From Peskin-Schroeder, p.212:
The term $$ \langle \Omega | \phi(x) | \Omega \rangle \langle \Omega |\phi(y) | \Omega \rangle$$ is usually zero by symmetry; for higher-spin fields, it is zero by Lorentz invariance.
Where $|\Omega \rangle$ is the ground state and $\phi(x)$ a scalar field.
How can I prove this?
Here is another take making use of the symmetry, as hinted in the text.
The vacuum state is by definition symmetric with respect to any Lorentz transformation and translation. Let $U(\Lambda)$ be a unitary representation of the Lorentz transformation $\Lambda$ acting on the states, and let $$T(a) \equiv \exp ( -i Pa) \equiv \exp (- i P_\mu a^\mu )$$ be the unitary spacetime translation operator. Then by Lorentz invariance we have $$ \tag{A-1} |\Omega \rangle = U(\Lambda) |\Omega\rangle, $$ $$ \tag{A-2}\langle\Omega | = U(\Lambda)^{-1} \langle \Omega |,$$ and by translational invariance we have $$ \tag{B-1} | \Omega \rangle = T(a) | \Omega \rangle, $$ $$ \tag{B-2} \langle \Omega | = \langle \Omega | T(a)^{-1}. $$ Moreover, $U(\Lambda)$ and $T(a)$ act on a scalar field $\varphi(x)$ as $$ \tag{C-1} U(\Lambda)^{-1} \varphi(x) U(\Lambda) = \varphi(\Lambda^{-1}x), $$ $$ \tag{C-2} T(a)^{-1} \varphi(x) T(a) \equiv e^{iPa}\varphi(x) e^{-iPa} = \varphi(x-a). $$ Using these relations we have $$ \tag{D-1} \langle \Omega | \varphi(x) | \Omega \rangle = \langle \Omega | U(\Lambda)^{-1}\varphi(x)U(\Lambda) | \Omega \rangle = \langle \Omega | \varphi(\Lambda^{-1}x) \Omega \rangle, $$ $$ \tag{D-2} \langle \Omega | \varphi(x) | \Omega \rangle = \langle \Omega | \varphi(x+a) | \Omega \rangle, $$ for any Lorentz transformation (whose vector representation is) $\Lambda$ and for any 4-vector $a$. In particular from (D-2) you can see that $$ \tag{E} \langle \Omega | \varphi(x) | \Omega \rangle = c,$$ where $c$ does not depend on the space-time variable $x$. The answer of @lionelbrits explains why this constant can often be considered to be 0.
For a spinor field $\psi_\alpha(x)$ (or a vector field for that matter) you can use the same argument as above to obtain $$ \tag{F} \langle \Omega | \psi_\alpha(x) | \Omega \rangle = \langle \Omega | \psi_\alpha(0) | \Omega \rangle \equiv c_\alpha,$$ but now you have to remember the non trivial action of the Lorentz group on a spinor field: you have $$ \tag{G} U(\Lambda)^{-1} \psi_\alpha(x) U(\Lambda) = S(\Lambda)_\alpha^{\,\,\,\beta} \psi_\beta(\Lambda^{-1} x), $$ where $S(\Lambda)$ is the spinor representation of the Lorentz transformation. Using (G) in (F) you obtain: $$ \tag{H} c_\alpha \equiv \langle \Omega | \psi_\alpha(0) | \Omega \rangle = S(\Lambda)_\alpha^{\,\,\beta} \langle \Omega | \psi_\alpha(0) | \Omega \rangle \equiv S(\Lambda)_\alpha^{\,\,\beta} c_\beta, $$ and due to the arbitrariness in the choice of the Lorentz transformation this implies that $c_\alpha=0$, i.e. remembering (F), $$ \langle \Omega | \psi_\alpha(x) | \Omega \rangle = 0.$$
It's a bit more involved than creation and annihilation operators. What is meant is that the field has no vacuum expectation value, and in a way this is by construction. You may think about why $\langle x \rangle = 0$ for a harmonic oscillator (or better yet, an an harmonic one). It's because $x\to -x$ is a symmetry of the Hamiltonian. It's not true for cubic potentials, though, but those theories can be sick.
In many situations, when we quantize a field, we perform perturbation theory around it's classical solution, and the dynamical field is the total field minus is vev, I.e. we write $\Phi = \langle \Phi \rangle + \phi$, and then work in terms of $\phi$. Thus the vev of the dynamical field vanishes by construction.
Edit:
To clarify, the reason you can't rely on creation and annihilation operators is because $\left\langle \Omega | a |\Omega\right\rangle$ is only zero if $\Phi$ has a zero vev, (circular argument). Imagine trying to naively quantize the Higgs field, which has a "Mexican hat" style potential where the minimum is located away from $\Phi = 0$. In that case, you do not have $\left\langle \Omega | \Phi |\Omega\right\rangle$ until you redefine the field to $\Phi = \langle \Phi \rangle + \phi$.
