Exercise 2.12

$\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 2.12}$

a) Show that the surface of the Bloch sphere can be parametrized in terms of two real-valued parameters, the angles $\theta$ and $\phi$ illustrated in above Figure. Make sure your parametrization is in one-to-one correspondence with points on the sphere, and therefore single-qubit quantum states, in the range $\theta\in [0, \pi]$ and $\phi\in [0, 2\pi]$ except for the points corresponding to $\ket 0$ and $\ket 1$.

b) What are $\theta$ and $\phi$ for each of the states $\ket +$, $\ket -$, $\ket{\i}$, and $\ket{-\i}$?

page revision: 0, last edited: 12 Nov 2011 05:36

a) A qubit can be represented as $\vert \psi \rangle = \alpha \vert 0 \rangle + \beta \vert 1 \rangle$, where $\vert \alpha \vert^2 + \vert \beta \vert ^2 = 1$. With $z = x + iy$ on the Bloch sphere, we can convert the qubit representation s.t. $\vert \psi \rangle = r_{\alpha} e^{i \phi_{\alpha}} \vert 0 \rangle + r_{\beta}e^{i \phi_{\beta}} \vert 1 \rangle$ since $z = x + iy = r e^{i \theta}$. Since we're concerned with the relative angle, we can substitute $\phi = \phi_{\beta} - \phi_{\alpha}$, so that $\vert \psi \rangle = r_{\alpha} \vert 0 \rangle + r_{\beta} e^{i \phi} \vert 1 \rangle$, which is equivalent to $\psi = r_{\alpha} \vert 0 \rangle + (x + iy) \vert 1 \rangle$. Before converting to polar coordinates its important to see that $0 \leq \theta \leq \pi$ and $0 \leq \phi \lt 2 \pi$, and that the angle separating two bases is maximum at $\frac{\theta}{2}$. Using the map: $(x, y, z) \to (r sin(\theta) cos(\phi), r sin(\theta) sin(\phi), r cos(\theta))$, we can convert to polar coordinates, so that we have that $\vert \psi \rangle = cos(\frac{\theta}{2}) \vert 0 \rangle + e^{i \phi} sin(\frac{\theta}{2}) \vert 1 \rangle$

b) For $\vert + \rangle$, $\theta$ and $\phi$ are found by first setting $cos(\frac{\theta}{2}) = \frac {1}{\sqrt{2}}$ and $e^{i \phi} sin(\frac{\theta}{2}) = \frac{1}{\sqrt{2}}$, with the result that $\theta = \frac{\pi}{2}$ and $\phi = 0$.

The same method can used to find phi and theta for the other bases in question.

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