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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 $$ \hat\Psi^+(f) = \int d^3x \, \hat\Psi^+(x) f(x)$$

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    $\begingroup$ Hi Mattia, you've just discovered the dirty secret that quantum fields are not operators. Instead, they are "operator-valued distributions". I'm not able to enlight you about the mathematical details, but roughly it means that they are properly interpreted as linear maps $f\mapsto \Psi (f)$, where $f$ lives in some space of regular funcions. $\endgroup$ – pppqqq Nov 24 '16 at 18:39
  • $\begingroup$ Are your $\Psi^+,\Psi$ the creation/annihilation operators more commonly written as $a^\dagger,a$, the Fourier modes of a free field? Where do you get the expression $\Psi^+(x)$ from? Also, why exactly are you surprised by this result? Try calculating $\langle p \vert p\rangle$ or $\langle x\vert x\rangle$ in standard non-relativistic one-particle QM, you get the same sort of non-sensical result. $\endgroup$ – ACuriousMind Nov 24 '16 at 18:49
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    $\begingroup$ @ACuriousMind No, $\Psi$ in his post are usual field operators, not creation/annihilation operators. Substitute $u_k=\exp(i p_k x)$ and you will understand what's happening. And NO, it's not the same as $\langle x|x\rangle$ - $\hat{x}$ in QM acting on normalizable states gives normalizable states (and simply don't have eigenstates in Hilbert space), but field "operator" doesn't. The reason was explained by pppqqq, and his comment deserves to be turned into real answer. $\endgroup$ – OON Nov 24 '16 at 21:48
  • $\begingroup$ Thank you guys, you have for real helped me a lot :) anyway, I'm sorry maybe I put too much things too much badly in the question, which actually can be simply stated as " why is <0| delta|0> = infinity?". I'm sure I'm missing some basic result.. $\endgroup$ – MattiaBenini Nov 25 '16 at 0:45
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The state $\Psi^+(\mathbf{r})|0\rangle$ means the particle created in the point $\mathbf{r}$, it is not very convenient to work with because wave function of this particle is the Dirac delta.

So if you need a practical recipe for calculation, you can try to smear out this state in space creating a particle with the wave function $f(\mathbf{r})$, i.e. working with the state $\int d\mathbf{r}\:f(\mathbf{r})\Psi^+(\mathbf{r})|0\rangle$ (in the end of calculations, you can turn $f(\mathbf{r})$ back into a Delta function). Its norm squared is $$ \left|\left|\int d\mathbf{r}\:f(\mathbf{r})\Psi^+(\mathbf{r})|0\rangle\right|\right|^2=\int d\mathbf{r}d\mathbf{r}'\:f^*(\mathbf{r}')f(\mathbf{r})\langle0|\Psi(\mathbf{r}')\Psi^+(\mathbf{r})|0\rangle\\ =\int d\mathbf{r}d\mathbf{r}'\:f^*(\mathbf{r}')f(\mathbf{r})\langle0|\delta(\mathbf{r}-\mathbf{r}')|0\rangle=\int d\mathbf{r}\:|f(\mathbf{r})|^2=1. $$

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