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I believe QFT represents a single free (stationary) electron as a an excitation of the electron matter field which then couples to the EM field to create a local 'attached' EM field - if this is accurate, what keeps the electron matter field excitation localized instead of dispersing out through the extended field. In an atom one can imagine that the electron matter field excitation (and associated attached EM field) stays put as a standing wave pattern due to attraction between the attached EM field and the positively charged nucleus. But for a single free electron what mechanism keeps the electron matter field excitation localized?

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There's three levels of approximations that eventually allow for the QFT electron to be described by the point particle. I will give them in order, but my suspicion is that you're looking for the 3rd one (coherent states).

Approximation #1: the adiabatic hypothesis, interacting QFT $\rightarrow$ free QFT. In the regime of weak interactions, an isolated system of electrons can be described by perturbation theory. The space of states of the QFT then is approximately described by the space of states of the corresponding free QFT, and the matrix elements of the Hamiltonian are given by a formal series in the coupling constant $\alpha$ (the fine structure constant).

This hypothesis is actually dead wrong, an important result known as the Haag's theorem. However, it turns out that adiabatic hypothesis works extremely well in practice, provided an important caveat: the interacting QFT and the corresponding free QFT have different values of field normalization constants and masses. Moreover, usually one of the two values is actually divergent.

The correct way to apply approximation #1 is therefore to:

  1. Build the Hilbert space of the QFT in question (QED in your example) as the Fock space of the corresponding free QFT with different "renormalized" values of particle masses and wavefunction normalization constants.
  2. Express physical observables such as e.g. the scattering matrix elements in terms of formal power series in the coupling constants (in your example, the fine structure constant $\alpha$).

This means that the leading (in $\alpha$) contribution to the behavior of an isolated electron comes from the free QFT with a renormalized (physical) electron mass.

Approximation #2: the non-relativistic limit, QFT $\rightarrow$ quantum mechanics. Now that we've approximated the original problem by a free QFT of non-interacting electrons and photons with the physical electron mass, let's consider its nonrelativistic limit.

One characteristic quantity that depends on $c$ is the (reduced) Compton wavelength of the electron

$$ \lambda = \frac{\hbar}{m c}. $$

In the formalism of the free QFT, the Compton wavelength plays a specific role. Say we've confined a particle into a region (in one dimension) of length $\Delta x$. Due to the uncertainty principle, this implies an uncertainty in the momentum. Due to the kinematic relation between energy and momentum, this implies uncertainty in energy. The characteristic scale of $\Delta x$ for which the uncertainty in energy is comparable to the rest energy of the electron is exactly $\lambda$.

In other words, try confining an electron to a region smaller than $\lambda$, and you'll get enough energy coming from the quantum uncertainty to create more electrons and positrons, so you don't have an isolated electron anymore.

All of the above was just a back-of-the-envelope calculation, of course. To describe what happens in the free QFT we should look at the behavior of the propagator for small spacelike separation. It turns out that the propagator decays as (numeric coefficients dropped for simplicity)

$$ S(x, y) \sim e^{-\left| x - y \right| / \lambda}, $$

which means that the quantum field values taken at spacelike separated points aren't orthogonal in the Hilbert space, unless the separation is much larger than $\lambda$, in which case they are orthogonal.

That is exactly the regime of approximation #2. It is easy to see that $\lambda \rightarrow 0$ corresponds to $c \rightarrow \infty$, that's why it is also the non-relativistic approximation.

Approximation #3: coherent states, quantum mechanics $\rightarrow$ classical mechanics. So now we've reduced our initial QED to a theory of a single isolated electron, and we now think of the delta-function basis of wavefunction as of an orthogonal basis in the (appropriate extension of the) Hilbert space, because in the non-relativistic limit $S(x, y)$ vanishes.

This is actually exactly equivalent to Pauli's theory of a non-relativistic electron. We've lost antimatter and other relativistic effects on approximation 2.

But even here the electron is just a mist of quantum magic, the wavefunction. What ensures that it doesn't spread?

Indeed, there are quantum-mechanical states for which the wavefunction rapidly spreads in time. Think delta-function-like distributions (but keep in mind that they have to be at least as wide as the Compton wavelength, or else the electron will explode in jets of QFT particles).

There's however a special subset of states which are states of minimal spread, or semiclassical states, aka coherent states. These are wave packets, given by Gaussian wavefunctions peaked at the classical position of the packet multiplied by a phase given by the classical value of the (non-relativistic) momentum.

Whenever the system is in a coherent state, the peak's spread in time is minimal (for free noninteracting theories like the idealisation here it vanishes). The classical values of position and momentum (determining the shape of the wavepacket) obey the classical equations of motion, which is the real reason why classical mechanics is a limiting case of quantum mechanics.

This approximation corresponds to $\hbar \rightarrow 0$, because the width of the wavepacket is proportional to $\hbar^{1/2}$. Hence, the classical approximation is valid when the scale of observation is large compared to the width of the wavepacket.

Conclusion. It is when the three approximations above are applied in sequential order that the elementary particles can be thought of as billiard balls moving through space. Note that this only works for massive particles, because for massless particles there’s no Compton wavelength and approximation #2 isn’t valid. Instead, one would typically take the classical limit of the relativistic QFT which leads to the classical field theory, eg electromagnetism. There's another notion of coherent states of the relativistic QFT, which is the eigenstate of the annihilation operator and describes the behavior of the classical field (and not a particle), which is the tool that highlights this correspondence.

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    $\begingroup$ This is incredibly well explained. $\endgroup$ – JPattarini Jan 26 at 5:04
  • $\begingroup$ Many thanks for a very clear and concise answer to my question! I think I should follow-up by further reading on the formation of Coherent States. $\endgroup$ – CSnowden Jan 28 at 19:01

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