Particles vs fields

Question

I've been reading the book "The Standard Model in a Nutshell" by Dave Goldberg and I'm confused by the notion of a particle.

Case 1: Suppose that $\phi$ solves the Klein-Gordon equation, i.e. $(\square + m^2 )\phi=0$. On p33, the author writes that $\phi$ represents the "dynamics of a particle of mass m". The author also refers (e.g. on p37) to $\phi$ as a "particle".

Case 2: To motivate the wave equation (e.g. p29), one thinks of space as being made up of an infinite array of point particles which vibrate as a harmonic oscillator. So in this case $\phi(t,x)$ describes how much a particle at the point $(t,x)$ deviates from its equilibrium position.

The second case makes good sense to me, unlike the first case. Are they both supposed to be the same? If not, how is the $\phi$ in case 1 a "particle"?

I feel like I am missing something obvious and I would very much appreciate help understanding what's going on!

Answer 1

Both cases are simultaneously correct, but the author is being quite sloppy. If you're a mathematician, the absolute last thing you want to do is to try to learn physics from a book like this.

Here's conceptually how it works in the case of phonons.

• Start with a grid of lattice ions of mass $M$ connected with springs with spring constant $k$. Describe their deviation from their equilibrium by a classical field $\phi(x)$. Vibrations in these fields correspond to classical sound waves.
• Working classically, we find solutions to the equation of motion for $\phi(x)$ which oscillate with frequency $\omega$, called normal modes. Since the solid has translational invariance, they will have the form of plane waves, with some wavenumber $k$.
• Now focus on a single mode. Upon quantization, it corresponds to a quantum harmonic oscillator whose energy level spacing is $E = \hbar \omega$. If this oscillator is in the state $|n \rangle$, we say that the mode contains $n$ particles, called phonons.
• The mass of the phonon depends on the relationship between $\omega$ and $k$, because by the de Broglie relations it yields a relationship between $E$ and $p$. In particular, it does not have anything to do with $M$. The only thing the parameters $M$ and $k$ do in this case is provide a characteristic frequency $\omega_0 \sim \sqrt{k/M}$. Phonons are not lattice ions. They are quanta of the excitations of lattice ions.
• The field $\phi(x)$ is promoted to a field operator $\hat{\phi}(x)$. It obeys the exact same classical equation of motion that $\phi(x)$ does.
• The field operator $\hat{\phi}(x)$ plays the same role as the operator $\hat{x}$ does in the case of a quantum harmonic oscillator. For example, applying this operator to the state, in the case of a free field, will increase or decrease the particle number by one. And you can get the "expected field value", just like the expected position, by taking expectation values of this operator.
• The field operator does not represent the state of a single particle in any sense. However, by a terrible coincidence, the equation of motion for the field is the exact same as the equation of motion for a single particle wavefunction. This led to much historical confusion which survives today in many textbooks.

The relativistic case is typically interpreted a bit differently.

• In a relativistic quantum field theory, like the Standard Model, we do not typically think of the field excitations as emerging from some underlying lattice. It would be possible to do this, but it's not conceptually necessary. Instead the field is elementary.
• Often, we work in the Heisenberg picture, where the field operators depend on time and do specify the state of the system.

Answer 2

In the first case, the Klein-Gordon equation describes the dynamics of a single particle of mass m, with $\phi$ being its wavefunction. Just like the Schrodinger equation in non-relativistic, undergraduate, quantum mechanics.

In the second case, we think of space as consisting of point-like quantum harmonic oscillators. These are not "particles", they are just abstract quantum systems that provide quantitative qualities to the points in space. This gives us the field $\phi(x,t)$ describing said properties at each point in spacetime. In this case we think of the waves in this field as particles. Well, technically, as khzhou explained, we think of the normal modes' quanta as particles.

Case 2 is more fundamental. Case 1 emerges from it as the effective dynamics of a particle (of a normal mode, if you will).

Answer 3

" one thinks of space as being made up of an infinite array of point particles which vibrate as a harmonic oscillator."

This is a hand-waving description of quantum field theory, an "aether" but a Lorenz invariant one, so the Michelson Morley data are not violated.

Both views become clear if one understands that quantum mechanical solutions are giving the dynamics of the probability for what the particle is doing.

The Klein Gordon, in addition to energy spectra for a given potential, gives the probability distributions of how a cumulative distribution of particles with the same boundary condition will react to the potential. Even though a single particle enters the equation, to see the dynamics, like crossections, one has to have cumulative distributions .

The quantum field theoretical framework is based on single particle solutions of the corresponding quantum mechanical equations, with zero potential as a method of calculating many body interactions, at each point in space.Creation and annihilation operators on this basis propagate the particles and describes their interactions. These can be summarized with feynman diagrams that are a recipe for calculating measurable quantities.

The whole objective of quantum mechanical calculations is to be able to calculate measurable quantities, and check them against experimental values.