1
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • $\begingroup$ 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 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 at 10:35
2
$\begingroup$

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

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

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.