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?


1 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.

  • $\begingroup$ Well for part 2, it should be zero, by Fock space commutation relation. I just wanted to make sure. $\endgroup$ Commented May 10, 2018 at 10:38
  • $\begingroup$ @BrionBrion Yes. But it is not true that this is the reason we use momentum space representation. Objects like correlation functions are easy to keep track of in momentum space. $\endgroup$
    – Bruce Lee
    Commented May 10, 2018 at 10:42
  • $\begingroup$ I already accepted the answer, but one more question: suppose we do scattering experiment. We prepare some particle in some position (OK, realistically sharp localization is not possible, but that's same with momentum as well) and run some scattering experiment. I thought we technically can talk about some particle at some position though only asymptotically. Is this correct understanding, or there is simply no localization even asymptotically? $\endgroup$ Commented May 10, 2018 at 10:49
  • $\begingroup$ Let's assume we don't have any interaction Hamiltonian at all, just free scalar theory. The solution of that is a sum over momentum modes. The quanta of the theory are defined associated with these modes, their coefficients (upon quantization, creation/annihilation operators) create or destroy particles acting on states in the Hilbert space. If you act this on the ground state, you get 1-particle states. Now in scattering processes, you start asymptotically with these 1-particle states, and then make them interact with some interacting Hamiltonian. $\endgroup$
    – Bruce Lee
    Commented May 10, 2018 at 11:08

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