My Professor in QFT did a move which I cannot follow:
Given the state $$\hat\phi|0\rangle = \int \frac{d^3p}{(2\pi)^3 2 E_p} a^\dagger_p e^{- i p_\mu x^\mu}|0\rangle,$$ he wanted to show that this state is an eigenstate of the position operator. Therefore he applied the position operator in the momentum representation which is $$\hat X^\mu = i\frac{\partial}{\partial p_\mu}.$$
Then a miracle for me appears as he interchanges the derivative and the integral hat hence gets $$\hat X^\mu \hat\phi|0\rangle= \hat X^\mu \int \frac{d^3p}{(2\pi)^3 2 E_p} a^\dagger_p e^{- i p_\mu x^\mu} = i \frac{\partial}{\partial p_\mu}\int \frac{d^3p}{(2\pi)^3 2 E_p} a^\dagger_p e^{- i p_\mu x^\mu} = x^\mu \hat \phi |0\rangle.$$
I cannot see why he is allowed to interchange the integral and the derivative.
Answer
Here is the answer (I will not consider the constants on the denominator of your Fourier transform for simplicity, however they are there ;-) ). When you write the operator $\hat{\phi}$ you have to be careful. I will drop the hats, because it will be clearer I think (maybe here the hat stands for an operator and not for the fourier transform). Your operator is not in the momentum representation, since you have the integration over $p$. That operator depends on $x$, and can be written as $\phi(x)$.
Denote the fourier transform by $\mathscr{F}$. It is a unitary transformation on $L^2$, so when acting on an operator $A$ of that space it is often written as $\mathscr{F}A\mathscr{F}^{-1}$. Because, by unitarity, $$\langle\psi_1, A\psi_2\rangle=\langle\mathscr{F}\psi_1,\mathscr{F}A\psi_2\rangle=\langle \mathscr{F}\psi_1,(\mathscr{F}A\mathscr{F}^{-1})\mathscr{F}\psi_2\rangle\; .$$ Thus the correct writing is that the position operator in the fourier transform representation is the derivative w.r.t. $p$: $$\mathscr{F}X^\mu\mathscr{F}^{-1}=+i\partial/\partial p_\mu\; .$$ In the same spirit, your operator $\phi(x)$ becomes in the Fourier transform, and acting on the vacuum (to follow your notation, the momentum vacuum $\lvert 0\rangle=\mathscr{F}\lvert 0_x\rangle$, while $\lvert 0_x\rangle$ is the position vacuum): $$\mathscr{F}\phi(x)\lvert 0_x\rangle=a^{\dagger}_p\lvert 0\rangle\; .$$ Therefore: $$X_\mu\phi(x)\lvert0_x\rangle= \mathscr{F}^{-1}(\mathscr{F}X_\mu\mathscr{F}^{-1})(\mathscr{F}\phi(x)\mathscr{F}^{-1})\lvert 0\rangle=i\mathscr{F}^{-1}\frac{\partial}{\partial p_\mu}a^\dagger_p \lvert 0\rangle\\=i\int dp \,e^{ip_\nu x^\nu}\frac{\partial}{\partial p_\mu}a^\dagger_p \lvert 0\rangle\; .$$ Now if you "integrate by parts" on the last term (in the sense of distributions) you get: $$X_\mu\phi(x)\lvert0_x\rangle=-i\int \, dp \Bigl(\frac{\partial}{\partial p_\mu}e^{ip_\nu x^\nu}\Bigr)a^\dagger_p\lvert 0\rangle = x^\mu \phi(x)\lvert 0_x\rangle\; .$$
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