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You can simplify "produce a state" problems by breaking them into three parts: Prepare the collection of magnitudes you'll need, without worrying about phase or which state has which magnitude. Fix the phases. Fix the ordering. Now consider the Hardy state. What are the magnitudes that we need to make? We need one instance of $3/\sqrt{12}$ and three ...

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With help of the Mariia's comment, here is my attempt to reiterate the steps as a self-check/ notes for the Hardy-state (which I'll call $\left| Hardy \right>$): $\left| Hardy \right> = \frac{1}{\sqrt{12}} (3\left| 00 \right> + \left| 01 \right> + \left| 10 \right> + \left| 11 \right>)$ Following the accepted answer, the starting state $\... 5 As DaftWullie pointed out, the question about$W_n$has an excellent collection of answers here. For the Hardy state question (and a lot of other tasks like it), you can approach it as follows. Start with the$|0...0\rangle$state. Start by putting the first qubit "in the right state", which is a state$(\alpha |0\rangle + \beta |1\rangle) \otimes |0...0\...

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AHusain's answer is absolutely correct, but perhaps lacks some detail. The circuit that you want to implement is Basically, the key is to realise that you want to apply phase $e^{i\alpha}$ to the basis elements $|00\rangle$ and $|11\rangle$, and $e^{-i\alpha}$ otherwise. In other words, you care about the parity of the two bits. If you can compute that ...

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You cannot force a superposition to collapse in a particular direction. When you perform a measurement that removes a superposition, that 'collapse' is random, and you cannot choose which way it collapses. However, if you know what superposition you have, you can always convert it into any other state that you want to via unitary evolution (at which point ...

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Conjugate by a CNOT, you'll see a controlled unitary of multiplying by $e^{\pm 2i a}$ depending on which CNOT you do and an overall phase of $e^{\mp i a}$.

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i think you have done entaglment $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ so when you measure the first qubit the second qubit forces to be collapses to the same state as the first qubit state.

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I guess what you may need is the Pauli's operators https://qiskit.org/documentation/autodoc/qiskit.quantum_info.operators.pauli.html?highlight=pauli%20operator#module-qiskit.quantum_info.operators.pauli https://www.sciencedirect.com/topics/engineering/pauli-operator B.R Parfait

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\begin{align} \Pr(\text{“0"}) & = \frac{1}{4} ( 2 + \left\langle \psi , \phi \mid \phi, \psi \right\rangle + \left\langle \phi , \psi \mid \psi , \phi \right\rangle ) \\ & = \frac{1}{4} ( 2 + \left\langle \psi \mid \phi \right\rangle \left\langle\phi \mid \psi\right\rangle + \left\langle \phi \mid \psi \right\rangle \left\langle\psi \... 4 In[13]:= H = 1/Sqrt[2]*{{1, 1}, {1, -1}}; T = {{1, 0}, {0, Exp[I*Pi/4]}}; CNOT = {{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 0, 1}, {0, 0, 1, 0}}; KroneckerProduct[IdentityMatrix[2], T].CNOT. KroneckerProduct[T,ConjugateTranspose[T]].CNOT // MatrixForm {"1", "0", "0", "0"}, {"0", "1", "0", "0"}, {"0", "0", "1", "0"}, {"0", "0", "0", "I"} So that red boxed ... 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(... 1 Quantum algorithms provide a computational speedup by orchestrating constructive and destructive interference of the amplitudes. It is as if there must be a "minus" sign somewhere in the matrices - otherwise we merely work in the classical world, and would not see a computational speedup. Let's consider the following gates as controlled Pauli matrices: \... 2 With ancillas, you could construct a gate that is controlled on the n qubits being 0 to apply X to the ancilla. This can be done with polynomially many gates and ancillas. Once you have that, you can apply contolled e^{-i \theta} on any of the original n. Then do all the uncomputation. 1 The terms of expression do not cancel out in the balanced function case. We start with\frac{1}{2} (|0\rangle|0 \oplus f(0)\rangle - |0\rangle|1 \oplus f(0)\rangle + |1\rangle|0 \oplus f(1)\rangle - |1\rangle|1 \oplus f(1)\rangle)$$If f(0) \neq f(1), consider the first two terms (the only ones which can cancel with each other, since the state of the ... 1 If your using the latest qiskit version then it is qiskit.aqua 1 I think you agree that if you start with the state (a|0\rangle+b|1\rangle)|0\rangle, the cnot produces a|00\rangle+b|11\rangle. The issue is why is the state of the first qubit not the same as a|0\rangle+b|1\rangle. The answer is if you only look at that one qubit and you only look in the standard, Z basis, then they do look the same. But those are ... 1 I'm a little confused about which gates operate on which qubits and how, but following the linked question, I think I understand that you are wondering why, given a single qubit in the state in a|0\rangle+b|1\rangle and preparing two qubits in a state a|00\rangle+b|11\rangle does not qualify as cloning the first bit, especially because the probabilities ... 4 Start No control equals each control \forall U : U = C(U) \cdot \bar{C}(U) Opposite controls commute \forall U, V : [C(U), \bar{C}(V)] = 0 No control equals each control \forall U : U = C(U) \cdot \bar{C}(U) Self-inverse operations self-cancel Done More generally, for any "V conjugated by U" operation of the form U_a \cdot C_b(V_a) \cdot U_a^{-... 2 There is a way to simplify it down slightly - for some controlled unitary C\left(U\right)=I\oplus U and some arbitrary unitary V (of the same dimension as U),$$\left(I\otimes V\right).C\left(U\right).\left(I\otimes V^\dagger\right) = \begin{pmatrix}V&0\\0&V\end{pmatrix}\begin{pmatrix}I&0\\0&U\end{pmatrix}\begin{pmatrix}V^\dagger&0\\...

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