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Consider any purification of $\rho$: $|\psi\rangle \in \mathbb{H}_A \otimes \mathbb{H}_B$. This purification necessarily admits a Schmidt decomposition of the form: $$|\psi \rangle=\sum_i \sqrt{\lambda_i} |\alpha_i\rangle_A |\beta_i \rangle_B $$ with the state $\rho$ being of the form: $$\rho=\sum_i \lambda_i |\alpha_i\rangle \langle \alpha_i |$$ This ...


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This topic seems to me as functions of operators, so I'll explain this issue and will use "n-th root NOT gate" as example. Any function you apply on the operator - is applied on it's eigenvalues. If the operator is diagonal - all the eigenvalues are on the diagonal and applying the function is simply apply it to any element on the diagonal. $$A=D=\Sigma_i ...


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The answer depends on the implementation, as is often the case when asking practical questions about quantum computing. To give you an example of the state of the art in trapped ions, the Lucas group in Oxford can achieve less than one error in 1 million single-qubit gates, which they claim is less than the fault-tolerance threshold. Their error rate for ...


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Since $|H\rangle= \frac{1}{\sqrt2}(|R\rangle + |L\rangle)$ and $|V\rangle= \frac{1}{i\sqrt2}(|R\rangle - |L\rangle)$ the state $|\psi\rangle = \frac{1}{\sqrt2}(|H\rangle_1|H\rangle_2|H\rangle_3 + |V\rangle_1|V\rangle_2|V\rangle_3)$ and the state $\frac{1}{\sqrt2}(|H\rangle_1 \frac{1}{\sqrt2}(|R\rangle_2 + |L\rangle_2) \frac{1}{\sqrt2}(|R\rangle_3 + ...


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If you want to hard code a specific set of qubits you could measure the spin (pass it through an inhomogeneous magnetic field) of a spin 1/2 system and then if you get the wrong spin, flip it (tune a photon to a wavelength based on a given uniform magnetic field strength). If you want to read an output you could do the same thing, measure the spin of each ...


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I'm guessing that by $n^{\text{th}}$ root you mean some gate which if you pass a signal through $n$ of them the output will be identical to the output of the gate you want to "root". For odd $n$ it's pretty obvious that the classical NOT gate is it's own $n^{\text{th}}$ root. In that an odd number of NOT gates one after the other are identical to a single ...


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The NOT gate can be written as $$ \sigma_x \propto \exp[i\pi/2\,\sigma_x] $$ (up to an irrelevant global phase). Its roots are therefore of the form $$ \exp[i\phi\,\sigma_x] = \cos(\phi)\,I + i\sin(\phi)\,\sigma_x $$ with $n\phi = \pi/2 + 2\pi k$ with $k\in\mathbb N$.


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The reason a measurement outcome on the second qubit forces a particular state for the first qubit is that the two qubits are entangled. You are correct about the probability of measuring state $\psi_b$, but the issue is that $\psi_b$ is under-determined, since $j$ can take any value in $\{0,1\}^n$. So when you perform the measurement on the second qubit, ...



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