# Do real particles start out as virtual particles?

I know that QFT states that space is composed of various quantum fields and that particles are just excitations of those field.

As far as I know: virtual particles are excitations that exist for only a tiny moment, because they don't have enough energy to become real particles. On the other hand, though, if enough energy is absorbed by an excitation to sustain itself, then that's what a real particle is. What I want to understand is, according to this theory, does every real particle "begin" as a virtual particle? Because technically it had to get that energy to become real, right? Or am I misunderstanding something? Thanks.

I've also read the other posts regarding VPs and gotten some useful info, but I think answers to this question can help me better. Thanks.

• Commented Aug 29, 2021 at 23:45
• To clarify, the "virtual particle" thing is a method, not a theory. Analogy: We can use hand tricks as a method for doing arithmetic, but arithmetic isn't about hand tricks. Similarly, we can use virtual particles as a method for doing (some) calculations in quantum field theory, but quantum field theory isn't about virtual particles. Commented Aug 30, 2021 at 0:05

The idea that a virtual particle doesn't have enough energy to become real is deeply misguided. It's imposing far too much reality on a virtual particle.

Consider the virtual photon in deep inelastic scattering (DIS):

It has a lot of energy. At SLAC it my be a 40 GeV beam scattering into the famed 8 GeV spectrometer. That's 32 GeV of energy going from electron to proton. What's the momentum? Well, it's more. For any real beam/scattered electron:

$$q^2=(k'-k)^2 < 0$$

Of course there is a frame in which the energy is 0, and the square momentum is negative... that's just the nature of $$t$$-channel scattering.

What about $$s$$-channel annihilation? Also at SLAC's Z-factory, 40 GeV positrons and electrons were collided:

Now that virtual photon has $$\nu=80\,$$ GeV and zero momentum in the lab. It can't be real. It has the energy, but it's stationary, and light needs to go at $$c$$.

What is the energy of an electron in a vacuum polarization loop:

It's anything you want it to be. Evaluation of that in a scattering diagram, or the ground state of a hydrogen atom, requires integrating over all energies positive and negative (renormalization notwithstanding).

I know the idea comes from popularized quantum field theory where the virtual particle is always introduced with hand wave to the Heisenberg uncertainty principle....borrowing, popping etc.

This does more harm than good. It contradicts, or at least obscures, a lot of more important quantum concepts, such has: eigenstates are stationary states, they have no time varying observables. Also: the vacuum is Lorentz boost invariant. (Don't start me on the Higgs-as-molasses analogy). It's hard to imagine a bubbling sea of particles that is time invariant and boost invariant.

So when a state, e.g. and electron in a beam and proton in a liquid hydrogen target transition to a final state: an electron in the 8 GeV spectrometer and proton sprayed about end-station A (see the 1st diagram), the EM and quark/gluon fields take all possible configurations (coherently, and some not e.g., external radiative corrections), and an analytic solution is intractable.

The problem is solved by perturbation theory in which the field configuration/evolution is described by diagrams with virtual particles over ever increasing complexity. That is it.

Nevertheless, they can be very instructive. The DIS virtual photon has a longitudinal and transverse polarization defined by $$k^{\mu}$$ and $$k'^{\mu}$$ that affects how it scatters from the target. The kinematics of the $$\gamma^*$$ can tell you when you're scattering from charges or from magnetic moments.

If the beam or target is polarized, the virtual photon has circular polarization that behaves as expected with respect to scattering off spin initial and final states. (It can also interfere with $$Z^0$$-exchange as if we're talking about two paths on a laser table).

Virtual particle are very useful in that respect.

There is one fact that remotely resembles the "not enough energy" quip: as

$$k'^{\mu} \rightarrow k^{\mu}$$

(forward elastic scattering):

$$q^2 \rightarrow m^2_{\gamma} = 0$$

which is no interaction. In the limit the particles don't interact, the exchange photon get closer to being "on-shell", or real.

• Does DIS mean deep inelastic scattering? Would be better to make it clear in the answer, since not everyone who reads it knows all the abbreviations. Commented Aug 30, 2021 at 11:12

Quantum Field Theory (QFT) developed as a theoretical model when experiments in the quantum micro world became complicated and could not be modeled with a simple potential and a corresponding wave function. It is based on the postulates of quantum mechanics, i.e. it uses the plane wave (free of potentials) wave functions of the equations corresponding to the particles, (Dirac, KleinGordon, Maxwell) as a basis on which many particle states can be modeled. With the use of Feynman diagrams for the calculations it has been very successful in modeling the data and predicting data.

Virtual particles are a direct artifact of the Feynman diagrams, see this simple case:

What is real, i.e. the four momenta measurable in the laboratory, are the incoming particles and the outgoing particles. The connecting line is called a " virtual photon" because it carries the quantum numbers of a photon, in order that there should be conservation of quantum numbers,BUT the four vector assigned to the line is not on mass shell.

The virtual lines in complicated diagrams carry the quantum numbers but are off mass shell. The mass varies between the limits of the Feynman integrals used to calculate the process.Virtual particles are mnemonic place holders for the quantum numbers which get very complicated in many outgoing particles diagrams.

For your question it is important to understand the propagator function connected with each virtual line. If you see the examples, it has the mass of the assigned name particle as a pole in the function representing the Feynman diagram. Real particles will give a singularity for this function.

does every real particle "begin" as a virtual particle? Because technically it had to get that energy to become real, right?

In theoretical principle, the whole universe could be described by one quantum wavefunction, all its particles described as virtual. Reality comes from the width of how far from the mass pole of the particle the four vector describing it is. That is what distinguishes real particles from virtual. The incoming protons in the LHC are real because their mass is fixed within tiny, unmeasurable, experimental errors, and we live in the classical mechanics world because of this. The virtual by construction of the Feynman diagram describing the interaction, cannot be called real.

Look at this diagram for Compton scattering:

Here the virtual particle is the electron. There is an outgoing electron, that is on mass shell, the energy is gained from the incoming photon energy.

It is misleading to just think in QFT terms, imo, one should be looking at what the model is modeling.

space is composed of various quantum fields and that particles are just excitations of those field.

A free real particle cannot be modeled by an excitation of a quantum field, because the probability of finding it would be all over space (plane wave). One needs to use the wavepacket solution to do so. (Fortunately Feynman diagrams do not need it)