It seems that in LSZ formalism approach, or just Feynman diagram approach, we can compute scattering amplitude of $\langle x_{out} | y_{in}\rangle$ (position space) and $\langle p_{out} | p_{in}\rangle$ (momentum space). However, I read that it is possible that $\langle x,t | y,t\rangle \neq 0$ for $x\neq y$, which suggests that there does not really exist position state. (Assume in scalar QFT that $|x,t\rangle = \phi(x)|0\rangle$)
Can $\langle x,t | y,t\rangle \neq 0$ be possible for $x\neq y$? If so, why can we consistently define position space asymptotically?
Isn't $\langle p_a,t |p_b,t\rangle = 0$ for $p_a\neq p_b$ where $p_a$ and $p_b$ refer to momenta? (Assume $a_p^{\dagger}|0\rangle = |p\rangle$, where $a_p^{\dagger}$ is creation operator) If so, is this the reason why we use momentum space for most QFT calculations?
Both momentum and position space are not properly normalizable in Hilbert space. How do we really deflect away these problems in practice?