**Abstract** : We propose a solution for the apparent paradox
of the double-slit experiment within the framework of our reconstruction of quantum mechanics (QM), based on the geometrical meaning of spinors.
We argue that the double-slit experiment can be understood much better
by considering it as an experiment whereby
the particles yield information about the set-up rather than an experiment whereby the set-up
yields information about the behaviour of the particles.
The probabilities of QM are conditional, whereby the conditions are
defined by the macroscopic measuring device.
Consequently, they are not uniquely defined by the local interaction probabilities in the point
of the interaction. They have to be further fine-tuned in order to fit in seamlessly
within the macroscopic probability distribution, by complying to its boundary conditions.
When a particle interacts incoherently with the set-up the answer to the question
through which slit it has moved
is experimentally decidable.
When it interacts coherently the answer to that question
is experimentally undecidable.
We provide a rigorous mathematical proof
of the expression $\psi_{3}= \psi_{1} + \psi_{2}$
for the wave function $\psi_{3}$ of the double-slit experiment,
whereby $\psi_{1}$ and $\psi_{2}$ are the wave functions of the two
related single-slit experiments. This proof
is algebraically perfectly logical and
exact, but geometrically flawed and meaningless for wave functions.
The reason for this weird-sounding distinction is that the wave functions are representations of
symmetry groups and that these groups are curved
manifolds instead of vector spaces.
The identity $\psi_{3}= \psi_{1} + \psi_{2}$ must
therefore be replaced in the interference region
by the expression
$\psi'_{1} + \psi'_{2}$, for which a geometrically correct meaning can be constructed in terms
of sets (while this is not possible for $\psi_{1} + \psi_{2}$).
This expression has the same numerical value as
$\psi_{1} + \psi_{2}$ , such that $\psi'_{1} + \psi'_{2} = \psi_{1} + \psi_{2}$,
but with $\psi'_{1} = e^{\imath \pi/4 } (\psi_{1} +\psi_{2})/\sqrt{2} \neq \psi_{1}$ and
$\psi'_{2} = e^{-\imath \pi/4 } (\psi_{1} +\psi_{2})/\sqrt{2} \,\neq \psi_{2}$.
Here $\psi'_{1}$ and $\psi'_{2}$ are the correct (but experimentally unknowable)
contributions from the slits to the total wave function $\psi_{3} = \psi'_{1} + \psi'_{2}$.
We have then $p = \vert \psi'_{1} + \psi'_{2}\vert^{2} = \vert\psi'_{1}\vert^{2} + \vert\psi'_{2}\vert^{2} = p'_{1}+p'_{2} $ such that the apparent paradox
that quantum mechanics would not follow the traditional rules of probability calculus for mutually exclusive events disappears.