Why are orthogonal quantum states represented as collinear in the Bloch sphere?

Question

We know that the angle between two orthogonal qubit states is 90 degrees. Why then, when we use the Bloch sphere, the angle becomes 180 degrees?

Answer 1

The Bloch representation of states is what you get when you decompose density matrices in terms of a basis of orthonormal Hermitian operators. That is, you write $\rho=\sum_k \sigma_k \langle\sigma_k,\rho\rangle$ where $\langle\sigma_k,\rho\rangle\equiv\operatorname{Tr}(\sigma_k\rho)$ and $\{\sigma_k\}_k$ is some basis of Hermitian operators. For single qubits, the standard choice is to use Pauli matrices, and write the decomposition as $$\rho=\frac{I + c_x \sigma_x+ c_y \sigma_y + c_z \sigma_z}{2}.$$ Note that the coefficient of the identity matrix $I$ has been fixed to $\frac12$. This comes from the normalisation constraint on states, as $\langle I,\rho\rangle=\operatorname{Tr}(\rho)=1$.

Now, how does the inner product (and thus the geometry) between quantum states relate to the inner product between their Bloch representations? Given pure states $|\psi_1\rangle,|\psi_1\rangle$, and corresponding density matrices $\rho_1\equiv|\psi_1\rangle\!\langle\psi_1|$ and $\rho_2\equiv|\psi_2\rangle\!\langle\psi_2|$, with Bloch representations $$\rho_i = \frac{I + \vec r_i\cdot \vec\sigma}{2},$$ we can see that $$|\langle\psi_1|\psi_2\rangle|^2 = \langle\rho_1,\rho_2\rangle = \langle\vec v_1,\vec v_2\rangle,$$ where $\vec v_i=(\frac12,\frac12 \vec r_i)\in\mathbb{R}^4$, and thus $$\langle \vec v_1,\vec v_2\rangle = \frac12(1+\langle\vec r_1,\vec r_2\rangle).$$ From this we can see that orthogonality of quantum states corresponds to orthogonality of the vectors $\vec v_i$, which corresponds to the translated vectors $\vec r_i$ being collinear, that is, to $\langle\vec r_1,\vec r_2\rangle=-1$. The vectors $\vec r_i$ are what people typically mean when talking about the "Bloch sphere representation" of a state.