New answers tagged nielsen-and-chuang
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TL;DR
Given a process matrix $\chi$ expressed in a normalized, ordered Pauli basis $\{P^{n}_{k}\}$, find its eigendecomposition $\{\lambda_{i},\mathbf{v_{i}}\}$, where $0 \leq \lambda_{i} \leq 1$ and $\mathbf{v_{i}} = (v^{i}_{P^{n}_{k}})$ is the eigenvector for eigenvalue $\lambda_{i}$ and contains 'coefficients' for every Pauli $P^{n}_{k}$. $\chi$ is ...
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The linear map $\mathcal{E}$ is what characterizes a quantum process,
$$\rho \rightarrow \mathcal{E}(\rho),$$
but $\mathcal{E}$ can be determined by $\chi$. Using the operator-sum representation,
$$\mathcal{E}(\rho) = \sum_i A_i \rho A^\dagger_i = \sum_{i}\sum_{m}\sum_{n}a_{im}\tilde{A}_m \rho a^*_{in}\tilde{A}^\dagger_n,$$
where the $a_{ij}$ are some set of ...
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Due to $\rho$ being a positive and adjoint, you can always spectrally decompose it into a convex combination of pure states, it's eigenvectors. However, since the measurement statistics, or the action of a unitary operator, as you observed, will not change if you do not represent it diagonally, a lot of papers or books will choose not to represent them as ...
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All your working is correct. Of course, if you want to define a density operator/matrix, you cannot define it in terms of another density operator/matrix which is why you will not find an initial presentation of density matrices going in this way. That said, you may want to look up the definition of a mixed separable state. This is typically done in the form
...
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I don't understand the idea of this question? I need some explanation of this question.
You're supposed to show that the operation is the same in both representations. That applying a rotation matrix to the state vector gives the same result as converting the state to a bloch vector, rotating the bloch vector, then converting back.
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Imagine you have a state $|\psi\rangle$ with Bloch vector $\vec{m}$. Now apply this rotation $R_n(\theta)$ to $|\psi\rangle$. Can you show how the vector $\vec{m}$ is changed by this unitary?
Without wanting to do too much of the calculation for you, I find it instructive to write
$$
|\psi\rangle\langle\psi|=\frac{1}{2}(I+\vec{m}\cdot\vec{\sigma})
$$
and, ...
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It's been a while since the question was posted but since I have a solution that differs from the one above, I would like to put it forth. It's mathematically gory but satisfactory in my opinion. We start by considering the state $|n, m\rangle$ as some basis state in the combined excited state space of $a$ and $b$ respectively.
We realize that $H$ and ...
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