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Without additional assumptions or context, there is no fundamental difference between an "$2^n$-dimensional qudit" and "$n$ qubits". Any "qudit system" over $2^n$ modes for some integer $n$ can be thought of as a system of $n$ qubits. Equivalently, an $n$-qubit system is nothing but a $2^n$-dimensional qudit system. The difference is in the fact that if you ...

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The Quantum Reality group at the Centre for Quantum Technologies (National University of Singapore) https://qreality.quantumlah.org/

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I'll cover a slightly more general case. Let $P_k$, $k=1,...,N$ a complete set of orthogonal projectors: $\sum_k P_k=I$ and $P_j P_k=\delta_{jk}P_j$. Consider the map $\mathcal E(\rho)=\sum_k P_k \rho P_k$. We want to find a set of unitaries $\mathcal U_k$ and probabilities $p_k$ such that $\mathcal E(\rho)=\sum_\ell p_\ell\mathcal U_\ell\rho\,\mathcal U_\... 4 This question is actually entirely about the basics of measurement on a quantum system, and nothing to do with secret sharing. Let's state the measurement postulate of quantum mechanics as it applies to projective measurements: A measurement is described by a set of projectors$P_i$satisfying$\sum_iP_i=1$. If a state$|\psi\rangle$is being measured, ... 5 If$P$is (ortho)projector, that is$P^2=P=P^\dagger$, then we can define unitary$U = I - 2P$. You can verify $$UU^\dagger = U^2 = (I-2P)(I-2P) = I-4P+4P = I$$ Now we can express$P=\frac{1}{2}(I-U), Q=\frac{1}{2}(I+U)$and calculate $$P\rho P + Q\rho Q = \frac{1}{4}(I-U)\rho(I-U) + \frac{1}{4}(I+U)\rho(I+U)=$$ $$= \frac{1}{4}(\rho - U\rho - \rho U +... 2 Recall the measurement postulate (as applied to projective measurements): A measurement is described by a set of projectors P_i satisfying \sum_iP_i=1. If a state |\psi\rangle is being measured, the outcome i is obtained with probability p_i=\langle\psi|P_i|\psi\rangle, and if outcome i is obtained, the state after measurement is P_i|\psi\rangle/... 5 You just need to do a bit more algebra: Note that$$ \sum_{i=0}^n (\overline{x_i+y_i})(x_i+y_i)=\langle x+y|x+y\rangle$$and then you can distribute the right-hand side to get$$\langle x|x\rangle+\langle x|y\rangle+\langle y|x\rangle+\langle y|y\rangle.$$Since$| x\rangle$and$| y\rangle$are normalized, we know that$\langle x|x\rangle=\langle y|y\...

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An application of an oracle does not return a value; rather, it modifies the state of the system in a non-collapsing way. The oracles are a bit similar to controlled gates in this respect (in fact, a lot of oracles rely on controlled gates for their implementation). Consider, for example, CNOT gate: it does not measure the control qubit and apply an X gate ...

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It's hard to know what are you referring to without context, but a "quantum oracle" is just a type of (generally unitary) gate. As such, it does not provide information about the system, and neither it induces collapse. You can find more details about oracles in this other question (and links therein). You cannot know the state of the system without ...

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Since this question seems to be in the context of Grover's search, I will explain using what happens in Grover's search, however, this is way more general. The oracle function $f$ itself can be thought of a specification as to what should happen given various basis states (in fact, just classical bit strings to classical bit strings). For example, $f$ can ...

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