I'm studying for an exam called Introduction to QFT. One of the main topics in this class is the quantized free fields.
I can now find the fields that solve the Klein-Gordon equation and the Dirac equation, but I don't really know what they do.
For example: if we are considering a Klein-Gordon scalar-free field, we would get something like this: $$ \phi(X) =\int \frac{d^3p}{(2 \pi)^{2/3} \omega_p} (e^{iP_\nu X^\nu} a_p + e^{-iP_\nu X^\nu} a^\dagger_p) $$ $$ P_0=\omega_p=\sqrt{m^2+p^2} $$
Now my problem is: what does this operator exactly do? I know what $a_p$ and $a^\dagger_p$ do, but I'm not really sure about $\phi(X)$.
We do also define $\phi(X) = \phi^+(X) + \phi^-(X)$, with $$ \phi^+(X) = \int \frac{d^3p}{(2 \pi)^{2/3} \omega_p} e^{iP_\nu X^\nu} a_p $$ $$ \phi^-(X) = \int \frac{d^3p}{(2 \pi)^{2/3} \omega_p} e^{-iP_\nu X^\nu} a^\dagger_p $$ My understanding is that $\phi^-(X)$ creates a particle in the space-time coordinate $X$, but is this true? If this was correct, $\phi^-(X)|0>$ would be an eigenfunction of the time-position operator.
Furthermore $\phi^{+\dagger}(X)=\phi^-(X)$, which means that $\phi^\dagger(X)=\phi(X)$, and so $\phi(X)$ is a self-adjoint operator. Does this mean that $\phi(X)$ is connected to some observable? What are its eigenvalues and eigenvectors?
My biggest concern here is on the effect of the operator $\phi(X)$: what does it do? does it define an observable?
I feel like these should be some very important questions for someone learning QFT, but I'm having a hard time finding the answers.
