The Bloch Sphere

1. 1
   Step 1:

   $$\alpha=r_0e^{i\gamma_0},\quad\beta=r_1e^{i\gamma_1},\qquad r_0,r_1\ge0,\ \gamma_0,\gamma_1\in\mathbb R$$

   Polar form of each complex amplitude.
2. 2
   Step 2:

   $$\lvert\psi\rangle=e^{i\gamma_0}\big(r_0\lvert0\rangle+r_1e^{i(\gamma_1-\gamma_0)}\lvert1\rangle\big)$$

   Factor out $e^{i\gamma_0}$ from both terms.
3. 3
   Step 3:

   $$\lvert\psi\rangle\simeq r_0\lvert0\rangle+r_1e^{i\phi}\lvert1\rangle,\qquad\phi:=\gamma_1-\gamma_0$$

   Discard the global phase $e^{i\gamma_0}$, which changes no measurement probability; only the relative phase $\phi$ survives. Reduce $\phi$ modulo $2\pi$ into $(-\pi,\pi]$.
4. 4
   Step 4:

   $$r_0^2+r_1^2=\lvert\alpha\rvert^2+\lvert\beta\rvert^2=1$$

   Normalisation, using $\lvert e^{i\gamma}\rvert=1$.
5. 5
   Step 5:

   $$r_0=\cos\tfrac\theta2,\quad r_1=\sin\tfrac\theta2,\qquad\tfrac\theta2\in\big[0,\tfrac\pi2\big]$$

   $(r_0,r_1)$ is a point of the unit circle in the closed first quadrant, so it equals $(\cos a,\sin a)$ for a unique $a\in[0,\pi/2]$; write $a=\theta/2$, hence $\theta\in[0,\pi]$.
6. 6
   Step 6:

   $$\lvert\psi\rangle\simeq\cos\tfrac\theta2\lvert0\rangle+e^{i\phi}\sin\tfrac\theta2\lvert1\rangle$$

   Substitute. At the poles $\theta=0$ or $\theta=\pi$ the factor multiplying $e^{i\phi}$ or the whole $\lvert0\rangle$ term vanishes, so $\phi$ is irrelevant there (like longitude at the geographic poles); the convention is $\phi=0$.

1. 1
   Step 1:

   $$\lvert\psi\rangle\langle\psi\rvert=\begin{pmatrix}\cos\frac\theta2\\e^{i\phi}\sin\frac\theta2\end{pmatrix}\begin{pmatrix}\cos\frac\theta2&e^{-i\phi}\sin\frac\theta2\end{pmatrix}$$

   The bra is the conjugate transpose of the ket, so $e^{i\phi}$ becomes $e^{-i\phi}$.
2. 2
   Step 2:

   $$=\begin{pmatrix}\cos^2\frac\theta2&e^{-i\phi}\sin\frac\theta2\cos\frac\theta2\\e^{i\phi}\sin\frac\theta2\cos\frac\theta2&\sin^2\frac\theta2\end{pmatrix}$$

   Column times row: entry $(r,c)$ is $\psi_r\overline{\psi_c}$.
3. 3
   Step 3:

   $$\cos^2\tfrac\theta2=\tfrac12(1+\cos\theta),\qquad\sin^2\tfrac\theta2=\tfrac12(1-\cos\theta)$$

   Add and subtract $\cos^2\tfrac\theta2+\sin^2\tfrac\theta2=1$ and $\cos^2\tfrac\theta2-\sin^2\tfrac\theta2=\cos\theta$ (double-angle formula), then halve.
4. 4
   Step 4:

   $$\sin\tfrac\theta2\cos\tfrac\theta2=\tfrac12\sin\theta$$

   Double-angle formula $2\sin\tfrac\theta2\cos\tfrac\theta2=\sin\theta$.
5. 5
   Step 5:

   $$\lvert\psi\rangle\langle\psi\rvert=\tfrac12\begin{pmatrix}1+\cos\theta&\sin\theta\,(\cos\phi-i\sin\phi)\\\sin\theta\,(\cos\phi+i\sin\phi)&1-\cos\theta\end{pmatrix}$$

   Substitute the three identities and Euler's formula $e^{\pm i\phi}=\cos\phi\pm i\sin\phi$.
6. 6
   Step 6:

   $$=\tfrac12\begin{pmatrix}1+r_z&r_x-ir_y\\r_x+ir_y&1-r_z\end{pmatrix}=\tfrac12\big(I+r_xX+r_yY+r_zZ\big)$$

   Compare with $\tfrac12(I+\vec r\cdot\vec\sigma)$ entry by entry: $r_z=\cos\theta$, $r_x=\sin\theta\cos\phi$, $r_y=\sin\theta\sin\phi$.
7. 7
   Step 7:

   $$\lVert\vec r\rVert^2=\sin^2\theta(\cos^2\phi+\sin^2\phi)+\cos^2\theta=1$$

   Pythagoras twice: every pure state lies on the unit sphere.

Quick check

Quick check

1. 1
   Step 1:

   $$R_{\hat n}(\alpha)=e^{-i\alpha\,\hat n\cdot\vec\sigma/2}=\cos\tfrac\alpha2\,I-i\sin\tfrac\alpha2\,\hat n\cdot\vec\sigma$$

   $(\hat n\cdot\vec\sigma)^2=I$ for a unit vector $\hat n$, so the power series of the exponential splits into even (cosine) and odd (sine) parts.
2. 2
   Step 2:

   $$U\,\tfrac12(I+\vec r\cdot\vec\sigma)\,U^\dagger=\tfrac12\big(I+(M\vec r)\cdot\vec\sigma\big),\qquad M_{jk}=\tfrac12\mathrm{Tr}\big(\sigma_jU\sigma_kU^\dagger\big)$$

   $U\sigma_kU^\dagger$ is Hermitian and traceless, so it expands in the Paulis with real coefficients $M_{jk}$ (found using $\mathrm{Tr}(\sigma_j\sigma_l)=2\delta_{jl}$). $M$ preserves $\lVert\vec r\rVert$ because purity is unitarily invariant, and $\det M=1$ by continuity from $U=I$, so $M\in SO(3)$.
3. 3
   Step 3:

   $$R_{\hat n}(\alpha)\ \longmapsto\ \text{rotation of }\vec r\text{ by angle }\alpha\text{ about }\hat n$$

   E.g. for $R_z(\alpha)=\mathrm{diag}(e^{-i\alpha/2},e^{i\alpha/2})$ the coherence $\rho_{10}$ picks up $e^{i\alpha}$, i.e. $\phi\mapsto\phi+\alpha$: the Bloch angle is the full $\alpha$ while the matrix uses $\alpha/2$.
4. 4
   Step 4:

   $$R_{\hat n}(2\pi)=\cos\pi\,I-i\sin\pi\,\hat n\cdot\vec\sigma=-I$$

   A full turn of the Bloch vector gives $-I$ on the Hilbert space; only a turn by $4\pi$ gives $+I$.
5. 5
   Step 5:

   $$(-U)\,\rho\,(-U)^\dagger=U\rho U^\dagger$$

   The signs cancel, so $U$ and $-U$ produce the same rotation $M$: the map $U\mapsto M$ is exactly 2-to-1. The factor $-1$ is a global phase, invisible on the sphere, which is why the amplitudes carry $\theta/2$ while the sphere shows $\theta$.

Quick check