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9

Well, it depends how you define "distinct fundamental particle". If you insist that Wigner's classification is what defines a particle, i.e. "particle = irreducible unitary representation of the Lorentz/Poincare group", then the photon is two particles, as you say. But, more commonly, we do not look at the particles like this - particles arise as the ...


1

All the other answers that "no, there is no triggering event, it just happens, quantum mechanics is like that" are perfectly right. But let's look at the experimental evidence for these answers. Yes indeed, there is considerable experimental evidence that heavily falsifies the idea that there is a triggering event. This evidence is the statistical ...


0

Consider a particle in a box, but where the box has a thin wall, and the energy level outside the box is lower than that inside. (This is, e.g., a neutron in an unstable heavy nucleus.) Follow the development of this wave in time. It will tunnel out eventually to the lower energy state and propogate away. From this, you can see that the decay is always ...


2

An " intuitive" approach is to consider that in QM, the exact location of particles doesn't exist. They're all probability waves, and you never have a 100% chance to find a particle in exactly one place. So for unstable nuclear atoms, the probability function of the protons and neutrons are smeared out even further. There's a significant non-zero ...


13

As the other answers state, the individual nuclei have a probability of decay and this happens randomly, as they sit there. You are correct though in wondering about a trigger, because at the atomic level that is exactly what happens with lasing, induced-emission = induced-decay. Spontaneous decay is random, controlled by the quantum mechanical individual ...


10

There really is none. Unstable elements (and unstable elementary particles) can decay into a less energetic state. However, each kind of decay depends on a quantum mechanical process, this is tunneling for $\alpha$, a virtual $W^\pm$ for $\beta$ or a transition from one nuclear shell to another for $\gamma$. Now these underlying processes can be strongly ...


6

Nothing happens! It's random! The nucleus is in an unstable state, and unstable states have a certain small probability to decay within a given amount of time (how small depends on the nucleus). There's not much else to it! Sometimes decay can be stimulated but the type of decay you're talking about is truly random.


0

Is this related to electroweak symmetry breaking and the Higgs field? Yes. There is a particular mixture of the $W^0$ and $B$ bosons that propagates freely in the Higgs field condensate; this freely propagating state is the photon. Why are mesons (hadrons) mentioned?? There was a time when the weak intermediate vector bosons were referred to as ...


2

Atoms are a small building blocks of matter, but not the smallest. Atoms are made up of electrons, protons, and neutrons. Electrons are (as far as anyone has been able to tell) honest-to-god point particles, i.e. they have no further internal constituents (I'm not saying they don't have spin or that they can't be described by wave functions, etc., just ...


1

Are you arguing that because particles come in pairs - particle and anti-particle - then there should be an even number of massless particles? If so, the argument fails because the photon is its own antiparticle i.e. there is no antiphoton.


0

Electrons, protons, neutrons as well as their antiparticles are able to receive and to emit photons. The photon exchange is possible between each of this particles and antiparticles and this does not change the properties of photons. Once emitted photons are the linear propagation of energy in the form of a oscillating electric and a oscillating magnetic ...



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