User:Drezet

Concerning me

1)I am Aurelien Drezet, a physicist living in Graz Austria and working on the fields of nano-optics and quantum physics. You can find a link to my current works [1].

2)I am curently involved into a dispute concerning the Afshar experiment. You can read my point of view in [2].

Theoretical discussion (i want just to save that here and to read it again latter on)

Wave-particle duality is considered to be one of the distinguishing characteristics of quantum mechanics, whose theoretical and experimental development has been honoured by more than a few Nobel Prizes for Physics. It has been discussed by prominent physicists for the last 100 years, from the time of Albert Einstein, Niels Bohr and Werner Heisenberg, onwards. On the basis of Bohr's principle of complementarity, it is indeed universally accepted that the observation of two properties, such as position and momentum, requires mutually exclusive experimental measurements.

Mathematically, a specific formulation of Bohr's complementarity can be obtained on the basis of the Englert-Greenberger duality relation. The wave function in the Young double-aperture experiment can be written as

$\Psi _{\mbox{Total}}(\mathbf {x} )=\Psi _{A}(\mathbf {x} )+\Psi _{B}(\mathbf {x} )\qquad \qquad (1)$.

Here A and B denote the two pinholes,

$\Psi _{A,B}(\mathbf {x} )=C_{A,B}\cdot \Psi _{0}(\mathbf {x} -\mathbf {x} _{A,B})$

are the wave function associated with both pinhole centered on

$\mathbf {x} _{A},\qquad \mathbf {x} _{B},$

and

$\mathbf {x}$

is a position in space downstream of the slits. We write

$C_{A},\qquad C_{B}$

the corresponding wave amplitudes and

$\Psi _{0}(\mathbf {x} )$

is the single hole wave function for a aperture centered on the origin.

To distinguish which pinhole a photon passed through, one needs some measure of the distinguishability between pinholes. Such a measure is given by

$D={\frac {||C_{A}|^{2}-|C_{B}|^{2}|}{|C_{A}|^{2}+|C_{B}|^{2}}}=|P_{A}-P_{B}|,\qquad \qquad$

where

$P_{A}={\frac {|C_{A}|^{2}}{|C_{A}|^{2}+|C_{B}|^{2}}},\quad P_{B}={\frac {|C_{B}|^{2}}{|C_{A}|^{2}+|C_{B}|^{2}}}$

are the probabilities of finding that the particle passed through aperture A or slit B.

We have in particular D=0 without which-path detection and D=1 for a perfect distinguishability. In the far-field of the two pinholes the two waves interfere and produce fringes. The visibility of the fringes is defined by

$V={\frac {I_{max}-I_{min}}{I_{max}+I_{min}}}\qquad \qquad (3)$

where max and min denote the maximum and minimum of intensity. This can be equivalently written

$V=2{\frac {|\Psi _{A}\cdot \Psi _{B}^{*}|}{|\Psi _{A}|^{2}+|\Psi _{B}|^{2}}}.$

In a good which path experiment we have V=0. Reciprocally we have V=1 without distinguishability. It is straighforward to see that the duality relation

$V^{2}+D^{2}=1\quad$

is always true. The present presentation was limited to a pure quantum state. For a mixture we have

$V^{2}+D^{2}\leq 1\qquad \qquad (4)$

The interpretation of this relation is the following: Consider the Young double slit experiment for photon with the lens and suppose first the experiment made without which path detector. If we detect the particles in the aperture plane we find statistically two narrow peaks with equal intensity (D=0). Now if we detect the photons in the focal plane (image) of the lens we find an interference pattern with visibility V=1. Naturally we then didn't record the same particles in the aperture plane since a photon can not be absorbed twice. We can deduce however that each photon has a probability equal to 1/2 to come from the pinhole A or B. It is important however to observe that in this experiment a photon detected directly in the apertures tell us where it is actually (which path information) and this even if D=0. The real meaning of the duality relation is then only in the logical inference : 'If the photon is actually detected in the fourier plane (momentum information) then we only know the probability of where it would have been before in the aperture plane'.

The introduction of a which-path detector does not change the story. Such which path detector is introduced in order to realize a non destructive measurement: we want to observe the photon in the Fourier plane of the lens and at the same time know where it was coming from. The photon must now be entangled with one other quantum system or with a internal degree of freedom like spin or polarization. When the detector fires (upper state) we have a joint-wave function like Eq. 1. If the photon is detected in coincidence, when it is still in the aperture, we find the two asymmetric peaks corresponding to :$D\neq 0\quad$. In the ideal case one of the two peaks has an intensity equal to zero e.g. D=1 which corresponds to a perfect which-path experiment. Now if we detect (still in coincidence with the detector in its upper state) the photon in the Fourier plane the same logical inference than previously can be done. Once again it is important however to observe that in this experiment a photon detected directly in the apertures tell us where it is actually (which path information).

The experiment of Afshar contains no detecting device using entanglement. In the formalism presented previously this means that D=0 and V=1. However in this experiment the photon are actually detected in the image plane of the lens and not in the Fourier plane (e.g in the focal plane). The logical inference is here inversed: 'If the photon is actually detected in the image plane (spatial information) then we only know the probability of where it would have been before in the focal plane plane (momentum information)'. The principle of complementarity should then be respected in this experiment. However this conclusion is still debated.