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I know that $b$ can be decomposed mathematically as $b= c_1u_1 + > \cdots + c_nu_n$ since these eigenvectors form an orthonormal basis. Why only consider the effect on $|u_j \rangle$? As you say, you know you can decompose $b$ in terms of the $|u_j \rangle$, so by linearity, if we know the effect on one basis state (which is pedagogically easier to ...
Note: the graphics have been generated with the LaTeX code available here. Credits to @Niel de Beaudrap. Yes it is possible! The HHL algorithm can be schematically depicted as Let's split down the parts: The first part aims at computing an approximation of the eigenvalues of $H$, $H = A$ if $A$ is hermitian, else $H = \begin{pmatrix} 0 & A \\ A^\... 3 Be careful! They don't apply$e^{i\pi A}$and$e^{i\pi A/2}$. They apply $$|0\rangle\langle 0|\otimes I\otimes I+|1\rangle\langle 1| \otimes I\otimes e^{i\pi A}$$ and $$I\otimes |0\rangle\langle 0|\otimes I+I\otimes |1\rangle\langle 1|\otimes e^{i\pi A/2},$$ i.e. controlled versions of the gates, controlled off two different qubits. So, consider the 4 ... 3 Define the states $$|\psi_t\rangle=\left\{\begin{array}{cc} |t\rangle\otimes(U_{t-1}U_{t-2}\ldots U_1|\psi\rangle) & t=1,2,\ldots T \\ |t\rangle\otimes(U_{T}U_{T-1}\ldots U_1|\psi\rangle) & t=T+1,T+2,\ldots 2T \\ |t\rangle\otimes(U_{3T+1-t}U_{3T-t}\ldots U_1|\psi\rangle) & t=2T+1,2T+2,\ldots 3T \end{array}\right.$$ Now let $$U=\frac{2}{T}\sum_{... 3 Let's assume that you have a Hermitian matrix$$ H=\left(\begin{array}{cc} 0 & A^\dagger \\ A & 0 \end{array}\right). $$Let |b\rangle be the state that we want to apply A to, extended to work on the space that H acts on. So, our aim is to implement H|b\rangle. Let X be the standard Pauli X matrix. If we implement a unitary evolution$$ ... 3 1) Are we not applying the conditional Hamiltonian evolution to$|\Psi_0 \rangle |b \rangle$? The operation $$\sum_{\tau = 0} ^{T - 1} |\tau \rangle \langle \tau| \otimes e^{iA\tau t_0 / T}$$ is a controlled operation. You can read it as:$\forall \tau \in [0, T-1]$, if the first register is in the state$\vert \tau \rangle$, apply$e^{iA\tau t_0 / T}$. ... 3 The simplest method to implement$e^{iA\theta}$for a small, Hermitian matrix$A$is to: Find the eigenvectors$|\lambda\rangle$and eigenvalues$\lambda$of$A$. Construct the unitary$U=\sum_i|i\rangle\langle\lambda_i|$. Implement the gate sequence:$Ue^{i\theta\sum_i\lambda_i|i\rangle\langle i|}U^\dagger$Now, for one qubit, you have the middle ... 3 Disclaimer: I'm the one that wrote the code of the 4x4 HHL. Controlling a quantum gate$U$can be achieved by controlling each of the$U_i$gates that are composing$U$. For the specific example you are considering, the implementation is available online. Some remarks about the code: I think the code is not up to date with the last version of Qiskit. I ... 2 Quick answer: You will not be able to fully recover$x$. Explanations: By design, the HHL algorithm stores$x$in the amplitudes of a quantum state. Because of how quantum mechanics works, the vector representing the quantum state (i.e. containing all the amplitudes of the quantum state) needs to be of unit-norm (according to the Euclidean norm). Because ... 2 (Remark: I have corrected a typo in the matrix exponentiation). Please notice that the matrix has the form: $$e^{i\alpha A} = \begin{pmatrix} a & b \\ b & a \end{pmatrix}$$ with $$|a|^2 + |b|^2 = 1$$ Now, please notice that this matrix can be expanded as:$$\begin{pmatrix} a & b \\ b & a \end{pmatrix} = \begin{pmatrix} e^{i \arg(a)} &... 2 There are plenty of methods that can be used to implement a given unitary matrix. Only for 1-qubit gates ($2\times 2$matrix): The algorithm described in https://arxiv.org/abs/1212.6253 seems to be the most efficient at the moment. It is restricted to 1-qubit quantum gates ($2\times 2$unitary matrices) and only decompose in the Clifford + T basis. See ... 2 Note that the ancilla qubit in your example has dimension 2, not 1, because it goes into a superposition of 2 basis states,$|0\rangle$and$|1\rangle$. For the HHL algorithm, you need an ancilla capable of recording the three options "nothing", "well" and "ill". As such, you need a Hilbert space of dimension 3 (also called a qutrit). Now, it is true that ... 1 I tend to find the bit about the geometric random variable a bit misleading. What they're trying to say is that if you apply$A^{-1}$to$|1\rangle|\psi\rangle$and you measure the first register, you'll get the answer$t$with a probability that goes like$e^{-t/T}$(approximately, assuming large$T\$), and that's a geometric distribution. Now, I think the ...