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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?

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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.

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@user1247 @LubosMotl please continue your discussion here, if you wish to do so. – Manishearth Feb 25 '13 at 14:57

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.

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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. – user1247 Feb 25 '13 at 11:34
@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. – Vladimir Kalitvianski Feb 25 '13 at 12:11

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