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.

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

  • $\begingroup$ Thank you, Adam. Consider the Fock space identity... Sorry, I'll edit the question. $\endgroup$ – Daniel Apr 20 '14 at 23:30
  • $\begingroup$ 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. $\endgroup$ – Adam Apr 21 '14 at 0:10
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    $\begingroup$ @Daniel: I've edited my answer to include the information given in my comment. If this answer is good, maybe you could accept it. $\endgroup$ – Adam Apr 21 '14 at 16:12
  • $\begingroup$ I just assumed that the configuration basis was orthogonal, it isn't. Problem solved. $\endgroup$ – Daniel Apr 21 '14 at 19:42
  • $\begingroup$ Oh, ok, done, I'm new at this. $\endgroup$ – Daniel Apr 22 '14 at 10:54

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