4

$\mathrm{X}$ is not equivalent to a $\mathrm{CNOT}$ gate. The former is a 1-qubit gate whereas the 2nd is a 2-qubit gate (in essence, a controlled-$\mathrm{X}$). The $\mathrm{X}$ basically flips the state of qubit B i.e., $|0\rangle_B\to|1\rangle_B$ and $|1\rangle\to|0\rangle_B$, and does not depend on the state of qubit A.


4

The Hadamard gate is: $$\frac{1}{\sqrt 2} \left(|0\rangle \langle 0 | + |0\rangle\langle 1| + |1\rangle \langle 0| - |1\rangle \langle 1|\right)$$ And since $|+\rangle$ is $\frac{1}{\sqrt 2}\left(|0\rangle + |1\rangle \right)$, you can work out that $H(|+\rangle) = |0\rangle$ So, $$CNOT(H|+\rangle \otimes |+\rangle)$$ $$= CNOT(|0\rangle \otimes |+\...


3

There is a more direct characteristic that makes the state of an entangled pair of qubits distinct from a non-entangled pair (which is also known as a separable state). When two qubits are not entangled, the state of the one qubit can be described without any knowledge of the state of the other qubit. Mathematically we can write: \begin{equation} |\psi\...


3

Suppose you have 2-qubit state $|\psi\rangle_{AB}$. Entangled or not, you can always write it as $$|\psi\rangle_{AB}=a_0|0\rangle_A|\psi_0\rangle_{B}+a_1|1\rangle_A|\psi_1\rangle_{B}$$ If you measure qubit $A$ in state $|0\rangle$, then the qubit $B$ is in state $|\psi_0\rangle$, and you can compute the probability of qubit $B$ being in state $|0\rangle$ as $...


3

I find it a little tough to understand your calculations directly. I am especially confused by the circuit diagrams in your question; why they are there and what you are using them for. If you are performing calculations on theoretical data (without noise), then I feel you can make do with an easier approach for quantum state tomography. As per my answer on ...


3

Q1) The qubits in the $|\Phi^+\rangle$ state are entangled - this means that (by definition) you can not represent the state of one of them individually without talking about the second one (mathematically this would be represented as tensor product of two single-qubit states). The best description of the individual qubits received by Alice and Bob is that ...


3

The four Bell states are $$ |\Phi_{\pm}\rangle=(|00\rangle\pm|11\rangle)/\sqrt{2}\qquad |\Psi_{\pm}\rangle=(|01\rangle\pm|10\rangle)/\sqrt{2}. $$ So, let's consider what happens then we try and measure in the Bell basis, i.e. project onto one of these four states. If we started with the state $|00\rangle$, then we can write it as $$ |00\rangle=\frac{1}{\sqrt{...


3

Simply start by writing out everything $$ |B_{00}\rangle_{13}|B_{00}\rangle_{24}=\frac12\left(|00\rangle_{13}|00\rangle_{24}+|00\rangle|11\rangle+|11\rangle|00\rangle+|11\rangle|11\rangle\right) $$ Let me rearrange each of these terms $$ \frac12\left(|00\rangle_{12}|00\rangle_{34}+|01\rangle|01\rangle+|10\rangle|10\rangle+|11\rangle|11\rangle\right). $$ Now ...


2

The most direct way to do this using normal notation is to simply write the Bell projections using the same convention for subscripts:$\def\ket#1{\lvert#1\rangle}\def\bra#1{\langle#1\rvert}\def\idop{\mathbf 1}$ $$\begin{aligned} \bra{\Phi^+}_{1,5} \;&=\; \tfrac{1}{\sqrt 2}\Bigl(\,\bra{0}_1\bra{0}_5 \,+\, \bra{1}_1 \bra{1}_5\,\Bigr), \\ \bra{\Phi^-}_{1,...


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