Is Calcite a parallel to the Stern-Gerlach experiment?

  • #1
Frigorifico9
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TL;DR Summary
I was thinking about how Calcite has birefringence and I realized it was very similar to the Stern-Gerlach experiment, but I'm not sure if it makes sense
When light passes through Calcite it is split into two beams opposite polarizations, doubling the image, and this sounds very similar to the Stern-Gerlach experiment where atoms are split into two beams with opposite polarizations

The difference is that with light the opposite polarizations are 90 degrees apart in physical space, while with electrons their polarization states are 180 degrees apart in physical space (but 90 degrees apart in amplitude space)

All of this is very pretty and makes a lot of sense, but there's one thing that makes me question how closely these two things are related

Is the Calcite crystal working as a filter of light, only letting through those photons with the right polarizations? Or is the Calcite crystal letting through all the photons, but changing their polarization in the process?

If the second thing is happening, then Calcite is indeed very similar to the Stern-Gerlach experiment, because this experiment changes the spin of the electrons instead of only letting through those whose spin match with the orientation of the experiment
 
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  • #2
Frigorifico9 said:
the Stern-Gerlach experiment, because this experiment changes the spin of the electrons instead of only letting through those whose spin match with the orientation of the experiment
What makes you think this? Remember we are talking about quantum physics here, not classical physics.
 
  • #3
PeterDonis said:
What makes you think this? Remember we are talking about quantum physics here, not classical physics.
In the Stern-Gerlach experiment 100% of the atoms you shoot through the magnetic field make it to the other side, except they are split 50/50 into two beams

Meanwhile, if you shine unpolarized light at a polarizer exactly half of the light will make ti through, the rest will be reflected

However Calcite isn't exactly a polarizer, so I'm not sure if it reflects some of the light or if 100% makes it through

Also, I have a masters in physics, I'm not a new at this, so feel free to explain this in as much high level as you can
 
  • #4
Frigorifico9 said:
In the Stern-Gerlach experiment 100% of the atoms you shoot through the magnetic field make it to the other side, except they are split 50/50 into two beams
Yes. But how does that justify the statement you made that I quoted? Again, we are talking about quantum physics here, not classical physics. You said the device changes the spin of the electrons. In classical physics that statement would be justified--but classical physics does not predict the actually observed result.
 
  • #5
Frigorifico9 said:
except they are split 50/50 into two beams
Note that this is not necessarily the case; it depends on how the electrons are prepared before they are sent through the device and what the device's orientation is.
 
  • #6
PeterDonis said:
Yes. But how does that justify the statement you made that I quoted? Again, we are talking about quantum physics here, not classical physics. You said the device changes the spin of the electrons. In classical physics that statement would be justified--but classical physics does not predict the actually observed result.
Look, I'm honestly sorry for how badly I expressed myself, I truly am and I am grateful for people like you who force me to be better. But I would like to know, are you just gonna point out my mistakes in phrasing this question or are you gonna answer it? Because it seems to me like you understand what I'm asking
 
  • #7
PeterDonis said:
Note that this is not necessarily the case; it depends on how the electrons are prepared before they are sent through the device and what the device's orientation is.
Dude, I know. I shouldn't have to write a question with perfect terminology to get an answer. This is not a research paper. I would like to think that the people here are smart enough to understand poorly phrase questions from stupid shitty people like me. I am stupid but I still want an answer to my question: How much unpolarized light does a clear calcite crystal reflect and how how much passes through? (Keeping in mind that by "unpolarized light" I don't mean light without polarization, that would be impossible. Rather I mean light in which there is no predominat polarization among the photons)
 
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  • #8
Frigorifico9 said:
I would like to know, are you just gonna point out my mistakes in phrasing this question or are you gonna answer it?
You're missing my point. Your question is based on an assumption that I think is false: the statement of yours that I quoted, that a Stern Gerlach device changes the spin of electrons passing through it. (Btw, you're making the same assumption in the second possibility you describe for the Calcite crystal.) So I need you to either justify that assumption or reframe your question to not use it. Until you do one of those two things, I can't answer your question because it's not well defined.
 
  • #9
PeterDonis said:
You're missing my point. Your question is based on an assumption that I think is false: the statement of yours that I quoted, that a Stern Gerlach device changes the spin of electrons passing through it. (Btw, you're making the same assumption in the second possibility you describe for the Calcite crystal.) So I need you to either justify that assumption or reframe your question to not use it. Until you do one of those two things, I can't answer your question because it's not well defined.
Forget about my original question. Thanks to you I have developed a new question which I know will not live up to your standards, but which I hope it is answerable in a way my original question wasn't because I am just that stupid. Here is the new question:

How much unpolarized light does a clear calcite crystal reflect and how how much passes through? (Keeping in mind that by "unpolarized light" I don't mean light without polarization, that would be impossible. Rather I mean light in which there is no predominat polarization among the photons)
 
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  • #10
Frigorifico9 said:
Thanks to you I have developed a new question which I know will not live up to your standards, but which I hope it is answerable in a way my original question wasn't because I am just that stupid.
Please stop with this attitude. It adds nothing to the discussion.

You say you have a master's degree in physics. That means it's perfectly reasonable for me to expect you to do a fair amount of the work in this discussion.

Frigorifico9 said:
How much unpolarized light does a clear calcite crystal reflect and how how much passes through?
As far as I know, it's possible to have a calcite crystal that lets all incoming light through.

However, that doesn't mean that the crystal is changing the polarization of incoming light, any more than a Stern Gerlach device changes the spin of incoming electrons.
 
  • #11
PeterDonis said:
Please stop with this attitude. It adds nothing to the discussion.

You say you have a master's degree in physics. That means it's perfectly reasonable for me to expect you to do a fair amount of the work in this discussion.As far as I know, it's possible to have a calcite crystal that lets all incoming light through.

However, that doesn't mean that the crystal is changing the polarization of incoming light, any more than a Stern Gerlach device changes the spin of incoming electrons.
Thanks for your answer. I hope we never interact again
 
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  • #12
Frigorifico9 said:
TL;DR Summary: I was thinking about how Calcite has birefringence and I realized it was very similar to the Stern-Gerlach experiment, but I'm not sure if it makes sense

When light passes through Calcite it is split into two beams opposite polarizations, doubling the image, and this sounds very similar to the Stern-Gerlach experiment where atoms are split into two beams with opposite polarizations
Indeed there's a very close analogy here, provided you send single photons through the crystal. If you prepare the single photon in a polarization state which is a superposition of H and V polarized em. waves (with H and V referring to the direction of the polarizations exiting the crystal in different directions), then you'll get randomly each single photon going to the one or the other direction with probabilities given by the normalized intensity of the two beams when using classical correspondingly polarized light (which quantum mechanically is described as a coherent state).
Frigorifico9 said:
The difference is that with light the opposite polarizations are 90 degrees apart in physical space, while with electrons their polarization states are 180 degrees apart in physical space (but 90 degrees apart in amplitude space)
That's because for the electron the "polarization" refers to a spin component, and the spin is 1/2 for the electron, while in the case of the photon the "polarization" refers to the helicity states of a massless spin-1 particle. Otherwise the math is pretty similar, because the polarization state is described by a 2D unitary space.
Frigorifico9 said:
All of this is very pretty and makes a lot of sense, but there's one thing that makes me question how closely these two things are related

Is the Calcite crystal working as a filter of light, only letting through those photons with the right polarizations? Or is the Calcite crystal letting through all the photons, but changing their polarization in the process?
The Calcite crystal is not a filter but it lets you sort to two polarization states, i.e., if the photon goes the one direction, the outgoing photon is H polarized, if it goes into the other direction it's V polarized. It's thus rather a "beam splitter" than a "filter", as is the magnet in the Stern-Gerlach experiment for atoms. The "action" of both the Calcit crystal on the photon state as well as the magnet on the atom's spin state is a unitary operation.

An example for a filter is a polarizer foil. There you absorb photons polarized in one direction (say H) and let the ones polarized in the perpendicular direction (in this example V) through. This is described by a projection operator rather than a unitary operator. That's why with a filter you always loose a part of the photons, i.e., it either lets a given photon through or completely absorbs it. The probability for that is given by Malus's Law for linearly polarized photons, i.e., if the photon is polarized in a direction making an angle ##\theta## wrt. the direction the polarizer lets the photon through with 100% probability is given by ##P=\cos^2 \theta##.
Frigorifico9 said:
If the second thing is happening, then Calcite is indeed very similar to the Stern-Gerlach experiment, because this experiment changes the spin of the electrons instead of only letting through those whose spin match with the orientation of the experiment
Right!
 
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  • #13
PeterDonis said:
What makes you think this? Remember we are talking about quantum physics here, not classical physics.
As I just wrote, I find the analogy for both experiments (with single atoms in the SGE and with single photons using a polarizing beam splitter) one to one.
 
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  • #14
PeterDonis said:
You're missing my point. Your question is based on an assumption that I think is false: the statement of yours that I quoted, that a Stern Gerlach device changes the spin of electrons passing through it. (Btw, you're making the same assumption in the second possibility you describe for the Calcite crystal.) So I need you to either justify that assumption or reframe your question to not use it. Until you do one of those two things, I can't answer your question because it's not well defined.
Of course the SGE changes the spin due to the interaction of the atomic magnetic moment with the magnetic field. The propagation of the atom through this field leads to an entanglement between the position (or momentum if you wish) and the spin component in direction of the magnetic field (or more precisely the large homogeneous part of the magnetic field).
 
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  • #15
vanhees71 said:
Of course the SGE changes the spin due to the interaction of the atomic magnetic moment with the magnetic field.
Not in QM, no. In QM the SG device entangles the two components of spin along the device's direction with the linear momentum in that direction. That's not the same as changing the spin.
 
  • #16
vanhees71 said:
The propagation of the atom through this field leads to an entanglement between the position (or momentum if you wish) and the spin component in direction of the magnetic field (or more precisely the large homogeneous part of the magnetic field).
Yes, this is correct, but, as I noted just now, it's not the same as changing the spin. The spin components of the atom and their amplitudes after passing through the device are precisely the same as they were before. They're just entangled with the linear momentum where they weren't before.
 
  • #17
PeterDonis said:
The spin components of the atom and their amplitudes after passing through the device are precisely the same as they were before.
If you send a beam of atoms that are in a spin up in the ##\hat x## direction through a Stern-Gerlach magnetic oriented in the ##\hat z## direction, you'll end up with beams oriented in up and down ##\hat z## states. I'd call that changing their spin orientation.
 
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  • #18
Doc Al said:
If you send a beam of atoms that are in a spin up in the ##\hat x## direction through a Stern-Gerlach magnetic oriented in the ##\hat z## direction, you'll end up with beams oriented in up and down ##\hat z## states. I'd call that changing their spin orientation.
No, it's, as I said, entangling the ##z## components of their spin with their linear momentum. But the spin components, in the ##z## basis, and their amplitudes, are exactly what they were before. So nothing about the spin degree of freedom by themselves has been changed. All that has changed is the entanglement of the spin degree of freedom with the linear momentum degree of freedom.

"Changing their spin orientation" would mean what the classical model says: the spin of every atom gets changed according to the magnetic interaction term in the Hamiltonian. But that is not what actually happens; that's precisely why the SG experiment was one of the most important experiments in the history of QM, because the result was so out of line with the classical prediction. And that was true even though the Hamiltonian in question is exactly the same as the classical one. Getting very different results from the same Hamiltonian seems to me to warrant a change of terminology.

For those who disagree, can you find a description of the SG experiment in a textbook or peer-reviewed paper that says it changes the spin? As an example of the description I am arguing for, see Ballentine, section 9.1 (top of p. 232 in my edition), where he says: "The effect of the interaction [in the SG device] is to create a correlation between the spin and the momentum of the particle." He says nothing about changing the spin. He says that, combined with a suitable detector to detect the output beams, an SG device can be used to measure the spin, but he never says the device changes the spin.
 
  • #19
PeterDonis said:
Yes, this is correct, but, as I noted just now, it's not the same as changing the spin. The spin components of the atom and their amplitudes after passing through the device are precisely the same as they were before. They're just entangled with the linear momentum where they weren't before.
No, the spin precesses around the magnetic field. The spin component along the magnetic field indeed doesn't change. But now one has to keep in mind that for the SGE for the spin-##z## component to work you need an inhomogeneous field, and because of ##\vec{\nabla} \cdot \vec{B}=0## the inhomogeneous part cannot be along the ##\vec{e}_z##.

A simple approximation for the magnetic field in a small enough region around the ##z##-axis is
$$\vec{B}=(B_0 + \beta z) \vec{e}_z -\beta y \vec{e}_y.$$
This means that there's always some probability of a "spin flip", i.e., if you start with a pure spin-up state, theres some probability that after running through the magnet you find it with spin down. A "good" SGE is constructed such that this probability is very small. This can be achieved by making ##B_0 \gg \beta d##, where ##d## is the maximal distance of the beam from the ##z## axis while traveling in the magnetic field. Then the contribution of the ##y##-component of the magnetic field to the Hamiltonian can be treated as a perturbation, and the apparatus works close to the idealized version of the experiment treated in almost all textbooks.

Also the entanglement between the ##s_z##-spin component and momentum or posisition is not ideal, but under the above given assumption it's also not too much violated since due to the large constant part ##\vec{B}_0=B_0 \vec{e}_z## the spin precesses with a quite high frequency around the ##z##-axis, while the motion in the magnetic field (i.e., typical changes the momentum) are much slower. So the force due to the inhomogeneous ##y##-component of the ##B##-field can also well be neglected, because over the slow time scales of the changes of the momentum this force averages to 0 due to the rapid precession of the spin around the ##z## axis.

Amazingly there are not too many fully quantum-theoretical treatments of the SGE. A nice one, including numerical studies for an exact solution of the Pauli equation can be found here:

https://arxiv.org/abs/quant-ph/0409206
https://doi.org/10.1103/PhysRevA.71.052106
 
  • #20
Thank you so much for helping me out. Could I ask you a follow up question?

I am struggling to think about the state of the wavefunction before the SG measurement

If the beam is moving in the $z$ direction we can choose to measure along the $y$ direction and we can represent the state of the wave function like this:

$$ \psi = \frac{1}{\sqrt(2)}\begin{pmatrix} 1 \\ 1 \end{pmatrix} $$

But then if we rotate the measurement direction 90 degrees clockwise then the we find the wavefunction was rotated 45 degrees clockwise in probability-space...

Does it make sense to say that the measurement apparatus rotates the probability vector even before the measurement has taken place?
 
  • #21
Frigorifico9 said:
Does it make sense to say that the measurement apparatus rotates the probability vector even before the measurement has taken place?
You may be confusing expressing the state in a different basis with a physical change. For example, if an object moves down a slope at an angle ##\theta##, then we normally choose to decompose the vertical gravity vector into components tangential and normal to the slope (to make the calculations easier). The gravity vector itself is unchanged by the angle of the slope.

Frigorifico9 said:
$$ \psi = \frac{1}{\sqrt(2)}\begin{pmatrix} 1 \\ 1 \end{pmatrix} $$
Assuming that is expressed in the z-basis, then the same spin state can be expressed relative to any other angle. The basis vector/state ##\ket {z+}##, is expressed in the z-basis as:$$ \ket{z+} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} $$And, in the basis at some angle ##\theta## to the z-axis as:$$ \ket{z+} = \begin{pmatrix} \cos(\frac \theta 2) \\ -\sin(\frac \theta 2) \end{pmatrix} $$In fact, the state ##\psi## above, expressed in the ##\theta## basis would be:$$\psi = \frac{1}{\sqrt 2}\begin{pmatrix} \cos(\frac \theta 2) - i \sin(\frac \theta 2)\\ -i\cos(\frac \theta 2) - \sin(\frac \theta 2) \end{pmatrix} $$And, in fact, you can see that the two basis components have equal weighting - as must be the case from a symmetry argument.

PS that expression, and the conclusion that was based on it, are wrong.

Frigorifico9 said:
But then if we rotate the measurement direction 90 degrees clockwise then the we find the wavefunction was rotated 45 degrees clockwise in probability-space...
No. We simply express the wavefunction in a different basis.
 
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  • #22
PeroK said:
You may be confusing expressing the state in a different basis with a physical change. For example, if an object moves down a slope at an angle ##\theta##, then we normally choose to decompose the vertical gravity vector into components tangential and normal to the slope (to make the calculations easier). The gravity vector itself is unchanged by the angle of the slope.Assuming that is expressed in the z-basis, then the same spin state can be expressed relative to any other angle. The basis vector/state ##\ket {z+}##, is expressed in the z-basis as:$$ \ket{z+} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} $$And, in the basis at some angle ##\theta## to the z-axis as:$$ \ket{z+} = \begin{pmatrix} \cos(\frac \theta 2) \\ -\sin(\frac \theta 2) \end{pmatrix} $$In fact, the state ##\psi## above, expressed in the ##\theta## basis would be:$$\psi = \frac{1}{\sqrt 2}\begin{pmatrix} \cos(\frac \theta 2) - i \sin(\frac \theta 2)\\ -i\cos(\frac \theta 2) - \sin(\frac \theta 2) \end{pmatrix} $$And, in fact, you can see that the two basis components have equal weighting - as must be the case from a symmetry argument.No. We simply express the wavefunction in a different basis.
I thought about this, but here's the thing: If you measure the spin of this state in different direction you will get different results, right? If you measured it in a direction that was aligned to it the beam would not split. But instead the beam ALWAYS splits, as if the vector was always written in the same way in the new basis, or in other words, as if it had been rotated
 
  • #23
Frigorifico9 said:
I thought about this, but here's the thing: If you measure the spin of this state in different direction you will get different results, right? If you measured it in a direction that was aligned to it the beam would not split. But instead the beam ALWAYS splits, as if the vector was always written in the same way in the new basis, or in other words, as if it had been rotated
It's the same vector, expressed in a different basis. Note, also, that the state vector is a vector in an abstract Hilbert space. Not a vector in 3D space. The same state vector can represent an equal weighting for measurements about all axes. In this case, equal weighting in the z-basis implies equal weighting in any other directional basis.
 
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  • #24
vanhees71 said:
the spin precesses around the magnetic field
Again, you're using a classical description, but we're not talking about classical physics, we're talking about quantum physics. In quantum physics, i.e., in actual experiments when they are actually run, the SG device does not change the amplitudes of the spin components at all. It just entangles them with linear momentum. Indeed, it would make no sense to use SG devices to measure the spin components (by measuring the relative frequencies of particles in the output beams) if the device changed them.

vanhees71 said:
there's always some probability of a "spin flip", i.e., if you start with a pure spin-up state, theres some probability that after running through the magnet you find it with spin down. A "good" SGE is constructed such that this probability is very small.
Yes, a probability of a spin flip. This flip would be a "change in spin" since in general it changes the spin components (although if a ##z## oriented device is being used to measure particles prepared in the spin ##x## up state, the flip actually does not change the relative weights of the spin components), but, as you note, it is considered an error and the devices are constructed to make this probability as small as possible. So the vast majority of the time, this spin flip does not happen and the spin does not change. And that vast majority of the time, when the SG device is functioning properly as part of a spin measurement, is what the post I originally responded to was talking about. In that case, the spin does not change, which was my point.

vanhees71 said:
the entanglement between the ##s_z##-spin component and momentum or posisition is not ideal
Yes, but again, this is another "error" in real devices, not a property of the ideal device. If we are going to get into all the messiness of real devices, we would also have to say that no real calcite crystal ever allows 100% of incoming light to pass through; there is always some small amount of absorption, which means that some photons are not split by birefringence. But I did not take the OP to be asking for a discussion of such complications, but about the ideal case (and also the most common case even with real devices) where the device does what it is designed to do.

The paper you reference also mentions other complications, which, again, I do not think the OP originally intended to discuss. Bringing them in does not change the "ideal case" answer already given; it just complicates it by having to consider the rare instances where some particles do not do what the ideal case describes. If the OP is interested in discussing those complications as well, then of course that's fine.
 
  • #25
PeroK said:
In fact, the state ##\psi## above, expressed in the ##\theta## basis would be:$$\psi = \frac{1}{\sqrt 2}\begin{pmatrix} \cos(\frac \theta 2) - i \sin(\frac \theta 2)\\ -i\cos(\frac \theta 2) - \sin(\frac \theta 2) \end{pmatrix} $$
This can't be right, because for ##\theta = 0## it gives $$\ket{\psi} = \frac{1}{\sqrt 2}\begin{pmatrix} 1 \\ -i \end{pmatrix} $$ not $$\ket{\psi} = \frac{1}{\sqrt 2}\begin{pmatrix} 1 \\ 1 \end{pmatrix} $$.

Also, the state $$\ket{\psi} = \frac{1}{\sqrt 2}\begin{pmatrix} 1 \\ 1 \end{pmatrix} $$ in the ##z## basis corresponds to spin ##x## up, so it cannot have "equal weights" at all angles.

In fact, a complete specification of all possible states for a qubit (i.e., a single spin-1/2 particle) requires two angles, not one. See here:

https://en.wikipedia.org/wiki/Bloch_sphere
 
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  • #26
PeterDonis said:
This can't be right, because for ##\theta = 0## it gives $$\ket{\psi} = \frac{1}{\sqrt 2}\begin{pmatrix} 1 \\ -i \end{pmatrix} $$ not $$\ket{\psi} = \frac{1}{\sqrt 2}\begin{pmatrix} 1 \\ 1 \end{pmatrix} $$.

Also, the state $$\ket{\psi} = \frac{1}{\sqrt 2}\begin{pmatrix} 1 \\ 1 \end{pmatrix} $$ in the ##z## basis corresponds to spin ##x## up, so it cannot have "equal weights" at all angles.

In fact, a complete specification of all possible states for a qubit (i.e., a single spin-1/2 particle) requires two angles, not one. See here:

https://en.wikipedia.org/wiki/Bloch_sphere
Sorry, there was a typo in my notes. For the SGE, I was trying to do things for the ##z-y## plane, but it's simpler to do things for the ##z-x## plane, where motion is in the y-direction. That would give simply:
$$\psi = \frac{1}{\sqrt 2}\begin{pmatrix} \cos(\frac \theta 2) + \sin(\frac \theta 2)\\ \cos(\frac \theta 2) - \sin(\frac \theta 2) \end{pmatrix} $$The answer to the OP's question is that if the state is prepared like this, then (of course) we don't have equal weighting in all directions. This is, as you say, spin-up in the x-direction.

Instead, in a SGE, the relevant electrons in the silver atom have a mixed random state. That's what provides the equal weighting about every angle. They do not all have a pure, equally weighted state in the z-basis.
 
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  • #27
PeroK said:
in a SGE, the relevant electrons in the silver atom have a mixed random state
Meaning, in an SGE done the way the original was done. AFAIK technical difficulties have prevented doing more complicated things with silver atom beams like putting the output beam of one SG device into a second SG device. But of course the corresponding experiments with photons are much simpler and have been done with all kinds of variations.
 
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  • #28
PeterDonis said:
Again, you're using a classical description, but we're not talking about classical physics, we're talking about quantum physics. In quantum physics, i.e., in actual experiments when they are actually run, the SG device does not change the amplitudes of the spin components at all. It just entangles them with linear momentum. Indeed, it would make no sense to use SG devices to measure the spin components (by measuring the relative frequencies of particles in the output beams) if the device changed them.
You have an oversimplified picture of the SGE, but let's discuss this idealized picture first, because as I tried to clarify in my previous posting, you can realize it at very good approximation. My arguments were indeed semiclassical, but since the equations of motion in the discussed approximation are linear, it's a very good picture.

Quantum mechanically, the indealized SGE is indeed an example for a von Neumann measurement. Let's assume that the (large homogeneous part of the) ##\vec{B}## field is along the ##z##-direction, i.e., we look at an SGE that measures the ##z##-component of the spin:
$$\vec{B}=(B_0 + \beta z) \vec{e}_z - \beta y \vec{e}_y.$$

Solving the Pauli equation for the approximate Hamiltonian,
$$\hat{H}_0=\frac{1}{2m} \hat{\vec{p}}^2 + \mu_B g_s (B_0+\beta z) \hat{s}_z$$
shows that you indeed entangle the ##s_z##-component with the position of the atom, i.e., if you arrange the initial conditions and the magnetic field properly you can achieve that one partial beam is prepared in a ##\sigma_1=+1/2## and the other partial beam in a ##\sigma_z=-1/2## eigenstate of ##\hat{s}_z##. Using the position as a pointer variable you have thus a perfect measurement of the spin-##z## component and by blocking one of the partial beams you even have a perfect preparation procedure for eigenstates of ##\hat{s}_z##, i.e., a "von Neumann filter measurement".

Now what do I mean "you change the spin" with this experiment. It, of course means, this preparation procedure, i.e., independent from which initial (mixed or pure) spin state you start, in each of the partial beams you have prepared an eigenstate of ##\hat{s}_z##. Only if the original particle was in such an eigenstate, this state remains unchanged. So in general "you change the spin state" with this experiment. That's the only meaning to say "the spin changes" that makes sense within QT.

You can also look at the expectation values ##\langle \vec{s} \rangle##. As long as the magnetic field is acting this expecation value precesses around the ##z## axis. Also this shows that of course within a magnetic field the spin changes, i.e., it's not a conserved quantity of the dynamics (which also is reflected in the operator equations of motion, ##\mathring{\hat{\vec{s}}}=-\mathrm{i} [\hat{\vec{s}},\hat{H}] \neq 0##.

You couldn't measure or prepare the spin-##z##-component if spin were conserved!
PeterDonis said:
Yes, a probability of a spin flip. This flip would be a "change in spin" since in general it changes the spin components (although if a ##z## oriented device is being used to measure particles prepared in the spin ##x## up state, the flip actually does not change the relative weights of the spin components), but, as you note, it is considered an error and the devices are constructed to make this probability as small as possible. So the vast majority of the time, this spin flip does not happen and the spin does not change. And that vast majority of the time, when the SG device is functioning properly as part of a spin measurement, is what the post I originally responded to was talking about. In that case, the spin does not change, which was my point.
As detailed above, what does not change in the "vast majority of time" is the spin-##z## component, because the approximate Hamiltonian ##\hat{H}_0## commutes with ##\hat{s}_z##, but it doesn't commute with the other spin components, i.e., the spin changes due to the interaction with the magnetic field even in this approximation of the Hamiltonian describing the ideal SGE, and that's the curcial dynamics that makes the SG magnet a measurement device for the spin-##z## component in the here discussed setup.

It's of course also important that the perturbation due to the "rest Hamiltonian"
$$\hat{H}_1=-\mu_B g_s \beta y \hat{s}_y$$
is indeed "small". This can be checked in first-order time-dependent perturbation theory, which also shows, how the magnetic field must be chosen to get a nearly ideal SGE for measurement of the spin-##z## component. What's shown by this calculation is the correctness of the semiclassical heuristic argument, usually given in the textbooks.
 
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  • #29
vanhees71 said:
Now what do I mean "you change the spin" with this experiment. It, of course means, this preparation procedure, i.e., independent from which initial (mixed or pure) spin state you start, in each of the partial beams you have prepared an eigenstate of ##s_z##. Only if the original particle was in such an eigenstate, this state remains unchanged. So in general "you change the spin state" with this experiment.
Yes, with the full experiment--i.e., including the filtering out of one output beam, so it is now a preparation procedure. But the SG device by itself has no filtering. That is done by an additional device. And the OP of this thread is not talking about the filtering, just about the SG device itself (and the calcite crystal itself). I stand by my statement that the SG device itself does not change the spin; only the full filtering process, which requires an additional device, does.

And note that, while, as I noted before, technical difficulties have precluded doing other kinds of experiments with SG devices, such as making the equivalent of a Mach-Zehnder interferometer, analogous experiments with photons have of course been done. So restricting attention to just the simple filtering process ignores all the other things that can in principle be done with an SG device or a calcite crystal.
 
  • #30
vanhees71 said:
You have an oversimplified picture of the SGE, but let's discuss this idealized picture
Yes, let's, because that's the subject of this thread. I have already acknowledged that no real experiment is 100% ideal.
 
  • #31
Yes, and this I did in the posting you are quoting. Obviously it's again a misunderstanding due to unclear language. You have to define what you mean by "the spin doesn't change". I gave several interpretations of change of spin, according to which spin changes during the motion of the atom in the magnetic field (also for the idealized approximate treatment).
 
  • #32
PeterDonis said:
Yes, with the full experiment--i.e., including the filtering out of one output beam, so it is now a preparation procedure. But the SG device by itself has no filtering. That is done by an additional device. And the OP of this thread is not talking about the filtering, just about the SG device itself (and the calcite crystal itself). I stand by my statement that the SG device itself does not change the spin; only the full filtering process, which requires an additional device, does.
Please define what you precisely mean that the SG magnet "does not change the spin". Clearly it changes the spin state, if the particle wasn't prepared in an ##s_z## eigenstate. That's crucial for the SGE being the measurement of the spin-##z## component (with the position or momentum of the particle behind the magnet as the "pointer observable").
PeterDonis said:
And note that, while, as I noted before, technical difficulties have precluded doing other kinds of experiments with SG devices, such as making the equivalent of a Mach-Zehnder interferometer, analogous experiments with photons have of course been done. So restricting attention to just the simple filtering process ignores all the other things that can in principle be done with an SG device or a calcite crystal.
Indeed, and the polarization state of photons letting them go through a birefringent crystal is pretty analogous to the SGE. Both can be said to be "polarizing beam splitters".
 
  • #33
vanhees71 said:
Please define what you precisely mean that the SG magnet "does not change the spin".
I've already done that multiple times in this thread.

vanhees71 said:
Clearly it changes the spin state, if the particle wasn't prepared in an ##s_z## eigenstate.
No, it doesn't. It entangles the spin degree of freedom with the linear momentum degree of freedom. It doesn't change the amplitudes of the spin components at all. I've already said this multiple times in this thread.
 
  • #34
You both are saying true things but committed to butting heads. (assuming the usual state leaving the source) It is true that after leaving the magnetic field the spin and momentum are entangled and the modulus of the coefficients in front of each of the spin components is the same. (assuming the usual state leaving the source) It is also true if you throw a half wall in after the magnetic field, the spin state of the beam making it past the wall is different than the spin state that left the source and the that the left the magnetic field. The former seems obtuse to me and the latter vacuous. Why die on this hill?
 
  • #35
Haborix said:
It is true that after leaving the magnetic field the spin and momentum are entangled and the modulus of the coefficients in front of each of the spin components is the same.
Yes, and that is what the OP of the thread was asking about.

Haborix said:
if you throw a half wall in after the magnetic field, the spin state of the beam making it past the wall is different than the spin state that left the source and the that the left the magnetic field.
Yes, and this is not what the OP of the thread was asking about. That's why I have emphasized the difference between the two cases.
 

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