The Quantum Spring Interferometer · AI Generator #GEN-1C8791FC

Physics · quantum-mechanics

Problem

A novel quantum interferometer uses the vibrational states of trapped ions to create interference patterns. In this device, a single ion of mass m=40  amum = 40\;\text{amu} is confined in a harmonic potential with frequency ω0=2π×106  Hz\omega_0 = 2\pi \times 10^6\;\text{Hz}. The ion is initially prepared in a superposition state ∣ψ⟩=12(∣0⟩+∣1⟩)|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle), where ∣0⟩|0\rangle and ∣1⟩|1\rangle are the ground and first excited vibrational states. To create the interference, a weak electric field is applied that shifts the harmonic potential by a small distance dd. This field acts as a perturbation H′=mω02dxH' = m\omega_0^2 dx where xx is the position operator. The field is applied for exactly one quarter of the oscillation period (t=π/(2ω0)t = \pi/(2\omega_0)), then turned off. (a) What is the energy difference between the vibrational levels in eV? (b) Using first-order perturbation theory, find the phase difference accumulated between the ∣0⟩|0\rangle and ∣1⟩|1\rangle components of the wavefunction due to the applied field, given that d=10  nmd = 10\;\text{nm}. (c) After the perturbation, what is the probability of measuring the ion in the ground state ∣0⟩|0\rangle? Take ℏ=1.055×10−34  J⋅s\hbar = 1.055 \times 10^{-34}\;\text{J·s}, 1  amu=1.66×10−27  kg1\;\text{amu} = 1.66 \times 10^{-27}\;\text{kg}, and e=1.602×10−19  Ce = 1.602 \times 10^{-19}\;\text{C}.

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