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 $(\alpha, \theta)$?


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.

  • $\begingroup$ Thanks. And what do you mean by "where σ⃗ is a vector containing the three Pauli matrices" how do I know what to include there? If I want to create : ρ=1/2(I+1⋅σz) from scratch. Which Pauli matrices I throw in σ ?? Ordo I first have to build the density matrix by taking ket by bra and only then can I infer what's σ ???? $\endgroup$ – bilanush Mar 3 '19 at 8:25
  • $\begingroup$ There are only three Pauli matrices and the vector $r$ is three dimensional. So it contains all three for the scalar product to match up. How each one contributes to the density matrix result depends of course on the components of the $\vec r$ vector. The formula works both ways, you can go from the state to the vector and from the vector to the state as I showed. $\endgroup$ – S.Move Mar 5 '19 at 13:38
  • $\begingroup$ Does it mean you multiply a coordinate of the vector by it's corresponding Pauli matrix? $\endgroup$ – bilanush Mar 5 '19 at 20:07
  • 1
    $\begingroup$ That is exactly right. The density matrix is the sum of the Pauli matrices, each with the weight of the corresponding component of the vector (plus the identity). $\endgroup$ – S.Move Mar 6 '19 at 14:28
  • $\begingroup$ This is very interesting and amazing $\endgroup$ – bilanush Mar 6 '19 at 17:03

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