Next, we can represent each $\ket{\psi_i}$ in this orthonormal basis, $U$. \end{pmatrix} &=& \left( \left\langle v | l \right\rangle \right)^* \left\langle w | m \right\rangle \\ $$,$$ \end{pmatrix} \\ $$,$$ \begin{eqnarray} \begin{pmatrix} \begin{pmatrix} J = \sqrt{A^{\dagger}A} &= \sqrt{\sum_{i,j}\lambda_i^{*} \lambda_j \left| i \right\rangle \left\langle i \middle| j \right\rangle \left\langle j \right|}\\ Nielsen Chuang Chapter 7 Solutions. \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}-\lambda &= \mathrm{tr} \left( \sum_i \lambda_i A^{\dagger}B_i \right) \\ &= \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 \\ =0 \begin{pmatrix} + v_3 A^{\dagger}A = \left| \Psi^+ \right\rangle &= \frac{\left( \left| 01 \right\rangle + \left| 10 \right\rangle \right)}{\sqrt{2}} \\ 1 & -1 & 1 & -1 \\ \left\langle \Phi^+ \middle| \Psi^+ \right\rangle &= \frac{1}{2}\left( \left\langle 00 \middle| 01 \right\rangle + \left\langle 00 \middle| 10 \right\rangle + \left\langle 11 \middle| 01 \right\rangle + \left\langle 11 \middle| 10 \right\rangle \right) = 0 \\ = I + \end{pmatrix} \end{pmatrix}. \begin{pmatrix} \end{align} \sigma_2^2 &= &= \frac{1}{2} \end{pmatrix} $$,$$ $$,$$ \end{pmatrix} \begin{eqnarray} \begin{pmatrix} \otimes \end{pmatrix} \begin{pmatrix} \left\langle X^2 \right\rangle &= \left\langle 0 \middle| X^2 \middle| 0 \right\rangle = \left\langle 0 \middle| I \middle| 0 \right\rangle = 1. \end{pmatrix} 1 & 1 & 1 & 0 \\ 03 - Entanglement. = i & 0 Finally, take $E_{m+1} = I - \sum_m E_i$. Then. 0 & 0 \begin{align} 0 \\ 1 \end{pmatrix} \end{pmatrix} \end{align} \\ $$,$$ $$,$$ \end{pmatrix} &A&\left| 0 \right\rangle = \left| 0 \right\rangle = 1\left| 0 \right\rangle + 0\left| 1 \right\rangle \\ $$,$$ i & 0 \left\langle \lambda_- \middle| \lambda_+ \right\rangle &= \frac{1}{\sqrt{10+2\sqrt{5}}\sqrt{10-2\sqrt{5}}}\left( 4+\left( -1-\sqrt{5} \right)\left( -1+\sqrt{5} \right) \right) = 0. \otimes \rho^1 &= \rho^2 = \frac{1}{3} = 0 \\ \end{pmatrix} \begin{pmatrix} $$,$$ \end{eqnarray} \end{align} $$,$$ $A = \sum_i a_i \left| i \right\rangle \left\langle i \right|,; B = \sum_j b_j \left| j \right\rangle \left\langle j \right|$. \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} = J = \sqrt{P^{\dagger}P} &= \sqrt{\sum_{i,j}\lambda_i \lambda_j \left| i \right\rangle \left\langle i \middle| j \right\rangle \left\langle j \right|}\\ \begin{pmatrix} \end{eqnarray} a_1 \\ b_1 1 & 1-\lambda Hence, $M_m = M_m^2$ so the measurement is projective. get the PDF. Furthermore, from spectral theorem, $U$ is diagonalizable as $U = V \Lambda V^\dagger$ where $V$ is unitary\footnote{Quick proof: $U$ can be written as $U=VTV^\dagger$ where $V$ is unitary and $T$ is upper triangular by Schur Decomposition. $$,$$ 1 & 0 \\ 1 & -1 Show that the average value of the observable $X_1Z_2$ ($X$ acting on the first qubit and $Z$ on the second) for a two qubit system measured in the state $\frac{\ket{00} + \ket{11}}{\sqrt{2}}$ is zero. \end{align} \begin{pmatrix} \end{pmatrix} \end{pmatrix}. Furthermore, $\bra{\beta_{00}}\ket{\beta_{00}} = \frac{\bra{00} + \bra{11}}{\sqrt{2}}\frac{\ket{00} + \ket{11}}{\sqrt{2}} = (\bra{00}\ket{00} + \bra{11}\ket{11}) /2 = 1$. So, $M_m$ are positive operators. &= \sum_j p_j \left|\psi_j\right\rangle\left\langle\psi_j\middle|\rho^{-1}\middle|\psi_i\right\rangle\delta_{i,j} \\ 1 & 0 \\ 0 & -1 \end{pmatrix} \rho^{B} &= \sum_i\lambda_i^2\left| i_B\right\rangle\left\langle i_B\right| . \end{pmatrix} \begin{pmatrix} Let H_1 and H_2 be Hermitian operators. \begin{align} \end{eqnarray} 2.4 of Nielsen and Chuang and Sec. &\left| v \right\rangle = \left( v_1\; v_2\; \cdots\; v_n \right)^T \\ \mathrm{tr} \left( X \right) &=& \mathrm{tr} \end{pmatrix} a_1 \\ b_1 2A^{-1}AB = A^{-1}0 = 0\;\longrightarrow\;B=0 0 & -i \\ \end{pmatrix} = \left| \lambda_+ \right\rangle. Reading: prime factor decomposition ( pdf ) ( ps ) problem, correction and appendix Homework 6 (graded): Fast Fourier Transform ( pdf ) ( ps ). \left( A, \sum_i \lambda_i B_i \right) &= \mathrm{tr} \left( A^{\dagger}\sum_i \lambda_i B_i \right) \\ 1 & 0 \\ 0 & 1 B\left( A\left| v_j \right\rangle \right) &=& B \left( \sum_i A_i {}_j \left| w_i \right\rangle \right) \\ \begin{pmatrix} \begin{pmatrix} &= -i\sum_j\log\left( e^{i\theta_j} \right) \left| j \right\rangle \left\langle j \right| \;\;\; \left( \because \mathrm{Exercise\; 2.18} \right) \\ \lambda_{\pm} = \frac{2}{\sqrt{5\pm\sqrt{5}}},\;\;\; \left|\lambda_{\pm}\right\rangle = \sqrt{\frac{2}{5\pm\sqrt{5}}} 3 & 3 \\ 3 & 3 , where $U_m := UV$. \begin{pmatrix} \left( \left( A^{\dagger} \right)^{\dagger} \left| v \right\rangle, \left| w \right\rangle \right) &=& \left( \left| v \right\rangle, A^{\dagger} \left| w \right\rangle \right) \\ \begin{align} To find the Schmidt decomposition of $\left|\psi\right\rangle\equiv\frac{\left| 00\right\rangle + \left| 01\right\rangle + \left| 10\right\rangle}{\sqrt{3}}$, we construct the density matrix $\rho$ and find the reduced density matrices $\rho^1$ and $\rho^2$. \begin{pmatrix} \end{pmatrix} \end{align} Now find the eigenvector $\left| \lambda_1 \right\rangle$. 1 & 0 & 0 & 0 \\ $$,$$ &=& \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) \\ $$,$$ \end{pmatrix} \frac{1}{3} & \frac{1}{3}-\lambda_- = 0 & 0 & 1 & 0 \end{vmatrix} $$,$$ v_1-iv_2 \\ \frac{v_1^2+v_2^2}{1+v_3} The states are only distinguishable if one can perform a measurement that acts on both qubits. + \sqrt{\frac{2}{3+\sqrt{5}}}\frac{1}{10-2\sqrt{5}} = H = \sum_i \lambda_i \left| i \right\rangle \left\langle i \right|. -1-\sqrt{5} & 3+\sqrt{5} U\left( t_1,t_2 \right)U^{\dagger}\left( t_1,t_2 \right) &= \exp\left[ \frac{-iH\left( t_2-t_1 \right)}{\hbar} \right]\exp\left[ \frac{iH\left( t_2-t_1 \right)}{\hbar} \right] \\ \end{eqnarray} \therefore \lambda = 1,\;7