# A Value of observable in statistical ensemble interpretation

1. Apr 11, 2017

### Demystifier

The statistical ensemble interpretation (SEI) is supposed to be a minimal interpretation of QM with the smallest amount of philosophy, vagueness and controversy. Yet it turns out not to be the case. For instance Ballentine, the inventor of SEI, interprets Bell theorem as a strong evidence of non-locality. By contrast, @vanhees71 , the strongest supporter of SEI on this forum, disagrees. Even though vanhees71 explained his position many times, many share the feeling that he never explained it sufficiently clearly.

In a more recent thread (which is now closed) it crystallized that one of the the most problematic issues in SEI is the concept of value of an observable. Since values of observables are numbers that are obtained in experiments, it's very important to have a clear interpretation of those values. In this thread I hope we can clarify them.

For that purpose let me start with a sharp question addressed to @vanhees71 . According to SEI, when does the value of an observable exist?

1) Value of an observable never exists.
2) Value of an observable always exists.
3) Value of an observable exists when and only when it is measured.
4) Value of an observable exists when and only when it can be predicted with certainty.
5) The question doesn't make sense.
6) Something else (what?).

From previous explanations by vanhees71 I suspect that he will choose answer 4). For that case I have additional questions. Predicted by who? What if one observer can predict it and the other can't? Does it mean that the existence of value is subjective? As a response to those questions, I expect that he would agree with the following refined version of 4):

4-refined) If there is at least one observer (say Alice) who can predict the value with certainty, then she has a right to tell that the value exists in an objective sense.

So far so good, but now I claim that 4) and 4-refined) contradict locality. Let the state initially at time $t_0$ be prepared in the EPR state
$$|A1\rangle|B1\rangle+|A2\rangle|B2\rangle$$
Then 4) implies that values of $A$ and $B$ do not exist at $t_0$. Next let Alice measure $A$ at time $t_1$ and let she obtain the value $A1$. Then, according to 4) and 4-refined), she can tell not only that $A$ has value $A1$, but also that $B$ has value $B1$. But when did $B$ started to has it's value? Since it did not have a value at $t_0$, it must have started to have the value at $t_1$. Since this value is objective by 4-refined), there must have been some objective change with the $B$-system at time $t_1$. But the only possible cause of that change can be the measurement of system $A$, which is spatially separated from system $B$. Therefore locality must be violated. Q.E.D.

Yet vanhees71 denies violation of locality in this case. I hope he can explain why exactly does he think that locality is not violated. He explained it many times before, but apparently he has never done it sufficiently clearly. If he doesn't agree with 4), the he should pick some other answer on the list 1)-6). If we agrees with 4) and 4-refined), the he should explain what exactly is wrong with my argument above that it violates locality. If he agrees with 4) but not with 4-refined), then he should refine 4) in his own terms.

vanhees71 likes to say that correlations between the observables existed even before the measurement. But that's not the issue here. The issue are the values of the observables, not their correlations.

Finally, let me mention that Einstein et al in the original EPR argument assumed something similar to 4) and locality, and from those assumptions they derived a contradiction. From the contradiction they concluded that QM was incomplete.

Last edited: Apr 11, 2017
2. Apr 11, 2017

### vanhees71

I think our discussion starts to get in circles. It starts right away with your first question: "According to SEI, when does the value of an observable exist?"

I've no clue, what could be meant by this. Perhaps it helps to explain, what I mean, again. Imho it's a very simple robust view on the meaning of the quantum formalism. One has to strictly distinguish between observables and values of observables to begin with:

An observable $A$ is defined as an equivalence class of measurment procedures. E.g., for the spin of a neutral particle, atom, etc. you can take the Stern-Gerlach apparatus as one realization of the observable "spin" (within non-relativistic QT). Outcomes of (idealized measurements) are values of observables.

In the formalism $A$ is represented by a self-adjoint operator $\hat{A}$, and quantum theory tells you that the possible values of the observable (i.e., outcomes of precise measurements of the observable) are given by the spectrum of this operator.

A quantum state is defined as an equivalence class of preparation procedures.

A quantum state is represented in the formalism by a statistical operator (self-adjoint positive semidefinite operator of trace 1) $\hat{\rho}$. Its meaning and its only meaning is that the probability to find the value $a$ (in the spectrum of $\hat{A}$) of the observable $A$ is given by
$$P(a)=\sum_{\beta} \langle a,\beta|\hat{\rho}|a,\beta \rangle,$$
where $\beta$ labels a complete orthonormal basis of the eigenspace of $\hat{A}$ to the eigenvalue $a$ (of course, you can generalize this to generalized "eigenstates" for the case of continuous spectra and continous labels $\beta$, but let's keep the formalism as simple as possible here).

Now there are two possibilities:

(1) There's a value $a$ in the spectrum of $\hat{A}$ such that $P(a)=1$: Then the state is such that the observable $A$ takes the determined value $a$.
(2) For all values $a$ in the spectrum of $\hat{A}$ one has $P(a)<1$: Then the state is such that the observable $A$ doesn't take any determined value, i.e., it's value is indetermined, and the probability to obtain the value $a$ when measuring $A$ is given by $P(a)$.

Values can thus be determined or indetermined according to QT. It doesn't make sense to say it exists or it doesn't exist. An observable by definition of course always "exists", because it's defined by a measurement procedure, that you can always apply to the system in question, no matter in which state it might be prepared. If you mean by "the value exists" in fact that "the value of $A$ is determined by the state preparation to be $a$", then the answer is of course (4). If understood in this sense, then "the value exists" or "the value doesn't exist" (i.e., the value is determined or it is not determined, respectively) not because of measuring it but because of the preparation procedure of the system in a state. Although state preparation often involves measurements, one must not mix up measurement and preparation (i.e., observables and states, respectively).

As to the question of locality, I don't know what Ballentine means, when he says that the successful prediction of the outcome of Bell tests by QT means the violation of locality. By locality I understand how it is defined in the usual successful relativistic QFTs, i.e., locality of interactions (Lagrangians/Hamiltonians built by local polynomials of field operators (including their derivatives) in a locally realized representation of the Poincare group) and microcausality (local observables like energy, momentum, angular-momentum densities etc. commute if the arguments of the corresponding operators (built from the field operators and their derivatives) are space-like separated). Then there cannot be any instantaneous interaction between any local measurement at A and the system measured at a space-like separated measurement event at B.

In the above discussed example of entanglement (let's think of spin components of two particles resulting from the decay of a $J=0$ state to simplify language) there are correlations for the outcomes of measurements for the spin components of each of the single particles, although no single-particle spin component has a determined value. These correlations are due to the preparation of the particles in the entangled state (in this case the preparation consists of putting an unstable scalar boson at rest with sufficient precision somewhere and wait until it decays) before any measurements at A and B are done. The correlations are not caused by measurement of one or both of the single-particle spins. That there is no cause-effect relation between such measurements is concluded due to the fact that (a) zillions of Bell tests have been done where with great care it was ensured that the measurment events ("clicks" in the detectors) where space-like separated and that there is not any hidden correlation between the measurement apparati and that (b) the outcomes are correctly described with local microcausal QFT (e.g., in the case of photons, where it's QED).

Of course, you cannot draw this conclusion from experiment alone, because you need the assumption that there is no action at a distance in the sense of the linked-cluster principle valid for local microcausal QFTs.

3. Apr 11, 2017

### Demystifier

You are probably right about that. As far as I can tell from your post above, you agree with my 4), you did not specify whether you agree with 4-refined) (which you probably do), and you did not explain what exactly is wrong (if anything) with my argument that 4) and 4-refined) lead to non-locality. You gave an independent argument for locality, but you did not discuss my argument for the opposite.

4. Apr 11, 2017

### vanhees71

I agree with 4-refined, and I think I've given a crystal-clear argument, why this does not violate any locality in the above explained sense. Alice (a) knows that the particle pair was prepared in the entangled state and (b) has measured here observable. Then it follows which value Bob must find in his measurement of this observable (if both measure in my example the spin component in the same direction). If B's particle is far away, then A's measurement cannot do anything to it for some time. All that changed is A's information about the value her particle's spin component and thus she knows what B will measure (or has already measured) due to the preparation of the particle pair in the entangled state, implying the 100% correlation, encoded in the entangled state.

5. Apr 11, 2017

### Demystifier

Clearly, Ballentine means violation of locality in a different sense. The local interactions you mention above do not describe the actual values of observables in a single run of an experiment. The local interactions describe the ensemble, not a single run. It is those actual single-run values that, according to Ballentine, must obey non-local laws. That's why I constantly insist on concentrating on the concept of values.

6. Apr 11, 2017

### Demystifier

But it cannot be all that changed, if 4) and 4-refined) are true. Namely, according to 4) and 4-refined), the B-value did not even exist before that, and now it suddenly exists. So it's not only information about the value that has changed, but also the value itself has changed (from non-existence into existence).

Last edited: Apr 11, 2017
7. Apr 11, 2017

### dextercioby

So how do you prepare an entangled quantum state with 2 observers one kilometer apart? I'm curious to know how this is physically done.

8. Apr 11, 2017

### atyy

vanhees71 means exist in the mind. So for example, the universe can exist and not exist simultaneously.

9. Apr 11, 2017

### bhobba

I think the ensemble interpretation is silent on if anything exists prior to observation or not. I don't think it says it doesn't exist, although it may be the case, or may not.

Witness Ballentines original paper:

Now somewhere in there, I cant actually recall where, it more or less implies its BM, or something similar, in disguise. Since then his textbooks have corrected the 'issue'.

Thanks
Bill

Last edited: Apr 11, 2017
10. Apr 11, 2017

### Demystifier

11. Apr 11, 2017

### Demystifier

Yes, the minimal version of ensemble interpretation would be silent about that. But vanhees71 is not silent about that. In one post he objected that something should exist before observation, because otherwise it would be solipsism (and I agree).

Last edited: Apr 11, 2017
12. Apr 11, 2017

### vanhees71

That's strange, because I thought that within the statistical interpretation you take Born's rule seriously and thus states refer to ensembles, because they provide nothing else than probabilities, and probabilities can be addressed experimentally only via ensembles.

13. Apr 11, 2017

### vanhees71

Again, I do not use this formulation "a value exist". It doesn't make sense! The observable exists, and it can take determined values or not, depending on the state the system is prepared in. The value has not changed. For Bob it's still as indetermined as before, while Alice knows the outcome of Bob's observation from her measurement. Thus only Alice's information has changed.

14. Apr 11, 2017

### Demystifier

Well, I can't imagine how seriously one should take the Born rule and ensembles to deny the existence of single runs of an experiment. Apparently, Ballentine does not take them that seriously. Do you?

Maybe you are really an adherent of the Mermin's Ithaca interpretation, the slogan of which is: Correlations exist, but not the correlata.

15. Apr 11, 2017

### vanhees71

Of course single runs of experiments exist. That's how experiments are done. I don't understand what this has to do with QT, given the definition of what an observable and a state is (namely real-world measurement and preparation procedures in the lab, not abstracta in Hilbert space).

16. Apr 11, 2017

### DrDu

If I throw a coin, but don't look at the result, I have some information about the outcome (namely a chance of 1/2 for each value). In the sense of statistics, the situation is the same if I only imagine having thrown a coin. I think you are trying to construct a difference between the two situations which isn't there even in classical statistics. From all experiments, we only get probabilities, although sometimes these are 100%, but we don't get a qualitative difference between existence and non-existence.

17. Apr 11, 2017

### Demystifier

Well, it's only a matter of language. Whenever I say "value exists" you can translate it as "the observable has a determined value". Whenever I say "value does not exist" you can translate it as "the observable does not take a determined value". My proof of non-locality can be repeated with that translation without any other changes.

18. Apr 11, 2017

### Demystifier

It looks as if you don't accept 4) in the first post. Instead, it seems that you accept something like 2). Fine, in that case the proof of non-locality is not so simple. But non-locality can still be proved, as done e.g. by Bell, or by GHZ, or (my favored proof) by Hardy.

19. Apr 11, 2017

### vanhees71

Ok, if you proved non-locality, then what's the observable consequence of it? Where do you have proven relativistic local microcausal QFT to be observably wrong? To the dismay of almost all HEP physicists there's no such observation in sight at the moment!

20. Apr 11, 2017

### Demystifier

Perhaps you are right that single runs have nothing to do with QT. But single runs definitely have something to do with Nature. So are you saying that there is something about Nature which is not contained in QT?

Anyway, the question is not whether QT is non-local. The question is whether Nature itself (which violates Bell inequalities) is non-local.