I understand from the internet that the Higgs particle is highly unstable! It decays as soon as it is created. If it is so unstable, how one can say that the Higgs field exist? Just like, if photons are unstable, there won't be electromagnetic field. I cannot imagine a field whose particle is highly unstable. Also, it is said that the mass depends up on how the matter interact with this higgs field. If the higgs bosons are highly unstable, how matter can interact with those unstable higgs bosons?

# [Physics] How the Higgs field exist if the Higgs boson is unstable

higgsparticle-physics

#### Related Solutions

The difficulty with Higgs boson is it's high mass, so in order to create it, you need lots of energy (125GeV, using $E=mc^2$).

What is important to give particles mass is s the Higgs field, not the Higgs boson (which is an excitation of the field).

The problem is that you have mixed the concept of real particles and "virtual" or "force carrier" particles. The latter can't be observed and can be created spontaneosly, because the energy requiered is "borrowed" via Heisenberg's Principle ($\Delta E\Delta t \geq \frac{\hbar}{2}$).

Comparing to the analogy you made: two charges will attract/repell, via EM interaction without photons being present. The EM is mediated by virtual photons, but these are not physically observable, unlike "light" photons.

The plot here is showing events with two photons, with the invariant mass $m_{\gamma\gamma}$ of the two photons plotted. The invariant mass is given by $m_{\gamma\gamma}=\sqrt{(E_1+E_2)^2-(\vec p_1+\vec p_2)^2}$. This particular quantity is of interest because if the two photons came from a decay $\rm X\to\gamma\gamma$, then by conservation of four-momentum, $m_{X}=m_{\gamma\gamma}$.

On the top plot, the lower red line is what we expect to see from Standard Model background events. That is to say, basically everything *except* $\rm H\to\gamma\gamma$. The lower plot shows the same data, but with these background events removed, so it shows only the Higgs peak and some statistical noise.

These peaks show that we have more events with two photons with a specific invariant mass than we would expect without the Higgs boson. In other words, they can be interpreted as a measurement of $\rm H\to\gamma\gamma$.

The Feynman diagram you show is one contributor to the result above. The $W$ boson can be replaced with any charged Standard Model particle and that will also contribute. But the Higgs boson couples more strongly to more massive particles, so the primary contributing diagrams are the one you show and a similar one with the W boson replaced by a top quark. The next heaviest charged particle in the Standard Model (the bottom quark) is much lighter and so doesn't contribute that much.

## Best Answer

A field and a particle are two different concepts and it is well that one should separate them.

We are in quantum mechanics when we are talking of the Higgs field.

The field in quantum mechanics is defined via operators.

So it is two separate concepts. The Higgs field exists from low to high energies, but it is only at high energies that the Higgs boson can be on mass shell and subsequently decay. The Higgs boson is not a gauge boson, it does not define the interactions possible within the standard model between particles, as the photon, W and Z or the gluons do. It is a manifestation of the underlying field everywhere, when the energy available allows it, and the vacuum expectation value of this field is 246 GeV..

Particles acquire mass through the symmetry breaking mechanism of the Higgs field, not interacting with the Higgs boson/particle which is just a manifestation that had to exist if the symmetry breaking model is correct.