= Solution
{c}
Interpret the unspecified Gaussian width $\Delta$ as the <standard deviation>, so
$$
p(\delta)=\frac1{\sqrt{2\pi}\Delta}
\exp\!\left[-\frac{(\delta-\delta_0)^2}{2\Delta^2}\right].
$$
The <characteristic function> of this <Gaussian distribution> is
$$
\langle e^{i\delta T}\rangle=e^{i\delta_0T}e^{-\Delta^2T^2/2}.
$$
Average the free-evolution <Bloch vectors> from part (c):
$$
\boxed{\langle\mathbf r(T)\rangle
=e^{-\Delta^2T^2/2}(-\sin\delta_0T,-\cos\delta_0T,0).}
$$
It starts at $-y$ and spirals towards the origin in the equatorial plane, with period $2\pi/\delta_0$ and shrinking length $e^{-\Delta^2T^2/2}$. Individual <pure states> retain unit-length <Bloch vectors>; only ensemble averaging produces the apparent loss of <quantum coherence>.
After the final pulse, the corrected return population is
$$
\boxed{\overline{\rho}_{11}(T)=\frac12\left[1-e^{-\Delta^2T^2/2}\cos(\delta_0T)\right].}
$$
The oscillations lie between $[1\pm e^{-\Delta^2T^2/2}]/2$, start at zero, and tend to $1/2$. The upper envelope has no implication of exponential energy relaxation: this is static inhomogeneous <dephasing>. If instead the intended Gaussian convention was $p\propto e^{-(\delta-\delta_0)^2/\Delta^2}$, every envelope above becomes $e^{-\Delta^2T^2/4}$. The printed-plus-sign convention would give the complementary fringe.
\Image[/past-exam-of-the-laser-science-and-quantum-information-processing-course-of-the-university-of-oxford/2021-ramsey-dephasing.png]
{title=Ensemble <dephasing> and the corrected <Ramsey interferometry> return fringe, with $\Delta$ a <standard deviation>}
Back to article page
