Einstein Relatively Easy

User Rating: 5 / 5

Star ActiveStar ActiveStar ActiveStar ActiveStar Active
 
Pin It

 

                            After having spent some time on the expectation value of the the sigma observable for a product state, let's try to do the same for an entangled state.

Let's choose our entangled state to be for instance[1]:

For simplicity, let's write each composite vector as |uu> and |dd>  - but still remembering that such vectors are nothing else than a tensor product of the basis vectors of each sub-space - and so simplify this equation from now on as :

Let's look first at the expectation value of Alice's σ of this state along the z-axis. We know already all the machinery we need to compute it

 

or

 so finally

 Let's look now at the expectation value of Alice's σ of this state along the y-axis

 

 

 

 Finally, if we focus on the x-axis of the expectation value

 and we are left once more with zero as result.

 We could have done the same calculation for σB and we would have found the very same result.

 So we get a very different result than for an product state, as all σ expectation values for any observer along any axis is zero.

 

An expectation value equal  to zero for an observable for which the eigenvalues are +1 or -1 means that there is an equal probability for each experimental outcome.

That's where lies the true weirdness of entanglement: we can know nothing a priori of the value of the subsystem, but however, if both observables are measured along the same axis, they are found to be correlated. This means that the random outcome of the measurement made on one particle seems to have been transmitted to the other, even if the distance and timing of the measurements can be chosen so as to make the interval between the two measurements spacelike, hence, any causal effect connecting the events would have to travel faster than light.

 In the next article, we will demonstrate the correlation between the two sub-systems, and introduce the concept of composite observables.

 

[1] We verify that our state is a normalized vector, as <entangled|entangled> = 1/2 + 1/2 = 1.

Language

Breadcrumbs

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

RSS Feed

feed-imageRSS

Who is online

We have 56 guests and no members online