Non-locality Quantum eraser experiment
figure 3. 2 intereference patterns , sum when alice measures photons polarization circular polarizer.
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{\displaystyle i(x)=i_{1}(x)+i_{2}(x)\propto {\frac {1}{d^{2}+(x+a/2)^{2}}}+{\frac {1}{d^{2}+(x-a/2)^{2}}}.}
the results of experiment summarized in fig.3. bob can distinguish 2 peaks in data after has had access alice s results: set of photons alice measured clockwise polarization, bob s subset of photons distributed according
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{\displaystyle i_{1}(x)}
, set of photons alice measured anti-clockwise polarization, bob s subset of photons distributed according
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{\displaystyle i_{2}(x).}
next, let alice use linear polarizer instead of circular one. first thing write down system s wave-function in terms of linear polarization states:
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{\displaystyle |\circlearrowright \rangle _{a}|\circlearrowleft \rangle _{b}+|\circlearrowleft \rangle _{a}|\circlearrowright \rangle _{b}={\frac {1}{\sqrt {2}}}\left(|h\rangle _{a}+i|v\rangle _{a}\right)|\circlearrowleft \rangle _{b}+{\frac {1}{\sqrt {2}}}\left(|h\rangle _{a}-i|v\rangle _{a}\right)|\circlearrowright \rangle _{b}}
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{\displaystyle \qquad ={\frac {1}{\sqrt {2}}}\left(|\circlearrowright \rangle _{b}+|\circlearrowleft \rangle _{b}\right)|h\rangle _{a}+{\frac {i}{\sqrt {2}}}\left(|\circlearrowright \rangle _{b}-|\circlearrowleft \rangle _{b}\right)|v\rangle _{a}}
so alice measures horizontally polarized photon. wave function of bob s photon in superposition of clockwise , anti-clockwise polarizations, means indeed can pass through both slits! after travelling screen, wave amplitude is
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{\displaystyle h(x)={\frac {1}{\sqrt {2}}}\left[f_{1}(x)+f_{2}(x)\right],}
and intensity is
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ϕ
12
{\displaystyle i_{h}(x)\propto |f_{1}(x)|^{2}+|f_{2}(x)|^{2}+f_{1}(x)f_{2}(x)^{*}+f_{1}(x)^{*}f_{2}(x)={\frac {i_{1}(x)+i_{2}(x)}{2}}+\cos \phi _{12}}
where
ϕ
12
{\displaystyle \phi _{12}}
phase difference between 2 wave function @ position x on screen. pattern indeed interference pattern! likewise, if alice detects vertically polarized photon wave amplitude of bob s photon is
figure 4. 2 intereference patterns , sum when alice measures photons polarization linear polarizer.
v
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{\displaystyle v(x)={\frac {i}{\sqrt {2}}}\left[f_{1}(x)-f_{2}(x)\right],}
and
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cos
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12
{\displaystyle i_{v}(x)\propto |f_{1}(x)|^{2}+|f_{2}(x)|^{2}-f_{1}(x)f_{2}(x)^{*}-f_{1}(x)^{*}f_{2}(x)={\frac {i_{1}(x)+i_{2}(x)}{2}}-\cos \phi _{12}}
and once again interference pattern appears, changed because of 180º phase difference between 2 photons traversing each slit. can used alice send message bob, encoding messages in changes between 2 types of patterns? no! remember that, before, if bob not told polarization alice measured, sees sum of both patterns. result therefore,
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h
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{\displaystyle i(x)=i_{h}(x)+i_{v}(x)=i(x)}
which again smudge. results given in fig.4.
so s odd experiment? correlations change according experiment conducted alice. despite fact total pattern same, 2 subsets of outcomes give radically different correlations: if alice measured linear polarization total smear subdivided 2 interference patterns whereas if alice measured circular polarization pattern sum of 2 other gaussian bell-shapes. how bob s photon know go in forbidden stripes of interference pattern when alice measuring circular polarization not when alice measuring linear one? can orchestrated global dynamic of system whole, cannot locally carried each photon on own. experiment demonstrates phenomenon of microscopic non-locality.
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