How does the thermal interpretation explain Stern-Gerlach?

[PeterDonis]

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:

https://arnold-neumaier.at/ms/foundIII.pdf
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?

 

[A. Neumaier]

See my post #135 and my post #290 in the thread on the thermal interpretation, maybe also #266 and #300.

 

[Demystifier]

@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]

@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]

See also my post #329 in the other thread.

 

[A. Neumaier]

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.

 

[Demystifier]

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?

 

[A. Neumaier]

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]

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]

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]

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]

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.

 

[PeterDonis]

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]

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.

 

[PeterDonis]

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]

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.

 

[PeterDonis]

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. 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]

Ok, then I guess I am confused about the math. 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.

 

[PeterDonis]

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]

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.

 

[PeterDonis]

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]

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.

 

[PeterDonis]

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]

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.

 

[charters]

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?

 

[PeterDonis]

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.

This exchange does raise a follow-up question that I didn't ask at the time (because I was focused on the case of a high intensity beam): the explanation given is basically spontaneous symmetry breaking. But spontaneous symmetry breaking occurs because a single equation that has a certain symmetry has multiple solutions that, taken individually, do not share that symmetry. For example, the classical bar under vertical pressure obeys an equation that is symmetrical about the bar's axis: but each individual solution of that equation describes a bar that is bent in one particular direction, i.e., not symmetrical. There is no individual solution that describes a bar bent in all directions at once.

However, in the case of the SG experiment for $N_e = 1$, we don't have that. We have a single solution of the equation describing the system that shares the equation's symmetry: it describes a superposition of a spot in the "up" position and a spot in the "down" position. We don't have two solutions, one of which describes a spot in the "up" position and one of which describes a spot in the "down" position. So how can spontaneous symmetry breaking explain why we only observe one spot?

 

[charters]

I don't see how the explanation can be both the instability of superpositions and inaccurate devices. The point of saying devices are inaccurate is so the superposition is actually stable - it is just that the detectors are bad, so detector 1 didn't register any of the incident (N=1) beam, while detector 2 registered too much of it. But the beam is ontologically at both detectors before and after each measurement or non-measurement, and in a single world.

The issues are A) how it can be reasonable for these bad detectors to be perfectly correlated in their inaccurate clicks and non-clicks and B) how is it plausible to think so many different types of detectors are bad in just the right way so that a sum over histories and collapse/MWI-branching explanation has coincidentally worked all along, given the infinite ways for a detector to be inaccurate. Superdeterminism is the only live option here, imo.

And I don't think one can invoke spontaneous symmetry breaking, given III.4.2, where the interpretation seems to be presented as a deterministic hidden variable theory, All SG results and bent metal bars should be functions of the pre-existing fine grained hidden variables.

 

[A. Neumaier]

The second reason is the need to explain correlations between detectors.

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.

See this post and the subsequent discussion.

 

[A. Neumaier]

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?

I am not doing the first. Regarding the second, I ask instead: assuming our measurements are telling the truth, what is the right way to interpret this truth?

 

[A. Neumaier]

This exchange does raise a follow-up question that I didn't ask at the time (because I was focused on the case of a high intensity beam): the explanation given is basically spontaneous symmetry breaking. But spontaneous symmetry breaking occurs because a single equation that has a certain symmetry has multiple solutions that, taken individually, do not share that symmetry. For example, the classical bar under vertical pressure obeys an equation that is symmetrical about the bar's axis: but each individual solution of that equation describes a bar that is bent in one particular direction, i.e., not symmetrical. There is no individual solution that describes a bar bent in all directions at once.

However, in the case of the SG experiment for $N_e = 1$, we don't have that. We have a single solution of the equation describing the system that shares the equation's symmetry: it describes a superposition of a spot in the "up" position and a spot in the "down" position. We don't have two solutions, one of which describes a spot in the "up" position and one of which describes a spot in the "down" position. So how can spontaneous symmetry breaking explain why we only observe one spot?

Instead of the continuous syymmetry breaking for the bar let us discuss the same issue in a sequence of technically simpler problems.

1. Symmetry breaking of a classical anharmonic oscillator in a double well potential. All positions are possible states. An initial state at the location of the local maximum of the potential is unstable under tiny random prturbations, and the small amount of dissipation associated with the randomness (fluctuation-dissipation theorem) forces the state to move intto one of the two minima. Note that the symmetry only needs to be local (in a neighborhood of the stationary point) for this to hold; the potential itself needs not be symmetric.

2. Classical symmetry breaking in a multimodal potential energy surface with two degrees of freedom, such as a little ball moving in a bowl with an uneven bottom. Again all positions are possible states. An initial state at the location of a stationary point is unstable and moves into one of the local minima of the potential energy surface. This clearly generalizes to any number of degrees of freedom.

3. Classical symmetry breaking in a multimodal potential energy surface on the 3-dimensional unit ball, with forces on the boundary pointing inwards or tangential. Now the positions are constrained to lie in the closed unit ball, but otherwise nothing changes, except that now local minima may occur on the boundary. It is easy to construct potentials that have just two local minima, located at the north pole and the south pole of the ball. As a result, tiny dissipative and stochastic prturbations force the dynamics to settle at one of the poles.

4. An isolated quantum spin in the thermal interpretation is described by a deterministic linear dynamics of its density matrix $\rho$. The latter can be mapped 1-to-1 onto the unit ball, with pure states being mapped to its boundary, the Poincare sphere or Bloch sphere. One can choose this mapping such that the up state and the down state are represented by the noth pole and the south pole. For a nonisolated quantum spin, the reduced density matrix is again represented on the Bloch sphere but has a nonlinear, dissipative dynamics obtained by coarse-graining. If the environment is such that it corresponds to a spin measurement with collapse to an up or down state, this dynamics is expected to have just two stable fixd points, both located on the boundary. local minima on the boundary and no interior minimum. Thus we have essentially a generalization of the siuation in case 3, except that the detailed dynamics is different. Again, tiny dissipative and stochastic prturbations force the dynamics to settle at one of the poles. Thus we obtain collapse of an arbitrary (pure or mixd) spin state to the up state or the down state.

 

[PeterDonis]

For a nonisolated quantum spin, the reduced density matrix is again represented on the Bloch sphere but has a nonlinear, dissipative dynamics obtained by coarse-graining.

This would appear to be the fundamental difference between the treatment you are giving and the "standard" QM treatment, which uses the linear dynamics of the Schrodinger equation combined with the obvious assumption about how the interaction Hamiltonian between the SG apparatus, with its inhomogeneous magnetic field, and the spin-1/2 passing through it acts on spin states that are parallel to the field inhomogeneity (e.g., +x and -x spins for an apparatus oriented in the x direction). In the standard treatment, the action of the Hamiltonian on a +z spin is given by simple superposition (which is allowed due to the linearity of the Schrodinger equation) of the actions on the +x and -x spins.

But the dynamics you are attributing to the system + apparatus is nonlinear, so the simple standard linear superposition picture does not work and there is no solution to the nonlinear dynamics that describes a superposition of a spot at the +x position and a spot at the -x position. Instead, there are two solutions, one describing a +x spot and one describing a -x spot, and which solution gets realized in a particular case depends on random and unmeasurable fluctuations.

Is this a fair description of what you are saying?

 

[A. Neumaier]

This would appear to be the fundamental difference between the treatment you are giving and the "standard" QM treatment, which uses the linear dynamics of the Schrodinger equation [...]

But the dynamics you are attributing to the system + apparatus is nonlinear, so the simple standard linear superposition picture does not work.

Standard quantum mechanics also has lots of nonlinear approximations. Prime examples are the Hartree-Fock equations and the quantum-classical dynamics discussed in Part III of my series of papers. Nothing is really new in the thermal interpretation except for the reintepretation of what a measurement means.

The instantaneous collapse in the standard interpretation mimicks the nonlinearities by ignoring details of what happens during the measurement, which in fact tkes a finite time. it is like mimicking the classical continuous dynamics of an electric switch by treating it as an instantaneous discontinuity.

Objective-collapse theories introduce such nonlinearities explicitly into the dynamics of the universe. But they are not needed since they arise automatically through the well-knon coarse-graining procedures.

there is no solution to the nonlinear dynamics that describes a superposition of a spot at the +x position and a spot at the -x position.

The solutions are time-dependent. The trajectories started anywhere (e.g., in a superposition) move within a very short time (the duration of the measurement) towards one of the distinguished up and down states.

 

[PeterDonis]

Standard quantum mechanics also has lots of nonlinear approximations.

Yes, but the standard description of what happens if you make a SG measurement oriented in the x-direction on a single +z spin-1/2 does not, as I understand it, use them. It uses the linear Schrodinger equation and superposition. As I understand it, you are saying that that is wrong. In fact, it would seem that the thermal interpretation implies that it is always wrong--that as soon as you include any kind of macroscopic apparatus to make a measurement (which in practice you always will), the dynamics are no longer linear and so the Schrodinger Equation is not correct.

 

[PeterDonis]

The solutions are time-dependent.

If this just means that the starting state is not preserved by the dynamics, of course this is true; but it's just as true for the linear Schrodinger Equation as for the nonlinear dynamics you appear to be using.

The trajectories started anywhere (e.g., in a superposition) move within a very short time (the duration of the measurement) towards one of the distinguished up and down states.

But that's not what happens with the linear Schrodinger Equation. The linear Schrodinger Equation says that if the spin-1/2 starts in the +z spin state, which is a 50-50 superposition of +x and -x spin (for a SG apparatus oriented in the x direction), the solution is a superposition of a +x spot and a -x spot. There is no such solution of the nonlinear dynamics, as you say; every solution ends up with either a +x spot or a -x spot, not a superposition of both.

 

[A. Neumaier]

If this just means that the starting state is not preserved by the dynamics, of course this is true; but it's just as true for the linear Schrodinger Equation as for the nonlinear dynamics you appear to be using.

Yes.

But that's not what happens with the linear Schrodinger Equation. The linear Schrodinger Equation says that if the spin-1/2 starts in the +z spin state, which is a 50-50 superposition of +x and -x spin (for a SG apparatus oriented in the x direction), the solution is a superposition of a +x spot and a -x spot. There is no such solution of the nonlinear dynamics, as you say; every solution ends up with either a +x spot or a -x spot, not a superposition of both.

Yes, but traditional interpretations claim the validity of the linear Schrodinger Equation only for isolated systems. A detector is never isolated, hence the linear Schrodinger Equation does not apply.

Traditional interpretations claim nothing during the measurement, which is usually thought of being instantaneous, and only claim that the state of the system after the measurement collapsed. This is obviously a nonlinear stochastic process.

The thermal interpretation explains this nonlinear stochastic process as an effect of the neglected environment.