# How do collisions of fundamental particles produce different fundamental particles?

When considering fundamental particles as waves in fields, it seems like any collision of two particles of some fundamental type could only create energy within that type's field. Why do we expect that this energy is transferred to a different field, creating a different fundamental particle? What is the mechanism of this energy transfer and what, if any, is the physical interpretation of the energy's "path"?

• We don't know! Why mechanical energy should be able to transform into heat? – Asphir Dom Mar 21 '14 at 11:51

A simple version of this is bremsstrahlung, i.e. an electron that decelerates and produces electromagnetic radiation / photons. By your reasoning the energy of the electron should only be able to go into other electrons: maybe it should radiate other electrons, maybe a single electron shouldn't lose energy as it travels.

But the electron can transfer some of its energy to the electromagnetic field, simply because it is coupled to the electromagnetic field.

It's the same in more general processes, the fields are all coupled together, and these couplings allow energy to flow from one kind of field to another.

EDIT: Added some new information below based on discussion in the comments.

First I just want to point out how coupling allows energy transfer at the level of classical equations of motion [ie, in terms of "ordinary massive particles"]. If I have linear equations of motion for $N$ particles of the form:

$$\ddot{x}_i + K_{ij} x_j = 0$$ where $i=1,...,N$ labels the particle and where $K_{ij}$ is some symmetric matrix. Physically this system amounts to having $N$ particles coupled to each other with springs with varying springs constants.

I can always perform a variable redefinition $q_i = T_{ij} x_j$ and work in terms of the normal modes of the system by cleverly choosing $T$ to diagonalize $K$. Then I have $N$ uncoupled differential equations for my normal modes $q_i$. These $N$ modes do not interact. It is completely trivial to solve these equations of motion for $q_i$ as a function of time, since each $q_i$ simply undergoes simple harmonic motion. If I set up initial conditions in terms of $q_i$, one mode will not evolve into another mode. If I compute the energy in each mode at the initial time, I can compute the energy in each mode $E_i$. If I evolve the system forward, and compute the energy in each mode at a later time, I will find that $E_i(t_{initial})=E_i(t_{later})$. Not only is the total energy conserved, but the energy distribution in each individual mode is conserved.

Let me now throw a nonlinear term into these equations of motion. Physically this would amount to having springs coupling the particles that were not exactly hooke's law springs. $$\ddot{x}_i + K_{ij} x_j + \lambda_{ijk} x_j x_k = 0$$ I can no longer solve this differential equation analytically because of the term proportional to $\lambda$. I am forced to solve the system numerically, or if I am lucky and $\lambda$ is small I can use perturbation theory to get some analytic understanding of what is going on. I can still write the solution as a superposition of the eigenmodes of the original system, however now the coefficients of the superposition will change with time. In other words, the total energy is conserved, but not the energy in each mode individually. Energy that was originally in the $j$-th eigenmode will be transferred into the other eigenmodes as time evolves, precisely because of the $\lambda$ term. In other words, let me set up some initial conditions. At these initial conditions, I decompose the solution into eigenmodes, and compute the energy in each eigenmode. Then let me evolve the system forward (say numerically, or using perturbation theory). Then at some later time, I will again decompose the system into eigenmodes and compute the energy in each eigenmode. The energy in each mode at this later time will not equal the energy in each mode at the earlier time. You can check all of these statements numerically by simply picking a specific example. Thus nonlinear terms in an equation of motion cause interactions between modes.

The lagrangian that gives rise to this equation of motion is $$\mathcal{L} = \frac{1}{2}\dot{q}_i \dot{q}_i - \frac{1}{2} q_i K_{ij} q_j - \frac{1}{3} \lambda_{ijk} q_i q_j q_k$$

So you see that the quadratic terms in the lagrangian give the 'free' (i.e. linear) pieces, and the cubic (and higher) terms give the interactions.

So that was classical mechanics of $N$ point particles. In order to really understand how you get quantum particles and interactions from a field theory lagrangian, you really need to go through a pretty involved field theory discussion about quantizing a free scalar field and then discussing how perturbation theory is done in quantum field theory. These lecture notes by Tong are a really excellent introduction if you are interested: http://www.damtp.cam.ac.uk/user/tong/qft.html.

However, I can try to give you some heuristic understanding. The lagrangian for QED (which I am taking to be that subset of the standard model involving only the photon and the electron/positron) is $$\mathcal{L} = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} -i \bar{\psi}( \gamma^\mu {\partial_\mu} + m )\psi + e \bar{\psi} \gamma^\mu A_\mu \psi$$ where $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. I might have screwed up a sign, but frankly I don't think that it matters for us here.

Here, $A_\mu$ represents a photon, and $\psi$ represents the electron and the positron (they are fundamentally inseparable in QFT). The $\gamma^\mu$ is essentially just a constant for our purposes, the point is that it doesn't represent a dynamical degree of freedom.

The first two terms are quadratic in the fields. Again these are free, and they just describe how an electron/positron or the photon propagate through space without interchanging energy.

However the last term (which contains the electron charge $e$) is exactly analogous to the $\lambda_{ijk} q_i q_j q_k$ piece from the classical mechanics example. It allows energy in the photon eigenmodes to leak into the electron eigenmodes as things evolve. The existence of this term is really no more surprising (mathematically) than the existence of non hooke's law springs. It doesn't mess up the total conservation of energy of the entire system, but it does allow energy to leak from one kind of mode to another. Quantum mechanically these modes represent particles, so the coupling term allows energy to leak from one kind of particle to another.

So mathematically field theory allows interactions in a very similar way as occurs in classical mechanics. Fundamentally your question is still, why should energy be allowed to leak from one mode to another at all? However the modern particle physics way of looking at things is precisely the opposite: the modern philosophy is that every possible term in a lagrangian allowed under certain conditions (symmetries, maybe renormalizability) should exist. It would require an explanation if the interactions were not there.

• So I guess my question is what is the cause and/or physical interpretation of these couplings (or are they too fundamental for the question to be meaningful)? – csabol Mar 21 '14 at 4:19
• But I would just say that interactions are facts. The term $A e e$ in the lagrangian [which describes the interaction of electrons with the photon] is simply postulated, and we derive physical results from it. These interactions are directly responsible for the ability of particles to scatter off each other. If things didn't couple, then you would have disjoint sets of objects that could never communicate. For example if I was made of electrons and electrons didn't couple to anything else, I could never do any experiment to discover protons. – Andrew Mar 21 '14 at 4:31
• @Andrew That's one of my favorite Feynman stories, mainly because his analogy went a long way towards helping me explain to my father at least. The sound of my voice when I speak wasn't somehow inside my voice box the entire time (it's not a literal box full of voice!), and it's not as if we have a stock of, say, 20 'ah's and once we use 'em up we have to go to the store and buy more. The vibrations of our vocal chords simply produce sounds on demand, and once a sound has completely dispersed it's gone for good, as opposed to having gone somewhere else. – David H Mar 21 '14 at 13:23
• Something else just occurred to me. That analogy basically is attempting to make the creation and annihilation seem less mysterious by pointing out that we're already used to waves exhibiting this behavior all the time. And QM says particles are also waves, right? So crisis averted. Historically though, wave-particle duality came about 30 years before matter creation and destruction. This must be one of the reasons why wave-particle duality gave people anxiety attacks in the beginning. – David H Mar 21 '14 at 15:53
• @csabol Hehe, the thing about a surprise is that eventually it stops surprising you if it keeps happening. We do get used things. It's happened before, though we're prone to forget it. For example, look at what you said: "both my vibrating vocal chords and the vibrating air particles are massive objects, and thus I would not be surprised by their interaction". Air is substance? This substance has mass? That mass is comprised of particles? The particles vibrate? That's four surprises right there that you apparently take for granted today. ;) – David H Mar 21 '14 at 18:05

When a type of quantum (elementary) particle is absent it just means that the field of that particle is in the quantum "vacuum state". But "vacuum state" does not mean absence of everything concerning that field. The vacuum state has various physical properties in spite of its name: It is nothing but a possible state or quantum configuration of the field, usually the most symmetric quantum state of that field with respect all possible symmetries.

Moreover, even if the quantum state of a certain quantum field is the vacuum one, the interaction with other fields, theoretically represented by an interaction Lagrangian does exist. Just the existence of that interaction is responsible for possible transition processes form the vacuum to other quantum states.

In relativistic quantum physics, also the number of particles of a certain field is a quantum observable exactly as the momentum is. So the number of particles follows the same rules of general QM. By means of a quantum transition, you can pass from a state with zero particles (the vacuum state which is however a state) to a state with some particles.

When some already existing particles make free an amount of momentum and energy, in principle, these quantities can be spent in changing the states into all states admitted by the conservation lows of the physical system. Each possibility has its own probability to become real in accordance with quantum rules described by the interaction Lagrangian. In particular the vacuum state of a field can be transformed into a different quantum state containing particles, in principle for every conceivable quantum field.

As an example consider the hydrogen atom (actually I will consider the electron only). The electron in the atom, in particular, interacts with the external (i.e. not produced by the proton) EM field, even if this EM field is in the vacuum state (absence of photons). The composite state of the complete system: electron in the ground state and EM field in the vacuum state is stable. But if you lift the electron to the first excited level of the atom and keep the EM field in the vacuum state, you have an unstable overall state in view of the particular form of the interaction Lagrangian. There is a non-negligible probability to transform the vacuum EM state into a state containing photons and the excited state of the electron into the ground state. In this case the energy content of the electron (the difference with respect to the energy of the ground state) passes to the state of EM field in terms of one or more photons.