I recently read an old physics news about the Higgs boson where it was observed to decay into 2 photons and I was wondering why it wouldn't have decayed into a single photon with the combined energy of 2 photons?
Particle Physics – Why Can a Particle Decay Into Two Photons but Not One?
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Related Solutions
Isn't the universe full of Higgs bosons, making up the Higgs field?
No. In particle physics, it is understood that the underlying (more fundamental) object is the field, not the particles. Particles are excitations of the fields that can be measured, and always carry certain properties like charge, mass, spin etc. The field that you are most familiar with is the electromagnetic field, its excitations being the photons. In another field the excitations are electrons, in another still there are gluons etc.
And there is a Higgs field, whose excitations are the Higgs bosons. The Higgs field, in contrast to the electromagnetic field has a non-zero value even if there are no Higgs bosons there.
To have an analogy in mind, think of a room full of air. When I speak, there are sound waves moving around the air. The air is the Higgs field, the sound waves are the Higgs bosons.
Why can't we detect the ones that are there already, like we can other bosons such as photons?
Higgs bosons are very massive, as particles go, so they require a lot of energy to be created in collisions. Additionally they have a number of decays pathways, so when they are created, they decay rapidly. So, even if Higgs bosons are created all the time in the atmosphere, or in supernovae or other events, they are rare and hard to detect. That is why we set up an experiment that can reproduce millions of collisions a second so to accumulate enough data.
When we've made our Higgs out of pure energy, why does it instantly decay into other particles?
This is kind of misguided. There is no clear meaning of "pure" energy. Energy is a quantity that is assigned to various phenomena, yet is common and interchangeable between them all. We speak of kinetic energy, potential energy, mass-energy, etc. but none of these forms is "purer" in any specific sense. In the particle collisions, the kinetic and rest mass energy of the protons is concentrated in a small part of spacetime, and can be redistributed in the kinetic, potential and mass energy of other particles.
Once a particle is formed, it does not really matter what way it has been formed. Just like a radioactive nucleus has the same probability of decaying in the next $10$ minutes irrespective of how long it has survived until now, a Higgs boson will decay with a certain probability into the particles it can decay to.
Does it actually directly decay into other particles, or is it rather the case that it just turns back into pure energy, and then that energy produces other, less massive particles?
Here we end up a little in metaphysics.You will have different answers depending on the interpretation of QM you choose. All we observe is the protons that go in the collision, and the shower of particles that comes out after the collision, together with their energy. That's all. Quantum theory will give you the statistics of these observations, but not what happens between the two observations; that is (for now) metaphysics, because it is unobservable.
Strictly speaking, no Higgs boson has been observed, in the sense that no Higgs boson has collided with the detectors. We have calculated how the existence of the Higgs field will affect the measurements, we found that it would affect them in a particular way, we did the experiments and indeed found that signature. The experiments and theory match so well that it is inescapable that there is a Higgs field, even though we have not "seen" (with our eyes) any Higgs bosons.
To speak about the exact way in which one particle comes into existence and decays is a bit beyond present physics (also worth exploring in other questions).
The photons have intrinsic spin (or, better, helicity) one, so the pair can have odd orbital angular momentum, still conserving total angular momentum (which has to be zero, as the pion is spinless).
Specifically, the spins of the two photon can combine to give total spin $S=1$. This, conmbined with an angular momentum $L=1$, has a $J=0$ component which permits the pion to decay into two photons. You can check from the Clebsch-Gordan table that the final two photon wavefunction is symmetric under particle permutation, as required by Bose statistics.
Best Answer
No massive particle can decay into a single photon.
In its rest frame, a particle with mass $M$ has momentum $p=0$. If it decayed to a single photon, conservation of energy would require the photon energy to be $E=Mc^2$, while conservation of momentum would require the photon to maintain $p=0$. However, photons obey $E=pc$ (which is the special case of $E^2 = (pc)^2 + (mc^2)^2$ for massless particles). It's not possible to satisfy all these constraints at once. Composite particles may emit single photons, but no massive particle may decay to a photon.