I put the question right now, writing some definitons later.
I want to calculate $ ||\hat\Psi^+(x)|0\rangle||^2$, where $$\hat\Psi^+(x) = \sum_{k = 1}^{\infty}u^*_k(x) \hat a^+_k $$ is a field operator. So I do $$ ||\hat\Psi^+(x)|0\rangle||^2 = \langle 0|\hat\Psi(x)\hat\Psi^+(x)|0\rangle$$ Using the field operator's commutation relation, this is equal to $$ \langle 0|(\delta(x-y) \pm \hat\Psi^+(x)\hat\Psi(x))|0\rangle$$ Thus, due to the fact that every annihilation operator destroys the vaccum state, only the delta term survives, $$ \langle 0|\delta(x-y)|0\rangle$$ But this results ends up being infinite, thus losing any physical sense. I can understand this but I'm not able to mathematically demonstrate it. Can someone enlighten me?
EDIT
For the sake of clarity, I'll add more information which I actually understand. In the course of statistical mechanics I'm following, we're facing the second quantization formulation. In particular, the ladder operators $\Psi$ and $\Psi^+$ have been introduced, and they've been defined as
\begin{equation}
\hat\Psi(f) = \sum_{k = 1}^{\infty} f_k \hat a_k\, ; \qquad \hat\Psi^+(f) = \sum_{k = 1}^{\infty} f^*_k \hat a^+_k
\end{equation}
They act by creating-destroying a particle in the state
$$ |f\rangle = \hat\Psi^+|0\rangle$$
This should be equivalent of saying that a particle is in the state $f(x)=\sum_{k = 1}^{\infty}f_k u_k(x)$, in the coordinates representation. In what I just wrote $k$ denotes the $k$-th state, $u_k(x)$, $u^*_k(x)$ are the $k$-th element (and its complex coniugate) of the orthonormal basis of the Foch space, $f_k$, $f^+_k$ are scalars, and $\hat a_k$, $\hat a^+_k$ are the annihilation and creation operators of a particle in a state $k$.
After that, the quantum field operators $\hat\Psi(x)$ and $\hat\Psi^+(x)$ have been introduced, and they've been defined as
\begin{equation}
\hat\Psi(x) = \sum_{k = 1}^{\infty} u_k(x) \hat a_k\, ; \qquad \hat\Psi^+(x) = \sum_{k = 1}^{\infty}u^*_k(x) \hat a^+_k
\end{equation}
It's been said that in a sense they are not true operatores, since every ladder operator is weighted by a function, and not by a scalar, like the creation-annihilation operators $\hat\Psi^+(f)$ and $\hat\Psi(f)$ defined before. Nevwertheless they are useful since every $\hat\Psi^+(f)$ can be written as
$$ \hat\Psi^+(f) = \int d^3x \, \hat\Psi^+(x) f(x)$$
