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In the context of scalable quantum computing, the polylog scaling needed for magic state distillation should not be a problem. Indeed, it is not the only polylog scaling we need to contend with. Using the $S$ and $T$ gates to approximate a general single qubit rotation can have a similar cost when using the Solvay-Kitaev algorithm (though this is no longer ...

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Disclaimer: these links do not all fit the criterion recent. First and foremost, it is important to realize that there is no 'one' magic state, and no 'one' magic state distillation. The term 'magic state' is, loosely speaking, used to refer to any state that can be used, in conjunction with fault-tolerant Clifford operations and a specific quantum code, to ...

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The short answer is that $(\rho^{\otimes N})^{T_B}=(\rho^{T_B})^{\otimes N}$. More explicitly, if $\rho=\sum_{ii'jj'}\rho_{ii',jj'}|i\rangle\!\langle i'|\otimes |j\rangle\!\langle j'|$, then we can write $$\rho^{\otimes N}=\sum_{I I' JJ'}\rho_{II',JJ'}\bigotimes_{k=1}^N \Big(|i_k\rangle\!\langle i_k'|\otimes |j_k\rangle\!\langle j_k'|\Big),$$ where $I\equiv(... 3 There are two different contexts where the term "entanglement distillation" is used, and are largely incomparable, even if they are conceptually extremely close (and I'm sure you'll be able to find papers that blur these boundaries). In the first, Alice and Bob share a known quantum state which is (usually) a pure state. They use this to make a maximally ... 3 Remember that the partial transpose condition is generally good for detecting entanglement, i.e. a bipartite state$\rho$is certainly entangled if the partial transpose is not non-negative. In other words, if there exists a state$|\psi\rangle$such that $$\langle\psi|I\otimes\text{T}(\rho)|\psi\rangle<0,$$ then the state is certainly entangled. If you ... 2 The entanglement entropy (what you call "von Neumann entropy") is a good measure for entanglement of pure states in the asymptotic setting, i.e. when one is dealing with many copies. However, it is not a good measure for mixed states. Distillable entanglement and entanglement cost are entanglement measures which apply to both pure and mixed states. ... 1 To see intuitively why this protocol increases the entanglement after each iteration, we can work out an example where our initial state is say$\lvert 00\rangle+\lvert 11\rangle$. Upon passing through a 50:50 Beam Splitter, we get:$\$|00\rangle+|11\rangle = |00\rangle+ a_{0}^{\dagger}a_{1}^{\dagger}|00\rangle \hspace{3mm} \text{transforms} \hspace{3mm}|00\...

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