1
$\begingroup$

How does one define phase and group velocities in QFT? More precisely, are they referring to the mode structure of the field or do they characterize excitations?

The question is motivated by question Speed of photon, which asks for a phase and group velocities of a single photon. Thus, if we take the definitions from classical electrodynamics, $$ v_{ph}=\frac{\omega}{k},\textrm{ }v_g=\frac{d\omega(k)}{dk}|_{k=k_0}, $$ then the two velocities characterize the mode structure of the field (plane waves in free space, but something else in a confined geometry), rather then actual excitation, and we could say that they are the same for a single photon, or any more complex state.

On the other hand, a single-photon state $a_{\mathbf{k},\lambda}^\dagger |0\rangle$ is not really any close to a plane wave (unlike a coherent state), so speaking about its phase and group velocities is a kind of meaningless (although it could be meaningful for a coherent state).

To restate the question:

  • Are there standard generalizations for phase and group velocities in QFT or Quantum Optics?
  • If not, what would be a meaningful way to extend these definitions to the quantum case?
$\endgroup$
2

1 Answer 1

1
$\begingroup$

A general single-photon state can be represented as $$ |\psi_1\rangle = \int \hat{a}_s^{\dagger}(\mathbf{k}) \psi_s(\mathbf{k})\ \frac{\text{d}^3 k}{(2\pi)^3} |\text{vac}\rangle , $$ where $\psi_s(\mathbf{k})$ is a normalized angular spectrum (or Fourier domain wave function), and $|\text{vac}\rangle$ is the vacuum state. As a result, the phase velocity and group velocity are governed by the properties of state are governed by $\psi_s(\mathbf{k})$, very similar to the way the angular spectrum determines these properties for a classical field.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.