When is there interference if two quantum systems are entangled?

Our derivation of the electron spin depends on the fact that the (hidden) spin rotation angles of an electron and its brother positron are entangled, so that destructive interference cannot happen in the electron rotation.

There certainly happens interference in the double slit experiment, though the passing photon is entangled in many ways with other particles. What exactly governs the appearance of interference?

In the double slit experiment, if we "can know" through which slit the photon passed, there is no interference pattern on the screen. What does it mean that we "can know"?

The passing photon always interacts with the walls of the slits. It is diffracted by the walls. The interaction does not destroy the interference pattern.

But if the photon collides with another particle close to either of the slits, then the interference pattern does not appear.

https://en.wikipedia.org/wiki/Weak_value

The weak measurement concept by Yakir Aharonov and others is of relevance here. A weak measurement does not destroy interference, though a strong measurement does.

The double-slit experiment

The wave of a single photon passes through both slits and forms the familiar interference pattern on a screen.

Let us try to detect through which slit the photon passes. That will destroy the interference pattern. We put lots of free floating electrons at one of the slits.

Classical electrodynamics explains why there is no longer an interference pattern. The electrons disturb the wave which comes through the slit. The inertia of the electrons makes them a source of new waves of various frequencies. If we have enough electrons, the wavefront through that slit is badly scrambled, and there is no interference pattern on the screen.

Free electrons make the wave equation nonlinear. They act as a source in the wave equation. We say that a photon has been scattered from an electron if we observe a photon which has a new frequency or whose path has been disturbed.

A produced pair of particles in a ring

Suppose that we have some process which produces two particles which circle a ring at the same speed to opposite directions. If there were just one particle, an integer number of the wavelength should fit in the ring. It is like the Bohr atom model. But if we have two particles?

It is easiest to handle the problem in cartesian coordinates, where we identify points

         n C,

where n is an integer and C is the length of the circle. The x axis is a covering space for the circle.

For a single particle, suppose the solution is of the form

       exp(i p x) * exp(-i E t).

The value of the first factor must return to its x = 0 value at x = C, so that there is no destructive interference. Setting t = 0, for the lowest energy state, we get the equation

       i 2π = i p C

      <=>

       p = 2π / C.

What if we have two particles? How are two particles born together? Suppose that we have a particle whose energy is 2E, momentum 0, and for which we do not know the position at all. Its wave function is

      exp(-i (2E t - 0 * x)).

The particle decays very fast into two particles, each of which has the energy E.  How do we determine the wave function of the two particle combination?

Suppose that the particles were born at x = 0. We may redefine the laboratory coordinates if not. The wave function of the whole configuration might be

       exp(-i (E t + p x_1)) * exp(-i (E t - p x_2)) * δ(-x_1, x_2)

where δ(y, z) is 1 if y = z, zero otherwise. We may define x_2 = x and x_1 = -x, and we get

       exp(-i (2E t - 2p x)).

The periodicity condition then gives p = π / C. That is, we can treat the whole system as a single particle who has the energy E and the momentum 2p.

What were our assumptions and what did we really prove if anything? We assume that when the particles are "born", they already have a stationary wave function which occupies the whole ring. We assume that the wave functions of both particles are symmetric at 0. We assume that we can get the wave function of the whole system by multiplication, and we assume that we can project out the solution with the δ() function.

From those assumptions we proved the Principle of combination 1 of our previous blog post. It is not at all certain that this "proof" is any more convincing than the Principle of combination 1 taken as an axiom.

What happens if we let the 2 particles to move to separate rings?

Is it essential that they are circling the same ring? Or is the periodicity condition satisfied even if one of them moves to another ring?

Since the electron and the positron can exist independently, the answer empirically is yes.

If we move the particles to separate rings, and let ring B move fast, then special relativity says that the clocks at the rings no longer agree. The phase of the particle in ring B has shifted, as measured by an observer moving with B. Does that mean that the Principle of combination no longer works and there is destructive interference in both rings?

Let A describe the particle in B with a wave function of type

      exp(-i (E' t' - m' v' x)).

A and B agree on x, but not on the quantities marked with the prime symbol '.

A sees the particle in B move slower, v', but that is compensated by the bigger mass m'.

A sees the time t' go slower in B, but that is compensated by a higher energy E'.

It might be that according to A, the phase of the rotation in B goes at the same rate, regardless of the translational speed of B. That would mean that A "knows" the phase of B, and the Principle of combination works even when the rings are separated from each other.

The reasoning depends on the fact that nothing can affect the absolute "rotation speed" of the wave function of B, as seen by A. The rotation axis can be turned, but the absolute value of the angular momentum stays the same.