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 $\ket{-\i}$?

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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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(1)a.We start from the general form $\ket{\psi} = \alpha \ket{0} + \beta \ket{1}$, where $\alpha, \beta \in \mathbb{C}$ and $|\alpha|^2 + |\beta|^2 = 1$. Since $\alpha$ and $\beta$ are complex numbers, we can separate them into the product of a real part and a phase: $\alpha = r_\alpha e^{i \phi_\alpha}$ and $\beta = r_\beta e^{i \phi_\beta}$. This gives:

We now define a global phase $\gamma = \phi_\alpha$ and a relative phase $\phi = \phi_\beta - \phi_\alpha$. By our normalization condition $\braket{\psi}{\psi} = 1$, we require $r_\alpha^2 + r_\beta^2 = 1$, and thus without loss of generality we can set $r_\alpha = \cos{\widetilde{\theta}}$ and $r_\beta = \sin{\widetilde{\theta}}$. So we have:

(2)Now, we want to choose $\widetilde{\theta}$ such that these states map onto the surface of a unit Bloch sphere using standard spherical coordinates (where $\theta$ is the polar angle and $\phi$ is the azimuthal angle), and such that $\ket{0}$ is at the top (north pole, i.e., $\theta = 0$) and $\ket{1}$ is at the bottom (south pole, i.e., $\theta = \pi$).

Using the expression we derived above, $\ket{\phi(\widetilde{\theta} = 0)} = e^{i \gamma} \ket{0}$ is at the north pole, and $\ket{\phi(\widetilde{\theta} = \frac{\pi}{2})} = e^{i (\gamma + \phi)} \ket{1}$ is at the south pole. So to achieve the correct result, we must set $\widetilde{\theta} = \frac{\theta}{2}$. This gives the desired expression:

(3)Note that the relative phase $\phi$ is irrelevant at the poles.

(4)b.Recall the definition of the states:

By identifying $\cos{\frac{\theta}{2}}$ as the coefficient of $\ket{0}$ and $e^{i \phi} \sin{\frac{\theta}{2}}$ as the coefficient of $\ket{1}$, we can derive the corresponding values of $\theta$ and $\phi$ for each state:

(5)Notice that all four of these states lie on the equator of the Bloch sphere and thus have $\theta = \frac{\pi}{2}$.

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