Exercise 4.1#

For a arbitrary single qubit state \(|\psi\rangle\), its Bloch sphere representation \((\theta, \varphi)\) can be obtained as follow,

(1)#\[ |\psi\rangle = \cos\frac{\theta}{2}|0\rangle + \sin\frac{\theta}{2}e^{i\varphi}|1\rangle \]

Here are eigenvalues and eigenstates of Pauli matrices.

  • For \(X\), the eigenvalues \(\lambda_1=1\) and \(\lambda_2=-1\), and their corresponding eigenvectors are

    (2)#\[\begin{split} |v_1\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}, |v_2\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ -1 \end{pmatrix} \end{split}\]

    We can obtain the Bloch coordinate of \(|v_1\rangle\) and \(|v_2\rangle\) via

    (3)#\[\begin{split} \begin{align} |v_1\rangle &= \frac{1}{\sqrt{2}} |0\rangle + \frac{1}{\sqrt{2}}|1\rangle \\ &= \cos\frac{\pi/2}{2}|0\rangle + \sin\frac{\pi/2}{2}e^{i\cdot0}|1\rangle \\ |v_2\rangle &= \frac{1}{\sqrt{2}} |0\rangle - \frac{1}{\sqrt{2}}|1\rangle \\ &= \cos\frac{\pi/2}{2}|0\rangle + \sin\frac{\pi/2}{2}e^{i\pi}|1\rangle \\ \end{align} \end{split}\]

    So the Bloch coordinate is \((\pi/2, 0)\) for \(|v_1\rangle\) and \((\pi/2, \pi)\) for \(|v_2\rangle\)

  • For \(Y\)​, the eigenvalues \(\lambda_1=1\)​ and \(\lambda_2=-1\)​, and their corresponding eigenvectors are

    (4)#\[\begin{split} |v_1\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix}, |v_2\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ -i \end{pmatrix} \end{split}\]

    We can obtain the Bloch coordinate of \(|v_1\rangle\) and \(|v_2\rangle\) via

    (5)#\[\begin{split} \begin{align} |v_1\rangle &= \frac{1}{\sqrt{2}} |0\rangle + \frac{1}{\sqrt{2}}i|1\rangle \\ &= \cos\frac{\pi/2}{2}|0\rangle + \sin\frac{\pi/2}{2}e^{i\cdot \pi/2}|1\rangle \\ |v_2\rangle &= \frac{1}{\sqrt{2}} |0\rangle - \frac{1}{\sqrt{2}}i|1\rangle \\ &= \cos\frac{\pi/2}{2}|0\rangle + \sin\frac{\pi/2}{2}e^{-i\pi/2}|1\rangle \\ \end{align} \end{split}\]

    So the Bloch coordinate is \((\pi/2, \pi/2)\) for \(|v_1\rangle\) and \((\pi/2, -\pi/2)\) for \(|v_2\rangle\)

  • For \(Z\)​, the eigenvalues \(\lambda_1=1\)​ and \(\lambda_2=-1\)​, and their corresponding eigenvectors are

    (6)#\[\begin{split} |v_1\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, |v_2\rangle = \begin{pmatrix} 0\\ 1 \end{pmatrix} \end{split}\]

    We can obtain the Bloch coordinate of \(|v_1\rangle\) and \(|v_2\rangle\) via

    (7)#\[\begin{split} \begin{align} |v_1\rangle &= |0\rangle = \cos\frac{0}{2}|0\rangle + \sin\frac{0}{2}e^{i\cdot 0}|1\rangle \\ |v_2\rangle &= |1\rangle = \cos\frac{\pi}{2}|0\rangle + \sin\frac{\pi}{2}e^{-i\cdot 0}|1\rangle \\ \end{align} \end{split}\]

    So the Bloch coordinate is \((0, 0)\) for \(|v_1\rangle\) and \((\pi, 0)\) for \(|v_2\rangle\).