# Quantum Information and Quantum Optics with Superconducting Circuits - Exercise Solutions (Chapter 6) : Part2

#### 6.10

(1)

Define as:

Then,

So, the interaction Hamiltonian is:

Since:

We have:

By dropping the non-RWA terms such as , we have:

(2) By dropping the third state, we have:

For

Hence, from the Schroedinger equation: , we have:

--- (1)

--- (2)

By combining them, we have:

, where

From the initial condition , , so,

From (2),

The initial condition for is:

So, if , at .

(3) The interaction Hamiltonian is:

In order to diagonalize the qubit space, introduce the states as:

Since ,

where,

We will apply an additional transformation with to a new interaction picture, then, the new interaction Hamiltonian is:

Since ,

For

So, from the Schroedinger equation , we have:

--- (1)

--- (2)

--- (3)

From (3),

By substituting (1), (2), we have:

, where

From the initial condition , , so,

Then

From the initial condition (where we suppose that ) and (3),

Hence, the excitation probability is:

When , to suppress the leakage below 1%,

(4)

Suppose that

The error rate is

To suppress the error rate ,

#### 6.11

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#### 6.14

Since the current rotates in the opposite directions in two loops, the external flux in the potential is replaced by .

#### 6.15

is (classically) determined by (6.52):

When

, and

Likewise,

For the inductor ,

Suppose that the wavefunction of and is approximately Gaussian:

Since , the integrand is an odd function.

The diagonal elements are:

Likewise,

Note that

Using the new basis:

we have:

#### 6.16

(1)

where,

Since ,

Hence, the interaction Hamiltonian is:

The first two terms oscillates two times faster than the energy gap , so as an approximation (of a weak interaction), we can drop these terms. Hence, getting back to the original Schroedinger picture, we have:

Using as basis, the matrix representation becomes:

By rearranging the order of basis as , we have:

So in the even space , and are the eigenvectors with eigenvalues and . In the odd space , are the eigenvectors with eigenvalues .

(2)(3) By applying the same matrix representation for the full Hamiltonian, we have:

In the odd space, we have the same eigenvectors (with the same eigenvalues) as (1). In the even space, the eigenvalues are .

The ground state with the eigenvalue is:

Likewise, the eigenstate with the eigenvalue is:

Hence, the probability of finding the exited state in is:

#### 6.17

(1) Using as basis, the matrix representation of non-interactive term is:

The interaction term is the same as in Problem 6.16. So, by rearranging the order of basis as , the full Hamiltonian becomes:

(2) The exact eigenvalues of full Hamiltonian are:

In the even space:

In the odd space:

On the other hand, if we apply the second-order non-degenerate perturbation in the odd space, the eigenvalues without interaction is , and the correction to the eigenvalue is:

Hence the approximate eigenvalues are:

This coincides with the exact result except . This approximation fails when (due to the degeneracy.)

(3) Without interactions, the state stays in the same one:

For ,

Hence, from the Schroedinger equation ,

From the initial conditions , the solution is:

Hence, the fidelity is:

To achieve 99% fidelity, , so .

This is achieved by, for example, .

(5) In the work by Barends et al. (2014), they used the system with coupling 30MHz and interaction time 43nsec to achieve around 99% fidelity.