$$, $$ 0 & 0 & 0 & 1 \end{align} \begin{pmatrix} So, this circuit implements a measurement of $U$. $$, $$ &= \frac{1}{2} Here we denoted c-$Z_{1,2}$ as a controlled-$Z$ gate which takes the first qubit as the controll qubit and the second qubit as the target qubit. Solutions: Quantum Computation and Quantum Information by Nielsen and Chuang Chapter 1: Introduction and Overview Chapter 2: Introduction to Quantum Mechanics Chapter 3: Introduction to Computer Science Chapter 4: Quantum Circuits Chapter 5: The Quantum Fourier Transform and Its Applications Chapter 6: Quantum Search Algorithms T &= \mathrm{e}^{i\pi/8} }\left( \frac{\theta}{2} \right)^2I+\frac{i}{3! \end{pmatrix} \end{pmatrix} \\ \end{align} }\left(\frac{iA\Delta t}{2}\right)^2+O\left(\Delta t^3\right)\right] \\ R_z\left( \frac{\pi}{2} \right)R_x\left( \frac{\pi}{2} \right)R_z\left( \frac{\pi}{2} \right) &= 0 & 0 & 1 & 0 \\ 0 & \mathrm{e}^{i\beta} \end{align} \mathrm{exp}\left(i\frac{\pi}{2}\right)R_{(1/\sqrt{2},0,1/\sqrt{2})}\left(\pi\right) &= i\left(\cos\left(\frac{\pi}{2}\right)I-i\sin\left(\frac{\pi}{2}\right)\left(\frac{X}{\sqrt{2}}+\frac{Z}{\sqrt{2}}\right)\right) \\ 0 & X &= \left(\left|0\right\rangle\left\langle0\right|-\left|1\right\rangle\left\langle1\right|\right)I_2 = Z_1 \\ -. 0 & 1 & 0 & 0 \\ \begin{pmatrix} So. $$, $$ &= \left(\left|0\right\rangle\left\langle0\right| I_2+\left|1\right\rangle\left\langle1\right| X_2\right)I_1\mathrm{e}^{-i\theta X_2/2}=CR_{x,2} X &= HZH, \\ \end{align} &= \mathrm{e}^{i\pi/8}\mathrm{e}^{-i\pi X/8} = \mathrm{e}^{i\pi/8}R_x\left(\frac{\pi}{4}\right) &= \mathrm{tr}_2(\rho) \begin{pmatrix} \begin{align} \begin{pmatrix} 0 & 1 & 0 & 0 \\ Then, we define |X as i √ p i|x i| i . &= \left(H\otimes H\right)\left|01\right\rangle = \left|+-\right\rangle\\ \begin{align} \begin{align} \end{pmatrix} \left|\psi_0\right\rangle &= \left(a_1\left|0\right\rangle+b_1\left|1\right\rangle\right)\otimes\left(a_2\left|0\right\rangle+b_2\left|1\right\rangle\right) \\ \end{pmatrix} \\ \end{align} a_{11}b_{00} & a_{11}b_{01} & a_{11}b_{10} & a_{11}b_{11} \begin{align} \end{pmatrix} \\ = X a_{01}b_{00} & a_{01}b_{01} & a_{01}b_{10} & a_{01}b_{11} \\ &+ a_1\left|0\right\rangle\otimes b_2\left|1\right\rangle\otimes\mathrm{e}^{\pi Y/8}X\left|\phi_3\right\rangle+b_1\left|1\right\rangle\otimes b_2\left|1\right\rangle\otimes\mathrm{e}^{\pi Y/8}\left|\phi_3\right\rangle \\ \mathrm{e}^{-i\pi/8} & 0 \\ The unitary matrix U acts non-trivially only on the state $\left|010\right\rangle$ and $\left|111\right\rangle$. Solutions 111 Problems of Chap. \end{pmatrix} \begin{pmatrix} HYH &= \left(\frac{1}{\sqrt{2}}\right)^2 &= \left|\left(\mathrm{e}^{-i\frac{\alpha}{2}\hat{n}\cdot\hat{\sigma}}-\mathrm{e}^{-i\frac{\alpha+\beta}{2}\hat{n}\cdot\hat{\sigma}}\right)\left|\psi\right\rangle\right| \\ $$, $$ \end{align} At the end of the circuit, the second qubit is either $a_2\left|0\right\rangle+b_2\left|1\right\rangle$ or $a_2U\left|0\right\rangle+b_2U\left|1\right\rangle$. V_7V_6V_5V_4V_3V_2V_1U &= 0 & 0 & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} CY_1C &= \left(\left|0\right\rangle\left\langle0\right| I_2+\left|1\right\rangle\left\langle1\right| X_2\right)Y_1\left(\left|0\right\rangle\left\langle0\right| I_2+\left|1\right\rangle\left\langle1\right| X_2\right) \\ &= \cos\left(\frac{\theta}{2}\right) I - i\sin\left(\frac{\theta}{2}\right)\left(n_xX+n_yY+n_zZ \right) \end{pmatrix} This and the previous lecture should cover 4.1 - 4.5.2. CZ_2C &= \left(\left|0\right\rangle\left\langle0\right| I_2+\left|1\right\rangle\left\langle1\right| X_2\right)Z_2\left(\left|0\right\rangle\left\langle0\right| I_2+\left|1\right\rangle\left\langle1\right| X_2\right) \\ \sqrt{\frac{2}{3}} & 0 & \frac{1}{\sqrt{3}} & 0 \\ 0 & 0 & 1 & 0 \\ Like many quantum computation researchers, back when I was first learning the basics of the field, I relied heavily on Nielsen & Chuang's "Quantum computation and quantum information" textbook. \begin{pmatrix} \end{align} E\left(R_{\hat{n}}\left(\alpha\right),R_{\hat{n}}\left(\alpha+\beta\right)\right) &= \left| \left[R_{\hat{n}}\left(\alpha\right)-R_{\hat{n}}\left(\alpha+\beta\right)\right]\left|\psi\right\rangle \right| \\ HTH &= \mathrm{e}^{i\pi/8}HR_z\left(\frac{\pi}{4}\right)H = \mathrm{e}^{i\pi/8}H\mathrm{e}^{-i\pi Z/8}H \\ $$, $$ \begin{pmatrix} 1 & -1 &-\frac{1}{2}\left[\left(A\Delta t\right)^2+\left(B\Delta t\right)^2+AB\Delta t^2+BA\Delta t^2+O\left(\Delta t^3\right)\right]\ \ \ \ \ \left(k:\ 2\right) \\ \begin{pmatrix} QCQI Exercise Solutions (Chapter 4) - めもめも. 1 & 0 & 0 & 0 \\ So, for all $m> 0$, $\left(3+4i\right)^m = 3+4i\ \left(mod\ 5\right)$. We are also doing 2.5 and 2.6, but this might spill over into the next class. So, $L$ is upper bounded by a polynomial in $n$. $$, $$ g_2:\ &\mathrm{0\ 1\ 1} \\ H &= X_1\otimes Y_2\otimes Z_3 \\ 1 & -i & -1 & i (JP, Chapter 4, revised 2001 version) Supplementary handwritten notes (scanned 3.6 ... (posted Tuesday 11 November 2008; due Wednesday 26 November 2008). 1 & 0 & 0 & 0 \\ \end{align} \end{align} \left|\psi_7\right\rangle &= a_1\left|0\right\rangle\otimes a_2\left|0\right\rangle\otimes\left|\phi_3\right\rangle+b_1\left|1\right\rangle\otimes a_2\left|0\right\rangle\otimes\mathrm{e}^{\pi Y/4}X\mathrm{e}^{-\pi Y/4}\left|\phi_3\right\rangle \\ ( &= \begin{pmatrix} \end{align} (2) Obviously, $3+4i=3+4i\ \left(mod\ 5\right)$. We use optional third-party analytics cookies to understand how you use GitHub.com so we can build better products. &= R_y(-\theta) 1 & 0 & 0 & 0 \\ \end{align} they're used to log you in. \end{pmatrix}^{\dagger} \\ &= \mathrm{e}^{iA\Delta t}\mathrm{e}^{iB\Delta t}\left[I+O\left(\Delta t^2\right)\right] + O\left(\Delta t^3\right) \\ \begin{align} $$, $$ H^2+HXH & H^2-HXH \\ \end{pmatrix} &= \sqrt{2-\cos\left(\frac{-\beta}{2}\right)-\cos\left(\frac{\beta}{2}\right)}\ \ \ \ \ \left(\because \mathrm{e}^{x}=\sum_j\frac{x^j}{j! \end{align} \begin{equation} Solutions: Quantum Computation and Quantum Information by Nielsen and Chuang Solutions: Quantum Computation and Quantum Information by Nielsen and Chuang $$, $$ \end{align} a_{00}b_{00} & a_{00}b_{01} & a_{00}b_{10} & a_{00}b_{11} \\ \begin{pmatrix} \end{pmatrix} 0 & 1 \\ \left|\psi_2\right\rangle &= \left|\phi_1\right\rangle\otimes a_2\left|0\right\rangle\otimes\mathrm{e}^{-\pi Y/8}\left|\phi_3\right\rangle + \left|\phi_1\right\rangle\otimes b_2\left|1\right\rangle\otimes X\mathrm{e}^{-\pi Y/8}\left|\phi_3\right\rangle \\ 2 & 0 & 0 & 0 \\
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