Post by: Eternal Student on 15/03/2024 15:01:23
Hopefully, some of you have heard of Stern-Gerlach apparatus.
(https://upload.wikimedia.org/wikipedia/commons/thumb/e/ee/Stern-Gerlach_experiment_svg.svg/300px-Stern-Gerlach_experiment_svg.svg.png)
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.
Post by: Halc on 15/03/2024 15:23:39
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.
Post by: Eternal Student on 15/03/2024 15:55:53
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.
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.
Post by: alancalverd on 15/03/2024 17:43:56
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
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!
Post by: Eternal Student on 15/03/2024 23:12:16
so you really need two SG2s to do the second splitSeems 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.
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.
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 246 times)or in practice a series of SGs each slightly rotated with respect to the previous oneThat 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.
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?
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.
Post by: Bored chemist on 16/03/2024 13:22:42
So we already know what happened.
Post by: paul cotter on 16/03/2024 13:49:27
Post by: Eternal Student on 16/03/2024 14:54:29
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
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.
Post by: hamdani yusuf on 18/03/2024 15:57:51
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.
Post by: 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.
Post by: Eternal Student on 19/03/2024 03:30:38
Thank you for your opinions @hamdani yusuf .
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.
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.
Post by: hamdani yusuf on 19/03/2024 04:52:01
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.
Post by: hamdani yusuf on 19/03/2024 04:57:03
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.
Post by: hamdani yusuf on 19/03/2024 05:53:02
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.
Post by: hamdani yusuf on 19/03/2024 06:44:43
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.
...
(https://upload.wikimedia.org/wikipedia/commons/thumb/3/35/Sg-seq.svg/1280px-Sg-seq.svg.png)
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.
Post by: 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, 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.
Post by: Eternal Student on 19/03/2024 17:41:06
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.
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.
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.
Post by: hamdani yusuf on 19/03/2024 21:13:02
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.
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.
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.
Post by: hamdani yusuf on 19/03/2024 21:19:32
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
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.
Post by: hamdani yusuf on 19/03/2024 21:36:19
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 evolutionYou 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.
Post by: Eternal Student on 20/03/2024 02:21:17
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.
Post by: varsigma on 20/03/2024 09:50:15
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.
Post by: hamdani yusuf on 20/03/2024 13:24:40
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.
Post by: hamdani yusuf on 20/03/2024 13:26:39
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?
Post by: hamdani yusuf on 20/03/2024 13:33:24
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?
Post by: hamdani yusuf on 20/03/2024 13:40:18
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.
Post by: hamdani yusuf on 20/03/2024 13:48:02
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.
(https://upload.wikimedia.org/wikipedia/commons/thumb/3/35/Sg-seq.svg/1280px-Sg-seq.svg.png)
Post by: hamdani yusuf on 21/03/2024 03:09:00
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
Post by: hamdani yusuf on 21/03/2024 04:31:13
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
Post by: Eternal Student on 21/03/2024 04:38:49
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.
Post by: hamdani yusuf on 21/03/2024 09:56:04
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.
Quote
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
Post by: hamdani yusuf on 21/03/2024 12:46:58
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.
Post by: varsigma on 25/03/2024 09:05:06
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.
Post by: varsigma on 25/03/2024 09:11:21
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.
Post by: hamdani yusuf on 25/03/2024 09:32:37
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.
Post by: varsigma on 25/03/2024 11:42:18
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?
Post by: hamdani yusuf on 26/03/2024 02:39:39
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-.
Post by: alancalverd on 26/03/2024 09:41:26
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 2N distinct streams each separated by X degrees.
Post by: alancalverd on 26/03/2024 09:45:05
Post by: Eternal Student on 26/03/2024 15:31:10
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 mz 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.
Post by: Eternal Student on 26/03/2024 17:25:30
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 ?
8) 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 mz = +(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.
Post by: alancalverd on 26/03/2024 18:06:07
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 Sx 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.
Post by: varsigma on 27/03/2024 01:23:47
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.
Post by: varsigma on 27/03/2024 01:30:29
Again, the diagram at the top doesn't say what you say.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.
(https://upload.wikimedia.org/wikipedia/commons/thumb/3/35/Sg-seq.svg/1280px-Sg-seq.svg.png)
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
Post by: Eternal Student on 27/03/2024 02:15:27
I should have written Sx and S-xI 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
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
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.
Post by: alancalverd on 27/03/2024 09:13:48
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.
Post by: hamdani yusuf on 27/03/2024 10:06:10
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.
Post by: Eternal Student on 27/03/2024 11:55:01
Gemini says that neutron can't be used in Stern Gerlach experimentI 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.
Post by: alancalverd on 27/03/2024 17:58:59
Post by: varsigma on 27/03/2024 19:38:52
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.
Post by: varsigma on 29/03/2024 20:41:00
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
Post by: varsigma on 29/03/2024 20:45:29
(https://upload.wikimedia.org/wikipedia/commons/thumb/3/35/Sg-seq.svg/1280px-Sg-seq.svg.png)
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.