Why quantum entanglement doesn't allow faster-than-light communication (2016)

Charlie doesn't see anything. He sends the electrons here and there. He's just produced the entangled electrons, he hasn't measured them. If he did he would destroy the entanglement and ruin the experiment (which is what secure quantum communication is about).

Unless he gets some reply (say a photon or electron sent by Bob), he doesn't know what either measure.

But if he does get a return particle then they're just communicating classically, so why not just pick up a phone?

Stated differently, the framing I saw was that the "spooky" action was somehow setting detector C to a specific setting would cause a different reading in detector B. And this was done in such a way that B could not know that C had changed. But, simply getting a new reading at B means that either A or C has changed, necessarily?

And again, going off old memory. Never my area of study, such that I assume I am misunderstanding. It is frustrating because most "pointing out my mistake" assumes I care about individual protons. I'm saying if we can agree to have A set to send with constant rate, then barring that getting broken, it seems you have a scheme whereby B and C can know what they are doing in aggregate.

Framing I saw was more of a truth table like where B/C have known states they can be in that each lead to a known distribution of outcomes. It was not clear that the known distribution was only an observed distribution.

> t^2 - x^2 - y^2 - z^2

Off by a constant factor (c).

> distance = [above]

distance^2.

This is still not quite right since we're interested in how coordinate changes affect the spacetime interval. So take distance -> S, the spacetime interval instead of the Euclidean distance, and then sprinkle in the deltas: dS^2 = c dt^2 - dx^2 - dy^2 - dz^2. That is the Minkowski metric (the flat metric of special relativity) in Minkowski coordinates.

The right hand side can be positive (for a timelike spacetime interval), negative (for a spacelike spacetime interval), or zero (for a null spacetime interval). Massless particles, when not accelerated, travel on null spacetime intervals, as do other solutions of massless wave equations. Light is rarely accelerated. However a null interval does not mean no distance, it only means that there is an exact balance between the spacelike portion of the spacetime interval and the timelike portion of the spacetime interval. An object moving against the t and x coordinates at speed "c" will (choosing units for clarity) tick off one second along the t-axis for every light-second it travels along the x-axis. A timelike trajectory ticks off more than one second per light-second along the x-axis; and a spacelike trajectory ticks off fewer seconds per light-second along the x-axis.

So, with light always in inertial motion, and always moving at "c", the spacetime interval for an element of light (classical or quantum) will always be zero. That does not imply in any way that the element of light does not travel at all. Also it does not imply that if we split the spacetime interval into timelike and spacelike components, that we find that the photon experiences "no time".

The confusion arises frequently because the usual way one splits a timelike spacetime interval (e.g. for a massive particle) into its timelike and spacelike components is to use the proper time interval (and that leaves us with proper lengths). Proper time is a specific affine time which can be caluclated for non-null spacetime intervals, but not for null spacetime intervals. We can use a different affine time for those instead (leaving us with affine lengths that are not "proper lengths"). But even so there are changes in coordinates (e.g. (t=0,x=0)->(t=10,x=10); the photon is ten seconds older and ten light-seconds further away along x).

[Experts who already know about invariance under coordinate transformations etc. should also know not to pick too hard at that last sentence. The point is that if we do c \def = 1, then 1dt^2 - dx^2 = 0 for arbitrary photon displacements along x and this is true at least up to rescaling.]

Refute what? I'm pointing out that classical (non-quantum) scratch-off lottery tickets can have the property you described. This is only thematically related to the CHSH game, and I'm rather confused as to what you're trying to say.

The Wikipedia articles about the CHSH inequality are IMO quite bad. The conclusion of CHSH has approximately nothing to do with measurement angles. Here's what CHSH is saying, based on one formulation, which is IMO far, far nicer than Wikipedia's and makes the same point.

1. There's a game that two players could play (with the help of a neutral third party). The neutral third party picks random bits x and y, tells player A the value of x and tells player b the value of y. Each player then chooses a play (A's play is a and B's is b) according to whatever strategy they like, except that they can't communicate. They are allowed to communicate with each other beforehand, though. You can look at sites like this for more details about the game works: https://circles.math.ucla.edu/circles/lib/data/Handout-2987-...

2. If A and B are only allowed to choose random numbers (jointly before the game and separately once it starts) and do local calculation, then they cannot win with more than a certain probability. This is the CHSH inequality, and it's a statement about classical statistics.

3. If A and B are allowed to produce an entangled state before the game and each hold on to one part of the state, then they can make measurements on their parts once the game starts and use those measurements as part of a better strategy. This strategy wins with probability higher than the CHSH inequality allows.