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I have heard it explained that a complex doublet scalar field produces the Higgs boson, and also three Goldstone bosons which are absorbed by $W$ and $Z$ bosons, giving them mass.

Does this mean that the $W$ and $Z$ absorb (and presumably emit) Goldstone bosons as they move through this scalar field?

My main question is: does this mean that their mass is the fluctuating result of an approximate energy equilibrium reached in a stochastic process?

A couple of tangential questions:

  • Does this mean that empty space is full of Goldstone bosons which are doing nothing in the absence of any W or Z bosons?

  • Does this mean that two very close W and Z bosons will have to share Goldstone bosons and hence lower mass?

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Your fantasy picture of the standard Higgs mechanism is off.

The field d.o.f. metaphorically "absorbed" by the Ws and Z would have been Goldstone particles, in the absence of coupling to would-be-gauge fields, but are "now" bonafide d.o.f. of massive vector fields, propagating as such in an asymmetric vacuum, without bonafide goldstons, massless scalars in the spectrum.

So there is no meaningful background "give-and-take" stochastic fluctuation process, as you are conjuring up. (Where did this fantasy come from? Stochastic quantization? Not there.) There are no such Goldstone degrees of freedom in the vacuum or the theory's spectrum any more: they have been solidly reassigned to the vector bosons. The only d.o.f. remaining out of the original Higgs doublet is the Higgs boson, a massive excitation. No massless d.o.f. left over, beyond the escapee photon.

While people find it instructive to consider the "duck in the wolf's belly", Nuc Phys B261 (1985) 379-431, the so-called Equivalence theorem relating in high energies the longitudinal components of vector bosons to their associated would-be Goldstone bosons, it is counterproductive to imagine this morphing as a real statistical process and infer spurious consequences.

Im' not sure if you are really asking about vector boson bound states in your last subquestion, but binding as a shortage of edible goldstons in a degenerate vacuum makes no sense. Individually, mass does not vary in interactions with other particles.

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    $\begingroup$ It is just a variable change that does not change the number of d.o.f. ;-) $\endgroup$
    – Vladimir Kalitvianski
    Commented Aug 27, 2020 at 15:39
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    $\begingroup$ Of course; but such d.o.f. s have permanently moved to the vector bosons, in whatever coordinate system, and do not rematerialize as massless goldstons in any statistical sense. This is the essence of my answer. People often confuse [unphysical] gauge choices with actual, physical, features of the spectrum. $\endgroup$
    – Cosmas Zachos
    Commented Aug 27, 2020 at 16:00

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