What is the difference between electro-optic sidebands and parametric down conversion sidebands?

Question

I am trying to understand the difference between the modes generated via electro-optic modulation (EOM) and parametric down conversion (PDC) in the context of quantum optics. Throughout the literature, PDC is the workhorse for the generation of quantum states, while EOM is never used. However, in a classical channel, particularly in integrated photonics, if I have some light coming through either a non-linear micro-ring or an electro-optical modulator, there will be coherent sidebands generated. Assume now that you have two photonic platforms:

Platform 1

Imagine a cavity in a material which has $\chi^{(3)}$ non-linearity, such as silicon nitride. Imagine now that such cavity has a free spectral range in the GHz range (say 20GHz), and we have no issues with phase matching due to dispersion. Then when light goes through it a number of non-linear processes takes place, particularly four-wave mixing effects, and generates sidebands at every 20GHz.

Platform 2

Imagine a cavity in every way similar to the one in platform 1, with the same FSR, but the material has a very low $\chi^{(3)}$, but a decent $\chi^{(2)}$. Therefore, we add an electro-optic phase modulator to the cavity and run a 20GHz (same as FSR) RF signal down the modulator. This will also generate coherent sidebands when light goes past it.

Question

Imagine we work on a single photon level and we send an single photon in an optical field with the same temporal profile on both platforms. How are the output quantum states different in each platform?

I truly want to understand this question and I welcome any literature on the subjects, particularly textbooks that go through detailed explanations.

Answer 1

There are several differences between the two processes. Spontaneous four-wave mixing (SPDC is a $\chi^{(2)}$ process) is a nonlinear process, while electro-optical phase modulation is an optically linear process. SFWM is a fundamentally quantum process due to its spontaneous nature, while phase modulation can be described classically. Additionally, the cavity scenario you describe is somewhat artificial, since a modulated EOM will generate sidebands even without a cavity, while a SFWM process will generally have a continuous spectrum.

You mention that EOMs are not used for single-photon generation. The reason for this is because they cannot generate single-photon states. As I mentioned, a phase shift is a linear-optical operation, and a time-varying (modulated) phase shift is as well. In linear optics, there is a formal equivalence between linear optics and coherent states, which can be summarised as the complex amplitudes of a multi-mode coherent state transforming the same as the amplitudes of a superposition of a single photon over the same modes.

If, for simplicity, we start from a classical picture and use our knowledge that a coherent state in the frequency mode $\omega_0$ is transformed by a periodic phase modulation as

$$ |\alpha\rangle_0 \to \Pi_{k=-\infty}^{\infty} |J_k(\beta) \alpha\rangle_k $$

where $J_k(\beta)$ are Bessel functions and $\beta$ is the modulation depth, then a single photon in the same frequency mode is transformed as

$$ |1\rangle_0 \to \sum_{k=-\infty}^{\infty} J_k(\beta) |1\rangle_k. $$

The reason electro-optical phase modulation cannot be used for single-photon generation is that the photon statistics in the sidebands are that of the input coherent state. For example, for low modulation depth the state would be

$$ |J_0(\beta) \alpha\rangle_0 \ |J_{-1}(\beta) \alpha\rangle_{-1} \ |J_{1}(\beta) \alpha\rangle_1. $$

Compare this with a SFWM process, which produces the state

$$ |\psi\rangle \approx C_0 |\alpha\rangle_0 |0\rangle_{-1} |0\rangle_{1} + C_1 |\alpha\rangle_0 |1\rangle_{-1} |1\rangle_{1}, \qquad C_0 >> C_1. $$

A single-photon state can be obtained by detecting a photon in for example the frequency mode $-1$, as this projects away the vacuum component and leaves a single photon in mode $+1$ due to the photon-number entanglement. In the phase-modulation process, the two coherent states in the modes $\pm 1$ don't have any photon-number correlation and photo-detection in one frequency mode does not herald the presence of a photon in the opposite mode.

If you want a full quantum description of electro-optical phase modulation, you can take a look at for example arXiv:1103.4747 and references therein.