Restricting the Cooper-pair box to the two lowest charge states and gives the Hamiltonian

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Quantum Information and Technology, Ch Superconducting qubits and circuit QED

Problem

Restricting the Cooper-pair box to the two lowest charge states

|n=0\rangle

and

|n=1\rangle

gives the Hamiltonian, in that basis,

H = \begin{pmatrix} 4E_C n_g^2 & -E_J/2 \\ -E_J/2 & 4E_C(1-n_g)^2 \end{pmatrix}.

Take

E_C/h = 0.30\,\mathrm{GHz}

fixed and consider three Cooper-pair boxes with

E_J/E_C = 0.5

1

and

2

, the regime where keeping two charge states is a fair approximation. ((a)) Diagonalise

H

to find the energy splitting (the

0

1

transition frequency) as a function of

n_g

, and evaluate it at the sweet spot

n_g = 1/2

and at

n_g = 0

for each of the three ratios. ((b)) For each ratio, compute the charge dispersion, defined here as the difference between the splitting at

n_g = 0

and at

n_g = 1/2

, and give the three values in

\mathrm{GHz}

in order of increasing

E_J/E_C

. ((c)) Show that for

E_J \gg 4E_C

this model's dispersion falls only as

8E_C^2/E_J

, and explain in one sentence why the two-state model therefore cannot describe a transmon with

E_J/E_C\approx50

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