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According to a popular definition

a system is called "entangled" if its quantum state cannot be factored as a product of states of its individual distinct ("local") constituents (e.g. individual particles).

The corresponding quantum state of an entangled system of two constituents would be expressed in general as

$$|\psi\rangle := \sum_{j = 1}^{N_A}~\sum_{k = 1}^{N_B}~c_{(j, k)}~|\phi_j^A\rangle \otimes |\phi_k^B\rangle,$$

where

(1): $|\phi_j^A\rangle$ denotes any one of the $N_A \ge 2$ pairwise ortho-normal eigenstates of operator $\hat A$ applied to identify one constituent (e.g. by "detector $A$"),

(2): $|\phi_j^B\rangle$ denotes any one of the $N_B \ge 2$ pairwise ortho-normal eigenstates of operator $\hat B$ applied to identify the other constituent (e.g. by "detector $B$", which is explicitly distinct from "detector $A$", for instance by detectors $A$ and $B$ being explicitly prescribed as separated from each other),

(3): state $|\psi\rangle$ is normalized, $\langle\psi|\psi\rangle = 1$; in particular there is at least one non-zero coefficient $c_{(j, k)}$, and

(4): there are no $N_A + N_B$ numbers $r_q$ to be found such that all coefficients could be expressed as factors $c_{(j, k)} := r_j~r_{(N_A + k)}$.
(Consequently there are at least two non-zero coefficients in the expression of $|\psi\rangle$.)

Now, considering only any one (valid) trial separately, there is a definite identification of the one constituent provided (by detector $A$) as, say, $|\phi_j^A\rangle$, and a definite identification of the other constituent provided (by detector $B$) as, say, $|\phi_k^B\rangle$.

This one trial might be considered one instance of an ensemble of $N$ trials whose quantum state is an entangled state as described above, with $c_{(j, k)} \ne 0$.

But this one trial might also (or instead) be considered one instance of an ensemble of $N$ trials whose quantum state is the same in all instances (perhaps aside from an incidental phase factor); thereby failing the consequence of condition (4).

Or this one trial might also (or instead) be considered one instance of an ensemble of $N$ trials whose quantum state is "mixed" but not an entangled state either, for failing condition (4) itself.

Therefore I wonder:
Is it appropriate to say, as apparently it is popular to do, that

a subatomic particle decays into an entangled pair of other particles

?

Is "having been entangled" a conclusive description of constituents in only any one (valid) trial?

Can the appropriateness (or lack thereof) of such a description referring to only one (valid) trial perhaps be quantified, e.g. in terms of numbers $N_A$, $N_B$ and $N$?

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This one trial might be considered one instance of an ensemble of $N$ trials whose quantum state is an entangled state as described above, with $c_{(j, k)} \ne 0$.

It not about considering. If $|\psi\rangle := \sum_{j = 1}^{N_A}~\sum_{k = 1}^{N_B}~c_{(j, k)}~|\phi_j^A\rangle \otimes |\phi_k^B\rangle,$ is the state it was in and then you did a measurement, its in a new state now (post measurement) and it was in that original (entangled) state prior to the measurement.

But this one trial might also (or instead) be considered one instance of an ensemble of $N$ trials whose quantum state is the same in all instances (perhaps aside from an incidental phase factor); thereby failing the consequence of condition (4).

This makes no sense. The entangled state is definite state of the system. And you can prepare systems in that state over and over again and do lots of different measurements.

If you are going to use one fixed measurement type over in A and one fixed measurement type in B then you won't be able to tell it is in that state. You might think someone just flips a coin and $\frac{|c_{(j, k)}|^2}{\sum_{j = 1}^{N_A}~\sum_{k = 1}^{N_B}~|c_{(j, k)}|^2}$ fraction of the time puts it into the state $|\phi_j^A\rangle \otimes |\phi_k^B\rangle.$

The way you tell it was in the state $|\psi\rangle := \sum_{j = 1}^{N_A}~\sum_{k = 1}^{N_B}~c_{(j, k)}~|\phi_j^A\rangle \otimes |\phi_k^B\rangle$ is by considering lots of potential different types of measurements.

Or this one trial might also (or instead) be considered one instance of an ensemble of $N$ trials whose quantum state is "mixed" but not an entangled state either, for failing condition (4) itself.

If you don't do the experiments to find out which state it is in and instead stick to one basis to measure on, then sure anything (entangled or not) could just be in a mixed state of eigenvectors of that operator.

Is it appropriate to say, [...], that a subatomic particle decays into an entangled pair of other particles?

Its as appropriate as aging anything is ever in any pure state. And in case that isn't clear. It's 100% totally appropriate and in fact the only way to agree with all possible measurements from a set of operators that don't commute. The pure state was designed to be the thing that gives the correct predictions for any possible measurement. And we would have used mixed states of definite values if that would have worked for all measurements. But it doesn't.

Is "having been entangled" a conclusive description of constituents in only any one (valid) trial?

Being entangled is about the state prior to measurement.

No one trial tells you the exact state prior to the measurement. All it tells you is that the state prior to measurement has a nonzero orthogonal projection onto the eigenspaces of that operator you just measured that one time. You don't know how big it until you do many. And even then if you just use that one operator you'll never be able to distinguish between a pure and a mixed state.

And the previous paragraph is a general fact, that has no bearing on entanglement except that an entangled state is a definite state and hence like all definite states no single operator can tell you it is in that state, not even with many measurements. Unless the state happens to be an eigenvector of that operator.

Can the appropriateness (or lack thereof) of such a description referring to only one (valid) trial perhaps be quantified, e.g. in terms of numbers $N_A$, $N_B$ and $N$?

No. Doing just one trial for one fixed operator doesn't require specifying the true state, just a list of possible eigenvalues. And its not about appropriate it like not looking at something and then saying it might be invisible. Your lack of effort to try to learn about something just tells us about how much effort you put into it. It doesn't tell us about the thing. Literally, it doesn't tell us about the thing.

"Being entangled is about the state prior to measurement." -- That's unacceptable. Even "preparing" an ensemble is about first measuring, and then discarding trials (as "invalid") which didn't match the prep. prescription.

Sometimes you do that, but that isn't the only way. Your entangled state could be a decay channel for an unstable system. If fact it is common to have decay products be in an entangled state. The word measurement is just a (rather silly) name for certain kinds of interactions. And it doesn't matter how you prepare the system. A minority say that a state merely or only represents an ensemble. And sure, a mathematical state is only a mathematical representation of a physical system.

But there is simply no empirical evidence that individual systems do not have a physical state, and that a physical state is exactly some physical thing that produces particular physical results with particular observable frequencies.

Its that highest levels idea that our mathematical models make predictions that agree with observations that is being tested. After all, you might not repeat that particular measurement but if every measurement on every system always produced results that your theory claimed were unlikely then the entire theory (QM) would be rejected even if you absolutely never did the exact same measurement twice (say for budget reasons).

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  • $\begingroup$ Timaeus: Thanks for your detailed and notably rapid answer; +1. "The way you tell it" ... "It": the (suitably numerous) ensemble ... "was in the [entangled] state [...] by considering lots of potential different types of measurements." -- Right (not least, to determine ratios between coefficients). Thus: just one trial is not enough for this characterization. "Being entangled is about the state prior to measurement." -- That's unacceptable. Even "preparing" an ensemble is about first measuring, and then discarding trials (as "invalid") which didn't match the prep. prescription. $\endgroup$ – user12262 Oct 29 '15 at 22:49
  • $\begingroup$ @user12262 Edited $\endgroup$ – Timaeus Oct 29 '15 at 23:25
  • $\begingroup$ Timaeus: "Edited. [...] common to have decay products be in an entangled state." -- Sure. Does this impede distinguishing the state attributed to the products of just one decay from the state atttributed to products of an ensemble of decays? "there is simply no empirical evidence that [...]" -- Claims of "evidence" (or lack thereof) can only be convincing if it's already agreed what constitutes (and how to gather) "evidence". $\endgroup$ – user12262 Nov 12 '15 at 19:53
  • $\begingroup$ Timaeus: "The word measurement is just a (rather silly) name for certain kinds of interactions." -- No (that'd be silly), but moreover: to record and to evaluate relevant interactions (or their absence); noting "with whom" and "in which sequence (or in coincidence)", and applying subsequent measurement/evaluation operators. "[Is it possible to quantify ...] No." -- Well, there seem to be some attempts after all, such as this (though admittedly I haven't figured out yet how this particular approach might be applied to my specific question). $\endgroup$ – user12262 Nov 12 '15 at 19:53

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