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4

While talking about knowing the position exactly is a nice theoretical ideal, in practice, you cannot do that. You'll really be asking: "In which 'bin' of width $\delta x$ where $x$ spans from $x_{\min}$ to $x_{\max}$ is the particle confined to?". This means that there's $(x_{\max}-x_{\min})/\delta x$ bins, and so you basically need $$\log_2\left((x_{\max}-... 4 Three outcomes amounts to more than one bit if the outcomes are all deterministic, and give you information about the original qubit. But suppose I have a coin (that is either heads or tails). I roll a dice, and if it comes 1 through 5, I tell you "H" or "T", depending on what the coin is. If it comes up 6, I tell you "6". There are three outcomes, but ... 4 Now If I chose the standard basis |0\rangle,|1\rangle what will be the result I will get? If I measure with respect of the projection operator \langle 0| I get \alpha and If I measure with respect of the projection operator \langle 1| I get \beta. This is wrong. First of all, arguably the most natural kind of measurement in QM consists in choosing ... 4 Look like a lot of misunderstanding. If you measure a state \alpha|0\rangle+\beta|1\rangle in computational basis, the state collapses either to |0\rangle or |1\rangle. In Quantum Information we say that we measured state |0\rangle or |1\rangle, or simply we measured 0 or 1. There is no chance to know \alpha and \beta from the measurement. ... 4 The question presupposes a misconception that the vector form of a state |\psi\rangle exists independently of its density operator form |\psi\rangle\langle\psi|, which is often described as secondary. In reality, the density operator of a state is all that truly exists --- and even then, it only exists as statistical information. In fact, you can ... 3 There isn't. A density matrix encodes all the knowledge available about a state, therefore if two states are described by the same density matrix, they are indistinguishable. Ket vectors differing by only a global phase have always the same density matrix, and represent the same physical state. 3 Preliminary I would like to rewrite the equation that you have in a slightly different manner. Since a density matrix can be written as a matrix, we can also write it down as a linear combination of elements from a basis for the space of density matrices. We can use essentially any basis to do this, but some are preferred: most notably, the Pauli basis. For ... 3 To measure, observe that you are simply projecting a quantum state onto some basis set of vectors. First, I will note that this state is not normalized. Let us first define the following quantum state.$$|\psi_i\rangle = \begin{pmatrix}1\\-1\\0\\0\end{pmatrix}.$$Then, calculating the corresponding probability yields:$$|\langle \psi_i|\psi_i\rangle|^2 = (...

3

As indicated by Danylo in his anwser, eq. (32) in arXiv: 1103.2030 presents the sixteen vectors ("ignoring overall phases and normalisation") \begin{equation} \left( \begin{array}{cccc} x & 1 & 1 & 1 \\ x & 1 & -1 & -1 \\ x & -1 & 1 & -1 \\ x & -1 & -1 & 1 \\ i & x & 1 & -i \\ i & x & -...

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You can find it here Symmetric Informationally Complete Quantum Measurements or here SIC-POVMs: A new computer study, in the appendix B. Update Given a single fiducial vector $v = (a_1,a_2,a_3,a_4)^T \in \mathbb{C}^4$ it's pretty easy to write down all SIC-POVM vectors. They are just $C^kS^lv$ for $k,l \in \{0..3\}$, where $C$ and $S$ are clock and shift ...

3

Given an arbitrary state $|\psi\rangle$, if it is expressed in the computational basis as $|\psi\rangle=\sum_k c_k |k\rangle$, then it will give the $k$-th result (when measuring in the computational basis) with probability $|c_k|^2$. Note that here by "computational basis" I simply mean the measurement basis under consideration. If you consider another ...

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Application of a unitary transformation $U$ on a state $|q\rangle$ really leads to a new state $U|q\rangle$. What you get after a measurement is a one particular outcome of $U|q\rangle$ state because the measurement leads to colapse of the state wave function. If you repeat measurement many times you will get probability distribution of possible outcomes of ...

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Thanks Peter for the clarification about information vs. outcomes. I accept his answer to acknowledge that, and want to add the possible construction of such measurement. In the same book section 2.2.8, a general method is described. In this case, one can add two qubits prepared as $|00\rangle$, apply a unitary on the three qubits and measure the two ...

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