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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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