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Oct 25, 2014 at 16:37 comment added yuggib with different boundary conditions.
Oct 25, 2014 at 16:29 history edited yuggib CC BY-SA 3.0
the previous statement was imprecise
Oct 25, 2014 at 16:22 comment added yuggib As I see it: physically you want states that have zero probability on the boundaries (or at the very least that have symmetric boundary conditions). On the other hand, you need an Hilbert space structure to do QM, so you need to consider the whole $L^2(0,1)$. Here the problem of boundary conditions affects the domains of operators; nevertheless once a (self-adjoint) operator is fixed (i.e. we fix its action and domain), it is ok. Then you may consider as physical only states with the desired boundary conditions, however in principle the action of an operator may change you to a state...
Oct 25, 2014 at 16:04 comment added Danu Good point. I have slightly updated my question, taking your response into account. The remaining question is essentially: Do we need to allow non-physical $\psi$ in order to define momentum properly? Maybe this is already addressed by your second paragraph, though.
Oct 25, 2014 at 16:01 history answered yuggib CC BY-SA 3.0