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6

We know that giving a single qubit starting in the state $|0\rangle$, which is a state one can initialize very fast with high fidelity, then we can put it in the superposition state $|\psi \rangle = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$ by applying a Hadamard gate. That is, $$H |0\rangle = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$$ And we can do for $N$-...

5

If you initialize three qubits to $|0\rangle$, apply a Hadamard gate to each, then measure each in the computational basis, the result will be an independent coin flip for each bit: that is, any of 000, 001, 010, ..., 111, each with probability 1/8. If you measure all three bits of the GHZ state in the computational basis, you'll get either 000 or 111, each ...

5

$\langle \psi|$ is not a quantum state, but a linear functional on the set of quantum states. $|\psi\rangle$ is a quantum state, any gates that you can apply to it can only take it to quantum states, so in particular not to $\langle \psi|$. Of course, mathematically it is trivial to transform $|\psi\rangle$ into $\langle \psi|$ and vice-versa, but this is ...

4

You can prepare equal superposition by application of Hadamard gate on each qubit. The result will be state $$|q\rangle=\frac{1}{\sqrt{2^n}}\sum_{i=0}^{2^n}|i\rangle,$$ i.e. the desired equally distributed superposition. As Hadamard gate belongs to Clifford group, it acts quickly. It can be even simulated on classical computer in polynomial time. But the ...

3

I think that asking for an exact solution is pointless, because quantum computers don't have infinite precision. You are limited, for example, by accuracy of pulses that control the gates. To implement the idea of the mentioned answer, you can refer to this paper which introduces a general method for constructing an efficient and highly accurate quantum ...

3

Let $\rho_{AB}$ be a quantum state shared between two parties, Alice and Bob. Suppose Alice performs a POVM measurement $\{M_i\}_i$ on her half of the state. Then the probability that Alice obtains outcome $i$ is given by the Born rule as $$p(i) = \mathrm{Tr}[\rho_{AB}(M_i \otimes I)].$$ But whenever we have the trace of a multipartite operator we can ...

2

The local state (described by density matrix) of each qubit in EPR state is \begin{equation} \rho=\frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{equation} It does not depend on basis, so both outcomes have 50% probabilities in every measurement basis. Proving directly the expression for qubit's local state is a little long. The EPR ...

2

Theorem: Suppose $U$ is a unitary operation on a single qubit. Then there exist real numbers $\alpha, \beta, \gamma, \delta$ such that $$U = e^{i\alpha} R_z(\beta) R_y(\gamma)R_z(\delta)$$ This is on page 175 of this textbook. Ignore the global phase, each of these gates can be implement directly on a quantum computer. So given a vector $\vec{x}$, you just ...

2

$\left<\psi\right|$ is not a state of a quantum system, it is a linear functional that takes a quantum state and returns a scalar. In terms of basic Linear Algebra, it is a row vector rather than a column vector, and the conjugate transpose of $\left|\psi\right>$. So $\left<\phi|\psi\right>$ is a scalar (inner product) while $\left|\psi\right>\... 2 I actually found the solution to my problem. Given$n$qubits initialized as$|0\rangle$, one can act with$R_y(\alpha_i)$and$R_z(\beta_i)$on all qubits$i$, excluding one of them (for me this will be the last one). This will put every qubit in an arbitrary superposition state$a|0\rangle + b|1\rangle$. Then one just needs to apply$n-1$CNOT gates with ... 1 As Martin Vesely mentioned in the comments, you can use the initialize function to perform such a task. For instance, to create the state you desire, you can do the following: import numpy as np from qiskit import QuantumCircuit from qiskit.transpiler.passes import Decompose qc = QuantumCircuit(2) initial_state = np.array([1, 1, 0, 1], dtype=float) ... 1 In fact, I figured out by myself. When I was reading this website, it posts a density matrix that I want(that is a long website, to find the corresponding part, just search the keyword 'identity'). Here comes the method. To produce a$2^n$dimensional density matrix, you need n qubits as ancilla and n qubits as the target. In the case of$n=1$, implement the ... 1 Ideal measurement Let$\rho_0$denote the state resulting from the$|0\rangle$preparation. The fidelity of$\rho_0$and$|0\rangle$is $$F: = F(|0\rangle, \rho_0) = \langle 0|\rho_0|0\rangle.$$ On the other hand, the probability of obtaining$|0\rangle$when measuring$\rho_0$in the computational basis is$\$ p(0|\rho_0) = \mathrm{tr}\left(|0\rangle\...

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