The abstract of the paper promises to show an experiment, in which it is possible to see interference on two slits with 100% visibility, and at the same time to know which of the two slits the particle went through. “The experiment makes use of a super-focused laser beam that is launched into only one of the two slits of the two-slit interference experiment.”
In reality the content of the paper is different:
Author believes that a particle in two slits experiment goes only through a single slit. However the particle always carries with it an empty wave. In the two slits experiment the particle carries half of the empty wave with itself through one slit A, and sends another half of its empty wave through another slit B. After the two slits screen empty waves unite and pull the particle to fit its trajectory to interference pattern.
To explain how it can be the author invented his own mathematics (section 2.3). Diffraction of the particle at a single slit A is described by wave Ψsingle,A. Diffraction on the two slit is described by the wave
function ΨA+ΨB. The product of two wave functions Ψsingle,A(ΨA+ ΨB) (see Eq. (1)) describes transmission of the particle through the slit A and at the same time produces interference on two slits with 100 % contrast.
Mathematics, and “invented” mathematics
The author does not discloses his “know how” for calculations of Ψsingle,Aand ΨA+ ΨB, so I shall do it instead of him.
In canonical science the wave field after two slits screen is calcu- lated according to Huygens-Kirchhoff recipe. If at a pointr0 (z0= 0) in a slit on the two slits screen the field of the incident wave isψ0(r0),
then the field at another screen at a pointr(z=D) is Ψ(r) = 1 4π Z S(z0=0) d2rk× × ψ0(r0) −→ d dz0 exp(ik|r−r0|) |r−r0| −ψ0(r 0) ←− d dz0 exp(ik|r−r0|) |r−r0| , (I1) where arrows over derivatives show which of the two functions in the brackets is to be differentiated.
To check this formula, let’s suppose thatψ0(r0) = exp(ik0r0) and
the first screen contains a single infinitely wide slit, i.e. the first screen at the point z0 = 0 is absent. Then at the point r we must obtain Ψ(r) = exp(ik0r) — the plain wave, which is identical to the
To do that it is sufficient to represent the spherical wave exp(ik|r− r0|)/|r−r0| by the two dimensional Fourier integral
1 |r−r0|exp(ik|r−r 0|) = i 2π Z d2p k pz
exp(ipk(r−r0)k+iDpz), (I2)
wherepz=
q
k2−p2
k, andpkare components of the vectorpparallel to the plane (x, y). To check (I1) it is sufficient to substitute (I2) into (I1) and to integrate over infinite planed2rk0.
The author, used a different approach — an approximate Huygens- Fresnel recipe, according to which
Ψ(r)∝
Z
S(z0=0)
dx0ψ0(r0) exp(ik|r−r0|). (I3)
He considers only two coordinatesxalong the screen perpendicular to slits, and z, which is perpendicular to screens, so r= (x, D) and r0 = (x0,0). He chosen the origin of the coordinate system at the middle point between centers of two slits, so coordinates in the slits arex0=−d/2 +ξ for slit A andd/2 +ξfor slit B, where variableξ varies in the interval−s/2≤ξ≤s/2, determined by the slits width s.
The function Ψsingle,A(r), according (I3), is
Ψsingle,A(r)∝ Z
A(z0=0)
dx0ψ0(x0) exp(ikRA), (I4)
whereRA=|r−r0A|. We can expand it overξ, and in linear approx- imation we obtainRA=|r−r0A| ≈D−ξ(x+d/2)/D. Substitution into (I4) gives
Ψsingle,A(r)∝
s/2 Z
−s/2
dξψ0(x0) exp(ikξ(x+d/2)/D). (I5)
This integral will give
Ψsingle,A(r)∝
sin(πs(x+d/2)/Dλ)
(see Eq. (11)) only if ψ0(x0) =const, i.e. only for single plane wave
incident normally on the first screen. We must point out that the parameterd is absolutely irrelevant to the single slit diffraction. If we choose the origin of coordinates not in the middle point between slits but at some distanceLaway from the slit A, then the function (I6) will look
Ψsingle,A(r)∝
sin(πs(x+L)/Dλ)
πs(x+L)/Dλ . (I6a) So diffraction on a single slit depends on choice of the coordinates origin. It cannot be.
The function ΨA(x) + ΨB(x) is calculated as a sum of two waves from point sources at the centers of slits A and B:
ΨA+ ΨB = exp(ikRA) + exp(ikRB), (I7) where 4RA= p D2+ (x−d/2)2≈D−xd/2D, and RB= p D2+ (x+d/2)2≈D+xd/2D.
So, substitution into (I7) gives
ΨA+ ΨB= exp(ikRA) + exp(ikRB)∝cos
πdx Dλ
. (I8) However we can obtain all that immediately without such an artificial separations of two functions. If we expandRA,Boverr0, then in linear approximation they becomeRA,B=D−x(ξ∓d/2)/D. Sinceψ0(r0)
is constant, it is the same constant over both slits. Therefore the integral (I3) becomes
Ψ(r)∝ s/2 Z −s/2 dx0exp(−ikξx/D) exp(ikdx/2D) + exp(−ikdx/D) = =sin(πsx/Dλ) πsx/Dλ cos πdx Dλ . (I9)
We see that diffraction on a single slit does not depend on choice of co- ordinates origin. We obtained the product, which author introduced artificially, and our diffraction pattern has 100 % contrast because we considered absolutely coherent plane wave normally incident on both slits.