152 S. Gogioso and C. M. Scandolo
Thanks to the graphical notation introduced above, we already have a first intuition of why density hypercubes display higher-order interference. However, a rigorous proof requires a complete set-up with states, projectors, measurements and probabilities for a d-slit interference experiment, so that is what we now endeavour to provide.
1. We choose a d-dimensional space H Cd, and we value our tensor indices in
=
the set X = {1, ..., d} (the same set that we use to label the d slits).
2.We fix an orthonormal basis (|x )x X , and we interpret |x to be the state in which the particle goes through slit x with certainty.
3.The initial state for the particle is the superposition state in which the particle
goes through each slit with the same amplitude. More precisely, it is the pure
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normalised density hypercube state ρ+ corresponding to the vector √ |
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ψ+ |
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√ |
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H |
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ρ+ := |
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Ψ+ |
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(27) |
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4.The particle goes through some non-empty subset U X of slits at random: afterwards, the experimenter knows which subset the particle passed through, but no more information than that is available in the universe.
5.The particle is measured at the screen, and the experimenter estimates the probability P[+|U ] that the particle is still in state ρ+ after having passed through the given subset U of the slits:
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Ψ+ |
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P[+|U ] := |
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P |
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(28) |
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+ |
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d2 Ψ+ |
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PU |
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It is immediate to see that the outcome probability P[+|U ] depends solely on the number of di erent pieces appearing in the decomposition of the projector
PU :
P[+|U ] = |
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· number of pieces in PU |
(29) |
d4 |
To count the number of pieces in PU , it is convenient to group them by shapes. If U is a subset of size k, standard combinatorial arguments can be used to obtain the number of pieces of each shape appearing in the decomposition (as a
convention, we set k = 0 for j > k):
j
(k1)·1! |
7 shapes, (k2)·2! each |
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(30) |
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6 shapes, (k3)·3! each |
(k4)·4! |