# Tag Info

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Josu, You might be mis-understanding something. Your equation for the Pauli channel says that when $t\rightarrow \infty$, all operators ($X,Y,Z,I$) have an equal probability (1/4) of transforming the density matrix $\rho$. You seem to be suggesting that $t\rightarrow \infty$, the probability of the $I$ operator acting on $\rho$ whould go to 0. Keep in mind ...

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Is it possible for quantum computers to exhibit behavior similar to flip errors in classical computers where a state |0⟩ becomes |1⟩ due to this error? Yes. It would be a very abrupt error if you're talking about errors on physical qubits. Usually, we'd think of an error as being a little bit of an X rotation (for example). However, the effect of performing ...

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If the measurement is an irreversible process then the probability of the resultant state is towards the initial quantum state it was in, that is, if the resultant state is $1$ with a probability of, say, $63\%$, that means that is has a $63\%$ probability that the initial state was $|\psi \rangle= |1\rangle$. This is a fundamental misinterpretation....

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Imgine that there is a bit string $\vec{k}\in\{0,1\}^n$. We use this to specify sites (bit value 1) where an error has occurred, and sites (bit value 0) where no error has occurred. The number of 1s in the bit string is $k$. The probability of this particular error arising is then $p^k(1-p)^{n-k}$ because there are $k$ sites with an error and $n-k$ sites ...

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Decoherence is the very general term which, more or less, is anything resulting in a loss of purity during the evolution of a system. Sometimes, when people are being a bit non-specific, they might be thinking of a particular type of decoherence such as dephasing (or perhaps depolarising) when they use the term decoherence. Relaxation and dephasing are two ...

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It's fundamentally similar to/the same as Baker–Campbell–Hausdorff (BCH). Generally, in quantum physics, this is most often used (or at least taught) with commuting Hamiltonians: $$e^{-i\left(H_1+H_2\right)t} = \sum_{n=1}^\infty\frac{\left(-it\right)^n}{n!}\left(H_1+H_2\right)^n = e^{-iH_1t}e^{-iH_2t}e^{\frac{1}{2}\left[H_1, H_2\right]t^2}\cdots$$ where the ...

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The state you've given, $|\psi\rangle=a|0\rangle+b|1\rangle$ is a pure state. It has not been decohered. Decoherence is a process which turns pure states into mixed states. We usually write these in terms of density matrices. A pure state can be written in this form:  |\psi\rangle\langle\psi|=\left(\begin{array}{cc} |a|^2 & ab^\star \\ a^\star b & |...

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