I'll use QED as an example, but my question is relevant to any quantum field theory.

When we have a particle in QED, where is its charge contained in the field? Is the field itself charged? If so, wouldn't the whole universe be filled with a constant charge density? If not, where does the charge come from, a separate field? If a particle is a ripple in the QED field, how can the ripple be charged but not the field itself, and what determines how the charge is distributed among the field amplitude?

This question is answered in quantum mechanics: we have point particles. But I find "particles as quanta of fields" confusing, because if the charge is contained in the field, and particles are just ripples in that field, then how is the charge of the ripple mapped to the ripple itself? Would a spread-out enough ripple have some small charge density filling a large region of space, or do we somehow map the charge of the ripple to a single point in space?


3 Answers 3


We say that the charge particle's field is charged itself but that's just a different way of saying that its excitations – the particles – are charged (mathematically, it means that the field gets multiplied by $\exp(iQ\lambda)$ under electromagnetic $U(1)$ gauge transformations) and this is a totally different statement from the statement about the location of the charge in a particular situation with a particular particle (or many particles).

If one considers a charged particle such as an electron (and it doesn't matter than a particle is an excitation of the field – a particle in QFT is defined as an excitation of a field, after all), the location of the charge is the very same thing as the location of the particle itself.

So it's an observable and just like any observable in any quantum theory, it is represented by an operator (matrix) acting on the Hilbert space. The observable may have various values – the eigenvalues $\vec x$ are the a priori allowed values – and we may measure it but the results of the measurements may only be predicted probabilistically. The probabilities are the usual squared absolute values of the inner products of the state vector and an eigenstate.

One may also define the charge density $\rho(\vec x)$ for a point in space, $\vec x$. This is an observable, too. So what I wrote for the position holds for this observable – and any other observable – as well. So in any quantum theory, $\hat \rho(\vec x)$ is an operator that has some eigenvalues and that may be predicted probabilistically.

In a system with one particle whose charge is $Q$, the density may be easily calculated as $$ \hat\rho(\vec y) = Q\cdot \delta(\vec y - \hat{\vec x})$$ where $\hat{\vec x}$ is the operator of the position of the charged particle but $\vec y$ is just an argument of $\hat\rho$ so $\vec y$ has no hat above it! It may be unfamiliar to calculate a function (even Dirac's delta-function) whose arguments depend on operators but it's possible.

We may calculate the expectation value of the charge density $\hat\rho(\vec y)$ in a particular state – that will be simply $$\langle \rho(\vec y)\rangle = Q\cdot |\langle \vec y|\psi\rangle|^2 $$ but the expectation value of an observable isn't the same thing as the observable. The observable itself has many possible values that may be obtained by measurements and probabilities of different outcomes are predictable. The expectation value is just the weighted (by the probabilities) average of the possible eigenvalues.

We may extract the "ordinary non-relativistic limit" of quantum field theories so that the formalism above is directly relevant for QED and other quantum field theories. This approximation is "non-relativistic", as I said, so it starts to be ill-behaved when the speeds of the particles (or the average speeds etc.) start to approach the speed of light. This nightmare scenario becomes inevitable by the uncertainty principle if we try to localize the particle within the Compton wavelength (of the electron, in the case of an electron) $\lambda = h/mc$. When one wants to localize the electron more accurately than these $2.4\times 10^{-12}$ meters, the uncertainty of the momentum becomes of order $mc$ due to the uncertainty principle which guarantees that there is a significant probability that the speed is close to the speed of light. When it's so, the non-relativistic approximation breaks down. Let me mention that the Compton wavelength is still 100 times smaller than the hydrogen atom so there's still a long interval of length scales for which the non-relativistic quantum mechanics is an OK approximation to QED.

When the particles are located to a better accuracy than the Compton wavelength, the extra kinetic energy $E_k$ becomes comparable to the latent energy $E=2 m_e c^2$ of an electron-positron pair and it becomes rather likely that the compression does create the electron-positron pair. Quantum field theory including loop processes etc. becomes the only viable description and the non-relativistic approximation has to be abandoned.

  • $\begingroup$ @user1247 @LubosMotl please continue your discussion here, if you wish to do so. $\endgroup$ Feb 25, 2013 at 14:57

(What started as an attempt to write a simple answer has turned into a page-long ramble about the nature of quantum field theory and observables. However, I think the basic message is still there and worth reading.)

Background: scalar fields and charge conservation

We can actually uncover the salient features of how to properly define and interpret the "charge" of a quantum field by just considering a complex scalar field theory. The Lagrangian for such a theory is


where the extra terms are possible interactions or mass terms which are irrelevant for our discussion. This theory contains a global $U(1)$ symmetry $\phi\to e^{i\alpha}\phi$, $\phi^*\to e^{-i\alpha}\phi^*$. The conserved current for this symmetry is easily obtained from the Noether procedure as


As we know from classical field theory, the charge density associated with this global symmetry is given by $j^0$, and the total conserved charge can be expressed as a functional in the field variables as


Once an electromagnetic field is turned on, this turns out to be the electric charge associated to the electromagnetic interaction. In this sense, your question is very easy to answer classically:

The charge associated to a classical field is spread everywhere that field has support, as a classical field carries a charge density $j^0$ associated to its Noether current.

Quantum field ontology

So what does this have to do with particles and other nonsense? How can we move from the interpretation of a classical field as carrying a classical charge density to that of a particle carrying a fundamental charge?

In order to answer this question, it's important to remember what we're doing when we do field theory. We have to remember the following:

Quantum field theory is not the study of fundamental particles, but, rather, the study of the quantum mechanical analogues of classical field theories.

That fundamental particles arise in the study of quantum field theory should, for the sake of this argument, be seen as a happy historical accident. What is most important is that we study the quantization of classical field theories.

To quantize a classical field theory is very simple, but somewhat subtle. We continue the same way that we quantize classical mechanical particles. We identify our canonical variables and associate to those variables a wavefunction that obeys a Schrodinger equation. In field theory, the canonical variable is $\phi(\textbf{x})$ -- the field configuration at any given point in time is the observable variable. Thus, it is natural to define a wave-functional $\Psi[\phi]$ which associates to each field configuration a quantum-mechanical amplitude for that configuration.

After we introduce the wave-functional, we wish to solve the Schrodinger equation


We do this the same way we do in standard quantum mechanics -- we break up $\Psi$ into a sum of energy eigenfunctions, then let the diagonalization take care of the time evolution.

We can (and should) also introduce a Dirac-type notation where $\Psi[\phi]=\langle\phi|\Psi\rangle$, where $|\phi\rangle$ is a state of definite field configuration (that is, a state such that a measurement of $\phi(\textbf{x})$ at any point will yield a definite result). This helps to clean up the notation.

Also, in direct analogy to the definition of momentum in standard quantum mechanics, we associate the conjugate momentum of $\phi$ (in our case, $\dot{\phi}$) with the field-momentum operator, which takes the following form when acting on wave-functionals:


Thus, the operator corresponding to the electric charge is given by


Back to defining charges

Let us say that we have diagonalized the Hamiltonian of our particular theory. We know that since $Q$ is conserved, when it is promoted to an operator, we will have $[\hat{H},\hat{Q}]=0$. Thus, the Hamiltonian and electric charge can be simultaneously diagonalized. That is, we can choose an energy eigenbasis such that each basis wave-functional will have a definite energy and electric charge.

Let's say that $|E,q\rangle$ is some state such that


for some energy $E$ and some charge $q$. Where would you say that the charge $q$ is located? This is not a question that has an answer in the quantum case, as the state $|E,q\rangle$ is in general a specific superposition of states of different charge density distributions. The only thing one can say is that, once you measure $q$, you know it must be conserved in time.

Asking where the charge is located in a specific state of a quantum field theory is sort of analogous to asking where the electron is located in an electronic orbital (ie, asking the value of an observable in a state which is not an eigenstate of that observable). It isn't quite right to say that the electron is "everywhere", because that's not how superpositions work. Indeed, the question of "where" only makes sense upon measurement.

In the same sense, the question of what the charge density at a certain point in a general state in a QFT is constitutes an equally nonsensical question. It is only when one observes the charge density at that point does the question make sense.

The question of "What is the charge density configuration?" for a given state in a QFT is just as nonsensical as asking the position of a particle in standard quantum mechanics. It only makes sense upon observation.

Particles (finally)

So far I haven't made any mention of how particles fit into all of this. The reason is that "particles" are simply special states in translation-invariant field theories. Namely, a "particle" is a state of definite energy, momentum, and charge (others will have a definition of "particle" in QFT that differs from mine, but the argument remains the same). Whether we're talking about particles or more generic states, the conclusion above remains the same. Asking about the charge density distribution in QFT is an ontologically nonsensical question until measurement.

I hope this helped! It's certainly confusing and hard to think about these things, and perhaps even harder to write about them. So if anything is confusing or just wrong, feel free to ask/correct me!

  • $\begingroup$ Nice answer. One aspect of my question you didn't quite get at, is if "upon observation" we ask the question "what is the charge density configuration," why is it that we only end up with charge localized to point particles? $\endgroup$
    – user1247
    Jan 5, 2019 at 16:17

Charge is a property of waves/particles to interact with each other in a certain manner. This understanding is the only one which is correct. Imagining charge existing in an empty space is useless. It is always meant to get into equations of motion to describe interaction with something else.

In a limited sense the charge of a wave/particle is the wave function "squared" ($|\psi|^2$) with the right meaning of the wave function itself as a probability amplitude, not the particle density in a classical sense. For example, if you scatter a charged projectile from an atom elastically (the initial and the final atom states are the same), then the wave function "squared" is involved as if it described the charge density. But if you scatter the projectile inelastically, the initial and the final wave functions are involved on equal footing and they are different, so it is actually the probability amplitude which is of the primary sense in describing interactions.

  • $\begingroup$ But my understanding is that there is no "wave function" in QFT. I understand the QM case; my issue is with QFT. Instead of a wave function corresponding to a single particle, we have a field that has an amplitude at many points in space, which is not interpreted itself as a probability amplitude. $\endgroup$
    – user1247
    Feb 25, 2013 at 11:34
  • $\begingroup$ @user1247: That field is quantum (quantized), not classical. In other words, it has operator amplitudes, not numerical ones. It encodes all plane waves with their wave functions in such a way that S-matrix (not the field itself) gives the probability amplitudes for transitions from one state to another. Quantized field is an auxiliary construction in the formalism of occupation numbers. $\endgroup$ Feb 25, 2013 at 12:11
  • $\begingroup$ @user1247 In QFT there is indeed a wave function $\Psi[\phi(\vec{x})]$, which takes in a field configuration $\phi(\vec{x})$ and returns a probability amplitude for the field to have that configuration. $\endgroup$ May 25, 2020 at 12:44

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