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\begin{pmatrix} \end{pmatrix} \begin{pmatrix} = I Blog Projects Solutions About Nielsen and Chuang -- Chapter 6 (6.2) Show that the operation $(2 \ket{\psi}\bra{\psi} - I)$ (where $\ket{\psi}$ is the equally weighted superposition of states) applied to general state $\sum_k \alpha_k \ket{k}$ produces \lambda_{\pm} = \frac{2}{\sqrt{5\pm\sqrt{5}}},\;\;\; \left|\lambda_{\pm}\right\rangle = \sqrt{\frac{2}{5\pm\sqrt{5}}} Hence, First, $v\cdot \sigma$ is Hermitian so it’s spectral decomposition is given by $v \cdot \sigma = U \Lambda U^\dagger$ for some unitary $U$, diagonal matrix $\Lambda$. $$, $$ \left| v_1 \right\rangle &=& \frac{\left| w_1 \right\rangle}{\left| \left| w_1 \right\rangle \right|}\\\ \begin{equation} \end{pmatrix}. \end{pmatrix} $$, $$ \end{pmatrix} &= \sum_k \frac{ \left[ -iH\left( t_2-t_1 \right)/\hbar \right]^{k}} {k!} 0 & 1 \\ $$, $$ \begin{eqnarray} H = \sum_i \lambda_i \left| i \right\rangle \left\langle i \right|. &= \mathrm{tr}\left( \sum_{j,k}p_jp_k \delta_{j,k} \left| j \right\rangle \left\langle k \right| \right) \\ \end{bmatrix} \Big\}. \end{equation} $$, $$ \begin{pmatrix} + v_2 \begin{pmatrix} = \frac{1}{\sqrt{2}} \left( \left| 0 \right\rangle + \left| 1 \right\rangle \right). \end{pmatrix} $$, $$ \rho^A = \left| i_A\right\rangle\left\langle i_A\right| . &=& \cos\theta \left( \left| \lambda_1 \right\rangle \left\langle \lambda_1 \right| + \left| \lambda_{-1} \right\rangle \left\langle \lambda_{-1} \right| \right) + i\sin\theta \left( \left| \lambda_1 \right\rangle \left\langle \lambda_1 \right| - \left| \lambda_{-1} \right\rangle \left\langle \lambda_{-1} \right| \right) \\ &= \frac{1}{2}\left[ \left\langle 0 \middle| I \middle| 0 \right\rangle + \left\langle 0 \middle| \vec{v}\cdot\vec{\sigma} \middle| 0 \right\rangle \right] \\ \begin{eqnarray} a_{11}B^T & \cdots & a_{N1}B^T \\ &=& \lambda^2 -\left( v_1^2 + v_2^2 + v_3^2 \right) \\ \rho^{\Psi^,2} &= \frac{\left| 0\right\rangle\left\langle 0\right| + \left| 1\right\rangle\left\langle 1\right|}{2} = \frac{I}{2} \\ \begin{align} \end{pmatrix} \begin{pmatrix} \begin{pmatrix} 0 & 1 \\ P_- = \left| \lambda_{-1} \right\rangle \left\langle \lambda_{-1} \right| &= \frac{1}{2\left( 1+v_3 \right)} \begin{pmatrix} a \\b = \lambda^2-1 = 0\;\;\;\;\therefore \lambda = \pm 1 Here \left\langle \Psi^+ \middle| \left( E\otimes I \right) \middle| \Psi^+ \right\rangle &= \left\langle \Phi^+ \middle| \left( X\otimes I \right)^{\dagger}\left( E\otimes I \right)\left( X\otimes I \right) \middle| \Phi^+ \right\rangle \\ $$, $$ \end{equation} \det \left( \vec{v}\cdot\vec{\sigma} - \lambda I \right) &= \end{pmatrix} = \sigma_3 \\ \end{pmatrix} \left\langle \Phi^- \middle| \Phi^- \right\rangle &= \frac{1}{2}\left( \left\langle 00 \middle| 00 \right\rangle - \left\langle 00 \middle| 11 \right\rangle - \left\langle 11 \middle| 00 \right\rangle + \left\langle 11 \middle| 11 \right\rangle \right) = 1 \\ \sigma_1^{\dagger} &=& Next, we can represent each $\ket{\psi_i}$ in this orthonormal basis, $U$. &= \left\langle \Phi^+ \middle| \left[ \left( c_I I - c_x X - c_y Y + c_z Z \right) \otimes I \right] \middle| \Phi^+ \right\rangle \\ v_1-iv_2 \\ 1-v_3 \end{align} v_3 & v_1-iv_2 \\ \end{pmatrix} \\ &=& \left( \left| w \right\rangle,\, \left| v \right\rangle \right)^{*}. \begin{pmatrix} \frac{2}{3}-\lambda_- & \frac{1}{3} \\ = \frac{1}{\sqrt{2}} \left( \left| 0 \right\rangle - \left| 1 \right\rangle \right). \left\langle \lambda_j \middle| H^{*} \middle| \lambda_i \right\rangle &=& \left\langle \lambda_j \middle| H \middle| \lambda_i \right\rangle = \lambda_j^{*} \left\langle \lambda_j \middle| \lambda_i \right\rangle = \lambda_j \left\langle \lambda_j \middle| \lambda_i \right\rangle \;\;\;\; &(4)^{'} 0 & 1 \\ \end{align} = -\left( \lambda^2 -1 \right) = 0\;\;\;\;\therefore \lambda = \pm 1. \end{equation} \end{pmatrix}. i & 0 \begin{equation} P \left| \lambda \right\rangle &=& \lambda \left| \lambda \right\rangle \\ \sigma^{\dagger}_3 &=& \begin{pmatrix} Let $A_1$ and $A_2$ be positive operators. \begin{equation} 1 \end{pmatrix} \begin{pmatrix} \begin{align} \end{eqnarray} $$, $$ 1 & 1 & -1 & -1 \\ \begin{pmatrix} &= \lambda^2-1 = 0\;\;\; \therefore \lambda = \pm 1. $$, $$ \\ Reading: prime factor decomposition ( pdf ) ( ps ) problem, correction and appendix Homework 6 (graded): Fast Fourier Transform ( pdf ) ( ps ). &= \frac{1}{2} \left( \left\langle 00 \middle| X_1Z_2 \middle| 00 \right\rangle + \left\langle 00 \middle| X_1Z_2 \middle| 11 \right\rangle + \left\langle 11 \middle| X_1Z_2 \middle| 00 \right\rangle + \left\langle 11 \middle| X_1Z_2 \middle| 11 \right\rangle \right) \\ \left| \lambda_1 \right\rangle = $$, $$ + v_3 $$, $$ \begin{pmatrix} &= \sum_j \theta_j \left| j \right\rangle \left\langle j \right| \\ = 0 \\ \begin{pmatrix} \end{pmatrix} Therefore, $\Lambda$ must have diagonal entries $\pm 1$. \end{eqnarray} \begin{equation} \end{eqnarray} \end{pmatrix} $$, $$ Therefore, Eve cannot infer anything about the information Alice sent. \begin{align} \begin{pmatrix} &=& \left( \left\langle v | l \right\rangle \left| w \right\rangle,\, \left| m \right\rangle \right) \\ 0 & 0 \\ \left| \psi'_j \right\rangle = \left| \psi_j \right\rangle - \sum_{k=1,k\neq j}^m\frac{\left\langle \psi_j \middle| \psi_k \right\rangle \left| \psi_k \right\rangle}{\left| \left| \psi_k \right\rangle \right|^2} $$ 0 \\ 0 \begin{align} Chapter 2.1 pp 61-79 Homework 1 (not graded): Nielsen Chuang, All exercises of. \begin{pmatrix} &= \frac{2i\sum_{l=1}^3\epsilon_{jkl}\sigma_l + 2\delta_{j,k}\mathbb{I}}{2} \\ \left| \lambda_1 \right\rangle = &= \sum_{N=0}^{\infty}\sum_{k=0}^{N} \frac{\left[ -iH\left( t_2-t_1 \right)/\hbar \right]^{N-k}}{\left( N-k \right)!} \end{equation} \sigma_1 &=& \left| 0 \right\rangle \left\langle 1 \right| + \left| 1 \right\rangle \left\langle 0 \right| \\ \begin{pmatrix} Chapters 10.6 pp. \end{pmatrix} \beta_a\alpha_b &= 0. \end{pmatrix} = 2iY \rho^{A} &= \sum_i\lambda_i^2\left| i_A\right\rangle\left\langle i_A\right|\\ When $N>0$, remembering the Binomial theorem, which is. From SVD, we have that $M_m = UDV$ for $U,V$ unitary and $D$ real, diagonal. So, $C=0$. 1 \\ 1 = \frac{1}{\sqrt{2}} = \left( 4-\lambda \right)^2 - 9 = 0 \end{pmatrix} \\ \begin{pmatrix} 2-\lambda & 1 \\ 1 & 1 \begin{eqnarray} $$, $$ $$, $$ = \begin{pmatrix} \begin{vmatrix} \end{pmatrix} So, $p_j^2 \le p_j$ for all $p_j$. Solution 2: \left\langle v \middle| U^{\dagger} U \middle| v \right\rangle = \left\langle v \middle| v \right\rangle = \lambda^* \lambda \left\langle v \middle| v \right\rangle\;\;\;\; \therefore \left| \lambda \right|^2 = 1 \rightarrow \lambda = \mathrm{e}^{i\theta} A^{\dagger}A = \end{pmatrix} &= -i\sum_j i\theta_j \left| j \right\rangle \left\langle j \right| \\ $$, $$ &= \sum_i \exp\left(a_i + b_i\right) \left| i \right\rangle \left\langle i \right| \\ \end{pmatrix}. \end{pmatrix} \\ &=& 0 \;\;\;\;\;\; \therefore \lambda=\pm 1 \end{pmatrix} 0 & -i \\ 1-\lambda & 0 \\ -2-2\sqrt{5} & 6+2\sqrt{5} &= \frac{1}{2} \left(I-\vec{v}\cdot\vec{\sigma} \right). $$, $$ \end{pmatrix} \;\;\; \therefore \left| \lambda_{-1} \right\rangle = \frac{1}{\sqrt{4+2\sqrt{2}}} H \left| \lambda_i \right\rangle &=& \lambda_i \left| \lambda_i \right\rangle \;\;\;\; &(1)\\ \left( Z\otimes I \right) \left| \Phi^+ \right\rangle &= \frac{1}{\sqrt{2}}\left( \left| 00 \right\rangle - \left| 11 \right\rangle \right) = \left| \Phi^- \right\rangle \\ Nielsen And Chuang Solutions Manual Rating: 3,5/5 2991 votes Solutions for problems in Nielsen and Chuang, but I was not able to find except for few problems in chapter. &= \frac{1}{\sqrt{2\left( 1+v_3 \right)}}\frac{1+v_3}{v_1-iv_2}

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nielsen chuang chapter 2 solutions