0
$\begingroup$

enter image description here

Equation 7.7 is given below: $$\hat\rho = \frac12(I +n_x(\hat X)+n_y(\hat Y)+n_z(\hat Z)) $$

Where $I$ is the identity matrix and $\hat X,\hat Y,\hat Z$ are Pauli matrices.

Now my attempt of this was to first square every term inside the bracket to get $$\hat\rho^2$$ When doing this, all Pauli matrices then convert into identity matrices. So if you were to add all terms together you would get a 2x2 matrix: \begin{bmatrix}4&0\\0&4\end{bmatrix} But I'm unsure of how the matrices will cancel out to give the final result?

$\endgroup$
2
  • 1
    $\begingroup$ I'm not quite sure why you are only squaring every term in the bracket. For example, $(A+B)^2 = (A+B)(A+B) = AA + AB + BA + BB$. $\endgroup$
    – Rammus
    Nov 16 at 14:06
  • $\begingroup$ Have you squared the $1/2$ out front? $\endgroup$ Nov 16 at 14:51
2
$\begingroup$

Using the fact that any pair of distinct non-identity Pauli matrices anti-commutes, we have

$$ \begin{align} \hat\rho^2 &= \left[\frac12\left(I +n_x\hat X+n_y\hat Y+n_z\hat Z\right)\right]^2 \\ &= \frac14\left[(1 + n_x^2 + n_y^2 + n_z^2)I + 2n_x\hat X + 2n_y\hat Y + 2n_z\hat Z\right] \end{align} $$

so

$$ \begin{align} \mathrm{tr}\hat\rho^2 &= \mathrm{tr}\left[\frac14(1 + n_x^2 + n_y^2 + n_z^2)I + 2n_x\hat X + 2n_y\hat Y + 2n_z\hat Z\right] \\ &= \frac12(1 + n_x^2 + n_y^2 + n_z^2) \\ &= \frac12(1 + \|\textbf{n}\|^2) \end{align} $$

where we used the fact that the trace of every non-identity Pauli matrix is zero.

$\endgroup$

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