# I Confusion about the thermal interpretation's account of measurement

#### A. Neumaier

(b) The description in the thermal interpretation papers seems to claim that, in fact, if I had a good enough description of the details of the initial state of the microscopic system and the measurement apparatus, I would be able to deduce which of the two possibilities "really happened", and that I could do this, again, using only ordinary unitary quantum mechanics with no collapse.
Correct.
Since both stories use the same initial condition and the same rule for evolving in time, these seem to be two different claims about the exact same mathematical object --- the density matrix of the final state of the system. If that's true, then one of them ought to be wrong.
No. Decoherence tells the same story but only in the statistical interpretation (using Lindblad equations rather than stochastic trajectories), where ensembles of many identically prepared systems are considered, so that only the averaged results (which must feature all possibilities) can be deduced. The thermal interpretation refines this to a different, more detailed story for each single case. Averaging the latter recovers the former.
My understanding of the thermal interpretation (remember I'm not its author so my understanding might not be correct) is that the two non-interfering outcomes are actually a meta-stable state of the detector (i.e., of whatever macroscopic object is going to irreversibly record the measurement result), and that random fluctuations cause this meta-stable state to decay into just one of the two outcomes. An analogy that I have seen @A. Neumaier use is a ball on a very sharp peak between two valleys; the ball will not stay on the peak because random fluctuations will cause it to jostle one way or the other and roll down into one of the valleys.
Correct.
However, the dynamics of this collapse of a meta-stable detector state into one of the two stable outcomes can't be just ordinary unitary QM, because ordinary unitary QM is linear and linear dynamics can't do that. In ordinary unitary QM, fluctuations in the detector would just become entangled with the system being measured and would preserve the multiple outcomes. There would have to be some nonlinear correction to the dynamics to collapse the state into just one outcome.
I explained how the nonlinearities naturally come about through coarse graining. An example of coarse graining is the classical limit, where nonlinear Hamiltonian dynamics arises from linear quantum dynamics for systems of sufficently heavy balls. This special case is discussed in Section 2.1
of Part IV
, and explains to some extent why heavy objects behave classically but nonlinearly.

Exactly, that's why I'm confused! My impression is that @A. Neumaier is somehow denying this, and that somehow the refusal to describe macroscopic objects with state vectors is related to the way he gets around this linearity argument, although I don't see how.

If we're supposed to be positing nonunitary dynamics on a fundamental level, then that would obviate my whole question, but from the papers I understood A. Neumaier to be specifically not doing that.
As you can see from the preceding, this is not necessary.

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#### A. Neumaier

Ordinary unitary QM with decoherence can give you two non-interfering outcomes. It can't give you a single outcome; that obviously violates linearity.
This is according to the traditional interpretations, where outcome = eigenvalue, which must be statistical. But in the thermal interpretation, outcome = q-expectation, which is always single-valued. This makes a big difference in the interpretation of everything! See Chapter 4 of Part IV.

#### vanhees71

Gold Member
No it goes in circles again. The identification of expectation values with meaurement outcomes had to be abandoned very early in the history of the development of QT. E.g., it contradicts the fact that the absorption and emission of electromagnetic radiation by charged-matter systems is in discrete "lumps of energy" $\hbar \omega$. That's in fact how the whole quantum business started with Planck's analysis of the black-body spectrum.

#### A. Neumaier

Now it goes in circles again. The identification of expectation values with measurement outcomes had to be abandoned very early in the history of the development of QT.
The observable outcome is a property of the detector, e.g., a photocurrent. This is not quantized but a continuous burst, for each single observation of a detection event. The derivation of macroscopic electrodynamics from QED by QED shows that the measured currents are q-expectations. No experiment in the history of quantum mechanics contradicts this.

The relation between observed outcome and true result is in general approximate (especially when the spectrum spacing is of the order of the observation error or larger) and depends of what one considers to be the true (unobservable) result. This is not observable hence a matter of interpretation.

Here tradition and the thermal interpretation differ in what they postulate to be the true result, i.e., how to split the observed result (a left spot and a right spot) into a true result (eigenvalue or q-expectation) and an observational error (the difference). See Sections 4.1 and 4.2 of Part IV.

Since this doesn't change the experimental record it is cannot be contradicted by any experiment.
E.g., it contradicts the fact that the absorption and emission of electromagnetic radiation by charged-matter systems is in discrete "lumps of energy" $\hbar \omega$. That's in fact how the whole quantum business started with Planck's analysis of the black-body spectrum.
The black body spectrum was explained by Bose 1924 by the canonical ensemble of a Bose-Einstein gas, although Planck had derived it from quantized lumps of energy. Only a quantized spectrum is needed, no discrete lumps of radiation energy.

Just as the photoeffect could be explained by Wentzel 1928 with classical light (no lumps of energy), although Einstein had originally explained it in terms of quantized light. Only quantized electrons are needed, no discrete lumps of radiation energy.

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#### vanhees71

Gold Member
Well, Planck's derivation was in terms of a canonical ensemble of a Bose-Einstein gas. Of course, at this time it wasn't known as such.

The "q-expectation" value in general does not reflect what's measured by an actual device. For this you'd have to put information on the device into the description. This of course always have to be done when data from a real detector are evaluated, but it cannot be part of the general description of a system.

Nowadays it's no problem to prepare single photons, and all experiments show that an "entire photon" is registered (if it is registered at all) but not some fraction of a photon. So obviously at this (today indeed technically realized!) resolution, you measure discrete photon energies $\hbar \omega$ and not some expectation value.

#### A. Neumaier

Well, Planck's derivation was in terms of a canonical ensemble of a Bose-Einstein gas.
For the equilibrium thermodynamics of a Bose-Einstein gas one doesn't need anything more than the maximum entropy state corresponding to the q-expectation of a Hamiltonian with discrete eigenvalues. The according to you necessary interpretation as discrete lumps of energy nowhere enters.

The "q-expectation" value in general does not reflect what's measured by an actual device.
Any measured current is the q-expectation of a smeared version of the QED current; only the details of the smearing depend on the actual device. This is the q-expectation that is relevant for the thermal interpretation.

Nowadays it's no problem to prepare single photons, and all experiments show that an "entire photon" is registered (if it is registered at all) but not some fraction of a photon. So obviously at this (today indeed technically realized!) resolution, you measure discrete photon energies $\hbar \omega$ and not some expectation value.
No. What is measured in each single photodetection event (called a photon by convention) is a magnified current of energy much larger than $\hbar \omega$.

#### vanhees71

Gold Member
Indeed, the Hamiltonian, representing the energy of the system (in this case an ensemble of non-interacting harmonic oscillators, representing the em. field), takes discrete values, which are the possible outcomes of precise measurement of energy, while the expectation values can take all continuous values $\geq 0$.

#### A. Neumaier

the Hamiltonian, representing the energy of the system (in this case an ensemble of non-interacting harmonic oscillators, representing the em. field), takes discrete values, which are the possible outcomes of precise measurement of energy
Well, only energy differences are measured, and in general many at the same time (through emission or absorption spectra). These all have widths and do not give exact values, and recovering from a spectrum the energy levels is a highly nontrivial process.

Thus the connection between measured values (always on a continuous scale, a q-expectation of something macroscopic) to the theoretical true values is always somewhat indirect, and therefore the designation of something (eigenvalue or q-expectation) as true measurement value is a matter of interpretation. Tradition and the thermal interpretation differ in the choice made in this.

#### PeterDonis

Mentor
in the thermal interpretation, outcome = q-expectation, which is always single-valued.
Yes, you're right, I left that part out. But for the benefit of @nicf, it might be worth spelling out how this works in the case of a simple binary measurement such as the Stern-Gerlach experiment. Say we are measuring a spin-z up electron using a Stern-Gerlach apparatus oriented in the x direction. Then we have the following account of what happens according to ordinary QM vs. the thermal interpretation:

(a) Ordinary QM: the measurement creates an entanglement between the spin of the electron and its momentum (which direction it comes out of the apparatus). When this entangled state interacts with the detector, decoherence occurs, which produces two non-interfering outcomes. How this becomes one outcome (or the appearance of one) depends on which interpretation you adopt (where "interpretation" here means basically collapse vs. no collapse, something like Copenhagen vs. something like MWI).

(b) Thermal interpretation: The q-expectation of the measurement is zero (an equal average of +1 and -1), but each individual measurement gives an inaccurate result because of the way the measurement/detector are constructed, so only the average over many results on an ensemble of identically prepared electrons will show the q-expectation. For each individual measurement, random nonlinear fluctuations inside the detector cause the result to be either +1 or -1.

#### nicf

(b) Thermal interpretation: The q-expectation of the measurement is zero (an equal average of +1 and -1), but each individual measurement gives an inaccurate result because of the way the measurement/detector are constructed, so only the average over many results on an ensemble of identically prepared electrons will show the q-expectation. For each individual measurement, random nonlinear fluctuations inside the detector cause the result to be either +1 or -1.
I might be misunderstanding you, but I actually think (b) is not what @A. Neumaier is saying, at least if by "random nonlinear fluctuations" you mean that there's some change to the unitary dynamics underlying quantum mechanics. Rather, he's saying that the nonlinearity comes from coarse-graining, that is, from neglecting some details of the state, which would actually evolve linearly if you could somehow add those details back in.

This was my reading from the beginning, and is the source of my question. I feel quite stupid right now and that I must be missing something obvious, but I'm going to press on anyway and try to be more specific about my confusion.

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Let's start with the setup in Section 3 of the fourth TI paper, where we've written the Hilbert space of the universe as $H=H^S\otimes H^E$ where $H^S$ is two-dimensional, and assume our initial state is $\rho_0=\rho^S\otimes\rho^E$. We have some observable $X^E$ on $H^E$, and we're thinking of this as being something like the position of the tip of a detector needle. Using thermal interpretation language, we can say that we're interested in the "q-probability distribution" of $X^E$ after running time forward from this initial state; this can be defined entirely using q-expectation values, so I think @A. Neumaier and I agree that this is a physically meaningful object to discuss. If, after some experiment, the q-probability distribution of $X^E$ has most of its mass near some particular $x\in\mathbb{R}$, then I'm happy to say that there's no "measurement problem" with respect to that experiment.

Consider two state vectors $\psi_1$ and $\psi_2$, and suppose they are orthogonal, and pick some initial density matrix $\rho^E$ for the environment. Suppose that:
(i) starting in the state $\psi_1\psi_1^*\otimes\rho^E$ and running time forward a while yields a q-probability distribution for $X^E$ with a single spike around some $x\gg 0$, and
(ii) similarly with $\psi_2$ around $-x$.

The question then is: what does the q-probability distribution of $X^E$ look like if we instead start with $\frac12(\psi_1+\psi_2)(\psi_1+\psi_2)^*\otimes\rho^E$?

(a) It will be bimodal, with a peak around $x$ and a peak around $-x$
(b) It will be unimodal, concentrated around $-x$ or around $x$, with the choice between the two depending in some incredibly complicated way on the exact form of $\rho_E$. (In this story, maybe there's a choice of $\rho^E$ that will give something like (a), but it would require a ludicrous amount of symmetry and so there's no need to worry about it.)

The reason I'm confused, then, is that I thought that the decoherence story involves (among other things) deducing (a) from (i) and (ii). In particular, I thought it followed from the linearity of time evolution together with the whole business with decaying off-diagonal terms in the density matrix, but I don't understand the literature enough to be confident here.

Am I just wrong about what the decoherence story claims? Is it just that they assume enough symmetry in $\rho^E$ to get (a) to happen, but actually (b) is what happens for the majority of initial environment states? I can see that this would be a sensible thing to do if you think of $\rho^E$ as representing an ensemble of initial environments rather than the "true" initial environment.

There is also the separate claim that, if the environment starts in a pure state (a possibility which TI denies but many other interpretations don't) then the linearity alone should leave the whole universe in a superposition of "what would have happened if $\rho^S$ were $\psi_1$" and the same with $\psi_2$, which feels to me like it ought to imply an outcome like (a), and it seems like I could then extract (a) for an arbitrary $\rho^E$ by writing it as a convex combination of pure states. I assume this paragraph also contains a mistake, but I would be interested to know where it is.

#### PeterDonis

Mentor
he's saying that the nonlinearity comes from coarse-graining, that is, from neglecting some details of the state, which would actually evolve linearly if you could somehow add those details back in.
I think he's saying that (sort of--see below), but remember that he's also saying that, in the thermal interpretation, the "beable" is not the eigenvalue; it's the q-expectation. So the reason there is only one result is that there is only one beable. There aren't two decoherent possibilities that both exist; there is only one result, which is an inaccurate measurement of the single beable, the q-expectation.

In other words, he is not interpreting a wave function that has two entangled terms that decohere to what look like two measurement results, as actually describing two real possibilities. He's just interpreting them as a tool for calculating the q-expectation, which is what is real. So in order to understand the TI, you have to unlearn much of what you learned from other QM interpretations, since all of them focus on eigenvalues instead of q-expectations.

The reason I said "sort of" above is that, if the "beable" is q-expectations, not eigenvalues, then I'm not sure there is an underlying linear dynamics; the linear dynamics is the dynamics of the wave function, which gives you eigenvalues. I don't know that the dynamics of the q-expectations is always linear even for wave function dynamics that are always linear.

(a) It will be bimodal, with a peak around $x$ and a peak around $−x$
(b) It will be unimodal, concentrated around $-x$ or around $x$, with the choice between the two depending in some incredibly complicated way on the exact form of $\rho_E$. (In this story, maybe there's a choice of $\rho^E$ that will give something like (a), but it would require a ludicrous amount of symmetry and so there's no need to worry about it.)
I think neither of these are correct; I think the TI prediction is that the q-expectation will be peaked around $0$. The $+x$ and $-x$ are eigenvalues, not q-expectations, and eigenvalues aren't "real" in the TI.

#### nicf

I think neither of these are correct; I think the TI prediction is that the q-expectation will be peaked around $0$. The $+x$ and $-x$ are eigenvalues, not q-expectations, and eigenvalues aren't "real" in the TI.
This part I actually think I understand well enough to explain. The q-probability distribution isn't the same object as the q-expectation; it's a measure on $\mathbb{R}$, whereas the q-expectation is just a number. (In fact, the q-expectation is the mean of the q-probability distribution.) You can extract the q-probability distribution from q-expectations of various observables --- the q-probability that $X$ lies in some set $U$ is $\langle\chi_U(X)\rangle$ where $\chi_U$ is the characteristic function of $U$ --- and so it's also "real" in the TI. (See (A5) on p. 5 of the first paper.) But there's no need to interpret it as a probability of anything; hence the "q". Eigenvalues don't enter into it at all.

The density matrix is also a completely legitimate physical object in the TI, since you can extract it if you know the q-expectations of all observables, and he's not changing the usual picture of how they evolve. (You can phrase the time evolution in terms of either the density matrix or the q-expectation, but they're mathematically equivalent. This happens on p. 6 of the second paper.) He does deny that we can always describe the universe with a pure state, that is, the density matrix doesn't have to have rank 1. But this isn't really all that big a deal for the question of the dynamics, since again it's possible to phrase ordinary unitary quantum mechanics in terms of density matrices without ever mentioning pure states.

I could, of course, be wrong about everything I just said, but I'm more confident about this part than about anything in my post two up from this one!

#### vanhees71

Gold Member
Yes, you're right, I left that part out. But for the benefit of @nicf, it might be worth spelling out how this works in the case of a simple binary measurement such as the Stern-Gerlach experiment. Say we are measuring a spin-z up electron using a Stern-Gerlach apparatus oriented in the x direction. Then we have the following account of what happens according to ordinary QM vs. the thermal interpretation:

(a) Ordinary QM: the measurement creates an entanglement between the spin of the electron and its momentum (which direction it comes out of the apparatus). When this entangled state interacts with the detector, decoherence occurs, which produces two non-interfering outcomes. How this becomes one outcome (or the appearance of one) depends on which interpretation you adopt (where "interpretation" here means basically collapse vs. no collapse, something like Copenhagen vs. something like MWI).

(b) Thermal interpretation: The q-expectation of the measurement is zero (an equal average of +1 and -1), but each individual measurement gives an inaccurate result because of the way the measurement/detector are constructed, so only the average over many results on an ensemble of identically prepared electrons will show the q-expectation. For each individual measurement, random nonlinear fluctuations inside the detector cause the result to be either +1 or -1.
Another nice example for the reason for abandoning the identification of expectation values with values of observables but using the standard one, namely that the possible values an observable can take is given by the spectrum of the representing self-adjoint operator of this observable. Thus, for the SGE "ordinary QM" provides the correct prediction, namely that a spin component of an electron can take two values $\pm \hbar/2$ when accurately measured. If the electron is "unpolarized", i.e., in the spin state $\hat{\rho}_{\text{spin}}=\hat{1}/2$ the expectation value is $\langle \sigma_z \rangle=0$, which according to QT is not in the spectrum of $\hat{\sigma}_z$ and not what should be measured when accurately measuring the spin component. Guess what: All experiments measuring the spin of an electron accurately (as was first done by Stern and Gerlach for the valence electron in a Ag atom) only the values $\pm\hbar/2$ have been found and not the expectation value $0$. The expectation value can be measured by measuring accurately the spin component on an ensemble of unpolarized electrons, which then of course gives $0$ (together with the correct statistical and usually also systematic error analysis of course).

#### timmdeeg

Gold Member
My understanding of the thermal interpretation (remember I'm not its author so my understanding might not be correct) is that the two non-interfering outcomes are actually a meta-stable state of the detector (i.e., of whatever macroscopic object is going to irreversibly record the measurement result), and that random fluctuations cause this meta-stable state to decay into just one of the two outcomes.
Does this mean that if an EPR experiment is performed "random fluctuations" at detector A are entangled with "random fluctuations" at detector B such that the outcomes are correlated?

#### A. Neumaier

Guess what: All experiments measuring the spin of an electron accurately (as was first done by Stern and Gerlach for the valence electron in a Ag atom) only the values $\pm\hbar/2$ have been found and not the expectation value $0$.
What is measured accurately are the positions of the spots in the Stern-Gerlach experiment. That these spots mean an accurate spin measurement is already an interpretation.

It is this interpretation that the thermal interpretation calls into question. It replaces it by the claim that it is an inaccurate measurement of a continuous particle spin with an error of the order of $O(\hbar)$ (as expected for nonclassical measurements, with the correct classical limit). This error is magnified by the experimental arrangement to a macroscopic size.

This is fully compatible with the experimental record and produces precisely the same statistics.

#### A. Neumaier

This part I actually think I understand well enough to explain. The q-probability distribution isn't the same object as the q-expectation; it's a measure on $\mathbb{R}$, whereas the q-expectation is just a number. (In fact, the q-expectation is the mean of the q-probability distribution.) You can extract the q-probability distribution from q-expectations of various observables --- the q-probability that $X$ lies in some set $U$ is $\langle\chi_U(X)\rangle$ where $\chi_U$ is the characteristic function of $U$ --- and so it's also "real" in the TI. (See (A5) on p. 5 of the first paper.) But there's no need to interpret it as a probability of anything; hence the "q". Eigenvalues don't enter into it at all.

The density matrix is also a completely legitimate physical object in the TI, since you can extract it if you know the q-expectations of all observables, and he's not changing the usual picture of how they evolve. (You can phrase the time evolution in terms of either the density matrix or the q-expectation, but they're mathematically equivalent. This happens on p. 6 of the second paper.) He does deny that we can always describe the universe with a pure state, that is, the density matrix doesn't have to have rank 1. But this isn't really all that big a deal for the question of the dynamics, since again it's possible to phrase ordinary unitary quantum mechanics in terms of density matrices without ever mentioning pure states.

I could, of course, be wrong about everything I just said, but I'm more confident about this part than about anything in my post two up from this one!
This is correct.
Let's start with the setup in Section 3 of the fourth TI paper, where we've written the Hilbert space of the universe as $H=H^S\otimes H^E$ where $H^S$ is two-dimensional, and assume our initial state is $\rho_0=\rho^S\otimes\rho^E$. We have some observable $X^E$ on $H^E$, and we're thinking of this as being something like the position of the tip of a detector needle.
This is correct.
Using thermal interpretation language, we can say that we're interested in the "q-probability distribution" of $X^E$ after running time forward from this initial state;
Here is the origin of your confusion. We are interested in the q-expectation of $X^E$, and in its distribution when $\rho_E$ has a random part that cannot be controlled in the experiment under consideration. This distribution has a priori nothing to do with the q-probability distribution of $X^E$. To relate the two constitutes part of the measurement problem - namely the part solved under suitable assumptions by decoherence.
Consider two state vectors $\psi_1$ and $\psi_2$, and suppose they are orthogonal, and pick some initial density matrix $\rho^E$ for the environment. Suppose that:
(i) starting in the state $\psi_1\psi_1^*\otimes\rho^E$ and running time forward a while yields a q-probability distribution for $X^E$ with a single spike around some $x\gg 0$, and
(ii) similarly with $\psi_2$ around $-x$.

The question then is: what does the q-probability distribution of $X^E$ look like if we instead start with $\frac12(\psi_1+\psi_2)(\psi_1+\psi_2)^*\otimes\rho^E$?
Because of the mixed terms in this expression, the q-probability distribution of $X^E$ cannot in general be deduced from the result of the two cases (i) and (ii). Thus the superposition arguments break down completely, due to the nonlinear dependence of expectations on the wave function.

To proceed and conclude anything, one therefore needs to make additional assumptions.
(a) It will be bimodal, with a peak around $x$ and a peak around $-x$

The reason I'm confused, then, is that I thought that the decoherence story involves (among other things) deducing (a) from (i) and (ii).
In each particular experiment, the q-expectation of the pointer variable is fully determined by $\rho_E$ and will in each case be close to one of $\pm x$. The q-probability distribution of the pointer variable is irrelevant for the TI but is centered close to the q- expectation.

But in each experiment, $\rho_E$ will be different, and the sign of the q-expectation depends chaotically on the details, hence appears random.

On the other hand, the q-probability distribution deduced by decoherence is based on assuming $\rho_E$ to be an exact equilibrium state, which is the case only in the mean over many experiments. Therefore it gives the mean of the q-probability distributions of the pointer variable in all these experiments, which is bimodal.
I think the TI prediction is that the q-expectation will be peaked around 000. The +x and -x are eigenvalues, not q-expectations, and eigenvalues aren't "real" in the TI.
No. You are probably thinking of the q-expectation of the measured variable, which in the symmetric case will have q-expectation zero, and for identically prepared qubits, this will be the case in all realizations. But for the interaction with the detector, the whole state of the measured system counts, not only its q-expectation; thus its q-expectation is fairly irrelevant (except when discussing errors).

Under discussion is, however, not the distribution of the q-expectation of the measured variable but the distribution of the q-expectations of the measurement results. According to the thermal interpretation, the q-expectation of the pointer variable will be essentially random (i.e., depending on the details of $\rho_E$) , with a strongly bimodal distribution reflecting the binary q-probability distribution of the variable measured.

This is indeed what decoherence claims as average result. However, decoherence cannot resolve it into single events since according to all traditional interpretations of quantum mechanics, single events (in a single world) have no theoretical representation in the generally accepted quantum formalism.

Note that this even holds for Bohmian mechanics. Here single events have a theoretical representation, but this representation is external to the quantum formalism, given by the additionally postulated position variables.

Only the thermal interpretation represents single events within the generally accepted quantum formalism, though in a not generally accepted way.

#### vanhees71

Gold Member
What is measured accurately are the positions of the spots in the Stern-Gerlach experiment. That these spots mean an accurate spin measurement is already an interpretation.

It is this interpretation that the thermal interpretation calls into question. It replaces it by the claim that it is an inaccurate measurement of a continuous particle spin with an error of the order of $O(\hbar)$ (as expected for nonclassical measurements, with the correct classical limit). This error is magnified by the experimental arrangement to a macroscopic size.

This is fully compatible with the experimental record and produces precisely the same statistics.
If the expectation value was the "true spin component" as you claim, you'd expect one blurred blob around $\sigma_z=0$, but that's not what's observed in the SGE, and that was one of the "confirmations" of "old quantum theory", though ironically just 2 mistakes cancel (gyro factor 1 but using orbital angular momentum and somehow discussing why one doesn't see three lines (which would be natural for $\ell=1$) but rather only two).

With the advent of half-integer spins and spin 1/2 with a gyrofactor of 2 all the quibbles were solved. The gyrofactor of 2 was, by the way, known from the refinements of the Einstein-de Haas experiment. De Haas himself had already measured a value closer to the right value 2, but was dissuaded by Einstein to publish it, because "of course the gyro factor must be 1" according to classical Amperian models of magnetic moments (ok, in 1915 you can't blame Einstein, but de Haas should have insisted to publish all experimental results, not the ones Einstein liked better, but that's another story).

#### A. Neumaier

If the expectation value was the "true spin component" as you claim, you'd expect one blurred blob around $\sigma_z=0$
You'd expect that assuming a Gaussian error.

But there is no necessity for a Gaussian error distribution. If you measure something with true value 0.37425 with a 4 digit digital device the error distribution will be discrete, not Gaussian.

The arrangement in a Stern-Gerlach experiment together with simple theory accounts for the discreteness in the response. Thus one must not attribute the discreteness to the value of the spin but can as well attribute it to the detection setup.

#### nicf

Here is the origin of your confusion. We are interested in the q-expectation of $X^E$, and in its distribution when $\rho_E$ has a random part that cannot be controlled in the experiment under consideration. This distribution has a priori nothing to do with the q-probability distribution of $X^E$. To relate the two constitutes part of the measurement problem - namely the part solved under suitable assumptions by decoherence.

Because of the mixed terms in this expression, the q-probability distribution of $X^E$ cannot in general be deduced from the result of the two cases (i) and (ii). Thus the superposition arguments break down completely, due to the nonlinear dependence of expectations on the wave function.

To proceed and conclude anything, one therefore needs to make additional assumptions.

In each particular experiment, the q-expectation of the pointer variable is fully determined by $\rho_E$ and will in each case be close to one of $\pm x$. The q-probability distribution of the pointer variable is irrelevant for the TI but is centered close to the q- expectation.

But in each experiment, $\rho_E$ will be different, and the sign of the q-expectation depends chaotically on the details, hence appears random.

On the other hand, the q-probability distribution deduced by decoherence is based on assuming $\rho_E$ to be an exact equilibrium state, which is the case only in the mean over many experiments. Therefore it gives the mean of the q-probability distributions of the pointer variable in all these experiments, which is bimodal.
Excellent! I think this gets at the heart of what I was asking about. Specifically, I think I was failing to appreciate both
(a) the obvious-once-you've-stated-it fact that mixed terms like $\psi_1\psi_2^*\otimes\rho^E$ prevent us from deducing the $\frac{1}{\sqrt{2}}(\psi_1+\psi_2)$ result from the $\psi_1$ and $\psi_2$ results, and
(b) the fact that the decoherence arguments depend on $\rho^E$ being an exact equilibrium state, and so in particular we shouldn't expect the "bimodal" result from a generic $\rho^E$. That is, I thought the decoherence argument was contradicting your analysis of the particular case, whereas they are in fact merely silent about the particular case and only making a claim about the average.

With respect to q-expectations vs. q-probability distributions, I think we actually agree and I was just speaking imprecisely. I'm happy to replace every instance of "the q-probability distribution is tightly centered around $x$" with "the q-expectation is $x$ and the q-variance is small"; I didn't mean to imply that the q-probability distribution necessarily has anything to do with the probability that the pointer needle is in a particular place. But it seems like you do want the q-variance to be small in order for $\langle X^E\rangle$ to be an informative measure of "where the pointer needle actually is". This is my reading, for example, of the discussion on p. 15 of the second paper.

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I think my question has basically been answered, but out of curiosity I'm interested in drilling deeper into the bit about the mixed terms in the q-expectation. Specifically, how important is it to your picture of measurement that we can't assign pure states to macroscopic objects? In a modified TI where the whole universe can always be assigned a pure state, does everything break in the story we're discussing? It seems to me like it might break, although my reason for thinking this could just be a repetition of the same mistake from before, so if you'll indulge me once again I want to lay out my thinking.

Suppose again that the state space of the universe is $H^S\otimes H^E$ with $H^S$ one-dimensional, but suppose now that we can represent the state as a vector $\psi_S\otimes\psi_E$. Fix $\psi_E$ and an observable $X^E$, and let $U$ be the unitary operator which runs time forward enough for the measurement to be complete. Suppose that $\psi_1$ and $\psi_2$ are orthogonal, and that
(i') in the state $\phi_1:=U(\psi_1\otimes\psi_E)$, we have $\langle X^E\rangle=x$ and the q-variance is very small, and
(ii') similarly for $\phi_2:=U(\psi_2\otimes\psi_E)$ and $-x$.

Then $U(\frac{1}{\sqrt{2}}(\psi_1+\psi_2)\otimes\psi_E)=\frac{1}{\sqrt{2}}(\phi_1+\phi_2)$, and we can ask what $\langle X^E\rangle$ is in this state. The answer is $\frac{1}{2}\langle\phi_1+\phi_2,X^E(\phi_1+\phi_2)\rangle$, and we again have "mixed terms" which prevent us from deducing the answer from (i') and (ii').

But in this case, I think the smallness of the q-variance might get us into trouble. In particular, in the extreme case where $\phi_1$ and $\phi_2$ are eigenvectors of $X^E$, the orthogonality kills off the mixed terms (since $\phi_1$ and $\phi_2$ are still orthogonal), and it seems like an extension of this argument should mean that the mixed terms are very small when the q-variances are very close to 0; I think I have most of a proof in mind using the spectral theorem. If this is right, it seems like, in this state, $\langle X^E\rangle$ should be close to 0, meaning it can't be $\pm x$.

Is this right, or am I spouting nonsense again? If so, what is it about only using density matrices that "saves us" from this?

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#### DennisN

Hi @A. Neumaier, @PeterDonis,
I read this post regarding the thermal interpretation and the Stern–Gerlach experiment:
For each individual measurement, random nonlinear fluctuations inside the detector cause the result to be either +1 or -1.
Considering these words, I wonder if the magnets in the experiment are considered part of the detector, or the "detector" is only the screen, so to say.

#### PeterDonis

Mentor
I wonder if the magnets in the experiment are considered part of the detector, or the "detector" is only the screen, so to say.
It's the latter. The magnets are part of the internals of the experiment; all they do is implement a reversible unitary transformation on the quantum state of the electron. There's no decoherence in that part of the experiment. Only the screen has decoherence and random fluctuations, etc.

#### DennisN

It's the latter. The magnets are part of the internals of the experiment; all they do is implement a reversible unitary transformation on the quantum state of the electron. There's no decoherence in that part of the experiment. Only the screen has decoherence and random fluctuations, etc.
Thanks!

#### DennisN

It's the latter. The magnets are part of the internals of the experiment; all they do is implement a reversible unitary transformation on the quantum state of the electron. There's no decoherence in that part of the experiment. Only the screen has decoherence and random fluctuations, etc.
Hmm, that got me thinking... if I remember correctly, sequential measurements of spin in one direction yield the same result (+1 and then +1 again, or -1 and then -1 again). How does this fit in with the thermal interpretation if it is random nonlinear fluctuations in the screen that cause decoherence*?

* From the previous quote:
For each individual measurement, random nonlinear fluctuations inside the detector cause the result to be either +1 or -1.
I'm confused.

#### PeterDonis

Mentor
sequential measurements of spin in one direction yield the same result (+1 and then +1 again, or -1 and then -1 again)
The term "sequential measurements" is a misnomer here, because in the process you're describing, there are not multiple detector screens, there's only one. The electron just passes through multiple Stern-Gerlach magnets before it gets to the one detector screen. So there is only one measurement being made in this process.

"Sequential measurements" properly interpreted would mean passing an electron through one Stern-Gerlach magnet, then have it hit a detector screen, then somehow take the same electron and pass it through another Stern-Gerlach magnet and have it hit a second detector screen. But that's not actually possible because once the electron hits the first detector screen it gets absorbed and you can't manipulate it any more.

How does this fit in with the thermal interpretation if it is random nonlinear fluctuations in the screen that cause decoherence*?
Because there's only one screen. See above.

"Confusion about the thermal interpretation's account of measurement"

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