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In Chapter 4 of his recent book Quanta and Fields, Sean Carroll uses a scalar free field to show how particles seem to emerge from field quantization.

On P.94, he talks about combining field modes with different wave numbers but all in their first excited (one particle) states to make a wave packet. He then writes:

What we're seeing is that a quantum state constructed from a superposition of free-field modes in their first excited states looks and acts like the wave function of a single particle of mass m.

Isn't the wave packet constructed by adding field modes just a field configuration for which you could compute an amplitude? Then I don't see how it could act like the wave function of a particle.

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    $\begingroup$ That is how wave packets emerge from plane waves. There is no such thing as a "particle" in nature. The minimal physical interaction in quantum mechanics is a quantum of energy, momentum, angular momentum and charge. Quantum theory can predict the average frequencies at which quanta of energy are being exchanged between systems and the correlations between such exchanges. It doesn't predict "particles". Many theorists like Sean Carrol are, unfortunately, too vetted to poorly conceived language from the early 20th century to give you the correct physical intuition for quantum mechanics. $\endgroup$
    – FlatterMann
    Commented Jul 18 at 22:38

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There is a mathematical equivalence between two different physical situations.

The first situation is a wavepacket in a classical field. This describes some localized packet of energy in an ordinary field, that will generically tend to dissipate if the field is obeying a normal wave equation with asymptotic boundary conditions that the field decays at infinity. The physical interpretation of the wave packet in the field is that you could measure the field profile by taking a "field meter" that measured the value of the field and poking it around the wave packet. The wave packet evolves according to the field equations. If you want to frame it quantum terms, this would be a state of the field with many non-interacting bosons in the same 1-particle state.

The second is a wavefunction for a single quantum particle, as a superposition of energy eigenstates, forming a wavepacket localized at some point in space. This describes a single particle that is not localized in space (until we perform a position measurement). The physical interpretation of the wavefunction is that it is the probability amplitude to find the single particle at any point in space. The equation obeyed by this single particle wavefunction, is the same equation as the classical field above. The fact that the two different physical situations are described by the same equation, is an endless source of confusion. However, it is no coincidence, as you can obtain the classical field case as the limit of many independent non-interacting quantum particles in the same state.

I also talked about this in the following Q&A: https://physics.stackexchange.com/a/816166/27732

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Before the sentence you quoted, Caroll writes:

Think of a state in which just one mode is involved, the mode with $k = 0$. Since $k = 2\pi/\lambda$, that’s a mode with “infinite wavelength”—basically the field has a constant value everywhere. Classically, that value can oscillate with time; quantum mechanically, it will have a wave function with some profile $\Psi(a)$. Imagine that this mode is in its first excited state. From (4.12) the frequency of the associated harmonic-oscillator potential is just $\omega = m$. The energy of the state is therefore $E = m$, since $E = \omega$, and the momentum is $p = k = 0$. (Remember, $\hbar = 1$.) That’s interesting: it’s just the energy and momentum of a single particle of mass $m$ sitting at rest.

Now consider a single mode with some other wave number k, also in its first excited state. It will have energy $E=\omega=\sqrt{k^2+m^2}$, exactly as we expect for a single particle with momentum $p = k$. Its spatial profile is not constant but looks like a plane wave, $e^{ikx}$.

What if we’re interested in field configurations that are not as simple as a single plane wave mode? No problem: we can combine the modes of different $k$’s, all of them in their first excited states, to get any field profile we like. A Fourier transform takes an arbitrary field profile and expresses it as a sum of plane waves, but we can also go backward, combining a collection of plane waves into whatever shape we care about. So for example, we can make a wave packet, which is what physicists call a localized wave that oscillates near some particular position in space but goes to $0$ far away.

What we’re seeing is that a quantum state constructed from a superposition of free-field modes in their first excited states looks and acts like the wave function of a single particle of mass m.

The second paragraph quoted contradicts your claim that Carroll claims the field is real valued. Nor can I find a place where he clearly states the field is real valued. Could you provide a quote and a page number if I am wrong?

The sense in which the wavepacket acts like a particle of mass $m$ is that (1) it is a superposition of fields with a minimum energy of $m$ just as a relativistic particle of mass $m$ has a minimum energy of $m$ and (2) the field is concentrated near a particular point.

If you would like to understand QFT better there are some good books that explain it more deeply such as "Quantum field theory for the gifted amateur" by Lancaster and Blundell and "Quantum field theory in a nutshell" by Zee.

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    $\begingroup$ Wave packets don't act like particles. They act like wave packets. We have never found a single particle in our experiments. What we are observing are quantized energy, momentum, angular momentum and charge exchanges between systems. Trying to pontificate "particles" out of the mathematical framework of quantum mechanics has, unfortunately, become somewhat of a past-time of textbook authors who haven't done a single actual quantum measurement in the lab. It does a great disservice to students who should be learning the correct phenomenology, even from theorists. $\endgroup$
    – FlatterMann
    Commented Jul 18 at 22:41
  • $\begingroup$ @alanf I think you're right about the field not being real valued (I must have read that elsewhere), so I edited my post. Note that Carroll writes that the wave packet constructed by superposing modes acts 'like the wave function' of a particle, not 'like a particle'. $\endgroup$
    – courno
    Commented Jul 18 at 23:14
  • $\begingroup$ @FlatterMann: while the "no-particle" point of view is reasonable for fundamental particles like photons, electrons and quarks, it does not hold for composite particles. The proton is associated with the confinement scale, which implies a region of space in which quarks and gluons are confined. This region now acts like a particle with a finite size. One can still have superpositions of such particles in a wave function. $\endgroup$
    – flippiefanus
    Commented Jul 19 at 3:26
  • $\begingroup$ @flippiefanus We aren't measuring "protons" or "baseballs" in quantum experiments. We are measuring energy, momentum, angular momentum and charges. That's why one can do diffraction experiments with anything. Light, electrons, neutrons, alphas, buckyballs. It doesn't matter. The basic behavior is always the same. The reason for that is because we aren't interested in the internal structure of the scattered field at all. We are only interested in where it deposits a quantum of energy. You are correct that the phase space that this energy can be distributed over is rather complex in QCD. $\endgroup$
    – FlatterMann
    Commented Jul 19 at 5:00

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