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.

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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.

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Dear Luboš, does the Higgs condensate exist everywhere in vacuum (in empty space) or it only surrounds particles whith which Higgs field interacts? – Vladimir Kalitvianski Nov 27 '12 at 12:43
Everywhere in the vacuum. – Luboš Motl Dec 4 '12 at 8:35
"...One must emphasize that the coherent state construction above may be expanded in terms of real particles..." But if it is so, then how to deal with both the dispersion relation $p^{2} = m_{h}^{2}$ for the Higgs boson one-particle state and the requirement of translational invariance? The nonzero dispersion relation provides requirement that VEV $$v = \langle 0|\hat{H}|0\rangle \sim cos(m_{h}t + \varphi),$$ while translational invariance requirement says us that the $v$ hust be coordinate independent. – Name YYY Apr 24 at 13:27
In my opinion, the case is similar to the case of Coulomb law, where photons are off shell, i.e., something which is called "virtual particle". – Name YYY Apr 24 at 18:46
Or to the case of constant magnetic field. – Name YYY Apr 24 at 18:59

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.

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Sorry but this is simply not the case. The basic formula for the new vacuum is cutoff-independent and only depends on low-energy physics. It's just a damn simple coherent state. It's also wrong to say that the Higgs condensate is made out of "virtual particles". Virtual particles, by definition, have finite life-time but the Higgs "ocean" is permanent. They're damn real particles that may be added to each scattering process. The Higgs vev is what adds "tadpoles" - vertices with 1 external leg which may be attached to Yukawa and other vertices just like external lines. – Luboš Motl Nov 27 '12 at 10:12
@LubošMotl: My answer is correct. In field theory, Bogoliubov transformations corresponding to field shifts are not unitarily realizable (except with cutoffs). They are therefore not well-defined. The coherent space you are talking about in you answer does not live in the Fock space generated by the unshifted vacuum. – Arnold Neumaier Nov 27 '12 at 10:18
There are cutoffs everywhere in QFT but it's wrong to dismiss the question as one having a cutoff-dependent answer because the cutoff dependence only affects the loop corrections (proportional to higher powers of the Planck constant), much like in the case of any other quantity calculable in a QFT. However, in this case, much like in most others, the leading term in the number of particles is cutoff-independent and calculable from the classical field theory. – Luboš Motl Nov 27 '12 at 10:22
Your comment that it's a completely wrong question because of "cutoff dependence" is equally wrong as if you dismissed a question about the electron-positron annihilation cross section or any other question in QFT because it is "cutoff dependent". To properly calculate higher-order corrections, one always needs to introduce the cutoffs and deal with the cutoff-dependent objects correctly. But in all cases, there's also the leading classical term which isn't cutoff-dependent, and the Higgs condensate is really a low-energy, in fact $p=0$, infrared physics. – Luboš Motl Nov 27 '12 at 10:25