Tag Info

After the first Hadamard gate you have the state $\frac{1}{\sqrt{2}}(|000\rangle+|100\rangle)$ (qubit zero being the leftmost). After the controlled $O_1$ gate, you get $\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle)$. And the next Hadamard leaves you with: \begin{align} |\psi\rangle &= \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}(|000\rangle+|100\rangle)+\... 0 2 assumptions for my answer (because I did not really understood your setup): Let's assume a pure state like |0010110> (not superposition) Assume you are doing the same experiment again and again (because in each time you collapse and project your state) The N Zs operator is actually counting if there is an odd or even number of ones in the state. Using ... 3 Not in general, no. Consider the state |\psi\rangle=\frac{1}{\sqrt{2}}\left(|0x\rangle+|1\bar x\rangle\right)|+\rangle $$for x\in\{0,1\}. We have \langle Z_i\rangle=0, and \langle Z_1Z_2Z_3\rangle=0 (to see this most trivially, look at the third qubit). However, the first two qubits are an eigenstate of Z_1Z_2 of eigenvalue (-1)^x. From the ... 1 You can use QuantumCircuit.decompose() for shallow decomposition circuit.decompose().draw('mpl') The result will be For further decomposition you can use Unroller transpiler pass from qiskit.transpiler.passes import Unroller from qiskit.converters import circuit_to_dag, dag_to_circuit # You can specify the target basis gate set: unroller = Unroller(basis=[... 3 Let \mathcal{H}_A denote the Hilbert space of qubits in partition A and similarly for \mathcal{H}_\bar{A}. Define the operator P:=Q|0^n\rangle\langle 0^n|Q^\dagger and write its Schmidt decomposition$$ P = \sum_a R_a\otimes Q_a\tag1 $$where R_a are operators on \mathcal{H}_A and Q_a are operators on \mathcal{H}_\bar{A}. See for example the ... 5 I think the key fact you're missing is that Z_2 \otimes Z_3 \otimes Z_4 = Z_2 \otimes Z_3 when you know qubit 4 is in the |0\rangle state; in the +1 eigenstate of Z. I'm not sure why that paper is using six CNOT gates instead of four. The ancilla isn't helping. Maybe it's just supposed to be an example for a more general case where it is helpful. 2 If you know the circuit that corresponds to the unitary U_{zz} for Z_1Z_2, then you also can also be sure that the same circuit will correspond to U_{zz} for Z_2 Z_3, except that the circuit will apply to qubits 2 and 3 instead of to qubits 1 and 2. This generalizes to any two qubits in your n-site Ising model, for example U_{zz} for Z_4Z_{13} ... 0 You are creating a gate from a manually-inputed unitary, and Qiskit does not know how such unitary can be decomposed out of the box. You can either tell Qiskit transpiler not to decompose your unitary, or extend the Equivalence Library (basically, class that stores information on how gates are decomposed) to know how to decompose your gate. The answer to ... 3 A canonical reference for gate decompositions is Barenco et al., Elementary gates for quantum computation. In particular, it also contains recipes to decompose an arbitrary n-qubit unitary into elementary gates (which, by parameter counting, requires about 4^n gates, assuming each gate has one real parameter.) 2 I believe this Q&A answers your question about decomposition in detail: Minimum number of 2 qubit gates to build any unitary In short, you are correct that the lower bound for a number of 2-qubit gates necessary to implement an arbitrary unitary U is \Omega(4^n) where n is the number of qubits. I am not entirely sure what authors meant, but perhaps ... 1 In superconducting quantum computers, gates are implemented with microwave pulses. So, the gates do not have input/output nodes (wires) in classical sense. It it true, that qubits are connected among each other (although not always fully) with coupling capacitors in order to apply two-qubits gates, however, again they are microwave pulses. Another physical ... 4 In quantum circuits, the qubits are represented with the "wires" (the horizontal lines on which gates are chained) rather than "input" and "output" nodes. They usually start in the |0\rangle state, their state changes as the gates are applied, and the classical output is the readout of the measurement nodes. 2 It's just Grover's algorithm. You don't need to mark the state manually, at least in Grover's algorithm. If you mark them manually, you might lose the essence of the oracle. If you mark the target state with an ancilla, the oracle can be easily build by using a control gate. In your case, the oracle should be$$\left( e^{i\theta}|00\rangle \langle 00|+e^{i\...