I Correlation setup between past and present

[entropy1]

Suppose A is an ensemble of measurement events in the past.
Suppose B is an ensemble of measurement events in the present.
Suppose there is a correlation between A and B that stays the same over time.
Suppose we can manipulate the outcomes of B (for example by choosing the orientation of the measurement basis or some other theoretical possibility).

If we manipulate B, and the correlation stays the same, can we then influence the outcomes of A?

If not, which of the following two reasons is the main reason:
1) Such a manipulation of B is not possible, OR:
2) There can be no influence to the past.

 

[phinds]

Correlation is not causation. We can't influence the past.

 

[PeterDonis]

Suppose there is a correlation between A and B that stays the same over time.

This doesn't make sense. Each set of events only happens once; there is nothing to "stay the same over time".

Suppose we can manipulate the outcomes of B

Then the correlations between B and A don't exist yet, because the B outcomes haven't happened yet.

If we manipulate B, and the correlation stays the same

This doesn't make sense. See above.

 

[entropy1]

I have this from Peter in another thread:
A measurement in Unitary manner can (according to Peter) be described by:

$$\left( |1> + |2> \right) |R> \rightarrow |1>|U> + |2>|D>$$

The measuring device gets entangled with the observable.

If we are measuring a specific result, we get either $|1 \rangle|U \rangle$ or $|2 \rangle|D \rangle$, right?
So, is the result determined by the eigenstates of the operator ($|1 \rangle$ and $|2 \rangle$) or could the outcome also be determined by the perception of the measured result ($|U \rangle$ or $|D \rangle$) (one goes with the other)?

If the observer gets entangled with the measuring device, and the measuring device got entangled with the observed particle, doesn't the observer get entangled with the observed particle? And if so, who is to say which determines the behaviour of which?

 

[PeterDonis]

I have this from Peter in another thread

This describes a single measurement, not an ensemble of them. It has nothing to do with what you were describing in your OP, which doesn't make sense, for reasons I've already given.

The measuring device gets entangled with the observable

The measuring device gets entangled with the measured system.

If we are measuring a specific result, we get either $|1 \rangle |U \rangle$ or $|2 \rangle |D \rangle$, right?

I don't understand what you mean by "measuring a specific result". The state after the interaction has both terms. But each term describes, not just a different state of the measured system, but a different state of the measuring device, plus everything else that interacts with it, including our brains. (I went into this in a later post in the other thread.) So as far as we observers are concerned, one result happened--but how that comes about depends on which interpretation of QM you adopt.

is the result determined by the eigenstates of the operator ($|1 \rangle$ and $|2 \rangle$) or could the outcome also be determined by the perception of the measured result ($|U \rangle$ or $|D \rangle$) (one goes with the other)?

The two are entangled, so this question doesn't make sense.

 

[entropy1]

@PeterDonis: I was referring to the manipulation of B which may come about by the perception of the outcome of B.

If the observer gets entangled with the measuring device, and the measuring device got entangled with the observed particle, doesn't the observer get entangled with the observed particle? And if so, who is to say which determines the behaviour of which?

 

[PeterDonis]

I was referring to the manipulation of B which may come about by the perception of the outcome of B.

What do you mean by "the perception of the outcome of B"? Where does that happen in the interaction I wrote down?