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

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?

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.
• The coherent state basis is overcomplete, and two different coherent state are never orthonormal ($\langle\phi|\phi'\rangle\neq 0$). The vacuum is just the coherent state $\phi(x=)0$ for all $x$. As I said, look at the case of a single harmonic oscillator. Then the (free) field theory is just the case of an infinite number of harmonic oscillator. – Adam Apr 21 '14 at 0:10