Why isn't the $\lvert\phi(x)=0\rangle$ Fock eigenstate the vacuum state?

Question

If, in a QFT of a scalar field $\phi$, a Fock space $n$-particle position eigenstate $\lvert x_1\cdots x_n\rangle $ is given by $$ \lvert x_1\cdots x_n\rangle =\hat\phi^\dagger(x_1)\cdots\hat\phi^\dagger(x_n)\lvert 0\rangle \,, $$ where $\lvert 0\rangle $ is the vacuum state, then we have $$ \langle x_1\cdots x_n\lvert\phi\rangle =\phi(x_1)\cdots\phi(x_n)\langle 0\lvert\phi\rangle \,, $$ with $\hat\phi(x)\lvert\phi\rangle =\phi(x)\lvert\phi\rangle $.

Now, the question is what is the value of $\langle 0\lvert\phi\rangle $?

If there's a vacuum configuration state $\lvert\phi_0\rangle = \lvert0\rangle$, then $\lvert 0\rangle $ is orthogonal to any other configuration in the basis, and we would have $$ \langle x_1\cdots x_n\lvert\phi\rangle = 0 $$ except for the case $\langle 0\lvert 0\rangle $ and thus a joined basis $\{\lvert x_1\cdots x_n\rangle ,\lvert\phi_0\rangle \}$ for a space of higher dimension than the Fock space. So, I conclude there's no vacuum configuration. But the zero configuration, $\phi(x)=0$, is anyway orthogonal to every Fock space state $\lvert x_1\cdots x_n\rangle $ different from $\lvert 0\rangle $, and thus excludes the presence of any particle. In other words, $$ \lvert\phi(x)=0\rangle=\lvert0\rangle\langle0\lvert\phi(x)=0\rangle $$ So, the zero field configuration corresponds to the vacuum state, a contradiction; what am I missing? What's wrong in all this?

Answer 1

The state $|\phi\rangle$ is a coherent state, which has a non-zero overlap with the vacuum state $|0\rangle$.

The vacuum state is defined by $\hat \phi (x)|0\rangle=0$ for all $x$. The vacuum state is also given by the coherent state $|\phi(x)=0\rangle$ for all $x$. Furthermore, one should keep in mind that the coherent state basis is overcomplete, and therefore $\langle \phi|\phi'\rangle\neq0$ for any $\phi$ and $\phi'$. All this properties solve the OP's problem.

To see that, one can look at the simpler case of the harmonic oscillator, and everything should be clear.