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TL;DR: Single-shot circuit of this sort would enable FTL comms, so is not possible. Multiple-shot variant can be realized using for example Quantum State Tomography or (more efficiently) Direct Fideli …

The most general pure state of a qubit can be written as $|\Psi\rangle=a|0\rangle+b|1\rangle$ where $a,b\in\mathbb{C}$. The amplitudes $a$ and $b$ can be written in polar form as $a=re^{i\alpha}$ and …

Many classical programming languages are equipped with a construct known as the conditional statement if (condition) { u(); } where condition is a boolean expression, i.e. an expression that eval …

Special case I: Product states A bipartite quantum state$^1$ $\rho_{AB}$ is said to be a product state if it can be written as $\rho_{AB}=\rho_A\otimes\rho_B$ for some quantum states $\rho_A$ and $\rh …

The following construction works for a pair of qubits and small perturbations with $\|H\|_2\le\frac15$. Remarkably, in this case we can choose operators $A_k$ and $B_k$ independently of $H$. Basis Beg …

Note that $$ \mathrm{tr}_B\left(|0\rangle \langle1|\otimes|0\rangle \langle1|\right) =|0\rangle \langle1| \mathrm{tr} (|0\rangle \langle1|) = |0\rangle \langle1| \, \langle1|0\rangle = |0\rangle \lang …

It is a variant of the formula for the average $\langle A\rangle_\rho$ of an operator $A$ measured on a state $\rho$ $$ \langle A\rangle_\rho = \mathrm{tr}(A\rho).\tag1 $$ In this case we are measurin …

Three real parameters are sufficient due to the constraint that $$ |\alpha|^2 + |\beta|^2 = 1\tag1 $$ where $\alpha$ and $\beta$ are the two components of a 2D complex vector describing the qubit stat …

Begin by writing down the density matrix $$ \begin{align} \rho_{AB} &= \frac{1}{4}\left[(|00\rangle+|11\rangle)(\langle00|+\langle11|)\right] + \frac{1}{4}\left[(|01\rangle+|10\rangle)(\langle01|+\lan …

Yes, with probability $\frac{1}{d}$. Begin by computing the output state $$ \begin{align} |\psi\rangle &= (I\otimes QFT^{-1} \circ CNOT \circ QFT \otimes I)|i\rangle|0\rangle \\ &= (I\otimes QFT^{-1} …

Qubits have more than three distinct states. Here are six examples of such states: \begin{align} |0\rangle\tag{1}\\ |1\rangle\tag{2}\\ |+\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}\tag{3} \\ |-\r …

Normalized states By the measurement postulate (see e.g. 2.2.3 on page 84 in Nielsen & Chuang), a measurement is described by a collection of operators $M_m$, indexed by the measurement outcomes $m$ s …

Let $U$ denote the desired unitary, i.e. $U$ is such that $UT|\sigma\rangle = TH|\sigma\rangle$ for any $|\sigma\rangle$. If two operators yield the same result on every input, then they are equal. Th …

TL;DR You have rediscovered the Bloch sphere! :-) Interesting special cases Before we prove that the map defined by equation $(1)$ is the representation of single-qubit quantum states as points on the …

TL;DR: This is impossible in general, because the space of two-qubit mixed states is significantly more complicated than the Bloch sphere. Dimension The space $D_N$ of all mixed states of an $N$-level …