The blockade section built the sequence : drive atom 1's qubit-to-Rydberg transition…

PHYSICS · Multi-part · Fermi University Physics · Quantum Information & Technology · Physical realizations and quantum technologies · Physical realizations and quantum technologies

Quantum Information and Technology, Ch Trapped ions, neutral atoms, photons, NMR and spins in solids

Problem

The blockade section built the sequence Uπ,2π,πU_{\pi,2\pi,\pi}: drive atom 1's qubit-to-Rydberg transition by a π\pi pulse, then atom 2's by a 2π2\pi pulse, then atom 1's by a second π\pi pulse, with only ∣1⟩|1\rangle coupled to the Rydberg state ∣r⟩|r\rangle, in the idealised limit of an infinite blockade. ((a)) Run the sequence on each of the four basis states ∣00⟩,∣01⟩,∣10⟩,∣11⟩|00\rangle,|01\rangle,|10\rangle,|11\rangle in this idealised limit, and confirm the map ∣q1q2⟩→(−1)q1+q2+q1q2∣q1q2⟩|q_1q_2\rangle\to(-1)^{q_1+q_2+q_1q_2}|q_1q_2\rangle. Show that a ZZ gate on each atom afterwards removes the factor (−1)q1+q2(-1)^{q_1+q_2} and leaves the controlled-ZZ, (−1)q1q2(-1)^{q_1q_2}. ((b)) The blockade is never truly infinite: two atoms both excited to the Rydberg state carry a real interaction energy V=C6/r6V=C_6/r^6, the same quantity that set the blockade radius Rb=(C6/ℏΩ)1/6R_{\rm b}=(C_6/\hbar\Omega)^{1/6}. Only the ∣11⟩|11\rangle input is affected, at the instant atom 1 sits in ∣r⟩|r\rangle and atom 2's 2π2\pi pulse is applied. Treat the two states ∣1r⟩|1r\rangle (atom 1 excited, atom 2 still in ∣1⟩|1\rangle) and ∣rr⟩|rr\rangle (both excited) as a driven two-level system of the kind built for a qubit, with Rabi frequency Ω\Omega and the blockade energy VV playing the part of a detuning Δ=V/ℏ\Delta=V/\hbar. Diagonalise this two-level system exactly, and expand the lower eigenvalue for V≫ℏΩV\gg\hbar\Omega to find the energy by which the finite blockade shifts the ∣1r⟩|1r\rangle level. ((c)) Hence show that, over the pulse duration T=2π/ΩT=2\pi/\Omega, the ∣11⟩|11\rangle input collects an extra phase δφ≈πℏΩ/(2V)\delta\varphi\approx\pi\hbar\Omega/(2V) beyond the ideal sign flip, and that the probability left in ∣rr⟩|rr\rangle once the pulse ends (population leaked outside the computational subspace) is of order (ℏΩ/V)2(\hbar\Omega/V)^2. ((d)) Evaluate δφ\delta\varphi for V=100 ℏΩV=100\,\hbar\Omega.

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