# Tag Info

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(Disclaimer: I don't know what's going on "under the hood" in Qiskit's initialize, my answer is based on Q# part of the question, but I suspect it's something similar.) The state of a quantum system is not something that can be directly initialized to the necessary state (quantum computing would've been a lot easier if it was possible!) Instead we ...

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The effect of noise is to give us outputs that are not quite correct. The effect of noise that occurs throughout a computation will be quite complex in general, as one would have to consider how each gate transforms the effect of each error. There is a very good article that shows Noise and its effect in practical using Qiskit https://qiskit.org/textbook/ch-...

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Aer simulators will support what you're looking for very soon: https://github.com/Qiskit/qiskit-aer/pull/834

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You can also create a Statevector, that can be directly initialized as follows: from qiskit.quantum_info import Statevector sv = Statevector.from_label('11') You can use sv.evolve(qc) to apply an operator/circuit to the state, where qc is the operator/circuit. sv.data gives you the numpy array, containing the actual implementation of the state. Check this ...

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may be you can define a function like below from qiskit import QuantumCircuit def qubitinitialize(n,n_init): #n = number of qubits #n_init = initial state of each qubit qc = QuantumCircuit(n) for i in n_init: if i == 1: qc.x(i) return(qc) This function will return a circuit with initial state as you provide in n_init n = 4 #...

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Classical information is stored the same way as it is for classical computers. You cannot use quantum states to store (in a retrievable way) more information that you would with classical bits.

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For interactions between non-nearest neighbour qubits, ancilla qubits are required, together with SWAP gates. The state of one of the (in this case) two qubits is swapped with the ancilla. This operation is repeated until the qubits are NN, and then the interaction can take place. After this is done, then the state of the ancilla is swapped back with the ...

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If you don't have the ability to perform the controlled-not directly between a pair of qubits, then you simply need to swap the qubits to place them onto a pair of qubits which can have a controlled-not applied to them.

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Qiskit assumes that initially each qubitt is set to the $|0\rangle$ state. So if you have $n$ qubits, the initial state is $|00..0\rangle$. If you want to flip the state of some specific qubits, you have to apply an $X$ gate to each of those specific qubits. For example, the following code sets the $|11\rangle$ state: qr = QuantumRegister(2) cr = ...

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The $\frac{1}{\sqrt{2}}$ is due to the normalization condition which says that sum of the squares of the amplitudes of the must be equal to one while the square of the amplitude refers to the probability of getting that particular state when the qubits are measured The vectors $|+⟩$ and $|-⟩$ are known as the eigenvectors for the Hadamard gate. When we apply ...

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The trick is that you don't need to calculate the inverse of $B$. What you really want to evaluate is $$(\langle 0|\otimes I)(B^\dagger \otimes I)(\text{select}(V))(B\otimes I).$$ So, the point is that you only need $\langle 0|B^\dagger$ which is the Hermitian conjugate of $B|0\rangle$, which you know.

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#1: the $1/\sqrt{2}$ is a normalization which ensures that the length'' of the vector is one. #2: The notation $|\pm\rangle$ is just a label for the two states defined above. Since the states $|0\rangle, |1\rangle$ are elements of a vector space, you can take linear combinations and therefore construct the states $|\pm\rangle$

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I think unlike the relative phase in the answer you reference, it is a global phase in your case: Your XHP-circuit where P=ID, prepares the state: [0.707+0j,-0.707+0j], where P=X, prepares the state: [-0.707+0j, 0.707+0j]. These states are differ by a global phase ${e}^{i\pi}=-1$. But the global phase is undetectable $|ψ⟩:={e}^{iδ}|ψ⟩$, also see the answer.

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On paper I think this a cool idea. Though the terms you are using like cost threshold, cost function, and weights will all have to be transformed. Classically this a well thought out idea and then to ask "If done on classical can I just tweak a few things and make them quantum?" is an awesome idea. The terms you used must be extended to the quantum ...

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Apparently, the specific question posed here has been answered in the affirmative--at least (first, we point out) through numerical means--by Arsen Khvedelidze and Ilya Rogojin in Table 2 of their 2018 paper, "On the Generation of Random Ensembles of Qubits and Qutrits: Computing Separability Probabilities for Fixed Rank States" ArsenIlya They ...

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No, there's not a lot you can say. Consider these two cases, both with $\epsilon=0$. First, the obvious one, $\rho=\sigma=|\psi\rangle\langle\psi|\otimes |\psi\rangle\langle\psi|$. Clearly $\rho_r-\sigma_r=0$. Second, let $|\psi^\perp\rangle$ be orthogonal to $|\psi\rangle$. You can have \rho=(|\psi\rangle\langle\psi|\otimes |\psi^\perp\rangle\langle\psi^\...

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What is "a single qubit state" in your question? If it is really a single qubit in a unique unknown state and there are no other qubits in the same state (for example, prepared in the same way), then in the answer it is already quite explained for this case. If it is a single qubit in a quantum program (circuit) designed to be executed on a NISQ ...

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