# Quantum Information and Quantum Optics with Superconducting Circuits - Exercise Solutions (Chapter 8)

#### 8.1

By expanding the tensor product, we have the all possible binary numbers.

As discussed in 6.1.4, for a single qubit, Hadamard gate can be implemented as:

using the coherent drive:

that's on resonance and .

To apply it to identical flux qubits, we have to have different gaps on neighboring qubits so that the drive doesn't affect neighboring qubits.

Since and ,

#### 8.3

where . Hence, we have:

Then,

So in the interaction picture with , the system is described with itself. Note that, in this picture, the photon has no energy. (Photon energy is absorbed in ).

We split as:

and take as a weak perturbation.

Now we apply the nondegenerate perturbation theory. In general, the transformation generator is given by:

We consider the states with minimum number of photons (or subspace of JC-Ladder), such as:

Since photons have no energy, their respective energy are:

Then, can be expanded as:

Note that, with similar calculations, you can confirm that the above expression for is valid for any subspace of JC-Ladder though we explicitly calculated in subspace.

So the perturbative correction to the Hamiltonian is given by:

Now,

Hence,

So, the effective Hamiltonian (in the interaction picture) is:

--- (1-1)

Since , we can go back to the original Schroedinger picture without modifying . So getting back to the original Schroedinger picture and dropping the photon energy , we have:

(2) Defining , we have:

Hence states are eigenstates of whereas states are mixed together via time-evolution.

As discussed in 8.3.2, when , the time evolution operator becomes:

So you can implement iSwap with up to local transformations:

--- (2-1)

(3)

From (1-1), the effective energy of qubits increase . This would influence the local transformation in (2-1). But I don't see how it affects the two-qubit exchange???

#### 8.4

We use the unit system with , and a consistent notation for energy levels (instead of ).

(1) The eigenvalues of subspace are:

where

The eigenvalues of subspace are:

where

(2) From the results of (1), energy levels degenerate at and , that is, and

(3) In , for large , we can ignore . So approximating as , we have:

They respectively correspond to states .

(4) Relabel the eigenvalues as:

Then, the phase changes are described as:

(5) Define:

Then,

If we modify the adiabatic change times slower, is modified as:

So, by adjusting the speed of the adiabatic change, we can achieve arbitrary value of .

Note that once we achieve the transformation with , by applying single qubit operations, we can achieve the transformation:

(6) N-bit Quantum Fourier transformation can be implemented with control-phase gates:

So we need .

Hence,

#### 8.6

(1) Asymmetric totally depolarizing channel:

This applies the phase flip (along the axe) with probability .

Since , by setting , we have:

In this symmetric case, with probability p, the system is replaced with the completely mixed status .

(2) Random flip channel is implemented as:

This is a special case of the asymmetric totally depolarizing channel .

(3) For ,

So applying it times, we have:

(4) Kraus operators for asymmetric totally depolarizing channel are:

Kraus operators for dephasing channel are:

Actually,

Since , for asymmetric totally depolarizing channel,

For other channels, you can confirm with similar calculations.

#### 8.7

Define and you can confirm with a direct calculation that:

.

SInce [tex|displaystyle {\rm tr}(A\otimes B)={\rm tr}(A){\rm tr}(B)], it's also true that:

.

So, under the inner product they are linear-independent and become orthobormal basis of dimension linear space.

As the matrix has elements, it can be expressed by a liner combination of .

And the coefficients are determined by .

#### 8.8

In the textbook, matrix representation and operator notation are somewhat mixed up. In the operator notation, we have:

--- (1)

where

--- (2)

where

From (1), we have:

.

From (2), we have:

Now we prove (8.16). For any ,

On both sides, we apply from left:

Since , we have:

--- (3)

Now we take , and using the relationship , we have:

#### 8.9

(1) Deduce and for .

For , from Exercise 8.8 (1), we have:

Applying and taking , we have:

For ,

Hence, from Exercise 8.8 (3),

(2) and for .

We use the Einstein summation convention in the following calculations.

For ,

where

If either or is , .

For other cases, from the relationships:

we have:

(1st term)

(2nd term)

(3rd term)

For ,

where

We use the convention that , then we have:

(3) and for CNOT .

Hence for ,

For ,

#### 8.10

The error mapping process is described as:

In the operator form, this can be described as:

Now the matrix elements of are calculated through the relationships:

For our particular case, using the notation :

Non-zero elements are:

By taking the limit , the error-free version is:

Hence, the process fidelity is:

The average fidelity is:

The matrix elements of can be calculated through the relationship:

#### 8.12

Define as a probability of depolarization, and the totally depolarization channel is:

Hence, for a pure state

So the fidelity for this particular state is:

Since this is independent of the state, the average over all pure states is given by:

Note that the depolarization parameter in the textbook (p.220) is a probability of not decaying: . So (8.24) should be: