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In the book Field Quantization of Greiner, in section 3.2 he introduces the field operators (for bosons), that are postuleted to satisfy the commutation relations $$[\hat{\psi}(\textbf{x},t), \hat{\psi}^{\dagger}(\textbf{x}',t)] = \delta(\textbf{x} - \textbf{x}')$$ $$[\hat{\psi}(\textbf{x},t),\hat{\psi}(\textbf{x}',t)] = [\hat{\psi}^{\dagger}(\textbf{x},t), \hat{\psi}^{\dagger}(\textbf{x}',t)] = 0.$$ Defining the vacuum state $| 0\rangle $ such that $\hat{\psi}(\textbf{x},t)|0 \rangle = 0$, he argues that the state $$|\textbf{x}_{1},...,\textbf{x}_{n}; t \rangle \equiv \frac{1}{\sqrt{n!}}\ \hat{\psi}^{\dagger}(\textbf{x}_{1},t)\cdots \hat{\psi}^{\dagger}(\textbf{x}_{n},t) |0\rangle$$ can be interpreted as a state of a system of $n$ (identical) particles localized at positions $\textbf{x}_{i}$ at the instant $t$. Thereby, it's easy to show that $\hat{\psi}^{\dagger}(\textbf{x},t)$ can be interpreted as an operator that creates a particle at position $\textbf{x}$ in the instant $t$, because $$\hat{\psi}^{\dagger}(\textbf{x},t) |\textbf{x}_{1},...,\textbf{x}_{n}; t \rangle = \sqrt{n + 1} \ |\textbf{x}, \textbf{x}_{1},...,\textbf{x}_{n}; t \rangle $$

My doubt is about the interpretation of the operator $\hat{\psi}(\textbf{x},t)$. I know that it annihilates a particle, but I would like to interpretated its action in the state above. Applying this operator and using the commutation relations, I have $$\hat{\psi}(\textbf{x},t) |\textbf{x}_{1},...,\textbf{x}_{n}; t \rangle = \frac{1}{\sqrt{n!}} \sum_{i} \delta(\textbf{x} - \textbf{x}_{i}) |\textbf{x}_{1},..., \textbf{x}_{i-1}, \textbf{x}_{i+1}, ...,\textbf{x}_{n}; t \rangle.$$ How can I interpretated this? To show that $\hat{\psi}(\textbf{x},t)$ is in fact a operator that destroy a particle at position $\textbf{x}$? If I take $\textbf{x} = \textbf{x}_{1}$, for exemple, the sum will diverge.

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The problem here is one of normalization. The state you've written down represents a state having been annihilated by the field operator only when the position of the field operator matches the position of one of the current particles in the state. Otherwise, the result vanishes. Of course, the state you have will not be normalized, and cannot be normalized without imposing some sort of ultraviolet cutoff into your theory (i.e., putting it on a lattice).

These reasons are why, when doing quantum field theory "rigorously," one tends to work with so-called "smeared" operators. Given a function $f$ on spacetime, the smeared operator $\psi_f$ is given by

$$\psi_f=\int\mathrm{d}^dx f(x)\psi(x).$$

Working with these operators intuitively "smooths out" the infinitely localized single-particle states created by the field operator alone, and makes dealing with problems of normalization much easier.

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