How many spots will a rotating Stern-Gerlach apparatus produce?

[Old Person]

TL;DR Summary

Pass electrons (along the x-axis) into a Stern-Gerlach apparatus (SG) aligned in the y-axis and you obtain two spots of electrons on a screen. Pass them through two SG, the first aligned in the y-axis and the second aligned the z-axis and you obtain 4 spots. What if you just passed them through one very long SG that was rotating so as to be aligned in the y-axis initially but aligned in the z-axis by the time the electrons emerge?

Hi.

Question as outlined in the summary. Note that there would be a Lorentz force if the particles were electrons. We can either live with that and take the consequence - or we can counter it by applying an E field across the SG (Stern-Gerlach) apparatus which will rotate with the SG. Assuming a suitable E field we can confine our attention just to the usual effects from spin - the basic theory just involving spin is what I'm mainly interested in. (The initial beam of electrons is in a suitably random mixed state, in an ordinary static SG we expect to get two spots of equal intensity on the screen).

I need a check on the answer please. I think there should be only 2 spots on the screen at the end. Provided you only send a short duration pulse of electrons, that's all. If it was a continuous stream of electrons then since the apparatus is rotating you trace out a circle on the screen.

Thank you for your time and consideration.

[Moderator's note: Off topic content deleted.]

 

[PeterDonis]

I think there should be only 2 spots on the screen at the end.

Why do you think that? Have you done any math?

 

[PeterDonis]

We can either live with that and take the consequence - or we can counter it by applying an E field

Or you can use what the original S-G experiment used, neutral silver atoms. The S-G magnet only acts on the outermost unpaired electron in each atom.

 

[Nugatory]

Provided you only send a short duration pulse of electrons

Short compared to what? The time it takes for the apparatus to make full rotation?

 

[Old Person]

Hi.

Have you done any math?
Yes. I thought it was sensible to keep the post short. Why would people have to read what I've done, it's boring. I am just asking the question: How many spots would there be on the screen?

Or you can use what the original S-G experiment used, neutral silver atoms

I know that's what was done. I don't want to use silver atoms. I want to use electrons. An E field should be sufficient to counter the Lorentz force.

[Moderator's note: Off topic content deleted.]

Short compared to what? The time it takes for the apparatus to make full rotation?

Yes. We need only enough electrons to arrive at the screen to get some discernible spots. If you leave the stream of electrons continuously pouring through the apparatus, the appartus isn't aligned with the y-axis for all of them when they entered and it won't be aligned in the z-axis for all of them when they exit.

Thank you for your time. This is not urgent, just of interest to me.

 

[PeterDonis]

Why would people have to read what I've done

To see what your claim in the OP, that there would be only two spots, is based on.

I am just asking the question

No, you're not. You claimed you already know the answer: two spots. What is that claim based on?

 

[PeterDonis]

I don't want to use silver atoms. I want to use electrons.

Why? Why introduce an extra complication that's irrelevant to the question at issue?

 

[Old Person]

Hi.

Why? Why introduce an extra complication that's irrelevant?

Electrons are fundamental particles and should be simple for most considerations. Silver atoms need heating in a furnace chamber and could then can be streamed through a hole, preferably all done in a vaccum. That's equipment I am unlikely to get. I would prefer to use electrons. However, if you would like to use Silver atoms, that's fine. It shouldn't affect the theory as concerns spin.

 

[PeterDonis]

That's equipment I am unlikely to get.

You mean you are proposing to run this experiment yourself?

 

[Old Person]

Hi.

Not today. The health and safety issues are worrying.
I'm mainly just interested in the theory. Electrons are fundamental and those are the particles I would like to consider. You can use Silver atoms if you wish (I shouldn't imagine vaporised Silver atoms are good for our health either, please use your own judgment if you were going to try the experiment).

 

[PeterDonis]

Not today. The health and safety issues are worrying.

Indeed.

I'm mainly just interested in the theory

Then, as I said, you are inviting needless complication by using electrons.

Electrons are fundamental and those are the particles I would like to consider.

"Fundamental" is not the primary theoretical issue here. The primary theoretical issue is having a simple spin-dependent response to an inhomogeneous magnetic field. That is what makes the theory simple. So, since you are mainly interested in the theory, you should be interested in whatever makes the theory simple.

(I shouldn't imagine vaporised Silver atoms are good for our health either, please use your own judgment if you were going to try the experiment).

I have no intention of trying any experiment.

 

[gentzen]

Electrons are fundamental and those are the particles I would like to consider.

Even if somebody would have done the experiment with electrons, the signal to noise ratio of the results would be much worse than for neutral particles. Speaking of neutral, would doing the experiment with neutrons be good enough for you? This has been done, and worked fine.

 

[Old Person]

Hi again.

No, you're not. You claimed you already know the answer: two spots.

This might be a fault in the phrases I used, sorry. I believe this is a USA based website and my English probably has differenet nuances. When I said:
I need a check on the answer please. I think there should be only 2 spots on the screen at the end.
That phrase was meant to imply: This is only what I think and not something I know to be true. Please can someone else consider the problem because I am not confident about my own assessment of the situation.

Why do you think that? Have you done any math?

If it's of any use. One approach you could take to the problem is to consider when a measurement is being made. The very long and rotating SG can be replaced with a series of many static SG, each slightly rotated relative to the one before.
The situation is (just to be clear I mean "it seems") comparable to having polarised light and sending it through a series of polarising filters, each slightly rotated relative to the previous filter. For the polarised light, we know that we can obtain light polarised in the y-axis from input light that was originally polarised in the z-axis provided we ask the question "is the photon polarised in this direction?" sufficiently often. Each time we ask the question there is wave function collapse. With only a small rotation in the axis on each measurement, we can get most of the light that came through the previous filter to go through the next filter. For example, two polarising filters, one aligned in the y-axis and one in the z-axis allow 0 light to pass through at the end. However, 3 polarising flters, each at 45 degrees to the previous one, will allow some light to pass through at the end. An infinite set of filters with infinitessimal rotation between each will let all the light that passes the first filter actually pass all the filters, so that we have effectively turned the polarisation of the light through 90 degrees with no loss of the light.

We could write down the mathematics for SG experiments and spin:
Suppose you start with a state |m _z =+1/2 >. If I can get LaTeX to work that will be
$| \uparrow >$
and put that through a SG aligned in the y-axis. We write the state on entry in terms of an appropriate basis for the SG it is going to enter.
$| \uparrow > = \frac{1}{\sqrt{2}} | \leftarrow > + \frac{1}{\sqrt{2}} | \rightarrow >$
So it collapses to the $|\leftarrow >$ or $|\rightarrow >$ states with probability = square of the magnitude of the coefficients = 1/2 for each possibility. (Doesn't look like the LaTeX did work on a "preview".... I hope the spirit is there).

Now we just need to consider a similar calculation where the next SG it enters wasn't orthogonal, it's just at a small angle to the earlier SG. I've found it hard enough to get the LaTeX to work for the above... and I don't want to prejudice any replies or the way anyone else might consider the problem. Considering the rotating SG as a series of measurements on the spin at slightly different alignments is only one way to consider the problem. There are other approaches but it's probably already been dull reading the above suggestion. I'd be more interested in seeing any opinions or ideas that others have.

Speaking of neutral, would doing the experiment with neutrons be good enough for you? This has been done, and worked fine.

Fine... just use whatever neutral particle you want. I'm being outnumbered here by people wanting to use a different particle. It's mainly just the effects from spin that I am intersted in, so it will be fine whatever you use.
People have actually done experiments with electrons and they did use E fields to counter the Lorentz force, I did try and spend a moment to see if it was possible. See
https://en.wikipedia.org/wiki/Stern–Gerlach_experiment#Experiment_using_particles_with_+1⁄2_or_−1⁄2_spin
"... If the experiment is conducted using charged particles like electrons, there will be a Lorentz Force..... This force can be cancelled by an electric field of appropriate magnitude......"

 

[Vanadium 50]

Have you done any math?

Have you?

It's hard to figure out what you want. It's electrons! No, it's theoretical! No, it;s electrons! The question is precessing faster than a charged particle in a magnetic field.

People have built polarized electron beams, usually not using a Stern-Gerlach mechanism, and have built spin-rotating magnets. This is a known technology and has been around for years. Is that good enough?

 

[Old Person]

I suspect there's been some overlap with the last reply I wrote and the recent one from @Vanadium 50 .

Let's go with any particle you like. It has spin 1/2 and no charge or the effect of Lorentz force cancelled.

 

[PeterDonis]

This is only what I think and not something I know to be true.

Why do you think it is true? You must have some reason.

If your reason is the analysis you give in the rest of your post, unfortunately, that's wrong. See below.

If it's of any use. One approach you could take to the problem is to consider when a measurement is being made.

That's easy: nothing that happens inside the SG magnet is a measurement. The operation the SG magnet performs is unitary: it entangles the linear momentum of the particles with their spins.

The measurement in the SG experiment does not happen until the particles hit the detector screen, which happens after they exit the SG magnet. So any analysis of the effects of the magnetic fields inside the SG magnet will not involve any kind of measurement.

The very long and rotating SG can be replaced with a series of many static SG, each slightly rotated relative to the one before.

Unfortunately, this viewpoint of yours is based on a false analogy. See below.

The situation is (just to be clear I mean "it seems") comparable to having polarised light and sending it through a series of polarising filters, each slightly rotated relative to the previous filter.

No, it is not, because, unlike an SG magnet, a polarizing filter filters: it absorbs particles that are not polarized in the direction it is oriented. An SG magnet does not absorb anything.

This invalidates everything you say based on this faulty analogy. Which basically means everything in your post.

 

[PeterDonis]

There are other approaches

What other approaches are there?

 

[Vanadium 50]

It has spin 1/2 and no charge

Electrons do not fit that description.

 

[Old Person]

Hi.

Could I just re-phrase the question again, please.

What distribution of electrons do you obtain on the screen?
If you don't know, that's fine. I don't know. I would be really grateful for some opinions. (I have to do some other things and may not reply as quickly - but will read anything left here).

Thank you.

 

[PeterDonis]

What distribution of electrons do you obtain on the screen?

The issue that makes this problem much more complicated than the standard S-G experiment is that the rotation of the SG magnet makes the interaction Hamiltonian time-dependent. Trying to analyze such a case is a fairly heavy lift.

 

[PeroK]

We could write down the mathematics for SG experiments and spin:
Suppose you start with a state |m _z =+1/2 >. If I can get LaTeX to work that will be
$| \uparrow >$
and put that through a SG aligned in the y-axis. We write the state on entry in terms of an appropriate basis for the SG it is going to enter.
$| \uparrow > = \frac{1}{\sqrt{2}} | \leftarrow > + \frac{1}{\sqrt{2}} | \rightarrow >$
So it collapses to the $|\leftarrow >$ or $|\rightarrow >$ states with probability = square of the magnitude of the coefficients = 1/2 for each possibility. (Doesn't look like the LaTeX did work on a "preview".... I hope the spirit is there).

This is wrong. The state does not collapse, but evolves within the magnetic field. It takes time for the evolving state to pick up lateral momentum, as it were. This factor is proportional to the time spent in the magnetic field. See, for example, Griffiths page 183.

Note also that any lateral momentum picked up from a SG magnet is not reversed by a subsequent orthogonal magnet. In the case of a z-oriented magnet following by a y-oriented magnet, the four components of the resulting superposition will have momentum in both the y and z directions.

Now we just need to consider a similar calculation where the next SG it enters wasn't orthogonal, it's just at a small angle to the earlier SG.

If the magentic field is continuously changing direction, then the state will have no time at any stage to evolve according to a typical SG experiment. I suggest you would end up with a circular smudge with no preferred direction.

 

[Old Person]

Hi.

I suggest you would end up with a circular smudge with no preferred direction.

That is an attempt at providing an answer or opinion. Thank you.
It's been a few days with no other opinions. I'm going to assume the chances of obtaining any further replies will now be quite low and therefore may not be following the thread for much longer.

 

[gentzen]

Speaking of neutral, would doing the experiment with neutrons be good enough for you? This has been done, and worked fine.

Fine... just use whatever neutral particle you want. I'm being outnumbered here by people wanting to use a different particle. It's mainly just the effects from spin that I am intersted in, so it will be fine whatever you use.

Doing the Stern-Gerlach experiment with neutrinos would be cool. Just a pity that neutrinos are so hard to detect.

 

[Vanadium 50]

Just a pity that neutrinos are so hard to detect.

And they are not affected by electromagnetic fields.

 

[gentzen]

And they are not affected by electromagnetic fields.

Can you elaborate? They are spin 1/2 particles, I guess we agree on that. But I admit that photons are spin 1 particles, and are not affected by electromagnetic fields (of reasonable field strength). So the absence of electric charge and inner structure also leads to the absence of a magnetic moment and hence to the absence of interaction with electromagnetic fields?

 

[PeterDonis]

Can you elaborate?

Neutrinos have no electric charge, so they are not affected by electromagnetic fields.

They are spin 1/2 particles

Which has nothing to do with whether they are affected by electromagnetic fields. EM fields don't couple to spin, they couple to electric charge.

the absence of electric charge

Is enough all by itself to not interact with EM fields.

and inner structure

"Inner structure" has nothing to do with it.

also leads to the absence of a magnetic moment

The absence of a magnetic moment is due to the absence of an electric charge. True, a charged scalar particle (spin 0) will also have no magnetic moment, even if it has an electric charge--but magnetic moment is not the only possible interaction with an EM field. A charged scalar particle still has an EM interaction. Again, the EM field couples to electric charge, so that is what you look at to see if a particle has any interaction with an EM field.

 

[gentzen]

Neutrinos have no electric charge, so they are not affected by electromagnetic fields.

That doesn't convince me. Neutrons have no electric charge either, but they do have a magnetic moment, and hence are affected by magnetic fields.

the absence of electric charge

Is enough all by itself to not interact with EM fields.

I have the impression that you simply don't understand why I asked that question. This is a thread about the Stern-Gerlach experiment, in this context the interesting interaction with the EM field is the one caused by the magnetic moment. Of course, if I had googled the magnetic moment of the neutrino, I would have seen that its magnetic moment is ridiculously small. My question is whether I could have known this (without googling), based on general principles.

"Inner structure" has nothing to do with it.

The quarks that constitute the inner structure of a neutron do have charge, but the neutron itself has none.

The absence of a magnetic moment is due to the absence of an electric charge. True, a charged scalar particle (spin 0) will also have no magnetic moment, even if it has an electric charge--but magnetic moment is not the only possible interaction with an EM field.

Well, but it is the relevant interaction in the context of the Stern-Gerlach experiment.

 

[PeterDonis]

Neutrons have no electric charge either

But the particles that compose them, quarks, do. Neutrinos are not composites of other particles.

I have the impression that you simply don't understand why I asked that question.

I have the impression that you have not thought through the problem carefully enough. See above.

if I had googled the magnetic moment of the neutrino, I would have seen that its magnetic moment is ridiculously small

No, you would have seen that the current experimental upper bound on the neutrino magnetic moment is ridiculously small. Which just means that any actual measurement has a finite resolution. We believe the actual value of the neutrino magnetic moment is zero, just as we believe the actual value of the photon rest mass is zero, even though experimentally the best we can say is that both of those quantities have ridiculously small upper bounds.

 

[gentzen]

I have the impression that you have not thought through the problem carefully enough.

I think I don't know enough QFT to think through this problem myself. What I can do (and should have done before posting my initial wrong comment) is google.

See above.

??? You are the one who wrote

"Inner structure" has nothing to do with it.

I am pretty sure that what I mean by "inner structure" is not the source of our misunderstanding.

No, you would have seen that the current experimental upper bound on the neutrino magnetic moment is ridiculously small.

I did google before I wrote that, and one of the results I found said:

Neutrinos "do" have a magnetic moment, in the sense that the Standard Model (with minimal additions to permit the neutrino to have mass) predicts that neutrinos acquire a very, very small magnetic moment. Something like $10^{−19}$ times the magnetic moment of the electron.

On the other hand, the magnetic moment of neutrinos has never been measured (only upper bounds have been set), so we don't know if neutrinos really do have a magnetic moment.

But this use of QFT and the Standard Model is not what I mean when I say "I don't know enough QFT to think through this problem myself". I rather mean that I am not sufficiently familiar with the general principles which would allow me to see what must be true (based on QFT principles), and what is experimental input.

 

[PeroK]

I'm not convinced that neutrinos are ideal candidates for the SG experiment!

 

[PeterDonis]

I am pretty sure that what I mean by "inner structure" is not the source of our misunderstanding.

If by "inner structure" you mean "composite of other particles", then no, we're in agreement on that point.

 

[PeterDonis]

I did google before I wrote that, and one of the results I found said:

Without a reference to back that statement up I'm not sure how reliable it is.

 

[Vanadium 50]

Lets tale a step back.

A spin-0 particle cannot have a dipole moment, either electric or magnetic, because there is no direction in which it can point. A similar argument can be made (more mathy) to show a spin-1/2 particle can have a monopole monent (charge), a dipole moment (electric or magnetic) but no higher moments. EDMs open a can of worms that is a big distraction - if you want to discuss them, that should be another thread. So we have MDMs.

The neutrino MDM is measured to be close to zero. As close as we can get. Since magnetic moments go as 1/m, and neutrinos are light, if there were any new physics effect, it should be large. Since we see no evidence whatsoever for this, we know it is very, very small. I do not know what the most stringent limit is, but if the neutrino had even a tiny MDM, the process $\gamma \rightarrow \nu + \overline{\nu}$ would go on all the time. Since we don't see it at all, it either does not happen, or happens way too rarely for us to see it.

In either case, the process is far, far too weak to be seen in a S-G experiment. If this were not the case, we would have seen it elsewhere.

Note that this is the same question as "How do we know that neutrinos are neutral and not almost neutral?" Same answer - "almost" is so close to reality that in virtually every case it does not metter.