Fields with non-vanishing vacuum expectation value In this post Arnold Neumaier wrote on necessity of vanishing vacuum expectation value if we want to have a particle interpretation of our field $\phi$:

As is easily checked, fields linear in creation and annihilation operators (and hence amenable to a particle interpretation) have zero vacuum expectation value. Thus the $\phi$ field with its non-vanishing vacuum expectation value cannot be given a particle interpretation.

I not see why it's necessary if we want have a meaningful interpretation as particle to demand that the vacuum expectation value should be zero. What is precisely running wrong if we deal with a field with non-vanishing expectation value physically and mathematically?
What are moreover in general the main drawbacks when we dealing with field theories with non-vanishing expectation values? I saw several times stated that this causes also troubles if we apply perturbative methods. Could somebody give a concrete example where perturbative calculations/reasonings going wrong if we apply them to such field theories?
 A: First, the only kinds of fields that can possibly have non-zero vacuum expectation values are scalar fields. This follows by Lorentz invariance.
Now let $\phi(x)$ be such a scalar field. Let us consider $\langle \phi(x)\rangle$. Let us first use translation invariance to argue that this must be constant. Indeed
$$\langle \phi(x)\rangle = \langle 0|\phi(x)|0\rangle = \langle 0|e^{-iP\cdot x}\phi(0)e^{iP\cdot x}|0\rangle=\langle 0|\phi(0)|0\rangle = \langle \phi(0)\rangle\tag{1}.$$
The main point here is that the vacuum is Poincaré invariant, and therefore $e^{iP\cdot x}|0\rangle = |0\rangle$.
Now we want to talk about particle interpretation. To do so we must look to the asymptotic region of spacetime and connect $\phi(x)$ properly with in/out fields $$\phi(x)\to \sqrt{Z}\phi_{\rm in/out}(x)\tag{2}$$
where we recall that $$\phi_{\rm in/out}(x)=\int \dfrac{d^3p}{(2\pi)^32\omega_p}(a_{\rm in/out}(p)e^{ipx}+a_{\rm in/out}^\dagger(p)e^{-ipx})\tag{3}.$$
Now because of (1) we can evaluate $\langle \phi(x)\rangle$ wherever we want. So push the point to the asymptotic region. There you can use (3) to write $$\langle \phi(x)\rangle = \sqrt{Z}\langle \phi_{\rm in/out}(x)\rangle\tag{4}$$
and now you can use (3). But now there is a tension. In fact, $a_{\rm in/out}(p)$ and $a^\dagger_{\rm in/out}(p)$ should annihilate the vacuum, when acting respectively to the right and to the left, and so the result should be zero. So what is wrong with the reasoning? Well, it is just that we are looking at the wrong variable. Define $$\varphi(x)=\phi(x)-\langle \phi(x)\rangle\tag{5}.$$
This field has zero vacuum expectation value and it does not run in that problem. So my take on this is that:

*

*Particle interpretation comes from (2) and (3) and you run into one problem if you try applying this to a field with non-zero VEV;


*Nevertheless there is nothing wrong. There is a particle interpretation associated to this field, provided you look to the right variable. Using (5) the appropriate variable is the deviation from the VEV. And in a sense this makes sense: particles are meant to be excitations about a ground state and that is exactly what $\varphi$ should capture.
