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Depending on your physical implementation, it may be that the two basis states are at different energies. Thus, when the state collapses, the energy changes. However, the average energy remains constant so you cannot use this for energy generation or similar. For example, assume we have a qubit with a Hamiltonian $$ H=E|1\rangle\langle 1|. $$ For any state $...


3

For me, generalised measurements cover everything (obviously, that's why they're generalised), with projective measurements being a simple case that covers what we usually want to be doing. So, yes, why introduce POVMs which are basically the generalised measurements but without the output state? Because they describe what actually happens in some ...


3

It is known that: $$ \left| 0 \right\rangle = \frac{1}{\sqrt{2}} \left( \left| + \right\rangle + \left| - \right\rangle \right) \qquad \left| 1 \right\rangle = \frac{1}{\sqrt{2}} \left( \left| + \right\rangle - \left| - \right\rangle \right) $$ By substituting in the initial state: $$ \left| \varphi \right\rangle = \frac{i}{\sqrt{3}}\left| 0 \right\rangle ...


3

Suppose you have 2-qubit state $|\psi\rangle_{AB}$. Entangled or not, you can always write it as $$|\psi\rangle_{AB}=a_0|0\rangle_A|\psi_0\rangle_{B}+a_1|1\rangle_A|\psi_1\rangle_{B}$$ If you measure qubit $A$ in state $|0\rangle$, then the qubit $B$ is in state $|\psi_0\rangle$, and you can compute the probability of qubit $B$ being in state $|0\rangle$ as $...


3

This can be done using the statevector_simulator provided with Qiskit Aer. It will return the statevector that describes the quantum state at the end of your circuit. It can be used in the same way as the qasm_simulator, only your circuit shouldn't have measurement gates at the end. There is more information about this simulator in this tutorial.


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It is not necessary to define the action of $U$ on other states. It is sufficient to know that the defined action is compatible with being a unitary (i.e. inner produces are preserved between all inputs) because there always exists an extension of the defined action such that it is unitary. So, all you really have to check is that $$ \left(\langle\psi|\...


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If first qubit is $|1\rangle$ there is no other possibility than second qubit to be $|1\rangle$ as well since probability of state $|10\rangle$ is zero. Hence probability of measuring $|1\rangle$ in second qubit is $1$. In case first qubit is $|0\rangle$ there are two possibe results: $|00\rangle$ or $|01\rangle$. Since probability of state $|00\rangle$ is ...


2

It depends on technology used for a qubit implementation. For example, here is an explanation provided by IBM Q team: We must perform the qubit measurements in a way that does not destroy the qubit quantum state. One method is to weakly couple each qubit to a microwave resonator whose resonance characteristics depend on the state of the qubit. Once the ...


2

There is a more direct characteristic that makes the state of an entangled pair of qubits distinct from a non-entangled pair (which is also known as a separable state). When two qubits are not entangled, the state of the one qubit can be described without any knowledge of the state of the other qubit. Mathematically we can write: \begin{equation} |\psi\...


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A matrix/operator is unitary if and only if it sends orthonormal bases into orthonormal bases. Therefore, to check whether $U$ is unitary, it is enough to check that it sends a subset of a basis (i.e. a non-complete set of orthonormal vectors) into a set of orthonormal vectors. In the present case, this means to prove that orthonormal states in the first ...


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If you are interested only in the measurements of the system that comprises only 2nd and 3rd qubits, then you could compute the reduced density matrix $\rho_{23}$ on those qubits from the total density matrix $\rho_{1234}=|\psi\rangle \langle\psi|$, i.e. $$ \rho_{23} = \text{Tr}_{14}(\rho_{1234}) $$ Density matrix carry all information that is need to ...


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Judging from your computations, the Bell states are $$|B_{00}\rangle = |\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$ $$|B_{10}\rangle = |\Phi^-\rangle = \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle)$$ $$|B_{01}\rangle = |\Psi^+\rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$$ $$|B_{11}\rangle = |\Psi^-\rangle = \frac{1}{\sqrt{2}}(|01\...


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