I am now reading the David Tong's lecture notes on quantum field theory.
http://www.damtp.cam.ac.uk/user/tong/qft/two.pdf
And I have some questions on whether there is some well-defined particle annihilation operator at position $\bf x$.
I already know that we can interpret $\phi ({\bf x}) \left| {0} \right\rangle$ as a single-particle state at position $\bf x$. However, it seems that we can not say that $\phi ({\bf x})$ is the creation operator for a particle at $\bf x$. This is because it is a real scalar field which satisfies $\phi ^{\dagger} ({\bf x}) = \phi ({\bf x})$. Therefore, if we try to interpret $\phi ({\bf x})$ as the particle creation operator then we will totally get lost of the corresponding particle annihilation operator at $\bf x$.
Therefore I am confused, is there any particle annihilation operator at position $\bf x$ in this rather simple real scalar field theory?
I would be grateful for any suggestion! Thanks!