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Consider the last part of your question: Why can't we have a query of the form $$O_{x}|i \rangle = (-1)^{x_{i}} |i \rangle$$ This transformation is certainly unitary. Andrew Childs notes [...] that we can't distinguish between $x$ and $\bar{x}$ (bitwise complement of $x$) if we exclude the $|b \rangle$ register. I don't see why this should be the case....
Let me augment the discussion by adding some insight into the derivation of the estimate provided. This will give you a good understanding of when the result is an approximation and when it is precise. After the algorithm has run, we are left with the following state on the first register: $$\frac{1}{2^{n}}\sum_{x=0}^{2^n - 1} \sum_{k=0}^{2^n - 1}e^{-\frac{... 1 Why not uncover the quantum computer—open the box—to reveal the mechanism? Well, we can’t. If we “watch” the computation happen, we expose the quantum computer to an environment and this will break the computation. The kind of things a quantum computer needs to do requires complete isolation from the environment. Just like a magician’s trick, if we reveal ... 1 Assuming you have n qubits then the domain of x is all bit-strings n. (2^n values) To encode the data, in many cases, we start with the maximum superposition state. Which means each qubit is in an equal superposition of |0> and |1> so the entire system of n qubits is in a massive superposition of all possible states (the entire domain of x). Then ... 1 You cannot force a superposition to collapse in a particular direction. When you perform a measurement that removes a superposition, that 'collapse' is random, and you cannot choose which way it collapses. However, if you know what superposition you have, you can always convert it into any other state that you want to via unitary evolution (at which point ... 1 i think you have done entaglment \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) so when you measure the first qubit the second qubit forces to be collapses to the same state as the first qubit state. 1 There is an operator P such that P^2 is the identity and commutes with the Hamiltonians. In this case P=\prod \sigma_x. This is a \mathbb{Z}_2 because P and the identity form a group isomorphic to the integers modulo 2 (odds and evens) Identity corresponds to evens, P to odds. P^2=1 corresponds to odd plus odd equals even. If there were ... 1 The terms of expression do not cancel out in the balanced function case. We start with$$\frac{1}{2} (|0\rangle|0 \oplus f(0)\rangle - |0\rangle|1 \oplus f(0)\rangle + |1\rangle|0 \oplus f(1)\rangle - |1\rangle|1 \oplus f(1)\rangle) If $f(0) \neq f(1)$, consider the first two terms (the only ones which can cancel with each other, since the state of the ...