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First, note that the Controlled-X gate can be written as: $$ CX = |0\rangle \langle 0| \otimes I + |1 \rangle \langle 1| \otimes X $$ This tells us that the first qubit is the controlled, and the second qubit is the target. And when the controlled qubit is $|0\rangle$ we do nothing hence the Idenity operator. When the controlled-qubit is a $|1\rangle$ we ...


5

There is the concept of controllability of a quantum system, i.e. do the given set of controls permit you to create any state or unitary? Usually this is computed by looking at the Lie Algebra of the system, and can be quite messy; you need to take the individual Hamiltonian terms that you can control, and calculate all their commutators to arbitrary orders. ...


5

Quantum control does not necessarily allow implementing just any gate. Imagine your control of the system is a time-dependent energy. That corresponds to the Hamiltonian $\hat{H}(t) = c(t) \hat{Z}$. Then you can only ever rotate about one axis of the Bloch sphere and your only choice is how fast to rotate when. This is obviously insufficient to even generate ...


4

It is true that these two circuits are equivalent: as the the controlled qubit $q_1$ is the same. So if $q_1$ is a $|1\rangle$ then you can see that it will apply two $X$ gates to $q_2$ and they will cancel each other out.


3

Control errors The term control error is generally used to refer to errors due to imperfections of the qubit control system. Hardware devices that control qubit evolution have a number of knobs that the control system sets to various values. In the process known as calibration we learn the settings of the knobs that correspond to each of the supported gates. ...


3

In general, an algebra $\mathcal{A}$ generated from a set $\{H_1, H_2,..., H_n\}$ by commutation refers to the algebra whose generators are $H_1,H_2,...,H_n$, all their first-order commutators $C_{ij} = [H_i,H_j]$, and all their second-order commutators $C_{ijk} = [[H_i, H_j],H_k]$ and so on.


2

You'll place the phase within the CRz gate. The approach you've taken essentially argues that: $$ e^{it H_3} \approx e^{it \alpha X_1 \otimes Y_2} e^{it \beta Z_1 \otimes Z_2} $$ So, when you're applying the Rz gate, you can select the $it \alpha$ coefficient to align with the necessary phase (likely you'll use $\theta = -2 t \alpha$, depending on the ...


1

The Qiskit units are unitless and require the drive Hamiltonian provided by the backend to link the program into a simulatable format. Keep in mind the Hamiltonian unit's are angular frequencies. Real hardware often has transfer functions and non-linearities that are not captured by the simplistic Hamiltonian that is returned, unless you account for these in ...


1

"Runtime" is not so easily quantified, it depends a lot on the compilation, the other operations in your circuit and whether you simulate or have a real backend. Generally, the different methods trade off circuit depth (more gates, but less qubits) against circuit width (more qubits, less gates). If we define the runtime by the number of gates we ...


1

You are correct that a controlled-gate should be written in the form $P_0\otimes I+P_1\otimes U$ for a unitary $U$. It certainly should not be of the form $I\otimes U$. I don't have a copy of your textbook to hand, but could you perhaps have misread it? Perhaps the author meant $I\oplus U$, which is an alternative way of describing the matrix structure? It ...


1

#This line creates the circuit instance circuit = cirq.Circuit() #These lines creates the cubits you want to use from a given length length = 3 qubits = [cirq.LineQubit(i) for i in range(length)] #Here you assign the control qubits (the first two) and then the target qubit controlled_rotation_on_Z = cirq.Z.controlled_by(*qubits[:-1]) circuit.append(...


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