Ensuring Clean Quantum Measurements in the Stern-Gerlach Experiment

[sokrates]

How do they make sure the "measurement" process in conventional Stern-Gerlach experiments are clean quantum measurements?

Spins dynamically precess in magnetic fields (uniform or non-uniform) and Stern-Gerlach (especially sequential SG setups) make precise predictions regarding the resultant beam.

You could easily end up rotating a z-beam making it an x-beam accidentally.

How is this difficulty avoided?

 

[sokrates]

Nobody has an idea... Everybody is an expert in the interpretation though...

 

[Doc Al]

You could easily end up rotating a z-beam making it an x-beam accidentally.

How would that happen "by accident"? You'd have to rotate your magnet--not something one does by accident.

Nobody has an idea... Everybody is an expert in the interpretation though...

 

[Vanadium 50]

Nobody has an idea

Based on not getting a response for two whole hours?

Doc Al is right - how do you rotate your magnet without noticing?

 

[sokrates]

How would that happen "by accident"? You'd have to rotate your magnet--not something one does by accident.

Rotating the magnet?

Spins, themselves WILL be rotating in the magnetic field because of the magnetic field)

how do you make sure there's no precession while the electron is traveling WITHIN the magnetic field?

 

[sokrates]

Based on not getting a response for two whole hours?

Doc Al is right - how do you rotate your magnet without noticing?

Based on not getting a response for a whole day. Did I say anything about rotating a magnet?
I am talking about the rotation of the individual SPINS while they are traveling inside the magnetic field.

 

[Doc Al]

Rotating the magnet?

Spins, themselves WILL be rotating in the magnetic field because of the magnetic field)

how do you make sure there's no precession while the electron is traveling WITHIN the magnetic field?

Again, I'm not quite sure what you're talking about. You had said:

You could easily end up rotating a z-beam making it an x-beam accidentally.

If the inhomogeneous magnetic field is aligned along the z-axis, then the beams will select for the z-component of spin. I don't see how a "z-beam" can accidentally turn into an "x-beam". (In a z-beam, the x-component of spin is random.)

 

[sokrates]

Let me try again: Electron spin responds to an external magnetic field, by making precessional movements when they are inside the field. I suppose you very well know this. torque ~ u x B (u= mag. moment, if there's no damping, such as in LLG equation)

If you write down the Hamiltonian for a z-beam, obviously you need to include the magnetic field, and independent of the force that's exerted on the electron that separates the up and down components, electrons would be precessing about the principal axis of the non-uniform magnetic field.

So when you try to measure the x-component of a z-beam, how do you know that the spins are not precessing around the x-axis?? Because that's exactly what would happen, a z-directed spin would precess about the x-axis, and if you carefully arrange the parameters (B-field and the magnet thicknesses etc..) you could end up with a y-beam just by rotation... or a minus z-beam.. or somewhere in between

Now the question is HOW is that taken into account?

Check this if you still don't see my point:
http://en.wikipedia.org/wiki/Larmor_precession

 

[Vanadium 50]

Oops...it was 14 hours, not 2.

But again, like Doc Al, I don't see how the beam can turn into an x-beam. Remember, once the beam splits in two, I don't have to preserve the spins, because the selection has already occurred.

 

[sokrates]

still very confusing, no convincing answers in two days.
I think details of this is important!

 

[humanino]

The components transverse to the magnetic field do oscillate at the Larmor frequency, while the longitudinal component can be taken constant. The Larmor frequency in a typical Stern-Gerlach is so large that it averages to zero. Still, the point of the experiment is that in the classical theory, any longitudinal component would be possible and the atoms would be deflected continuously according to their arbitrary component. However, in the quantum theory we obtain only J "jets" of width in agreement with Heisenberg.

See for instance Feynman vol2 35.2

 

[sokrates]

The components transverse to the magnetic field do oscillate at the Larmor frequency, while the longitudinal component can be taken constant. The Larmor frequency in a typical Stern-Gerlach is so large that it averages to zero. Still, the point of the experiment is that in the classical theory, any longitudinal component would be possible and the atoms would be deflected continuously according to their arbitrary component. However, in the quantum theory we obtain only J "jets" of width in agreement with Heisenberg.

See for instance Feynman vol2 35.2

How come "any arbitrary component" could NOT be zero if they are precessing wildly (in classical theory)? Because, if what you are writing here is correct, even in the classical theory, any non-longitudinal component of spin would precess with a large frequency (therefore average out to zero, as you say) while there will only be TWO and ONLY TWO components present that do not precess because they are collinear with the non-homogeneous (but unidirectional) magnetic field : + longitudinal and - longitudinal.

You just proved that even classically, only two discrete spots are expected. (Because all other directions precess and average out to zero)

 

[Cthugha]

Because, if what you are writing here is correct, even in the classical theory, any non-longitudinal component of spin would precess with a large frequency (therefore average out to zero, as you say) while there will only be TWO and ONLY TWO components present that do not precess because they are collinear with the non-homogeneous (but unidirectional) magnetic field : + longitudinal and - longitudinal.

I am puzzled about what you mean exactly. Classically you have the projection of spin on the longitudinal axis, which does not precess and the remaining spin components, which precess wildly and average to zero. In a "classical description of spin" only the total spin is given and the projection on the longitudinal axis can have any magnitude between - and + total spin and therefore there should be a line instead of two spots classically. Or did I miss your point somewhere?

 

[humanino]

You just proved that even classically, only two discrete spots are expected. (Because all other directions precess and average out to zero)

That's your understanding, but that's not what "I proved", that's not what's in any elementary introduction book to quantum mechanics, and that's most importantly not what this experiment tells us.

 

[conway]

How do they make sure the "measurement" process in conventional Stern-Gerlach experiments are clean quantum measurements?

Spins dynamically precess in magnetic fields (uniform or non-uniform) and Stern-Gerlach (especially sequential SG setups) make precise predictions regarding the resultant beam.

You could easily end up rotating a z-beam making it an x-beam accidentally.

How is this difficulty avoided?

I am trying to figure out which axis you are using for your magnetic field? Conventionally we talk about z but in this quote you appear to be taking it about y. Later you talk about z rotating to y, which would happen if the field is along x. What is your intended orientation?

 

[sokrates]

That's your understanding, but that's not what "I proved", that's not what's in any elementary introduction book to quantum mechanics, and that's most importantly not what this experiment tells us.

Why do you dodge your "own" remarks and your OWN explanation? Why do you have to senselessly attack instead of focusing on the physics?

YOU said that: ANY non-longitudinal component of spin WILL precess about the magnetic field direction. Rotation can be resolved in three coordinates clasically, so ANYTHING that does not coincide with the magnetic field axis will average out to zero, GIVING TWO DISTINCT directions.

This is what you say. And I am pretty sure, you understand this kindergarden logic, but now you are confused, you don't want to admit it. You don't know it any better than I do, you are just in a state of denial about it.

 

[sokrates]

I am trying to figure out which axis you are using for your magnetic field? Conventionally we talk about z but in this quote you appear to be taking it about y. Later you talk about z rotating to y, which would happen if the field is along x. What is your intended orientation?

No. Rotating Z to Y could happen in a Y-directed field as well, because of the precession...

A bicycle wheel hanging on a z-directed rope would make a 2*pi rotation ABOUT the z-axis, passing through both y-axis and the x-axis.

Depends on the rotation (if there's only precessional motion)

But since the field is non-uniform , I guess the motion is more complicated, still unknown to me.
Humanino's answer by the way, has to be wrong, he apparently solved the mystery.

 

[humanino]

Why do you dodge your "own" remarks and your OWN explanation? Why do you have to senselessly attack instead of focusing on the physics?

Sokrates, you are the one constantly attacking. You began with your nasty remark "everybody has an interpretation, but nobody to answer my question".

YOU said that: ANY non-longitudinal component of spin WILL precess about the magnetic field direction. Rotation can be resolved in three coordinates clasically, so ANYTHING that does not coincide with the magnetic field axis will average out to zero, GIVING TWO DISTINCT directions.

This is what you say. And I am pretty sure, you understand this kindergarden logic, but now you are confused, you don't want to admit it. You don't know it any better than I do, you are just in a state of denial about it.

I am not confused at all. As I am trying to explain calmly to you, classical theory does not predict only J possible projections along the magnetic field direction, but any possible projection. The deflection will be proportional to the component of the moment along the magnetic field direction. So in the classical setting, you get a continuum of possible deflections, including zero.

In the quantum setting, whatever the initial polarization can be written in a basis of proper states along any arbitrary direction. But whatever the choice you make in the direction, only J projection along this direction are possible. In particular, you can see if you take an electron and write it down as a superposition of two possible polarization projections along the magnetic field direction, you get only two possible discrete deflection, not a continuum. You will get different interpretation of what's going on if you decide to use another direction to write the state, but you will compute the same output possible BTW.

 

[sokrates]

I am puzzled about what you mean exactly. Classically you have the projection of spin on the longitudinal axis, which does not precess and the remaining spin components, which precess wildly and average to zero. In a "classical description of spin" only the total spin is given and the projection on the longitudinal axis can have any magnitude between - and + total spin and therefore there should be a line instead of two spots classically. Or did I miss your point somewhere?

You are right. "Magnitude" quantization is the key here. IT cannot be understood classically, but I wonder why all this complicated spin dynamics is avoided in discussing SG experiments?

 

[humanino]

A bicycle wheel hanging on a z-directed rope would make a 2*pi rotation ABOUT the z-axis, passing through both y-axis and the x-axis.

The point of the Stern Gerlach experiment is that it is probing a quantum spin, not a bicycle wheel. If the angular momentum is not large compared to hbar, then you can clearly distinguish the possible projections along the quantization axis. For a bicycle wheel, the number of projections is so large that it appears continuous.

Seriously Sokrates, a more civilized tone would be appropriate.

 

[sokrates]

Sokrates, you are the one constantly attacking. You began with your nasty remark "everybody has an interpretation, but nobody to answer my question".

I am not confused at all. As I am trying to explain calmly to you, classical theory does not predict only two possible projections along the magnetic field direction, but any possible projection. The deflection will be proportional to the component of the moment along the magnetic field direction. So in the classical setting, you get a continuum of possible deflections, including zero.

In the quantum setting, whatever the initial polarization can be written in a basis of proper states along any arbitrary direction. But whatever the choice you make in the direction, only J projection along this direction are possible. In particular, you can see if you take an electron and write it down as a superposition of two possible polarization projections along the magnetic field direction, you get only two possible discrete deflection, not a continuum. You will get different interpretation of what's going on if you decide to use another direction to write the state, but you will compute the same output possible BTW.

This is clear to me and has nothing to do with what I asked. I know the significance of the experiment, I have been talking about the spin dynamics, which involve aligning with the magnetic field, precession, etc...

 

[sokrates]

The point of the Stern Gerlach experiment is that it is probing a quantum spin, not a bicycle wheel. If the angular momentum is not large compared to hbar, then you can clearly distinguish the possible projections along the quantization axis. For a bicycle wheel, the number of projections is so large that it appears continuous.

Seriously Sokrates, a more civilized tone would be appropriate.

What are you talking about? I am civil to everyone who politely engages in my discussions, and let me remind you, you are the one who is constantly stalking my posts pointlessly attacking me. I don't have a problem with you, just discussing physics is my objective here.

You don't even understand my points before you post, the precession of an electron (IF YOU ASSUME IT HAS INTRINSIC ANGULAR MOMENTUM) could just as well be demonstrated by a bicycle wheel, it has nothing to do with the quantization... Same equations! and same interpretation

I won't conclude by saying: See for example, Feynman I, precession.

as you do.

 

[humanino]

This is clear to me and has nothing to do with what I asked. I know the significance of the experiment, I have been talking about the spin dynamics, which involve aligning with the magnetic field, precession, etc...

If this is so clear to you, then what is wrong with this simple answer : write down the wavefunction as a combination of proper states along the magnetic field and your precession is gone. It does not matter what the original direction of the spin is, I can always choose to write it down in the basis along this convenient direction.

 

[humanino]

What are you talking about? I am civil to everyone who politely engages in my discussions, and let me remind you, you are the one who is constantly stalking my posts pointlessly attacking me. I don't have a problem with you, just discussing physics is my objective here.

You already managed to have a mentor a scientific advisor not interested in this discussion anymore.

 

[sokrates]

If this is so clear to you, then what is wrong with this simple answer : write down the wavefunction as a combination of proper states along the magnetic field and your precession is gone. It does not matter what the original direction of the spin is, I can always choose to write it down in the basis along this convenient direction.

Look... You are simplifying things. What you say here is in every textbook. So I can just as well go read Sakurai, which I have already done.

Precession "averaging out" to zero is added here, WE KNOW that it averages to zero, I don't think it comes out of the SINGLE-PARTICLE Schrodinger equation.

If you want to write down the states, you have to include the magnetic field through the vector potential representation to the Hamiltonian, then solve this Hamiltonian assuming that you are only interested in the magnetic field region and then two distinct states, and electrons going up, etc... etc.. should come out. BUT:

I don't think precession averaging to zero comes out of such a naive picture.

But instead, if you say, I KNOW my states, and they have only TWO possible outcomes, of course the problem becomes much simpler and probably they yield the same results,

but it's just NOT that trivial, if you think about the details, and that's my whole point.

 

[sokrates]

You already managed to have a mentor a scientific advisor not interested in this discussion anymore.

Yeah too bad. And you already managed to make the thread irrelevant and unintelligible.

 

[humanino]

But instead, if you say, I KNOW my states, and they have only TWO possible outcomes, of course the problem becomes much simpler and probably they yield the same results, but it's just NOT that trivial, if you think about the details, and that's my whole point.

In any setting, there are many ways to do the calculation wrong and there are a few clever ways to solve the problem. I agree that it is not trivial that thinking in a unappropriate manner will lead to the same result as the simple and clever calculation.

 

[sokrates]

In any setting, there are many ways to do the calculation wrong and there are a few clever ways to solve the problem. I agree that it is not trivial that thinking in a unappropriate manner will lead to the same result as the simple and clever calculation.

You wouldn't know the simple and clever calculation if it wasn't in every QM textbook. It appears that it didn't even occur to you how to do the problem EXACTLY. And finally, you cannot justify WHY the states can be written so simply (for a system of utmost delicacy (spins) and with reasonable complexity (non-uniform B-fields)) without referring to your beloved textbooks.

It now is obvious that you are not the bit of an expert who should take this question, so please don't pollute the thread and prevent me from learning from all the other people here. Maybe there's still a chance that somebody will respond after all this unnecessary fight.

 

[humanino]

It now is obvious that you are not the bit of an expert who should take this question, so please don't pollute the thread and prevent me from learning from all the other people here. Maybe there's still a chance that somebody will respond after all this unnecessary fight.

Very nice and civilized manner indeed.If you like the gory details, you may read
Ph. A. Martin and M. Sassoli de Bianchi, "Spin precession revisited", Foundations of Physics Volume 24, Number 10 / October, 1994

Personally, I'd better get out of here.

 

[sokrates]

Very nice and civilized manner indeed.If you like the gory details, you may read
Ph. A. Martin and M. Sassoli de Bianchi, "Spin precession revisited", Foundations of Physics Volume 24, Number 10 / October, 1994

Personally, I'd better get out of here.

You could have given me this, 15 posts, and 4 hours earlier. But thank you anyway...
This might be it, "the gory details" ...
( I can't download it now, but I will. The first page gives an idea, but I still didn't see the non-uniform field)
Thanks anyway,

 

[sokrates]

Spin Precession Revisited

Abstract: The passage of a spin-1/2neutral particle through a region of *uniform magnetic field* and the corresponding precession mechanism is analyzed from the viewpoint of scattering theory, with particular consideration of the role of the field boundaries.

I hope they have also treated the SG setup here... OR did you just google this without really looking?

 

[transience]

The way I understand it is that the Stern-Gerlach experiment is a measurement device that splits the beam according to the projection of the spin component on the z-axis.

The quantum state of whatever makes up the beam isn't pure, meaning that the the z component of the spin can be expressed in terms of spins along the x or y axis, because of this it doesn't matter how you rotate the magnets you will still see the splitting.