Confusion in normalization of position space and momentum space

Question

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 $|x,t\rangle = \phi(x)|0\rangle$)

1. 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?

2. 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?

3. Both momentum and position space are not properly normalizable in Hilbert space. How do we really deflect away these problems in practice?

Answer 1

If you define $|x,t\rangle = \phi(x)|0\rangle$, then the inner product given by $\langle y, t|x,t\rangle$ gives you an equal time two point function, which isn't necessarily 0 if $x\neq y$. You are confusing this with position space inner product, so note that your definition of $|x,t\rangle$ isn't the position space eigenket, but it is a state obtained by acting the scalar field operator on the ground state. So parts 1 and 3 are resolved.

As for part 2, since $a_p^{\dagger}|0\rangle = |p\rangle$, I would ask you to calculate the mentioned quantity yourself as an exercise, and then we can discuss what you found.

The reason for using momentum space for many textbook calculations is because standard stuff, eg correlation functions, Feynman diagrams are easy to compute, instead of in position space.