What is the expectation value of the number operator when the vacuum has a VEV? The number operator N applied to a field whose vacuum has zero VEV gives $N|0>=0$. What if we apply it to the Higgs field?
The background of this question is that in popular scientific accounts, the Higgs field is sometimes described as a 'sea' of particles. I would like to clarify the meaning of this, and what is the physical interpretation of a non-zero VEV.
 A: The vacuum with $\langle 0|h|0\rangle=v$ has a shifted expectation value but it is given by a similar Gaussian as the would-be $h=0$ vacuum. To shift expectation values in this way, you have to construct the so-called "coherent state". The $h=v$ vacuum is a coherent state, too.
However, this coherent state still preserves the space and time translational symmetries, so only $p^\mu=0$ physical Higgs bosons are added to this condensate. The physical vacuum may be imagined as a coherent state similar to this one:
$$ |0\rangle = \exp(C\cdot a^\dagger_{p=0}) |h=0\rangle $$
Much like all coherent states in quantum physics (harmonic oscillators of any dimension), it refuses to be an eigenstate of the number operator $N$, of course. If you Taylor-expand the exponentials, you will see that the coherent state is a linear superposition of states with different values of $N$ – with different powers of the creation operator.
However, you may still calculate the expectation value of $N$. You will find out that because $a^\dagger,a$ are normalized to the 3D delta-function, the expectation value of $N$ will scale proportionally to the three-dimensional volume of the region. So $\langle N \rangle \sim V$. To calculate the coefficient, one needs to do some maths and the result won't be too meaningful anyway because the $h=0$ state isn't an energy eigenstate for the physical choice of the potential, anyway. And even if you allowed different potentials to be used for the two vacuum states, the two Gaussian "wave functional" will probably have a different width as well so that one will be not just coherent but "squeezed" state constructed out of the other.
But one may still "morally" assign an order-of-magnitude estimate to the coefficient. By dimensional analysis, a power of the electroweak scale is the only dimensionful constant that may enter. Therefore,
$$\langle N \rangle = K\cdot V\cdot m_{ew}^3 $$
where $K$ is a numerical constant of order one and $m_{ew}$ is of order of the Higgs vev $v$ or Higgs mass $m_h$. There is roughly one Higgs boson in the condensate per "electroweak volume"; the latter is roughly a cube with side $10^{-18}$ meters.
One must emphasize that the coherent state construction above may be expanded in terms of real particles. We are talking about states in the Hilbert space so of course all of them belong to the "Fock space" which is only made out of real particles. It makes no sense to say that we're adding "virtual particles". Virtual particles are just additions to a particular "process" (a Feynman diagram), not additions to a state in the Hilbert space! And they're short-lived while the Higgs condensate is static and infinitely long-lived.
The addition of the Higgs condensate that shifts the vev is equivalent to adding a new "linear" vertex to to the Feynman rules of the original theory. This linear vertex only has 1 external leg. It may be attached by Higgs propagators to other vertices, e.g. the cubic Yukawa vertex, and this propagator looks like an external leg and produces mass terms for the fermions (in the example of the Yukawa vertex) out of the Yukawa cubic vertex. Similarly for other interactions.
A: If a field $\phi(x)$ has a nonzero VEV $v$, the field whose Fourier components define the creation and annihilation operators is $\phi(x)-v$, which has a zero VEV, and again $N|0\rangle=0$. 
An interpretation in terms of the original $\phi$ is highly ill-defined and cutoff-dependent, and cannot be given a sensible physical meaning, although formally, it looks like a sea of indefinitely many virtual particles.
