In standard quantum mechanics, the wave function have discrete energy-values due to a potential. However, my very limited understanding of QFT is that electrons are excitation in the Dirac field, and the number of electrons is discrete even in free space. What is the reason for this, and why is there a minimum excitation?

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    $\begingroup$ The number of electrons in free space is discrete, but the spectrum of each is continuous, no? $\endgroup$ – Cosmas Zachos Aug 4 '20 at 14:48
  • $\begingroup$ For the energy? I would guess so, plus a constant mc^2. The energy spectra was ment as an analogy, I guess. I just have difficulties understanding how any representation of the states in free space become discrete. $\endgroup$ – Martin Johnsrud Aug 4 '20 at 15:34

I'll address what I understand to be your question, namely

Where is the discreteness of the number of excitations of QFT coming from, even in free propagation that apparently does not involve a potential and the corresponding compactness associated with discreteness?

(This is the magic of Fock space often referred to as "second quantization", an aggressively confusing term I and most avoid, truth be told...)

The short answer is

  • QFT is a repackaging of an infinity of quantum harmonic oscillators, each one with a discrete spectrum, each level of which corresponds to a particle.

This begs the question of where the harmonic potentials come from, if we are talking about free particles, no?

But this is a classical problem. Your classical continuum mechanics course describes, e.g., one-dimensional field theory, e.g. a "string" often discretized for computational and visualization convenience--the continuum limit is taken at the end of the day, and that discreteness is "fake". What is the crucial part is that the next-neighbor couplings lead to many coupled oscillators, an infinity at the end of the day, whose decoupling leads to normal modes in momentum space. The spectra of these oscillators, each one of them, are continuous before quantization.

But, upon quantization, the spectra become discrete. What you actually quantize is not x or its Fourier conjugate k, mere labels of the oscillators; but, instead, their displacement from equilibrium, the true dynamical variables.

Now each discrete excitation of each oscillator is a particle. The assembly has the built-in option of destroying as well as creating particles.

The ground state, $|0\rangle$, is the vacuum, with no particles. $a^\dagger_k|0\rangle$ is one particle with momentum k , and there is a continuous infinity of them, as many as there are momenta. $a^\dagger_k a^\dagger _l|0\rangle$ is two particles, one with momentum k and one with momentum l, suitably (anti)symmetrized, depending on the statistics of your field. That is, every excitation ("phonon") creator gives you a new particle, etc... You sum the different energies of all your particles to get your total energy of your multi-article state.

The packaging algorithm of quantized oscillators (a functor) is artful and encodes symmetries, Lorentz covariance, etc, but this is mere fine print. The above are naive states; if you want more realistic pictures involving wave packets beyond plane waves, you huff and puff a little, but that is an independent technical problem.

So to answer what I believe to be your question, "Where are the harmonic potentials coming from?", they come from nearest neighbor interactions of your coupled degrees of freedom, the elastic medium itself, even when your quantum fields themselves, and the particles, are free

Might like this question.

  • $\begingroup$ Thank you so much! I think this really answers the core of the question, though i guess I need to just get more familiar with the math to understand it better. $\endgroup$ – Martin Johnsrud Aug 4 '20 at 22:45
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    $\begingroup$ Here is a friendly summary. $\endgroup$ – Cosmas Zachos Aug 5 '20 at 0:04

In normal quantum mechanics, we consider an individual particle, turn its momentum and position into vector operators $\hat{P}_i$ and $\hat{X}_i$, and enforce the canonical commutation relations $[\hat{X}_i,\hat{P}_j]=i\hbar\delta_{ij}$. In Quantum Field Theory, we want to apply the laws of quantum mechanics to the field itself, and not to particles. A field $\phi(\mathbf{x})$ is treated as a generalized infinite-dimensional coordinate (each position $\mathbf{x}$ is one degree of freedom, or one dimension). Its conjugate momentum is defined, using the usual procedure from Hamiltonian classical mechanics, as $$\pi(\mathbf{x})\equiv\frac{\partial\mathcal{L}}{\partial\dot\phi(\mathbf{x})},$$ where $\mathcal{L}$ is the Lagrangian of the field. What we do in Quantum Field Theory is change the field $\phi$ and the conjugate momentum $\pi$ into operators. Then we need to figure out the eigenstates of the Hamiltonian. The way this is done depends on the field. For free Klein-Gordon fields, we do this using the commutation relations $[\phi(\mathbf{x}),\pi(\mathbf{x})]=i\hbar\delta^{(3)}(\mathbf{x}-\mathbf{y})$. We use the same trick that is often used in normal quantum mechanics to solve the quantum harmonic oscillator without invoking the Schrodinger equation. This is because the fourier-transformation the Klein-Gordon equation is of the same form as a harmonic oscillator equation. When we apply this method, we get creation and annihilation operators $a^\dagger(\mathbf{p})$ and $a(\mathbf{p})$ that create and destroy energy-momentum eigenstates out of the vacuum. You can choose any value of $\mathbf{p}$. The particles can have any momentum. However, the energy eigenvalues are $\sqrt{|\mathbf{p}|^2+m^2}$, so the particles always have mass $m$. There is no creation operator that gives you a "half" particle. There's the ground state $|0\rangle$, a state with one particle $a^\dagger(\mathbf{p})|0\rangle$, a state with two particles $a^\dagger(\mathbf{p}_1)a^\dagger(\mathbf{p}_2)|0\rangle$, but nothing in between. The reasoning that leads to discrete particles is the same as the reasoning that leads to discrete energy states in the quantum harmonic oscillator. You could certainly form a linear combination of a one particle and a two particle state, but when measured, it will collapse into an eigenvalue of whatever operator corresponds to that measurement. The procedure for electrons is different, since electrons obey the Dirac equation, but the procedure for the Klein-Gordon field should give you the general idea.


There are two facts to be distinguished. Electrons are what we loosely call particles, so they only ever occur in discrete numbers. Millikan demonstrated the discreteness of charge. Secondly, localised states in general have discrete energies. Examples of these are atomic and molecular states. Free electrons have continuous electrons. So called free electrons in metals for all practical purposes have continuous energies if the metal volume is macroscopic.

Why electrons are discrete particles or why quantum states are often discrete is not known.

A further question is how we account for these properties mathematically, but I believe this is not what you are asking.


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