A charged particle has uncertainty of position and momentum.

The electric field's value at a point due to a charged particle in space depends on the distance from that point to the particle. If the particle's position is uncertain, the distance to any point is uncertain, and so it would seem as if the electric field due to the charge would be uncertain.

A moving charge also produces a magnetic field whose magnitude is dependant on the particle's velocity. Given that momentum contains velocity and momentum is uncertain, so too would appear that the magnetic field induced by the moving particle is uncertain in magnitude.

Am I wrong? If I am, why are fields exempt from this uncertainty, and what does a field induced by an uncertain charged particle really look like? If I am right, are there implications, e.g. due to the fact that fields extend infinitely, a charge may emit a photon at arbitrary distance from itself.

  • $\begingroup$ Though I don't know quite enough QFT to officially answer your question, you are correct in thinking that electromagnetic fields have uncertainty. This is true even in the absence of a particle to generate them (fluctuations can and do happen in vacuum). $\endgroup$ Feb 7, 2018 at 0:45
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    $\begingroup$ Near duplicate. Yes, quantum fields are collections of quantum harmonic oscillators, so they have nontrivial constraints on variances. You are inching towards replicating the seminal analysis of N. Bohr and L. Rosenfeld, Kgl. Danske Vid. Sels. , Math. -fys. Medd. 12, No. 8 (1933), where they carried out your thinking, all but proving that if you quantize particles, you must also quantize fields both generated by them and interacting with them. $\endgroup$ Feb 7, 2018 at 2:05
  • $\begingroup$ ... Basically, vector potentials and electric fields are (generalized) conjugate variables, like x and p, so their variances are correlated.... $\endgroup$ Feb 7, 2018 at 2:06

1 Answer 1


You are right in thinking that fields are uncertain. In fact, Heisenberg's uncertainty principle can be generalized to canonical coordinates ($q_i$) and momenta ($p_j$) of the Lagrangian: $[q_i,p_j]= i\hbar\delta_{ij}$ with $[q_i,q_j]=[p_i,p_j]= 0$.Given field theories, such as electromagnetism, we can likewise find canonical coordinates and momenta associated with the Lagrangian (or Lagrangian density) of the system to find the corresponding Heisenberg uncertainty principle.

The easiest example to see this would be free scalar field theory with Lagrangian density \begin{equation} \mathcal{L}[\phi, \partial_{\mu}\phi] = \dfrac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\dfrac{1}{2}m^2{\phi}^2 \end{equation} that gives Klein-Gordon equation, ($\square +m^2)\phi=0$, as its equation of motion. Given the canonical coordinate ${\phi}$, canonical momentum gives

\begin{equation} \pi = \dfrac{\delta \mathcal{L}}{\delta \dot{\phi}} = \partial_t\phi \end{equation}

Therefore, we can apply Heisenberg's uncertainty principle to obtain the equal-time commutation relation:

\begin{equation} [\phi(\vec{x},t),\pi(\vec{y},t)] = [\phi(\vec{x},t),\partial_t\phi(\vec{y},t)] = i\hbar\delta^3(\vec{x}-\vec{y}) \end{equation}

If you know how to express the scalar fields $\phi(\vec{x},t)$ using creation and annihilation operators, you can use the commutation relation between the ladder operators to derive the above result as well. Note that the above equation implies that you cannot measure both the field and the time derivative of the field with infinite precision.

Likewise, you can find the canonical coordinates and momenta for the electromagnetism to obtain similar relation. It is a bit more complicated than the scalar field theory because we need to fix the gauge in order to remove unphysical degrees of freedom. If you are interested in how to find such canonical variables for electromagnetism, I highly encourage you to read Chapter 8, sections 1-3 of Weinberg's Quantum Theory of Fields, Volume I.


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