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1

Although it is not explained up to that point in the Qiskit textbook, the quantum toss is in reality applying the Hadamard gate, denoted $H$. In matrix form, this operator looks like: $$ H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} $$ Now, we express the basis states in column form as follows: $$ \begin{gather} |0\rangle = \...


0

The set $\mathrm L(\mathcal X)\equiv\mathrm L(\mathcal X,\mathcal X)$ is the set of linear maps from $\mathcal X$ to $\mathcal X$. In other words, $A\in\mathrm L(\mathcal X)$ iff $A$ is a linear function of the form $A:\mathcal X\to\mathcal X$. Note that $\mathrm L(\mathcal X,\mathcal Y)$ is itself a vector space. That means you can have, for example, $\...


1

Quantum (pure) states are, by definition, defined up to a scalar complex factors. That means that a state that we write as $|\psi\rangle$, should really be understood as the full set of vectors (an equivalence class if you will) $\{\lambda|\psi\rangle : \lambda\in\mathbb C\}$. The more formal way to put this is to say that quantum states are elements in the ...


1

First of all, if we write down $\left|\psi_1\right\rangle$, we get: $$\left|\psi_1\right\rangle=\frac{1}{\sqrt{2}^n}\sum_x|x\rangle\left[\frac{|0\rangle-|1\rangle}{\sqrt{2}}\right].$$ Applying $f$ on this state gives us: $$\left|\psi_2\right\rangle=\frac{1}{\sqrt{2}^n}\sum_x|x\rangle\left[\frac{|f(x)\rangle-|1\oplus f(x)\rangle}{\sqrt{2}}\right].$$ Note that ...


1

When we consider an uniform (or equal, as stated in Nielsen and Chuang) superposition, that is, a state that can be written as: $$|\psi\rangle=\frac{1}{2^n}\sum_x|x\rangle,$$ it is quite common not to write the normalisation constant $\frac{1}{2^n}$. Similarly, when the amplitutes of all vectors on the superposition are equal, we omit the normalisation ...


3

Let $M\in\mathrm{Lin}(\mathcal Y\otimes\mathcal X)$ be some linear operator whose input and output spaces are both $\mathcal Y\otimes\mathcal X$, for some pair of finite-dimensional Hilbert spaces $\mathcal X,\mathcal Y$. Moreover, suppose $M$ is positive semidefinite: $M\ge0$. It being positive semidefinite implies it admits a decomposition of the form $M=\...


1

An intuitive way to think about it is that $E[M]=E[X_1 \otimes Z_2]=E[X_1 \otimes \mathbb{1}]E[\mathbb{1} \otimes Z_2]$ If we only think about $E[\mathbb{1} \otimes Z_2]$, it is just the expectation value of $Z_2$ on the second qubit. Consider that our second Qubit in the entangled state $\frac{| 00\rangle + | 11\rangle}{\sqrt{2}}$ is measured to be $\frac{+\...


5

Choi operator of a linear map $\mathcal{E}$ is defined as $$ J(\mathcal{E}) = \sum_{ij} \mathcal{E}(|i\rangle\langle j|)\otimes |i\rangle\langle j|.\tag1 $$ Substituting $\mathcal{E}(\rho)=\sum_k E_k\rho E_k^\dagger$ into $(1)$, we have $$ \begin{align} J(\mathcal{E}) &= \sum_{ijk} \left(E_k|i\rangle\langle j| E_k^\dagger\right)\otimes |i\rangle\langle j|...


1

Taking the last two terms of last expression you gave, we can do the following $$ \begin{align} M \left(\frac{|00\rangle+|11\rangle}{\sqrt{2}}\right) &= X_1\otimes Z_2\left(\frac{|00\rangle+|11\rangle}{\sqrt{2}}\right) \\ &= \left(\frac{X_1|0\rangle \otimes Z_2|0\rangle+X_1|1\rangle \otimes Z_2|1\rangle}{\sqrt{2}}\right) \\ &= \left(\frac{|1\...


4

We know that $$ \begin{gather} |0\rangle = \frac{|+\rangle+|-\rangle}{\sqrt{2}} \\ |1\rangle = \frac{|+\rangle-|-\rangle}{\sqrt{2}} \end{gather} $$ Thus, we can rewrite the $GHZ$ state as $$ \begin{align} |GHZ\rangle &= \frac{1}{\sqrt{2}}\left(|0\rangle|00\rangle+|1\rangle|11\rangle\right) \\ &=\frac{1}{2}\left(|+\rangle|00\rangle+|-\rangle|00\rangle+...


3

Consider the state $|\Psi\rangle$. This has a Schmidt decomposition $$ |\Psi\rangle=U_A\otimes U_B\sum_i\lambda_i|ii\rangle. $$ Its reduced density matrix is $$ \rho_A=U_A\left(\sum_i\lambda_i|i\rangle\langle i|\right)U_A^\dagger. $$ It must be that if $|\Phi\rangle$ has the same reduced density matrix, the density matrices have the same spectrum and hence $|...


6

Given $\rho$ and a fixed ensemble $\{ |\psi_i \rangle \}$ it might not be possible to write $\rho$ as $\sum_i p_i |\psi_i \rangle \langle \psi_i |$. For example, let $| + \rangle = \frac{1}{\sqrt{2}} (| 0 \rangle + | 1 \rangle )$. Then the state $|+\rangle \langle + |$ cannot be expressed as a convex combination in the ensemble $\{ | 0 \rangle, |1\rangle \...


1

$P$ is acting on the space $V$, projecting onto the subspace $W$. Yes, if it only acted on the subspace $W$, it would be identity, but it is acting on a larger space. For example, think about a qubit, where $V$ is spanned by the basis states $\{|0\rangle,|1\rangle\}$. You can define a projector $P=|0\rangle\langle 0|$ which projects onto the space $W$ which, ...


2

As per N&C, fidelity is "analogous to the probability of doing the decompression correctly" (emphasis added). The goal is to do the operation correctly with 100% probability, which means the probability is 1. This is the desired limit of fidelity, so no error means the fidelity is 1.


2

There are many demos on https://pennylane.ai/qml/demonstrations.html. You could perhaps get some inspiration from there.


2

This is due to how the $\mathbf{A}$ matrix was defined; from that same tutorial page we have: $$\tag{1} \mathbf{A} = \sum_{n} c_n A_n $$ where each $A_n$ is unitary and $c_n$ is complex (and in the original VQLS paper they further impose $\lVert {\mathbf{A}}\rVert<1$ and bounded condition number) but $\mathbf{A}$ is never required to be unitary. Therefore,...


1

In a similar way to how the global phase difference of a state makes no physical difference, neither does amplitude of a state. We normalise states to have unit magnitude for mathematical convenience in the same way we don't carry around an $e^{i\phi}$ factor for arbitrary $\phi$ with all our states. This is because having unit vectors means we don't need to ...


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