How can a truly elementary particle decay into other particles?

Question

Consider (for example) the next particle decays:

The decay of a Higgs and the decay of a muon [in the diagram of which we also see the decay of (virtual) $W^-$ into an electron and its associated neutrino)].

If a particle (or an excitation of the associated quantum field) is truly elementary doesn't that imply that it can't change into other particles?
The case of a muon (which is considered elementary, i.e. not built up out of other particles) changing into an electron, and two neutrinos (which are all three also considered to be elementary) can very easily be described in the Rishon Model of Haim Harari (of which I'm a big fan), in which only two (!) truly elementary (apart from the photon, gluon and the $Z^0$ and $W^{+/-}$, of which the last three are considered to be composed particles which transmit not a truly elementary force, but a residual force of a deeper force, as once the $\eta$ was thought to be the transmitter of the strong force, which turned out to be the residual force of the strong force as it is known today) particles are said to exist: the T- and V-rishon (and their anti-particles).

$\bar T\bar T\bar T$ (the muon, in this model considered as an excited state of the electron) gives a $VVV$ (neutrino), $\bar V\bar V\bar V$ (the anti-neutrino, associated with the electron), and a $\bar T\bar T\bar T$ (the electron). So before and after the change, the same (net) combination of rishons exists.
The virtual $W^-$ is a short existent $\bar V\bar V\bar V\bar T\bar T\bar T$ combination, which in the rishon model obviously has an electric charge -1 because the $T$-rishon has electric charge +1, and the $V$-rishon has no electric charge. In this change, the (according to the rishon model) truly elementary particles keep their identity, so a $T$-rishon can't change into a $V$-rishon and vice-versa, and it only seems that what we consider elementary particles can change into other elementary particles.

The decays, in the rishon model, are nothing more than rearrangements of T-rishons and V-rishons (and their antiparticles), while at the same time virtual $T$ and $V$ rishons can become real in the decay and contribute to the process. Clearly, these $T$ and $V$ rishons can't change their identity if they are truly elementary (if so that would indicate that these two elementary particles would be composed of even more elementary particles, which is nonsense if you need only two elementary particles to explain the abundance of quarks and leptons; the model doesn't address the force-carrying particles, which are all massless in this model).

In the case of the changing Higgs (which in the rishon model isn't needed to give mass to particles, but it nevertheless exists because it has been detected so it can be considered as a boson particle), the change results in two pairs of $TTT$ and $\bar T\bar T\bar T$ combinations of T-rishons (and their anti-particles), the electron and it's anti-particle and a muon together with its anti-particle. The two $Z^0$ particles that appear shortly are both $TTT\bar T\bar T\bar T$ combinations (with obvious electric charges of zero). So the Higgs can be a combination of six $T$-rishons and six $\bar T$-rishons. Again the truly elementary particles (the $T$ and the $V$) don't change their identity.

So does the fact that what we consider as elementary particles (excitations of the associated field) can change into other particles mean they are not truly elementary?

Answer 1

Being elementary does not imply that a particle cannot decay. Decaying is not scaling some sort of scale of "elementariness" (I'm totally inventing this concept, it does not exist). Particles changing into one another is a matter of interaction.

To say it à la Weinberg [1], elementary particles are quantum states (yes, particles are states) that can be described by only specifying their impulse and another discrete set of quantum numbers (that is then interpreted, loosely speaking, as the particle spin state). Elementary particles are called "elementary" because you do not need to know their internal composition: to describe them, you just need an impulse and a spin, no internal composition needed.

Let's take your second diagram as an example. Note that everything that I'm saying is simplified. We have muons $\mu$, muonic neutrinos $\nu_\mu$, electorns $e$, electronic neutrinos $\nu_e$ and the $W$ boson. To each particle corresponds a field, and field dynamics is ruled by a Lagrangian. The relevant part of the SM langrangian governing this weak process can be split in two pieces: a part that is sum of terms that contain the fields that I have enumerated, in quadratic combinations: this is the free part of the Lagrangian, and determines things like the mass of those fields. The interacting part is composed by a sum of terms in which the fields appear more than once, and it regulate decays and Feynmann vertices.

In the case of the muon decay, we have as relevant interaction terms $$ V_{int}=g_1 \bar\mu W^-_\alpha\gamma^\alpha\nu_\mu+g_2 \bar e W^-_\alpha\gamma^\alpha\nu_e+h.c. $$ where h.c. means "Hamiltonian conjugate", and contains terms like $\bar \nu W^+\gamma\mu$ (neglecting all indexes). Notice that in $V_{int}$ $\alpha$ is a Lorentz index, and I won't use $\mu$ as index (damn muon). $g_1$ and $g_2$ could be related, but that's not the issue here.

By reading the interaction Lagrangian, you can understand what decays into what. Let's take the first term, that is involved in the first vertex. You can see any term as "creating" or "destroying" the corrispective particles (with creation and destruction operators, in accord with our interpretation of particles as states on which operators can act). Any field operator removes an antiparticle from an incoming state (the bra) or creates a particle in the outgoing state (the ket). As an example, in the first term you use $\bar \mu$ to destroy $\bar\mu$'s incoming antiparticle (the muon $\mu$), and you use the other two fields to create a $W^-$ and a $\nu_\mu$. In the second vertex, that is ruled by a term hidden into $h.c.$, you use a $W^+$ operator to destroy the incoming $W^-$, and $e$ and $\bar \nu_e$ operators to create the final particles. The amplitudes $g_1$ and $g_2$ rule the amplitude of the process.

To conclude, elementary particles can decay. An elementary particle is just a state that we can identify with as few informations as possible. Which particle can decay into which other is ruled by the Lagrangian, specifically by the interacting terms.

[1] The Quantum theory of fields. Vol. 1: Foundations, Steven Weinberg, right at the beginning of page 63

Answer 2

So does the fact that what we consider as elementary particles can decay mean they are not truly elementary?

Define elementary:

The Webster definition:

a : of, relating to, or dealing with the simplest elements or principles of something

b : of or relating to an elementary school

so it is the a: that is used in the elementary particle table of the standard model of physics.

This is an encapsulation of innumerable data gathered in the 80 years or so, and is a Lagrangian with the SU(3)xSU(2)xU(1) symmetry and the three interactions, strong weak and electromagnetic.

The table of elementary particles for the standard model is the simplest table possible within the above group symmetry, and thus adheres to the Webster definition.

Other theories may propose different particles as elementary and dirfferent interactions, but they are not mainstream, mainly because they cannoct accommodate/predict the plethora of data that the standard model does.

That many of the particles in the table decay is within the model, as the other answer by Salvatore explained. They are elementary in the meaning of a: above.

String theories are the only exception I know because they can embed the standard model. There is no standard string theory yet, there are thousands of possibilities. In string theories what is elementary is the string.