it's more complicated with electrons, or any charged particle, in a magnetic field, because their path will also be be deflected by the Lorentz force.
Their path can still be calculated.
Quote from: alancalverd on 19/03/2024 10:43:09
That said, if we start with an electrically neutral atom we know that application of a homogeneous field results in alignment, and a divergent field produces separation,
We've got the same conclusion here.

Quote from: alancalverd on 19/03/2024 10:43:09
so sequential application of N divergent fields, each slightly rotated with respect to its predecessor, would be expected to produce a circular distribution consisting of 2N opposing arcs.
But different here.
SG apparatus has two effects on streaming atoms. Twisting into alignment, and splitting towards either magnetic poles. In every stages, the atoms are re-aligned to the axis of the SG apparatus at that stage. In the end, they will be aligned with the axis of the final SG apparatus.
« Last Edit: 21/03/2024 03:11:58 by hamdani yusuf »
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Offline hamdani yusuf

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #18 on: 19/03/2024 21:19:32 »
Quote from: Eternal Student on 19/03/2024 17:41:06
That video seems to be presenting an alternative view of what gravity might be.   I only watched a minute, its relevance to the thread seemed limited.
You can skip to the experiment part at around 5:30.
The experiment shows that
Quote from: hamdani yusuf on 19/03/2024 06:44:43
Homogeneous field only twist dipole objects in it without moving them from their existing position. Whereas diverging field can move them from their existing position.

It also shows that the same set of observation can produce different conclusions from different observers. They depend on preexisting assumptions held by those observers.
« Last Edit: 19/03/2024 21:25:13 by hamdani yusuf »
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Offline hamdani yusuf

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Re: How many spots are produced from Stern-Gerlach apparatus that rotates?
« Reply #19 on: 19/03/2024 21:36:19 »
Quote from: Eternal Student on 19/03/2024 17:41:06
This (un-measured) component of spin along a different axis isn't something that affects anything.   I don't mean it doesn't affect the next SG apparatus and what happens in that, I mean it doesn't affect anything, not anything what-so-ever or at all.  A property that the particle has really should be something that matters to the particle.   It has got to affect something about its behaviour, future evolution
You wrote contradicting statements in the same paragraph. Precession or wobbling is a widely observed phenomenon in macroscopic spinning objects. Assuming its complete absence in microscopic objects is an extraordinary assumption, IMO.
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