# Interferences (with diagonal density matrix)

• I
Heidi
suppose that elecrons are in a state described by a diagonal density matrix for their spin (we are not interested in their spatial matrix). They are used in the double slit experiment. will we get fringes.
I ask the question because when Bob ans Alice share pairs of electrons (the total spin of the pair being null) then Bob will see no frange. can you confirm this point?
for him his electron is in a diagonal density matrix. Is this always like that? diagonal => no fringe?

Gold Member
2022 Award
It's not clear which experiment you are referring to. So it's impossible to answer your question. If you have an entangled pair of spin-1/2 in the singulett state
$$|\Psi \rangle=\frac{1}{\sqrt{2}} [|1/2,-1/2 \rangle -|-1/2,1/2 \rangle],$$
each of the electrons is unpolarized, i.e., the reduced density matrices of each of the electrons is $$\hat{\rho}_1=\hat{\rho}_2=\frac{1}{2} \hat{1}.$$
It's not clear to me, why there shouldn't be interference fringes in a double-slit experiment with unpolarized electrons.

• Heidi
Heidi
this question is important to me because i have this idea for a long time. Do you agree that when such a singlet pair is created it can be done such that the total momentum P is also null and that for each electron we have also a diagonal density matrix in the p basis?
this is the firsr question.
second question:
if Bob and Alice shares this pairs do you agree that whatever Alice can do, this will not change the interference pattern seen by Bob in his double slit experiment? he only have 2 slits and a screen behind. no polarizers or things like that . only the simplest Young experiment with the source in the orthogonal plane between the middle of the slits.
if you agree with these two points, we will discuss about an known experiment.

Gold Member
2022 Award
Yes, that sounds all plausible.

Heidi
the experiment is the Dopfer experiment.
https://www.researchgate.net/figure/color-online-The-Dopfer-experiment-of-the-Zeilinger-Group-Innsbruck-If-detector-D2-is_fig7_265787833
you can read her text in german here: https://www.univie.ac.at/qfp/publications/thesis/

pairs of singlet photon are emitted and send to a 2 slits plane (Bob)
the pattern is always without fringes but a coincidence counter enables to extract from the set of points on the screen points with fringes . what is important here is that the Bob photon has a I/2 density matrix which gives fringe visibility = 0.
it would be the same with singlet pairs of electrons.
I think that particles at the center of the Bloch (or Poincaré) sphere give a fringe visibility = 0 and par on the sphere give a visibility equal to one.
Do you agree?
I do not thind that the visibility is equal to the distance to the center of the sphere. Maybe a common formula can be found for visibility with photons and for electrons.

Gold Member
2022 Award
That's correct. It's characteristic for such delayed-choice experiments that you can choose partial ensembles of a full ensemble depending on a measurement protocol of coincidence measurements. Though the total ensemble does not show interference fringes, the partial ensembles do.

The thesis is great. Unfortunately there seems not to be an English translation or paper about it, although it's realizing the earliest idea about the uncertainty relation for position and momentum (the "Heisenberg microscope" as famously analyzed by von Weizsäcker ). Additionally there's of course this even more astonishing twist with the use of entangled photon pairs. While the single-photon double-slit experiment doesn't show specifically quantum effects concerning the interference fringes, because it's just the interference pattern you also obtain with classical light, this coincidence experiments with entangled photon pairs are specifically quantum ("non-classical").

• Heidi
Heidi
A point in the sphere has S1,S2,S3 as coordinates.
Its distance to the center is the Degree of Polarizamtion
I wonder if we can use them for the fringes.
I read this in wikipedia (polarization)

The coherency matrix contains all second order statistical information about the polarization. This matrix can be decomposed into the sum of two idempotent matrices, corresponding to the eigenvectors of the coherency matrix, each representing a polarization state that is orthogonal to the other. An alternative decomposition is into completely polarized (zero determinant) and unpolarized (scaled identity matrix) components. In either case, the operation of summing the components corresponds to the incoherent superposition of waves from the two components. The latter case gives rise to the concept of the "degree of polarization"; i.e., the fraction of the total intensity contributed by the completely polarized component.

in the second case we decompose the coherency matrix in a basis in which we have I/2 and a pure density matrix . the first one gives no fringe and the second a visibility = 1
Cannot we use this to calculate the fringe visibility?

Heidi
I am confident we will get fringe visibility as a function of P in the sphere in the case of more or less entangled pairs. but are photons entangled with something? why pure polarization would mean no path information? not obvious.

Heidi
I am confident we will get fringe visibility as a function of P in the sphere in the case of more or less entangled pairs. but are photons entangled with something? why pure polarization would mean no path information? not obvious

Heidi
I wonder how the density matrix of an internal property can be useful for a spatial thing like fringe visibility in the 2 slits experient. With a Stern Gerlach there is a magnetic field which entangles entamits spin and its path.
Maybe there is something like that with the slits. most of the particles do not pass throu the 2 slits. maybe it is why the other particles share the same density matrix for the paths and with the Bloch sphere.

I am confident we will get fringe visibility as a function of P in the sphere in the case of more or less entangled pairs. but are photons entangled with something? why pure polarization would mean no path information? not obvious.

Yes, one can take a similar approach. See "Duality between partial coherence and partial entanglement" by Saleh, Abouraddy, Sergienko and Teich, Phys. Rev. A 62, 043816 (2000). https://journals.aps.org/pra/abstract/10.1103/PhysRevA.62.043816

Unfortunately, I am not aware of a free version on the ArXiv.

• Heidi and vanhees71
Heidi
As mixed states in the middle of the sphere give no fringes in the Young experiment, I had in mind something that is false: that pure states on the poincaré sphere would all give fringes. It seems that they give fringes it they do not carry which parth information.
Let us take the Birgit Dopfer experiment.
the slits plane is orthogonal to the direction of the source and the detector D1 is on the axis of the lens in the plane where parallel light fronts converge.
D1 select also a light front parllel to the light with equal probability for each slit.
we have frinces.
Let us now put D1 on a circle centered on the lens axis (orthogonal to it)
it will select a light front arriving on the slits with a given direction.
the only directions with no "preferred slit" are those in the slits symmetry plane.
I think that in this case the fringe visibility is equal to 1
in the other cases it would decrease.
So all the D1 would give different fringe visibility on the circle.
Is this correct?