5

First note that, $$(|0\rangle + |1\rangle)(|00\rangle + |11\rangle) = |0\rangle |00\rangle + |0\rangle|11\rangle + |1\rangle |00\rangle + |1\rangle|11\rangle$$ then you can extend this to $|\psi_2\rangle$. That is: \begin{align} |\psi_2\rangle &= \frac{1}{2}\bigg[\alpha(|0\rangle+|1\rangle)(|00\rangle+|11\rangle)+\beta(|0\rangle-|1\rangle)(|10\rangle+|01\...


4

This is because we are using a spherical coordinate system. You can think of this coordinate system as specifying a radius, then two angular coordinates $\theta$ and $\phi$. Pure single-qubit states have a Bloch vector with radius equal to unity, so they only need two more coordinates to fully specify the direction in which the vector points. When you look ...


4

Note that the two bases in the Schmidt decomposition do not necessarily coincide (for reasons related to the fact that the two unitary matrices in the singular value decomposition do not necessarily coincide). It is therefore clearer to write $$ |\psi \rangle = \lambda_0 |\phi_0^A \rangle |\phi_0^B \rangle + \lambda_1 |\phi_1^A \rangle |\phi_1^B \rangle $$ ...


4

You have $$\newcommand{\ket}[1]{\lvert #1\rangle}U(\ket\psi\otimes\ket0) = \bigg(I\otimes \underbrace{\sum_\ell \ket\ell\!\langle\ell|}_{\equiv I}\bigg) U (\ket\psi\otimes\ket0) \\ = \sum_\ell (I\otimes \ket\ell\!\langle\ell|)U(\ket\psi\otimes\ket0) = \sum_\ell (U_{(\ell,0)}\ket\psi)\otimes\ket\ell $$ where $$U_{(\ell,0)} \equiv (I\otimes \langle\ell|)U(I\...


3

From Cauchy-Schwarz inequality $|\langle u|v\rangle| \le \|u\|\|v\|$, we have $$ |\langle\psi|U^\dagger M|\Delta\rangle| \le \|MU|\psi\rangle\|\||\Delta\rangle\|. $$ But $\|MU|\psi\rangle\| \le 1$, because $U$ is unitary and $M$ a POVM element. Therefore, $$ |\langle\psi|U^\dagger M|\Delta\rangle| \le \||\Delta\rangle\|. $$ Similar reasoning shows that $|\...


3

This is wrong for sure. And according to the book reviews on Amazon, this book is "unreliable", "riddled with errors", and "someone studying for the first time will get confused"


3

By far the easiest way to prove this is to realise that if a density matrix is pure, it is a rank 1 projector. Hence, to prove that a Hermitian $\rho$ is pure, you simply have to verify that $\text{Tr}(\rho)=1$ (thus showing the rank is 1 if it's a projector, although this is automatic for a correctly constructed density matrix) and $\rho^2=\rho$. Now, you ...


2

Because for a general state $|\phi\rangle = \alpha |0\rangle + \beta |1\rangle$ with $\alpha,\beta \in \mathbb{C}$, we want to describe states with $\langle \phi|\phi\rangle = 1$. There are 4 degrees of freedom, two for the real and imaginary components of $\alpha$ and $\beta$ each. The normalization condition $\langle \phi|\phi\rangle = 1$ reduces that to 3 ...


2

$X \otimes Z = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 \cdot \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} & 1\cdot \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\\ 1 \cdot \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} & 0 \cdot \begin{...


2

Personally, I would do the calculation a little differently. Start by writing $$ \langle\psi|X^0_1Z_2X^0_1|\psi\rangle+ \langle\psi|X^1_1Z_2X^1_1|\psi\rangle= \langle\psi|X^0_1Z_2X^0_1+X^1_1Z_2X^1_1|\psi\rangle $$ Next, think about a term like $X^0_1Z_2X^0_1$, but expand out the tensor product. This is just $(X^0X^0)\otimes Z$. But since $X^0$ is a projector,...


1

Where did the article say M/N is the probability of error? M/N in the article is only for the use of normalization. For example(2 qubits), after $H^{\otimes 2}$ acted on initial $\mid 0\rangle^{\otimes 2}$, the state becomes $\mid\psi\rangle\equiv1/2(\mid 00\rangle+\mid01\rangle+\mid10\rangle+\mid11\rangle)$ . If the answer is $| 01\rangle$, then the $\mid\...


1

Given a quantum state $|\psi\rangle$, you can performa a basis change on this state using the $I$, the Identity operator, which of course can be expanded as $\sum_{i}|i\rangle\langle i|$, where $|i\rangle$ is simply a particular basis from a complete set of basis states. Applying $I$ to $|\psi\rangle$, we get $$I|\psi\rangle=\sum_{i}|i\rangle\langle i|\psi\...


1

From this paper, suppose you have three bases $S_x$, $S_y$ and $S_z$. And suppose you define the states of the $S_x$ and $S_y$ bases on the $S_z$ basis as follows: $$ |+x\rangle = \frac{1}{\sqrt{2}}(|+z\rangle + |-z\rangle) \\ |-x\rangle = \frac{1}{\sqrt{2}}(|+z\rangle - |-z\rangle) \\ |+y\rangle = \frac{1}{\sqrt{2}}(|+z\rangle + i|-z\rangle) \\ |-y\rangle = ...


1

Dirac notations, as far as I have seen, concerns quantum states (kets) and their complex conjugates (bras). For gates, you could simply use the capitalized letter that represents the particular gate. For example, $H$ for Hadamard gate, $X$ for bit flip, and so on. Please note that, when you apply two quantum gates on the same quantum state, the notation ...


1

Why is the amplitude simply so without including the $x \dot z$ factor? When you calculate the amplitude of the $|0\rangle^{\otimes n}$ state, you have $z = 0$ (the integer representation of the state you're looking at), so $x \dot z = 0$ for any $x$. Why is the resulting amplitude for the constant case is ±1 and 0 for the balanced case? For the constant ...


1

It is postulate or axiom of quantum mechanics that if a state $|\psi\rangle $ that is a linear superposition of eigenstates $\{ |e_i\rangle\}$ of some observable, $$ |\psi \rangle = \sum_i \alpha_i |e_i\rangle $$ then upon making measurement with respect to this observable, the state is observed in the state $|e_i\rangle$ with probability $|a_i|^2$. That is, ...


1

The classical Fourier transform acts on a vector $(x_{0}, x_{1}, \ldots, x_{N-1}) \in {C} ^{N}$ and maps it to the vector $(y_{0}, y_{1}, \ldots, y_{N-1}) \in {C} ^{N}$. According to the formula: $y_{k}={\frac {1}{\sqrt {N}}}\sum _{n=0}^{N-1}x_{n}\omega _{N}^{-kn},\quad k=0,1,2,\ldots ,N-1$ Similarly, the quantum Fourier transform acts on a quantum state $|x\...


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