# Einstein Relatively Easy

### What's Up

User Rating: 5 / 5      So far we have described the entangled state as this very puzzling composite state where we ignore everything about each subsystem taken separately but where the measure of one subsystem gives us direct (or real in the EPR vocabulary) information about the measurement of the other subsystem.

The previous article Expectation value of an entangled state exposed the first part: the outcome of the measurement of each subsystem is completely random. It's time now to look at the second part, the correlation.

For this, we have to introduce a new kind of observable, wider than the ones that Alice and Bob can measure separately, by using only his own detector. The measurement of this new family of observables, the composite observables, requires both detectors.

More precisely, the composite observable is an observable that is mathematically represented by first applying Alice's observable and then Bob's observable.

To make it more concrete, let's take the example of the previous entangled state If we ask Alice to measure σAz, Bob to measure σBz and then to compare their results, that's what we have to calculate to predict the result: We have done half the work in the previous article: Applying now σBz, we get: or more simply: What does it mean is when Alice measure σAz on the entangled state with her detector and when Bob measures σBz on the same state with his detector and when they come together and compare their results, they find they've measured identical values.

Sometimes (randomly as seen in the previous article) Alice measures -1 and Bob measures -1 as well. Other times, Alice measures +1 and Bob measures +1 as well. The product of their two measurements is always +1. They are correlated.

We said in a previous article that this should be true every time Alice and Bob measure the observable along the same axis.

Let's verify it, by measuring by example the product σBxσAx Again, every time Alice and Bob measure their respective σx, they find they have the same value.

We could easily do the same calculation on the z-axis and we would definitely find the same correlation.

Welcome to the quantum weirdness ;-)

Remark 1: we get σBxσAx|entangled>=|entangled>  which means by definition that |entangled> is an eigenvector of the observable σBxσAx with eigenvalue +1. Everytime the observable is applied to |entangled> we will measure +1. Hence, necessarily, we deduce that the average value <σBxσAx> =1.

The statistical correlation between Alice's and Bob's observations is defined as

BxσAx> - <σBx> - <σAx>

When the statistical correlation is non zero, we say that the observations are correlated.

That holds perfectly true in our case: we have <σBxσAx> - <σBx> - <σAx> = 1 - 0 != 0

Alice's and Bob's observations are correlated.

Remark 2: we said that the correlation exists when the observables are done along the same axis. We will see that even if is perfectly doable for Alice and Bob to observe together in different directions,  say Alice along direction x and Bob along direction y (which is not true for the same observable, as the components x, y,  and z of σ dont commute with each other inside the same space), however the product operators won't commute anymore.

### Quotes

"The essence of my theory is precisely that no independent properties are attributed to space on its own. It can be put jokingly this way. If I allow all things to vanish from the world, then following Newton, the Galilean inertial space remains; following my interpretation, however, nothing remains.."
Letter from A.Einstein to Karl Schwarzschild - Berlin, 9 January 1916

"Quantum mechanics is certainly imposing. But an inner voice tells me that it is not yet the real thing. The theory says a lot, but does not really bring us any closer to the secret of the 'old one'. I, at any rate, am convinced that He is not playing at dice."
Einstein to Max Born, letter 52, 4th december 1926