Field operators in Klein Gordon field: creation of a particle at position $x$ Let's start with the quantised field operators $\phi(\mathbf{x})$ and $\pi(\mathbf{x})$, defined as usual, and the relativistic normalisation $|\mathbf{p}\rangle = \sqrt{2E_p}a_p^{\dagger}$.
I know that :
$$\phi(\mathbf{x}) \left|0\right> = \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_p}e^{-i\mathbf{p} \cdot \mathbf{x}} \left|\mathbf{ p}\right> , $$ which can be interpreted as $\phi(\mathbf{x})$ creating a particle at $\mathbf{x}$, according to Peskin and Schroeder, page 24.
How can we interpret $$\pi(\mathbf{x}) \left|0\right> = \int \frac{d^3p}{(2\pi)^3}\frac{-i}{2}e^{-i\mathbf{p} \cdot \mathbf{x}} \left|\mathbf{ p}\right> ?$$
 A: OK in the end I think I got somewhere on my own. I am just posting this here in case it is useful for someone else.
The reason $\phi(\mathbf{x}) \left|0\right>$ can be interpreted as creating a particle at position $\mathbf{x}$ is because: $$\phi(\mathbf{x}) \left|0\right> = \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_p}e^{-i\mathbf{p} \cdot \mathbf{x}} \left|\mathbf{ p}\right> ,$$ and if we take $\langle \mathbf{x}|\mathbf{p} \rangle  =  $$\langle {0}|\phi(\mathbf{x}) \left|\mathbf{p}\right>$, we get $e^{i\mathbf{p}\cdot\mathbf{x}}$ which is analogous to the quantum mechanical momentum eignestate in the position representation. 
If we look at $\langle {0}|\pi(\mathbf{x}) \left|\mathbf{p}\right>$, we get $-i\,E_{\mathbf{p}}\,e^{i\mathbf{p}\cdot\mathbf{x}}$.
Now recall that $\pi(\mathbf{x}) \equiv \frac{\partial \phi(\mathbf{x})}{\partial t} $ for the Klein-Gordon Lagrangian. We can extend $e^{i\mathbf{p}\cdot\mathbf{x}} $ to $e^{-ip_{\mu}\,x^{\mu}}$ in order to include time (using the $-+++$ signature). So taking the derivative of $\langle {0}|\phi(\mathbf{x}) \left|\mathbf{p}\right> = e^{i\mathbf{p}\cdot\mathbf{x}}$ with respect to time brings down a factor of $-i\,E_{\mathbf{p}}$. 
Basically what I am saying that it all makes sense, $\pi$ is still just the derivative of $\phi$. There is nothing to interpret really.
