What's the Cause of Quantum Entanglement?

Question

What is the cause of quantum entanglement?

When two particles become entangled what property of them basically changes as to establish a link between them and how the information is exchanged between them?

And can multiple particles be entangled with each other, if yes then suppose 10 particles are entangled to each other and some force is applied to one of the particle then all of the particles will reflect the same changes. But in another case if we apply different force on two of the particles then what would happen?

a) entanglement will be broken for all particles

b) only those 2 particles become untangled and others remain entangled

if case a) happens then doesn’t it pose a risk on quantum computers as a fault in a particle could bring down the whole system?

Answer 1

I.The cause of the quantum entanglement is that the wave-function (w.f.) of the involved particles doesn't have the form of the w.f. of independent particles.

So, let's begin by defining what is the w.f. of independent particles. The w.f. of a single particle, let's call it A, can look like

$$|A> = \frac {\lvert x_A \rangle + \lvert y_A \rangle.} {\sqrt(2)}$$

where x and y can be the possible polarizations of the particle if it has spin, or it is a photon. Similarly, for the particle B

$$|B> = \frac {\lvert x_B \rangle + \lvert y_B \rangle.} {\sqrt(2)}$$

So, the w.f. of these two particle should be

$$|A>|B> = \frac {(\lvert x_A \rangle + \lvert y_A \rangle)(\lvert x_B \rangle + \lvert y_B \rangle)} {\sqrt(2)}$$

Opening parentheses,

$$ |A>|B> = \frac {\lvert x_A \rangle \lvert x_B \rangle + \lvert x_A \rangle \lvert y_B \rangle + \lvert y_A \rangle \lvert x_B \rangle + \lvert y_A \rangle \lvert y_B \rangle} {2}$$

So, this is the w.f. of the two particles if they are independent i.e. what happens with one particle has no connection with what happens with the other. We use to say that this form is factorizable, i.e. can be represented as a product in which each factor refers to a single particle.

In an entangled pair of particles one of the products of states is missing, e.g.

$$|A, B> = \frac {\lvert x_A \rangle \lvert x_B \rangle + \lvert y_A \rangle \lvert x_B \rangle + \lvert y_A \rangle \lvert y_B \rangle} {\sqrt(3)}$$

We cannot factorize this state. This is what changes.

II. How the information is exchanged between A and B, we don't know, we only know the result of this supposed exchange of information. The result, see the non-factorizable w.f. is as follows:

If we measure the polarization of the particle A and find it along x, then a measurement of the polarization of B can be found only y. But if we find for A, polarization y, then for B we can find either x, or y. Similar things we can say about measuring the polarization of B and what happens with A.

III. One can entangle not only 10 particles, much more. Though, it will be difficult to work with that wave-function. People study much simpler entanglements, i.e. of 2, 3, 4, particles.

There is a theorem of no-communication that says

NO ACTION DONE ON ONE OF THE PARTICLES HAS AN EFFECT ON THE OTHER PARTICLES.

IV. Operations done on particles can be reversible, or non-reversible. An example of irreversible operation is to measure the polarization of one particle as in our example in part I. Yes, such an operation breaks the entanglement for all the particles entangled with the measured particle. On the other hand, a reversible operation, e.g. passing one particle through a classical field, has no implication on the entanglement. he entanglement remains entire.

Answer 2

There is no precise answer to the cause of entanglement is known (I did not come across any such answer). The superposition principle and the tensor product structure of Hilbert space of combined system leads to the possibility of entangled state. It is well understood in the case of pure state, but for the mixed state case it is quite involved.