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I am vexed by a particular derivation. Given a state $\psi$ and corresponding density matrix $\rho = |\psi\rangle \langle \psi|$, or $\rho = \begin{bmatrix} a & c \\ b & d \end{bmatrix}$, I can compute the Bloch sphere coordinates as the following (in Python):

  a = rho[0, 0]
  b = rho[1, 0]
  x = 2.0 * b.real
  y = 2.0 * b.imag
  z = 2.0 * a - 1.0

This code works, but - how? I fail to properly derive it. Any hints or pointers are appreciated.

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  • $\begingroup$ First, decompose the density matrix $\rho$ into the form $\rho = \dfrac{I + \vec{r} \cdot \vec{\sigma} }{2} $ where $I$ is the 2 by 2 idenity matrix and $\vec{\sigma} = \langle X, Y, Z \rangle$ where $X,Y,Z$ are Pauli matrices. The vector $\vec{r}$ is the coordinate you are looking for. $\endgroup$
    – KAJ226
    Apr 21 '21 at 6:50
  • $\begingroup$ see also quantumcomputing.stackexchange.com/a/4121/55. In the density matrix formalism, you just need to change $\langle\psi|\sigma_i|\psi\rangle$ into $\mathrm{Tr}(\rho \sigma_i)$ for $i=x,y,z$ $\endgroup$
    – glS ♦
    Apr 21 '21 at 10:35
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The point $(x, y, z)\in\mathbb{R}^3$ corresponds to the state

$$ \rho = \frac{I + xX + yY + zZ}2 = \frac12\begin{bmatrix} 1+z & x-iy \\ x+iy & 1-z \end{bmatrix}, $$

see also definition in Wikipedia. Therefore, if $\rho = \begin{bmatrix}a & c \\ b & d\end{bmatrix}$, then

$$ 2a = 1 + z \\ 2b = x + i y $$

and so

$$ x = 2\,\mathrm{Re}(b) \\ y = 2\,\mathrm{Im}(b) \\ z = 2a - 1. $$

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