Confusion about the thermal interpretation's account of measurement

[PeterDonis]

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.

The two competing answers I'm worried about are:
(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]

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]

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.

The two competing answers I'm worried about are:
(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]

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.

----------------------------------------------

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?

 

[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]

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]

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.

 

[microsansfil]

It's a property of the system like angular momentum in Classical Mechanics.

However, this is deducted from the calculation of a mathematical expected value !

http://www.cithep.caltech.edu/~fcp/physics/quantumMechanics/densityMatrix/densityMatrix.pdf

/Patrick

 

[A. Neumaier]

However, this is deducted from the calculation of a mathematical expected value !

The calculation of a q-expectation. This has in the thermal interpretation the meaning of a property of the individual system, not that of an ensemble expectation or a time average.

 

[DarMM]

However, this is deducted from the calculation of a mathematical expected value !

Yes, but this expectation is interpreted as a property itself not the long term average of observations.

 

[microsansfil]

Hi,

What is the reason for this interpretation? In quantum optics, a coherent state is called semi-classical because the expectation value of the quadrature (position, momentum) of the system oscillates like the position of a classical simple harmonic oscillator.

/Patrick

 

[A. Neumaier]

Hi,

What is the reason for this interpretation? In quantum optics, a coherent state is called semi-classical because the expectation value of the quadrature (position, momentum) of the system oscillates like the position of a classical simple harmonic oscillator.

/Patrick

The thermal interpretation is motivated from statistical thermodynamics (global and local equilibrium), where the observed macroscopic quantities are expressed microscopically as expectation values. There were several earlier discussions here on PF.

 

[A. Neumaier]

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?

That we can't assign pure states to macroscopic objects is a question of consistency.

The state of the universe determines every other state by taking partial traces. This doesn't preserve pureness; hence the state of a general quantum system is mixed. Indeed, in experimental practice one can make only those systems pure that have either very few degrees of freedom or widely separated energy levels. This is never the case for a macroscopic system.
Thus all large systems have mixed states. The only possible exception would be the whole universe, as its state is not a partial trace of something bigger. But why make an exception for the universe?

 

[akvadrako]

But why make an exception for the universe?

If there is no intrinsic randomness in nature then there can’t be any uncertainty about the state of the universe. So what would it mean for the universe to be in a mixed state?

 

[A. Neumaier]

If there is no intrinsic randomness in nature then there can’t be any uncertainty about the state of the universe. So what would it mean for the universe to be in a mixed state?

Subjective uncertainty about the state is not part of quantum theory, which is about what is objectively true, and not about which knowledge human beings have about what is objectively true.

In the thermal interpretation, a mixed state is therefore not a sign of uncertainty about the state. Indeed, the state of every system is determined by the state of the universe by taking the appropriate partial trace. Note that essentially every state is mixed. Even the states that we regard as pure are in fact only approximately pure. (For example, it is impossible to determine exactly the direction of a magnetic field in a Stern-Gerlach experiment, but it would be needed to prepare an exact spin up particle.)

Uncertainty about a state just means having an inaccurate state instead of the true one, just as uncertainty about the value of a measurement result (e.g., the number of atoms in a piece of matter).

 

[DarMM]

If there is no intrinsic randomness in nature then there can’t be any uncertainty about the state of the universe. So what would it mean for the universe to be in a mixed state?

Slightly related, it's actually an open question in QFT whether the global state can be a pure state. We already know states of matter in a finite volume must be mixed in QFT.

 

[akvadrako]

Subjective uncertainty about the state is not part of quantum theory, which is about what is objectively true, and not about which knowledge human beings have about what is objectively true.

In the thermal interpretation, a mixed state is therefore not a sign of uncertainty about the state. Indeed, the state of every system is determined by the state of the universe by taking the appropriate partial trace. Note that essentially every state is mixed. Even the states that we regard as pure are in fact only approximately pure. (For example, it is impossible to determine exactly the direction of a magnetic field in a Stern-Gerlach experiment, but it would be needed to prepare an exact spin up particle.)

Uncertainty about a state just means having an inaccurate state instead of the true one, just as uncertainty about the value of a measurement result (e.g., the number of atoms in a piece of matter).

Okay, but how does that apply to the universe? If it’s not uncertainty and there is nothing to trace over, how does one interpret the mixed state?

 

[DarMM]

Okay, but how does that apply to the universe? If it’s not uncertainty and there is nothing to trace over, how does one interpret the mixed state?

In the thermal interpretation what is denoted as a mixed state is just another possible state of the system no different than a pure state.

 

[nicf]

Thus all large systems have mixed states. The only possible exception would be the whole universe, as its state is not a partial trace of something bigger. But why make an exception for the universe?

Yes, I think I understood this part already; this motivation makes sense to me and I agree with it. I realize now that my last post was too long and not very clear, so I'll try to say it with fewer words :).

It is commonly argued that the measurement problem can't be solved with ordinary quantum mechanics. People who argue this find a situation where, by the linearity of time evolution, the universe has to end up in a superposition of both possible measurement outcomes, so the final state doesn't tell us "which one actually happened".

You reject this picture in favor of one where what's important is the q-expectation of the observable $X^E$ that records the macroscopic result of the experiment. The presence of mixed terms like (in our earlier notation) $\psi_1\psi_2^*\otimes\rho^E$ mean that we can't extract the result of measuring $\frac{1}{\sqrt{2}}(\psi_1+\psi_2)$ from the results of measuring $\psi_1$ and $\psi_2$.

This explanation depends on the fact that these mixed terms can contribute a large amount to the q-expectation of $X^E$. Furthermore, it seems like we would also like, in all three states, the q-variance of $X^E$ to be small. I'm worried these two desires might be incompatible, that requiring the q-variances to be small might also force the mixed terms to be small. The intuition I'm working from comes from pure states: in the extreme case where the final states are eigenstates of $X^E$, the mixed terms are zero, and for pure states that are merely close to being eigenstates we should be able to make the mixed terms small, which also seems bad.

The question I'm asking is more a mathematical one than a physical one; I realize we have good physical grounds to disallow pure states, but I'm interested in exactly how that prescription interacts with the picture of measurement you're advocating.
(a) If, hypothetically, the thermal interpretation did allow pure states, would the thing I just described be a problem, or did I make a mistake again?
(b) If the answer to (a) is yes, why does the problem go away when we disallow pure states?

I think what might help me is an explicit example. How hard is it to construct a $\rho^E$, $X^E$, $\psi_1$, and $\psi_2$ that behave this way after applying some unitary map?

 

[A. Neumaier]

Yes, I think I understood this part already; this motivation makes sense to me and I agree with it. I realize now that my last post was too long and not very clear, so I'll try to say it with fewer words :).

It is commonly argued that the measurement problem can't be solved with ordinary quantum mechanics. People who argue this find a situation where, by the linearity of time evolution, the universe has to end up in a superposition of both possible measurement outcomes, so the final state doesn't tell us "which one actually happened".

You reject this picture in favor of one where what's important is the q-expectation of the observable $X^E$ that records the macroscopic result of the experiment. The presence of mixed terms like (in our earlier notation) $\psi_1\psi_2^*\otimes\rho^E$ mean that we can't extract the result of measuring $\frac{1}{\sqrt{2}}(\psi_1+\psi_2)$ from the results of measuring $\psi_1$ and $\psi_2$.

This explanation depends on the fact that these mixed terms can contribute a large amount to the q-expectation of $X^E$. Furthermore, it seems like we would also like, in all three states, the q-variance of $X^E$ to be small. I'm worried these two desires might be incompatible, that requiring the q-variances to be small might also force the mixed terms to be small. The intuition I'm working from comes from pure states: in the extreme case where the final states are eigenstates of $X^E$, the mixed terms are zero, and for pure states that are merely close to being eigenstates we should be able to make the mixed terms small, which also seems bad.

The question I'm asking is more a mathematical one than a physical one; I realize we have good physical grounds to disallow pure states, but I'm interested in exactly how that prescription interacts with the picture of measurement you're advocating.
(a) If, hypothetically, the thermal interpretation did allow pure states, would the thing I just described be a problem, or did I make a mistake again?
(b) If the answer to (a) is yes, why does the problem go away when we disallow pure states?

I think what might help me is an explicit example. How hard is it to construct a $\rho^E$, $X^E$, $\psi_1$, and $\psi_2$ that behave this way after applying some unitary map?

All these problems are mathematically difficult. You conjecture a particular problem based on informal worries, I conjecture its absence based on seeing what the quantum formalism is known to achieve and the experimental record requires.

Note that by definition, the measured macroscopic value is the q-expectation of the pointer variable, independent of the value of its uncertainty. Whether the latter is small is secondary, though it seems to be automatically the case for macroscopic systems.

 

[akvadrako]

In the thermal interpretation what is denoted as a mixed state is just another possible state of the system no different than a pure state.

But doesn’t it mean it can be written as a mixture of several disconnected pure states? So you can consider each of those pure states as separate universes that evolve independently.

 

[A. Neumaier]

But doesn’t it mean it can be written as a mixture of several disconnected pure states? So you can consider each of those pure states as separate universes that evolve independently.

By the same argument you can postulate that the possibility of writing 5=2+3 can be considered to mean that there are different universes in which 5 is realized as 2 amd 3.

The thermal interpretation declares such statements to be nonsense.

 

[vanhees71]

That we can't assign pure states to macroscopic objects is a question of consistency.

The state of the universe determines every other state by taking partial traces. This doesn't preserve pureness; hence the state of a general quantum system is mixed. Indeed, in experimental practice one can make only those systems pure that have either very few degrees of freedom or widely separated energy levels. This is never the case for a macroscopic system.
Thus all large systems have mixed states. The only possible exception would be the whole universe, as its state is not a partial trace of something bigger. But why make an exception for the universe?

Well, you should be able to answer this within your thermal interpretation, because you claim to have a physically testable meaning about single systems. Within the standard statistical interpretation the idea of a "state of the entire universe" doesn't make sense, at least I've never understood the attempts to define such a mathematical fiction.

Maybe, one day there's an answer to that question, when we have a complete understanding of gravity and spacetime that is consistent with quantum theory.

As long as that, we have to take "cosmology" as what it is: an extrapolation of local observations and thereby derived natural laws, extrapolated to the "entire universe" (or to be more careful the piece of the universe that is at least in principle observable to us). This works indeed not too badly and there has been a lot of progress in the past few decades, but that's what it is: An extrapolation of locally discovered physical laws considered universal by hypothesis (cosmological principle).

 

[DarMM]

But doesn’t it mean it can be written as a mixture of several disconnected pure states? So you can consider each of those pure states as separate universes that evolve independently.

I think it would view this as no different to how in classical electromagnetism states of the electromagnetic field are sums of other states of the field.
Even within interpretations of QM modern forms of Copenhagen would be similar, viewing mixed states as simply states of higher entropy.

In QFT mixed states of finite volume systems aren't sums of pure states so viewing them like that is difficult to maintain.

 

[vanhees71]

The von Neumann-Shannon-Jaynes entropy is defined in QT (no matter whether it's formulated as a QFT or not) is given by (in SI units)
$$S=-k_{\text{B}} \mathrm{Tr}(\hat{\rho} \ln \hat{\rho}).$$
One can show that it is 0 if and only if $\hat{\rho}=|\psi \rangle \langle \psi|$, for some normalized vektor $|\psi \rangle$, and that's by definition representing a pure state.

I don't understand your last sentence.

Since $\hat{\rho}$ is a self-adjoint operator, it provides a (generalized) orthonormal eigenbasis and thus you can write it as sum (and/or integral) of the form
$$\hat{\rho}=\sum_{k} P_k |k \rangle \langle k|.$$
Again, it doesn't matter whether your QT is formulated as a QFT or not.

 

[A. Neumaier]

Well, you should be able to answer this within your thermal interpretation, because you claim to have a physically testable meaning about single systems.

Within the thermal interpretation it is considered to be in a mixed state. Indeed, the experimental record tells us that the universe has a positive temperature, whereas no temperature can be assigned to a pure state.

 

[DarMM]

Again, it doesn't matter whether your QT is formulated as a QFT or not.

It does. In QFT the local observable algebra involves type $III_{1}$ factors which have no pure states, thus density matrices for finite volume systems do not decompose like this.

 

[A. Neumaier]

It does. In QFT the local observable algebra involves type $III_{1}$ factors which have no pure states, thus density matrices for finite volume systems do not decompose like this.

For type $III_{1}$ factors and how this, at first counterintuitive fact, may arise in examples, see this post and the surrounding discussion there.