What's a vector in the Bloch sphere representation?

Question

The Bloch sphere isn't so intuitive for me. But I am not sure how you are supposed to manipulate it using vectors and matrices.

How do you actually represent a vector on it? Is it $(\cos(a) , e^i\sin(b))$, $(\alpha, \theta)$ or $(1,0)$ In which format do we work with?

On one hand why we would use $(\cos(a), e^i\sin(b))$ if we can simply use $(\alpha, \theta)$ which covers the whole sphere. So what do we use? I saw $(\cos(a), e^i\sin(b))$ mentioned but I don't understand then, why do we all the time manipulate vectors like $|1\rangle, |0\rangle$? Is the idea that $|1\rangle$, for example, is just $\cos a$? But what is then the basis? $|1\rangle$, $|0\rangle$ or $(\cos(a) , e^i\sin(b))$? What do we manipulate using gates and matrices?

I also don't understand how do $|1\rangle$, $|0\rangle$ or $(\cos(a), e^i\sin(b))$ go together physically in the Bloch sphere itself. I mean, what does $|0\rangle$ mean? Is it really the vector vertical with $z=1$? Then how would that be represented by $(\cos(a) , e^i\sin(b))$? What would the angle values be for $|0\rangle$, $|1\rangle $? I guess a is zero and the $b$ is free.

The third question is, using the rotational matrices, I understand them only if we use vectors like $(1,0), (0,1)$ and the like because then it's easy to see how $R_z$ is simply a rotation acting on both $|0\rangle$, $|1\rangle$, $R_y$ is the regular two dimensional rotation, and $R_x$ is simply a rotation like the previous one but also taking into account imaginary of $Y$. But if we use something like $(\cos(a), e^i\sin(b))$ or $(\alpha, \theta)$ as vectors, I wasn't able to see how $R_x$ or $R_y$ rotate. Maybe the claim is, that $1,0$ and $0,1$ are valid basis vectors so could be used to manipulate even if we use $(\cos(a), e^i\sin(b))$ or $(\alpha, \theta)$?

Answer 1

The components of the Bloch vector of a state are the expectation values of the X,Y and Z Pauli matrices in that state and it has to be a full three-dimensional vector to capture the interior of the Bloch sphere as well, which represents mixed states.

In general a state with density matrix $\rho$ of a single Qbit has Bloch vector $\vec{r}$ when $$ \rho = \frac12(I+\vec r \cdot\vec\sigma) = \frac12(I+r_x\hat\sigma_x+r_y\hat\sigma_y+r_z\hat\sigma_z) $$ where $\vec\sigma$ is a vector containing the three Pauli matrices and $I$ is the identity matrix. Looking at pure states for example the vector $(0,0,1)$ leads to: $$ \rho = \frac12(I+1\cdot\sigma_z) = \frac12\left(\begin{matrix}1 &0\\0 &1\end{matrix}\right) + \frac12\left(\begin{matrix}1 &0\\0 &-1\end{matrix}\right) = \left(\begin{matrix}1 &0\\0 &0\end{matrix}\right) = |0\rangle\langle0| $$ exactly as you would expect.

The utility it provides is a way to immediately visualize in an intuitive way how a state changes under various transformations. Unitary operators rotate the sphere around some axis (which leads to a very simple proof for the universal set of quantum gates) and decoherence channels shrink the sphere along some axis for example.

I hope that also clears up some of the confusion between the state vector and the Bloch vector which are two entirely different beasts.