[317070]

When you observe it, you collapse the wave-function of which color the ball has into a particular value (red or blue). Before you observed it, the ball was in a superposition of the two colors. And this collapse instantaneously also collapses the wave-function to the American ball.

Now, that is obviously not true for macroscopic objects like balls. Those are not in a superposition of colors until they are observed, but it is true for quantum objects like electrons.

[AnthonyMouse]

> Now, that is obviously not true for macroscopic objects like balls. Those are not in a superposition of colors until they are observed, but it is true for quantum objects like electrons.

But then what is it that can I do with two entangled electrons that I can't do with two literal billiard balls known to be different colors than one another?

[kgwgk]

You can prepare a pair of photons in a state such that when you measure the polarization of both of them along the same axis, for whatever direction you want to choose, you get the same result. But they are entangled, each photon considered separately is not in a well-defined state.

You can also prepare two photons in the same state, so the have the same polarization for some direction chosen at that time. But the measurements along other axis won't be perfectly correlated (if they are correlated at all).

The red/blue color example is too simple to be interesting.

[Koshkin]

You cannot paint a billiard ball with a superposition of colors.

[AnthonyMouse]

But what is the implication of that? What do those words cause to be different in practice?

[Koshkin]

The randomness in the observed color of the ball. (The observable represented by the "color operator" could have two or more distinct eigenvalues.)