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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}-...
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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 ...
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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.
...
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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.
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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 = (...
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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 ...
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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 ...
2
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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