The motional-bus section built the interaction H_I(t) = 2 ( _x^ (1) + _x^ (2) ) (a…

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Problem

The motional-bus section built the interaction

H_I(t) = \frac{\hbar\eta\Omega}{2}\left(\sigma_x^{(1)}+\sigma_x^{(2)}\right)\left(a\,e^{i\delta t}+a^\dagger e^{-i\delta t}\right)

and the displacement-composition rule

D(d\alpha)\,D(\alpha) = e^{i\,\mathrm{Im}(\alpha^*d\alpha)}\,D(\alpha+d\alpha)

, valid because

[a,a^\dagger]=1

. Held in an eigenstate of

S=\sigma_x^{(1)}+\sigma_x^{(2)}

with eigenvalue

s\in\{+2,0,-2\}

, the interaction drives the bus through a sequence of infinitesimal displacements

D(d\alpha_s)

, one every instant

dt

, with

d\alpha_s=-i\dfrac{\eta\Omega s}{2}e^{-i\delta t}\,dt

. ((a)) Integrate

d\alpha_s

to find the trajectory

\alpha_s(t)

traced by the bus for a general spin eigenvalue

s

. ((b)) Using

\mathrm{Im}(\alpha_s^*\,d\alpha_s)

as the phase added by each step, find the total phase

\Phi_s

built up once the loop first closes, at

t=2\pi/\delta

, and confirm that its magnitude is

2\pi(\eta\Omega s/2\delta)^2

. ((c)) Writing

S^2=2I+2\sigma_x^{(1)}\sigma_x^{(2)}

as an operator identity, show that the closed-loop unitary is

e^{-i\theta\sigma_x^{(1)}\sigma_x^{(2)}}

up to an overall phase, with

\theta=\pi\eta^2\Omega^2/\delta^2

, and find the detuning

\delta

that gives the maximally entangling value

\theta=\pi/4

. ((d)) Suppose instead the drive is switched off early, at

t=\pi/\delta

, half the loop. Find

\alpha_s

at that instant, evaluate

|\alpha_s|

for

s=\pm2

at the

\delta

found in part (c), and explain in a sentence why stopping there leaves the spin and the motional bus entangled rather than delivering a clean two-qubit gate.

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