A How does the thermal interpretation explain Stern-Gerlach?

  • Thread starter PeterDonis
  • Start date
24,390
6,026
I am posting this question separately from the ongoing thermal interpretation thread started by @A. Neumaier since it is a question about a specific experiment and how that interpretation explains it.

The experiment is the Stern-Gerlach experiment. For concreteness, I will specify that we are considering a beam consisting of a large number of electrons all prepared with spin-z up. The beam then passes through a Stern-Gerlach device oriented in the x direction. Experimentally we know that the beam splits in two: there is a left beam and a right beam (corresponding to the two possible spin-x eigenstates), with a low intensity region in the middle.

Part III of the series of papers by @A. Neumaier on the thermal interpretation deals with measurement:


On p. 4 of this paper, it is stated:

[T]he measurement of a Hermitian quantity ##A## is regarded as giving an uncertain value approximating the q-expectation ##\langle A \rangle## rather than (as tradition wanted to have it) as an exact revelation of an eigenvalue of ##A##.
Applying this to the Stern-Gerlach experiment, we would view the observed result as an uncertain measurement of the q-expectation of the spin-x operator as applied to the electrons in the beam. This q-expectation is easily seen to be zero. For this case we can use the usual formalism of state vectors and matrix operators: in the spin-z basis, the state of the beam after preparation is the vector ##\psi = \begin{bmatrix} 1 \\ 0 \end{bmatrix}##, the spin-x operator ##\sigma_x = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}## flips the state to ##\begin{bmatrix} 0 \\ 1 \end{bmatrix}##, and the expectation ##\langle \psi \vert \sigma_x \vert \psi \rangle## is therefore the inner product of ##\psi = \begin{bmatrix} 1 \\ 0 \end{bmatrix}## and ##\begin{bmatrix} 0 \\ 1 \end{bmatrix}##, which is ##0##.

So the thermal interpretation is telling us to interpret the experimental result, of a left beam and a right beam with low intensity in between, as an uncertain measurement of the q-expectation of ##0##, which would correspond to a point in the center. But this does not seem right: it seems like an uncertain measurement of a quantity whose q-expectation is ##0## would be a normal distribution about the value ##0##, i.e., an intensity peak in the center and decreasing to the left and right. (This is, of course, the classical EM prediction which was famously falsified by Stern and Gerlach.) So my question is, in the light of this seeming inconsistency, how does the thermal interpretation explain the Stern-Gerlach experiment?
 

Demystifier

Science Advisor
Insights Author
2018 Award
9,707
2,731
@A. Neumaier if I understood you correctly, in #290 you say that, in the thermal interpretation, the measurement device is treated classically, not in terms of a quantum state. For practical purposes, that probably resolves the problem that @PeterDonis addressed.

But it leads to another problem that the thermal interpretation shares with the Bohr's Copenhagen interpretation. How to know, in general, what is treated clasically and what is treated in terms of a quantum state? Where exactly the "cut" between micro and macro is?

In particular, what if the initial beam (you refer to in #135) contains only one silver atom? If the beam is interpreted classically (as you seem to suggest), then how can the beam containing only one atom split?
 

A. Neumaier

Science Advisor
Insights Author
6,052
2,206
@A. Neumaier if I understood you correctly, in #290 you say that, in the thermal interpretation, the measurement device is treated classically, not in terms of a quantum state.
This part referred to the Copenhagen interpretation only. In the thermal interpretation, the measurement device is always a quantum device.

what if the initial beam (you refer to in #135) contains only one silver atom? If the beam is interpreted classically (as you seem to suggest), then how can the beam containing only one atom split?
The beam is not interpreted classically but as a quantum field. This is the difference to Schrödinger's failed early attempts to give a continuum interpretation of quantum mechanics.

In this case, the quantum field representing the beam is in a state with sharp ##N_e=1##, but otherwise nothing changes. Unlike a particle (a fuzzy notion usually pictured as being a local object, which creates the mind-boggling situations that appear to make QM weird), a quantum field is always distributed; it has a density everywhere.
 

A. Neumaier

Science Advisor
Insights Author
6,052
2,206
See also my post #329 in the other thread.
 

A. Neumaier

Science Advisor
Insights Author
6,052
2,206
So the thermal interpretation is telling us to interpret the experimental result, of a left beam and a right beam with low intensity in between, as an uncertain measurement of the q-expectation of 000, which would correspond to a point in the center. But this does not seem right: it seems like an uncertain measurement of a quantity whose q-expectation is 000 would be a normal distribution about the value 000, i.e., an intensity peak in the center and decreasing to the left and right.
It is what one gets when one wants to measure a q-expectation ##\langle \sigma_3\rangle \in[-1,1]## by experiments that, by their very spatial QFT analysis, allow only approximate measurements at the end of the beams, i.e, possible values left spot (-1) or right spot (+1). This is an instance of the quantum bucket intuition.
The uncertainty of a single measurement is predicted to be ##\sigma=1##, and indeed in your case (true value 0), the error of both approximate measurement results ##\pm 1## is ##1##. To improve the accuracy one needs to average over multiple measurement, and gets better results that converge to the true value 0 as the sample size gets arbitrarily large.
 
Last edited:

Demystifier

Science Advisor
Insights Author
2018 Award
9,707
2,731
This part referred to the Copenhagen interpretation only. In the thermal interpretation, the measurement device is always a quantum device.

The beam is not interpreted classically but as a quantum field. This is the difference to Schrödinger's failed early attempts to give a continuum interpretation of quantum mechanics.

In this case, the quantum field representing the beam is in a state with sharp ##N_e=1##, but otherwise nothing changes. Unlike a particle (a fuzzy notion usually pictured as being a local object, which creates the mind-boggling situations that appear to make QM weird), a quantum field is always distributed; it has a density everywhere.
Fine, but I still don't understand the Stern-Gerlach experiment from the point of view of thermal interpretation. Consider the quantum field in a state with sharp ##N_e=1##. It is an experimental fact that a dark spot will appear at a relatively sharp position on the screen, as shown at the picture. This spot will appear at either up position (which we call spin up) or down position (which we call spin down). The point is that for ##N_e=1## only one spot will appear. How does the thermal interpretation explain this?

fa4_fig01.jpg
 

A. Neumaier

Science Advisor
Insights Author
6,052
2,206
Fine, but I still don't understand the Stern-Gerlach experiment from the point of view of thermal interpretation. Consider the quantum field in a state with sharp ##N_e=1##. It is an experimental fact that a dark spot will appear at a relatively sharp position on the screen, as shown at the picture. This spot will appear at either up position (which we call spin up) or down position (which we call spin down). The point is that for ##N_e=1## only one spot will appear. How does the thermal interpretation explain this?
Conservation of mass, together with the instability of macroscopic superpositions and randomly broken symmetry forces this, just as a classical bar under vertical pressure will bend into only one direction.
 

Demystifier

Science Advisor
Insights Author
2018 Award
9,707
2,731
Conservation of mass, together with the instability of macroscopic superpositions and randomly broken symmetry forces this, just as a classical bar under vertical pressure will bend into only one direction.
This explanation does not look very specific to the thermal interpretation. Is this explanation supposed to be similar to the explanation in some older interpretation, and if so, which one?
 

A. Neumaier

Science Advisor
Insights Author
6,052
2,206
This explanation does not look very specific to the thermal interpretation. Is this explanation supposed to be similar to the explanation in some older interpretation, and if so, which one?
No, it is specific to the thermal interpretation; I haven't seen it invoked anywhere else in quantum foundations. It needs a deterministic interpretation of quantum field theory with conserved currents and a dynamics which produces the necessary instability.

Perhaps it can be modified to apply to Bohmian mechanics, since this is also deterministic; but I don't care because Bohmian mechanics introduces irrelevant variables without providing more realism than the thermal interpretation. Thus Ockham's razor eliminates it. (Not to speak of the difficulties Bohmian mechanics has with quantum field theory.)
 

Demystifier

Science Advisor
Insights Author
2018 Award
9,707
2,731
No, it is specific to the thermal interpretation; I haven't seen it invoked anywhere else in quantum foundations. It needs a deterministic interpretation of quantum field theory with conserved currents and a dynamics which produces the necessary instability.
So how does the thermal interpretation explain the instability of macroscopic superpositions? It seems that here you have in mind a mechanism different from decoherence.
 

A. Neumaier

Science Advisor
Insights Author
6,052
2,206
So how does the thermal interpretation explain the instability of macroscopic superpositions? It seems that here you have in mind a mechanism different from decoherence.
The mechanism is more or less the same, but the details are quite different.

Decoherence looks at the dynamics of the reduced density matrix and produces an improper mixture that is diagonal in the preferred basis. But this mixture describes only an ensemble and says nothing about a single silver atom. It only proves that after many silver atoms arrived, both spots contain half the total amount.

The thermal interpretation looks instead at the reduced dynamics of the relevant macroscopic q-expectations and finds that it is bistable. Thus strictly speaking it is not the macroscopic superpositions that are shown unstable but nonzero amounts of deposit on both spots.
 
24,390
6,026
It is what one gets when one wants to measure a q-expectation ##\langle \sigma_3\rangle \in[-1,1]## by experiments that, by their very spatial QFT analysis, allow only approximate measurements at the end of the beams, i.e, possible values left spot (-1) or right spot (+1). This is an instance of the quantum bucket intuition.
I don't see how the quantum bucket intuition is relevant to my question in the OP, because I was specifically not talking about the case of ##N_e = 1## that @Demystifier is asking about. I'm not asking how we can view a single dot on the screen, at either left spot (-1) or right spot (+1), as an uncertain measurement of the expectation value of 0. I'm asking how we can view two spots, both appearing at the same time, because we are firing a continuous beam of as high intensity as you like through the apparatus, as an uncertain measurement of the expectation value of 0. The quantum bucket intuition doesn't help here because the flow is large, so I should be able to measure it accurately; I shouldn't have to accept the inherent uncertainty of taking a single bucketful at random intervals. So by your own description, for the case of a high intensity beam being fired at the apparatus, I shouldn't get two spots any more; I should get the classically predicted result (normal distribution about the center, value 0). But I don't.
 

A. Neumaier

Science Advisor
Insights Author
6,052
2,206
I don't see how the quantum bucket intuition is relevant to my question in the OP, because I was specifically not talking about the case of ##N_e = 1## that @Demystifier is asking about. I'm not asking how we can view a single dot on the screen, at either left spot (-1) or right spot (+1), as an uncertain measurement of the expectation value of 0. I'm asking how we can view two spots, both appearing at the same time, because we are firing a continuous beam of as high intensity as you like through the apparatus, as an uncertain measurement of the expectation value of 0.
Post #135 explained the Stern-Gerlach experiment as follows:
In the thermal interpretation, the Ag field is concentrated along the beam emanating from the source, with a directional mass current. The beam is split by the magnetic field into two beams, and the amount of silver on the screen at the end measures the integrated beam intensity, the total transported mass. This is in complete analogy to the qubit treated in the above link. Particles need not be invoked.
Thus what is measured by a spot is the intensity of the silver flow into the spot, not the spin of single electrons. The original Stern-Gerlach paper (and the early discussion about it) indeed talked about Richtungsquantelung (quantization of directions) and not of spin measurement. The fact that two beams appear is a consequence of the spin of the electron field, but has nothing per se to do with measuring the spin state. The latter is defined only for single electrons!

Things are different if you want to consider the Stern-Gerlach experiment as a spin measurement experiment. In this case one must - like every introductory text - treat the silver source as producing a large ensemble of single atoms and the quantum bucket becomes relevant. Each single atom deposited is one bucket event emptying the inflowing silver field. It approximates the true value 0 by either ##+1## or ##-1##, the only possible bucket results (nowhere else is positive field density). This holds for every single atom, and hence for all silver arriving in the two spots.

The quantum bucket intuition doesn't help here because the flow is large, so I should be able to measure it accurately; I shouldn't have to accept the inherent uncertainty of taking a single bucketful at random intervals. So by your own description, for the case of a high intensity beam being fired at the apparatus, I shouldn't get two spots any more; I should get the classically predicted result (normal distribution about the center, value 0). But I don't.
Taking the single buckets as measurement results each time in a binary measurement of the true (theoretically predicted) uncertain number ##0\pm 1## consistent with the measurement error of 1 in each case. This is completely independent of the flow rate!

To improve on the resolution one must consider a different operator, namely the mean spin ##s=N^{-1}(s_1+\ldots+s_N)##, where ##s_k## is the ##S_z## of the ##k##th silver atom in the ensemble measured. This mean spin operator has an associated (theoretically predicted) uncertain value of ##0\pm N^{-1/2}##, which is approximately measured by the mean of the bucket results. This mean is for large ##N## Gaussian distributed with zero mean and standard deviation ##N^{-1/2}##, matching the prediction.
 
24,390
6,026
what is measured by a spot is the intensity of the silver flow into the spot
That's fine, but it doesn't address my question. The expectation value of "intensity of silver flow" is still zero--it says the silver should be flowing into the center, not into two separated spots. So why is it flowing into two separated spots? And to be clear, I am not asking why in terms of how the math gives that result; I already know that. I'm asking in terms of interpretation: how does the thermal interpretation, as an interpretation, explain how an uncertain measurement of silver flow into the center (since that's the q-expectation, and therefore that's what the thermal interpretation claims is "real") shows up as two separated spots?
 

A. Neumaier

Science Advisor
Insights Author
6,052
2,206
The expectation value of "intensity of silver flow" is still zero--it says the silver should be flowing into the center, not into two separated spots
No. You are conflating it with the q-expectation of the spin of single particles, which is zero!

The intensity of silver flow is the function of the position on the screen defined by the q-expectation of the incident current integrated over a spot centered at this position. No other expectations are involved.
So why is it flowing into two separated spots?
Given the setup, the intensity is positive at the two spots predicted by the mathematics of the theory, and zero elsewhere. This is why the silver flows into these two spots and nowhere else.
I'm asking in terms of interpretation: how does the thermal interpretation, as an interpretation, explain how an uncertain measurement of silver flow into the center (since that's the q-expectation, and therefore that's what the thermal interpretation claims is "real") shows up as two separated spots?
The only q-expectation that matters is the intensity of silver flow as a function of screen position.
It is zero at the center, so nothing flows there.
 
24,390
6,026
The intensity of silver flow is the function of the position on the screen defined by the q-expectation of the incident current integrated over a spot centered at this position.
Ok, then I guess I am confused about the math. :wink: Mathematically, how is the q-expectation of silver flow calculated? (Since obviously the math I gave in the OP is not the right math for this.)
 

A. Neumaier

Science Advisor
Insights Author
6,052
2,206
Ok, then I guess I am confused about the math. :wink: Mathematically, how is the q-expectation of silver flow calculated? (Since obviously the math I gave in the OP is not the right math for this.)
See post #282 of the other thread, except that the electric current is replaced by a silver current (which means that the formula defining it is a complicated multibody current). And of course for the silver deposited you need to also integrate over the time of the experiment.
 
24,390
6,026
See post #282 of the other thread, except that the electric current is replaced by a silver current
I'm afraid this doesn't help much, because that example is for a galvanometer reading a current, i.e., a flow rate. This is a single number that will be subject to some uncertainty, i.e., in an actual measurement I expect a normal distribution about a single expected value. But we're talking here about two spots, i.e., two expected values, not one (or an expected distribution that has two peaks, not one). So it seems like there has to be some operator involved other than the one you gave in that post.
 

A. Neumaier

Science Advisor
Insights Author
6,052
2,206
I'm afraid this doesn't help much, because that example is for a galvanometer reading a current, i.e., a flow rate. This is a single number that will be subject to some uncertainty, i.e., in an actual measurement I expect a normal distribution about a single expected value. But we're talking here about two spots, i.e., two expected values, not one
No. The current is a field with values at each spacetime position. The galvanometer responds to a small region of spacetime defining ## \Omega ##. In the case of a screen, every pixel of the screen (in an appropriate resolution) corresponds to such an ##\Omega##. Most of the resulting integrals vanish, except for those at the two spots.
 
24,390
6,026
The current is a field with values at each spacetime position. The galvanometer responds to a small region of spacetime defining ##\Omega## . In the case of a screen, every pixel of the screen (in an appropriate resolution) corresponds to such an ##\Omega##. Most of the resulting integrals vanish, except for those at the two spots.
In post #282 of the other thread, you gave the current field as a q-expectation ##J^\mu (x) = \mathrm{Tr} (\hat{\rho} \hat{j}(x))##, where ##\hat{j}(x)## is the (normal ordered) electron current operator and ##\hat{\rho} = e^{-S / k_B}## is the density operator describing the galvanometer. To calculate the integral, I assume we would have to evaluate all of these in the spacetime region ##\Omega## that describes the galvanometer over the small range of times when it would be expected to see the electron beam pass by and register a current.

You appear to be saying that, for the case of the screen in the Stern-Gerlach apparatus, I have to do a separate integral for each spacetime region ##\Omega## corresponding to a pixel on the screen at the time when we would expect that pixel to see a silver current arrive. And there are only two such regions (corresponding to the two spots) where the integral does not vanish, because those are the only two regions where the current is nonzero. But that just pushes the question back to, why are those two regions the only ones where the current is nonzero? The normal ordered electron current operator you wrote in post #282 doesn't seem to do that; it's just the free electron propagator, which would be expected to be nonzero in a single "world tube", not one that split into two world tubes. So, again, it seems like there has to be some other operator involved, that splits the current world tube. What operator is that?
 

A. Neumaier

Science Advisor
Insights Author
6,052
2,206
In post #282 of the other thread, you gave the current field as a q-expectation ##J^\mu (x) = \mathrm{Tr} (\hat{\rho} \hat{j}(x))##, where ##\hat{j}(x)## is the (normal ordered) electron current operator and ##\hat{\rho} = e^{-S / k_B}## is the density operator describing the galvanometer. To calculate the integral, I assume we would have to evaluate all of these in the spacetime region ##\Omega## that describes the galvanometer over the small range of times when it would be expected to see the electron beam pass by and register a current.

You appear to be saying that, for the case of the screen in the Stern-Gerlach apparatus, I have to do a separate integral for each spacetime region ##\Omega## corresponding to a pixel on the screen at the time when we would expect that pixel to see a silver current arrive. And there are only two such regions (corresponding to the two spots) where the integral does not vanish, because those are the only two regions where the current is nonzero. But that just pushes the question back to, why are those two regions the only ones where the current is nonzero? The normal ordered electron current operator you wrote in post #282 doesn't seem to do that; it's just the free electron propagator, which would be expected to be nonzero in a single "world tube", not one that split into two world tubes. So, again, it seems like there has to be some other operator involved, that splits the current world tube. What operator is that?
Will reply in more detail tomorrow if needed.

But we do not have a free field because of the magnetic field in the experiment. This treats different components of the spinor field in opposite ways, turning a single beam into two..

Note also that the density operator is that of the whole apparatus, and the integration over a cell to which a piece of the equipment responds is done after the trace computation.
 
24,390
6,026
we do not have a free field because of the magnetic field in the experiment. This treats different components of the spinor field in opposite ways, turning a single beam into two..
Ok, so this would be the extra operator.

the density operator is that of the whole apparatus, and the integration over a cell to which a piece of the equipment responds is done after the trace computation.
Hm, ok. So schematically, the electron current operator ##\hat{j}(x)## gets split from one beam into two by the magnetic field, the operation ##\mathrm{Tr} (\hat{\rho} \hat{j}(x))## yields a current ##J## that is nonzero in two small spots, and integrating over each spot gives a total intensity of half of the original beam (before the apparatus) in each spot?
 

A. Neumaier

Science Advisor
Insights Author
6,052
2,206
Ok, so this would be the extra operator.
Well, it is a term in the Hamiltonian of the field theory that changes the dynamics. This is like in the traditional analysis of the Stern-Gerlach experiment in terms of single silver particles. There the dynamics is treated semiclassically for simplicity, and the two beams also appear as the only possible pathways.
Hm, ok. So schematically, the electron current operator ##\hat{j}(x)## gets split from one beam into two by the magnetic field, the operation ##\mathrm{Tr} (\hat{\rho} \hat{j}(x))## yields a current ##J## that is nonzero in two small spots, and integrating over each spot gives a total intensity of half of the original beam (before the apparatus) in each spot?
Yes. And integrating over other regions of the screen gives zero since the integrand is zero there.
 
No matter how accurately you measure a dot created by a single event in a Stern-Gerlach experiment, it always gives only a very inaccurate measurement of the incident field intensity, which is given by a q-expectation.
Under this account, it is not clear why there is never a simultaneous click in both the up and down detectors. It is difficult to understand how the SG detectors can be so inaccurate as to wrongly report +1 or -1 when the true value is 0, yet at the same time so reliable that A) one or the other detector always misreports on every single experimental run and B) they never both misreport at the same time.

Again, the only viable I can foresee is for you to say that there are superdeterministic correlations between the detectors that happen to mimic standard QM by virtue of their very special initial conditions.

Related, it is always an option to dismiss inconvenient and hard to interpret problems by claiming the theory and measurements are simply not really telling us the truth. It seems to me far more interesting to ask: assuming our measurements are telling the truth, what is the universe like?
 
Last edited by a moderator:

Want to reply to this thread?

"How does the thermal interpretation explain Stern-Gerlach?" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top