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The standard meaning of this notation is that you're using $n$-ary encoding of the indices. In this case $n=3$, so \begin{align}11 \sim 1\\ 12 \sim 2 \\ 13 \sim 3 \\ 21 \sim 4 \\ 22 \sim 5 \\ 23 \sim 6 \\ 31 \sim 7 \\ 32 \sim 8 \\ 33 \sim 9 \end{align} The matrix element $M_{1,6}$ is then represented as $M_{1,23}$ which is given by $M_{213}$.

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Yes. Any mixed state $\rho$ is a convex combination of pure states, that is $$\rho = \sum_i \lambda_i |\phi_i\rangle\langle\phi_i|$$ where $\lambda_i >0$, $\sum_i\lambda_i=1$. The partial trace is linear, so that $$\rho_A = \sum_i \lambda_i \text{Tr}_B(|\phi_i\rangle\langle\phi_i|).$$ Every $\lambda_i\text{Tr}_B(|\phi_i\rangle\langle\phi_i|)$ is a ...

2

Bob does not know the outcome of Alice's measurement. All he knows is that Alice would have obtained $|\psi\rangle$ with probability $\frac12$ and $|\psi^\perp\rangle$ with probability $\frac12$ for some orthonormal basis $|\psi\rangle$, $|\psi^\perp\rangle$. Therefore, his state is a mixture $$\rho_B = \frac12|\overline\psi\rangle\langle\overline\psi|+\... 2 Let us recall the Ekert-91 (E91) protocol. Initially Alice and Bob each independently receive respective elements of a plurality of Bell pairs (for example, generated by spontaneous parametric down-conversion; Upon deciding to generate a secret key, Alice and Bob each independently measure their pairs in one of three bases, and publicly announce their ... 0 In Quirk, you look at state vectors of qubits using the amplitude display. You can get conditional state vectors, such as the state vector implied by a particular measurement result, by combining the amplitude display with a control on the measurement bit. Here is an example: And another example: 1 Cirq has a function cirq.sub_state_vector which can extract a single qubit's state from a full state vector. It doesn't just do the single qubit case, it can do arbitrary subsets of qubits. It will raise an exception if the subset you pick is entangled with other stuff. It's unfortunately a bit picky about error tolerances and input shape. 0 Take a look at "entanglement of formation". It's a nice metric for how much is a state entangled. the "units" are (a log of) how many bell states required to get into this state. See here for more info, and also here. Example: from qiskit.quantum_info import entanglement_of_formation import numpy as np li = np.sqrt(0.5) #shortcut for 1/... 3 The following construction works for a pair of qubits and small perturbations with \|H\|_2\le\frac15. Remarkably, in this case we can choose operators A_k and B_k independently of H. Basis Begin by defining pure states$$ \begin{align} |\psi_0\rangle &= |0\rangle& |\psi_2\rangle &= \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac23}e^{\frac{2\pi ...

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That's an appropriate example, yes. The underlying reason this happens is that the set of entangled states is not convex, which roughly speaking means you sum entangled states and obtain (upon renormalisation) separable ones. As a slight generalisation of the example you already provide, you can consider any pair of entangled states with Schmidt ...

5

Yes, that's correct, and your example is exactly what I had in mind after reading the question title. More generally, the four Bell states form a basis in the space of all 2-qubit states, so you can express any state using a linear combination of them, including all unentangled states.

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In general, the problem of finding a separable decomposition (or even deciding if it exists) is NP-hard. I don't know if it's that hard for operators of the form $S = \mathbb{I}_\mathcal{X} \otimes \mathbb{I}_\mathcal{Y} - H$ where $\lVert H \rVert_2 \leq 1$, but if you want just examples it's easier to go backwards. You can start with a separable $S$, say, ...

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It is not true that the increase in von Neumann entropy of entanglement between the two partitions due to a single cross-partition two-qubit gate is at most one. See below for an example where the increase is two. However, we can use log Schmidt rank to justify a different asymptotically linear upper bound. Product states A two-qubit gate cannot produce more ...

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Another standard example are Werner states. These can be written as $$W_p\equiv p \frac{I}{4} + (1-p) \frac{|\Psi^-\rangle\!\langle\Psi^-|}{2},\qquad p\in[0,1],$$ and you can tune the entanglement by changing $p$. You can prove that the $W_p$ is entangled for $p<2/3$, and you go from a maximally entangled state at $p=0$ to a totally mixed state at $p=1$.

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