Englert-Greenberger duality relation

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The Englert-Greenberger duality relation relates the visibility V of interference fringes with the definiteness or distinguishability D of the photons' paths in quantum optics.^[1]^[2]^[3] as an inequality:

$D^{2}+ V^{2}\le 1 \,$

1 The mathematics of two-slit diffraction
2 Complementarity
3 See also
4 Further reading
5 References and notes

[edit] The mathematics of two-slit diffraction

This section reviews the mathematical formulation of the double-slit experiment. The formulation is in terms of the diffraction and interference of waves. The culmination of the development is a presentation of two numbers that characterizes the visibility of the interference fringes in the experiment, linked together as the Englert-Greenberger duality relation.The next will then discuss the orthodox quantum mechanical interpretation of the duality relation in terms of wave-particle duality. Of this experiment, Richard Feynman once said that it ``has in it the heart of quantum mechanics. In reality it contains the only mystery´´.

The wave function in the Young double-aperture experiment can be written as

Ψ Total (x) = Ψ A (x) + Ψ B (x)

The function

Ψ A (x) = C A Ψ 0 (x - x A)

is the wave function associated with the pinhole at A centered on $x A$ ; a similar relation holds for pinhole B. The variable $x$ is a position in space downstream of the slits. The constants $C A$ and $C B$ are proportionality factors for the corresponding wave amplitudes, and $Ψ 0 (x)$ is the single hole wave function for an aperture centered on the origin. The single-hole wave-function is taken to be that of Fraunhofer diffraction; the pinhole shape is irrelevant, and the pinholes are considered to be idealized. The wave is taken to have a fixed incident momentum $p 0 = h / λ$ :

$\Psi_{0}(x)\propto \frac{e^{ip_0\cdot|x|/\hbar}} {|x|}$

where $| x |$ is the radial distance from the pinhole.

To distinguish which pinhole a photon passed through, one needs some measure of the distinguishability between pinholes. Such a measure is given by^[4]

$\begin{matrix}D=|P_A-P_B|, \end{matrix}$

where $P A$ and $P B$ are the probabilities of finding that the particle passed through aperture A or aperture B respectively.

Since the Born probability measure is given by

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

and

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

then we get:

$D=\left|\;\frac{|C_A|^2-|C_B|^2}{|C_A|^2+|C_B|^2}\,\right|$

We have in particular $D = 0$ for two symmetric holes and $D = 1$ for a single aperture (perfect distinguishability). In the far-field of the two pinholes the two waves interfere and produce fringes. The intensity of the interference pattern at a point y in the focal plane is given by

$I(y)\propto 1+V\cos{(p_yd/\hbar+\phi)}$

where $p_y= h/\lambda\cdot \sin(\alpha)$ is the momentum of the particle along the y direction, $φ = Arg(C A) - Arg(C B)$ is a fixed phase shift, and $d$ is the separation between the two pinholes. The angle α from the horizontal is given by $\sin(\alpha)\simeq \tan(\alpha)=y/L$ where $L$ is the distance between the aperture screen and the far field analysis plane. If a lens is used to observe the fringes in the rear focal plane, the angle is given by $\sin(\alpha)\simeq \tan(\alpha)=y/f$ where $f$ is the focal length of the lens.

The visibility of the fringes is defined by

$V=\frac{I_\mathrm{max}-I_\mathrm{min}}{I_\mathrm{max}+I_\mathrm{min}}$

where $I max$ and $I min$ denote the maximum and minimum intensity of the fringes respectively. By the rules of constructive and destructive interference we have

$I_\mathrm{max} \propto ||C_A|+|C_B||^2$

$I_\mathrm{min} \propto ||C_A|-|C_B||^2$

Equivalently, this can be written as

$V=2\frac{|C_A\cdot C_B^*|}{|C_A|^2+|C_B|^2}.$

And hence we get, for a single photon in a pure quantum state, the duality relation

$\begin{matrix}V^2+D^2 = 1 \end{matrix}$

There are two extremal cases with a straightforward intuitive interpretation: In a single hole experiment, the fringe visibility will be zero (as there are no fringes); that is, $V = 0$ but $D = 1$ since we know (by definition) which hole the photon passed through. On the other hand, for a two slit configuration, where the two slits are indistinguishable with $D = 0$ , one has perfect visibility with $I min = 0$ and hence $V = 1$ . Hence in both these extremal cases we also have $V 2 + D 2 = 1$ .

The above presentation was limited to a pure quantum state. More generally, for a mixture of quantum states, one will have

$V^{2}+D^{2}\leq 1$ .

For the remainder of the development, the light source will be assumed to be a laser, so that $V 2 + D 2 = 1$ can be assumed to hold, following from the coherence properties of laser light.

[edit] Complementarity

The mathematical discussion presented above does not require quantum mechanics at its heart. In particular, the derivation is essentially valid for waves of any sort. With slight modifications to account for the squaring of amplitudes, the derivation could be applied to, for example, sound waves or water waves in a ripple tank.

In order for the relation to be a precise formulation of Bohr complementarity one must introduce wave-particle duality in the discussion. This means one must consider both wave and particle behavior of light on an equal footing. Wave-particle duality implies that one must A) use the unitary evolution of the wave before the observation and B) consider the particle aspect after the detection (this is called the Heisenberg-von Neumann collapse postulate). Indeed since one could only observe the photon in one point of space (a photon can not be absorbed twice) this implies that the meaning of the wave function is essentially statistical and can not be confused with a classical wave (like it exists in air or water).

In this context the direct observation of a photon in the aperture plane precludes the following recording of the same photon in (F). Reciprocally the observation in (F) means that we did not absorb the photon before. If both holes are open this implies that we don't know where we would have detected the photon in the aperture plane. $D$ defines thus the distinguishability of the two holes A and B.

A maximal value of distinguishability $D = 1$ means that only one hole (say A) is open. If now we detect the photon at (F), we know that that photon would have been detected in A necessarily. Conversely, $D = 0$ means that both holes are open and play a symmetric role. If we detect the photon at (F), we don't know where the photon would have been detected in the aperture plane and $D = 0$ characterizes our ignorance.

Similarly, if $D = 0$ then $V = 1$ and this means that a statistical accumulation of photons at (F) will build up an interference pattern with maximal visibility. Conversely, $D = 1$ implies $V = 0$ and thus, no fringes will appear after a statistical recording of several photons.

The above treatment formalizes wave particle duality for the double-slit experiment.

[edit] See also

[edit] Further reading

Berthold-Georg Englert, Marlan O Scully & Herbert Walther, Quantum Optical Tests of Complementarity , Nature, Vol 351, pp 111-116 (9 May 1991). Demonstrates that quantum interference effects are destroyed by irreversible object-apparatus correlations ("measurement"), not by Heisenberg's uncertainty principle itself. See also The Duality in Matter and Light Scientific American, (December 1994)

Aurelien Drezet, Complementarity and Afshar's experiment, University of Graz Institute of experimental physics, Austria, [1] for the details on the duality inequality.

[edit] References and notes

^ Gregg Jaeger, Abner Shimony, Lev Vaidman, "Two interferometric complementarities", Phys. Rev. A, Vol. 51, 54 (1995)
^ Berthold-Georg Englert, "Fringe Visibility and Which-Way Information: An Inequality", Phys. Rev. Lett., Vol. 77, 2154 (1996)
^ Daniel M. Greenberger, Allaine Yasin ,"Simultaneous wave and particle knowledge in a neutron interferometer", Phys. Lett. A 128, 391, (1988)
^ Actually, what is called "distinguishability $D$ " here is usually referred to as "predictability $P$ ".