The weak force transforms particles from one type to another. What can we say about the transition?

Take this as an example:

enter image description here

In the discrete view:

  • Before junction A there are 2 up quarks and one down quark
  • After junction A and before junction B, there are 2 down quarks, one up quark, and a w+ boson
  • After junction B there are 2 down quarks, one up quark, a positron, and electron-nutrino.

I am suspicious that this view is not completely accurate. It introduces discontinuities in the fields, so wouldn't that make the fields undifferentiable.

My intuition is that "counting particles" is itself not really a valid operation, and during the continuous time periods within the junctions, we could say that, for example, the W+ boson partially exists for a while, coming into existence gradually and continuously as the boson field gradually and continuously changes until eventually it reaches a temporarily-stable excited state which we artificially label as "one more boson".

Is my intuition correct? Are these transitions discrete or continuous? What can we say about what happens inside these junctions?

  • 1
    $\begingroup$ "instantaneous" doesn’t exists since this would mean 2 ≠ particles ( or quarks or… ) in the same position in space-time. $\endgroup$
    – dan
    Aug 29, 2020 at 9:31

2 Answers 2


The example you show is an effective Feynman diagram. Feynman diagrams are a method of representing the integrals that have to be calculated in order to predict observable values.When the weak force enters at a vertex, the calculations need the weak coupling constant in the integral, which gives smaller crossections and longer decay times in the end.

I call it effective feynman diagram because the Quantum Field algebra does not work for the strong interaction, because of the large coupling constant and asymptotic freedom. One needs different tools, as QCD on the lattice to do calculation, although to first order an estimate can be made.

Are weak force particle transformations continuous or instantaneous?

Nothing is instantaneous, since everything in quantum mechanical theories is Lorenz transformation consistent, so all times are limited by the velocity of light. In addition there is the Heisenberg uncertainty for time and energy which gives an envelope for the widths. As we are successful in calculating crossections and lifetimes, which needs calculus, one bets on continuity.

Quantum field theory has creation and annihilation operators, in a sense "counting" particles in a field ( an positron field an electron field a W field ..) and is the basis of the calculations of the Feynman method, so your intuition is off.

The whole space time is filled with the fields of the elementary particles and antiparticles in the table of the standard model , type of coordinate system on which particles and their interactions are modeled.


Your intuition about the particles partially existing is sort-of correct, but mostly an oversimplification of how the mathematical formalism describes the situation. A more developed description is that the various particle fields are more like a fixed basis for the state space of the system, and that it is the coefficients multiplied by these basis elements that you might need/want/hope to impose continuity constraints on. The number of $W$ bosons (on a particular spacelike hyperplane of spacetime, if you want to be rigorous about the special relative structure) is an observable of the system, but (as long as you don't disturb the system by measuring) what you get there is the expected value of this observable, which is not at all required to be an integer and therefore can continuously increase from zero to small positive values as you move your hyperplane forward in time.

All of this is however with the reservation that this is not at all what the formalism looks like, from which these Feynman diagrams pop out, and in the process of getting there the various quantities in the theory are subjected to enough transformations that the issue of whether the original quantities were continuous becomes rather esoteric — lots of the quantities you work with aren't even functions (though maybe distributions) when we reach the point that the Feynman diagrams appear, and we can start to actually calculate something. On the other hand, the results of those calculations are generally a good match for the experimental results, so there ought to be some sense to the whole thing.

One common point of confusion here concerns the role of the particle fields, because even though for example the photon field $A$ corresponds to the classical electromagnetic field (or more technically its $4$-vector potential, of which the electric field $E$ and magnetic field $B$ are derivatives), it is not like the classical EM field. The classical EM field carries information and has a state, but the photon field just sits there, showing photons how to behave. Whatever information there is in the system, such as the value of the classical electric field, is not in the photon field, though possibly of it. There are differential equations describing the particle fields — the Dirac equation, the Klein–Gordon equation, the wave equation, etc. — but the solutions are all just plane waves of varying wavelength that permeate all of spacetime, totally oblivious to any matter or energy found therein. (Well, I suppose spacetime curvature should affect them if these PDEs are what define them — that would be interesting to hear if anyone has done any work on — but in particle physics the particle fields are essentially taken to be fixed. (As a mathematician I would like to say that they are constants, but I fear that in a physics context that would be misinterpreted as saying they are independent of position and time, which these fields certainly are not, even if it is just the complex argument that varies.)) The information instead sits in the quantum state of the system, which we will somehow have to unravel if we are to actually predict the results of any experiment.

The Schrödinger equation says the quantum state $\lvert\Psi(t)\rangle$ evolves according to $i \hbar \frac{\partial}{\partial t} \lvert\Psi(t)\rangle = H \lvert\Psi(t)\rangle$, where the operator $H$ is the Hamiltonian of the system. If $H$ is independent of time (e.g. it encodes only the fundamental laws of nature), then this (at least formally) has the solution $\lvert\Psi(t)\rangle = e^{-i H t / \hbar} \lvert\Psi(0)\rangle$. To prepare the initial state $\lvert\Psi(0)\rangle$ of the system, one starts from the vacuum state $\lvert0\rangle$ (where no particles are present) and acts upon it with an appropriate sequence of creation operators to add the particles one wants (with the momentum, spin, etc. desired). Conversely, to find the amplitude of a particular outcome in $\lvert\Psi(t)\rangle$, one may act upon it by the sequence of annihilation operators that would remove the expected ensemble of particles, and finally compute the amplitude for then having arrived back at the vacuum state using the vacuum bra vector $\langle 0\rvert$; the amplitude for starting with two $a$ particles (whatever those may be) of momenta $\mathbf{p}_1$ and $\mathbf{p}_2$ but ending at time $t$ with two $a$ particles of momenta $\mathbf{p}_3$ and $\mathbf{p}_4$ would thus be

  • $\langle 0\rvert a(\mathbf{p}_3) a(\mathbf{p}_4) e^{-iHt/\hbar} a(\mathbf{p}_1)^\dagger a(\mathbf{p}_2)^\dagger \lvert 0\rangle$

(Confusingly in view of how a ✝ cross may be used to mark death of humans, the convention is that creation operators carry a dagger symbol $\dagger$, whereas the annihilation symbol do not.) Interestingly enough it is (at least in perturbative QFT) not necessary to construct an explicit representation for the creation and annihilation operators to act upon, because every creation operator combines with an annihilation operator elsewhere in the expression — either in the initial/final state, or in the expression for the Hamiltonian $H$.

Now these expressions undergo a number of transformations (not necessarily in the order mentioned, and I definitely fast-forward this, but my aim is to get the general gist across):

  1. The exponent of an operator $e^H$ (for simplicity) is expanded as a power series $\sum_{n=0}^\infty \frac{1}{n!} H^n$, so likewise the final state $\lvert\Psi(t)\rangle$ ends up as a superposition of the initial state having gone through $H$ zero times, or one time, or two times, or three times, etc.
  2. The Hamiltonian $H$ is reexpressed in terms of the Lagrangian (density) $\mathcal{L}$ of the theory. The Lagrangian is an expression in the particle fields of the theory.
  3. The Lagrangian $\mathcal{L}$ is semi-arbitrarily split up into free and interaction parts.
    • The free part gives rise to the equations for the various particle fields, so a trip through that term in an $H$ just contributes a factor $e^{i \omega (t_2-t_1)}$ for where the angular velocity $\omega$ depends on the details of the particle (e.g. mass, momentum).
    • The interaction part terms have the effect of annihilating some particles and creating (potentially different) particles. These are what become vertices in the Feynman diagrams.
    • The split is somewhat arbitrary. Most famously the mass terms for particles are usually put in the free part, but when you view mass as being an effect of interaction with the Higgs field, those terms rather belong in the interaction part.
  4. Integrals over all of spacetime (making an actual Lagrangian out of the Lagrangian density $\mathcal{L}$) are combined with the exponential factors from the evolution of the free field parts and reinterpreted as Fourier transforms, after which a product [piece of the annihilation side of an interaction term]·[trip through the free part]·[piece of the creation side of an interaction term] gets converted into a propagator that constitutes an internal edge in the Feynman diagram. In doing so, time and space as variables are replaced by four-momenta.

So you can't really (within the matematical formalism) say that the transition $u \to d + W^+$ in your diagram corresponds to an event that takes place at a particular time, it's more like a partial result at a particular point in a particular expansion of the expression for the final state vector.

In particular, once you've transformed to momenta as variables, you no longer have a time order on things; the vertex you interpret as $u \to d + W^+$ might just as well be $u + W^- \to d$, since a particle going forward in time is exactly the same as its antiparticle going backward in time. The common description of virtual particles as "borrowing energy" and then giving it back fast enough to stay below the shroud of Heisenberg's uncertainty principle doesn't seem to have much support in the actual math — it's more like "everything that can happen will happen" (with the probability determined by all the factors in the expression).

On the other hand, having momenta as variables is great for describing particle accelerator experiments, because the momenta are the things you primarily know: you prepare two beams of known particles with known momenta, aim them at each other hoping they'll interact, and then stand back while your detectors measure (in particular type and momenta of) stuff coming out. The microscale time and place of the interactions are not so important.


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