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Optimal quantum spectroscopy using single-photon pulses

Sourav Das, Aiman Khan, Francesco Albarelli, Animesh Datta

Abstract

We provide the ultimate precision attainable in spectroscopy of a quantum emitter using single-photon pulses. We find the maximum for estimating the linewidth to be independent of the details of the emitter's bare Hamiltonian while that for the detunings not to be so. We also identify optimal pulse shapes attaining these precisions.

Optimal quantum spectroscopy using single-photon pulses

Abstract

We provide the ultimate precision attainable in spectroscopy of a quantum emitter using single-photon pulses. We find the maximum for estimating the linewidth to be independent of the details of the emitter's bare Hamiltonian while that for the detunings not to be so. We also identify optimal pulse shapes attaining these precisions.

Paper Structure

This paper contains 2 sections, 18 equations, 2 figures, 1 table.

Figures (2)

  • Figure 1: The single-photon $\ket{1_{\xi_\text{in}}}$ excites the emitter from the ground state $\ket{g}$ to the single-excitation subspace (SES) of the emitter which is illustrated by the light blue shaded oval. Its elements $\{\ket{j}\}$$(j=1,2,..,N)$ (with energies $\omega_j$ relative to $\ket{g}$ which is defined to be zero) are illustrated with black solid lines in the SES. The detuning of the $j$-th singly-excited state is $\Delta_j = \omega_j - \omega_c$, where the thin horizontal dashed line along the path of the pulse indicates the carrier frequency $\omega_c$. $\Gamma$ denotes the coupling strength with the pulse and $\ket{\gamma}$ (thick red line) is a normalized superposition of the states $\{\ket{j}\}$ in the SES. The coupling encodes information of the emitter in the state of the scattered pulse $\ket{1_{\xi_\text{out}}}$.
  • Figure 2: The dimensionless QFI of estimating $\Gamma$ for a TLS using regularized optimal incident pulses as a function of the regularization parameter $\kappa/\Gamma$. The regularized delta functions are defined as the Lorentzian $\delta_\kappa(\omega)=\kappa/(\pi(\kappa^2+\omega^2))$, the Gaussian $\delta_\kappa(\omega)=e^{-\frac{\omega^2}{2\kappa^2}}/(\kappa\sqrt{2\pi})$, and, the rectangular $\delta_\kappa(\omega)=\Theta(\omega)\Theta(\kappa-\omega)/\kappa$.