I Confusion about the thermal interpretation's account of measurement

[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.

 

[vanhees71]

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.

What is considered to be in a mixed state? The entire universe? Well, why not?

 

[A. Neumaier]

What is considered to be in a mixed state? The entire universe?

yes, its state was under discussion. Assigning to the universe a temperature distribution (microwave background) means that the state of the universe is approximately a local equilibrium state and hence a mixed state.

 

[vanhees71]

Well, the microwave background temperature is a very much coarse grained observable. It's about the large-scale "average" structure of the universe, described by an FLRW metric using the above considered assumption of the cosmological principle to extrapolate it to "the entire (observable) universe". In this sense it's a paradigmatic example for the concept to consider coarse-graining to understand the classical behavior of macroscopic observables of a many-body system.