Physicist here: nope. The coupling between the Higgs field and other particle fields is what creates the phenomona of mass. The Higgs particle is just an excitation of that field.
Virtual particles aren't "real" in the same way that the first order Taylor series of sin(x) being x doesn't max x a real sinusoidal wave. All it does is let you make your calculations "close enough" for your accepted definition of "close enough."
Is it fair to say that both virtual Higgs and 'real' Higgs are two types of excitations of that field, except that virtual particles can't be directly observed?
I find the name virtual particles interesting, since they don't seem to be exactly the same kind of ripples as particles, and physicists frequently seem to disagree on how 'real' they are as a matter of interpretation
They are great for trying to help human brains make concrete the details of the process. But again they are just artifacts of the way we do our computations.
And the Higgs field is a particularly great example of this. If you shift a sin wave 90 degrees it becomes a cosine wave. If you don’t expand about the lowest energy of the Mexican top hot potential for the Higgs field it creates an entirely different description of the physical phenomena and virtual particles.
To save someone else the trouble, no, that wasn't an autocorrect mistake on "Mexican."
https://en.m.wikipedia.org/wiki/Spontaneous_symmetry_breakin...
> Also a countertop swivel chair physicist, my understanding is that Higgs bosons are recruited as virtual particles all the time to bestow other particles their mass.
I think you're mixing up two subtle ideas. One is virtual particles, which as lanza explains, is really a name for a step in a calculation. The other is the idea of vacuum expectation value, or vev.
As another commenter mentioned, a particle is (loosely) like a ripple on a pond. The vev tells you how much water there is, even at rest. With a simple harmonic potential (parabola shaped) the vev is the x-coordinate of the turning point (ie. the minimal energy), which we usually put at zero because of symmetry. However, the 'Mexican hat potential' respects the symmetry but has minima that aren't located at zero radius. The Higgs potential is like that, and so the Higgs has a nonzero vev.
The way the Higgs interacts with fermions implies that the masses are proportional to the Higgs vev.
Rather than it being virtual Higgses, it's more like the particles are swimming in a viscous liquid. Sleek particles aren't bothered and are hardly slowed down at all (these particles have not much mass) and some are not hydrodynamic and have a hard time.
The analogy is imperfect: the viscous soup doesn't provide friction in the way that water does (you can't give your momentum to the vacuum).
You can also create bona-fide Higgs particles, but you have to put in a lot of energy to do it (like the LHC does), and it will proceed to immediately disintegrate into lighter particles, which is what the LHC is observing in detectors
I think that when particles use the Higgs field to get their mass, the virtual particle is only a temporary ripple living on borrowed time, so it can interract with other fields but has to go back to nothing when the interaction is over, like a local variable in a function.
When you have a real Higgs, if it decays into say two photons, then yes you have a wave in the Higgs field dissipating, while equal energy/charge/<other conserved quantities...> waves replace it in other fields
(IANAP. This does not constitute physical advice.)