1
$\begingroup$

I'm relatively new to the subject of quantum computing, and I recently came across the idea of eigenvalues and eigenvectors. I believe I understand the relationship between the two, where eigenvalues determine the factor by which an eigenvector is stretched, and eigenvectors represent the direction that the transformation is pointing towards, but I was wondering how exactly eigenvectors and eigenvalues would be expressed in the Bloch sphere.

$\endgroup$
1
$\begingroup$

The eigenvectors for a one-qubit unitary are two orthogonal vectors. As such, on the Bloch sphere, they are visualised as a single axis (going through the origin). (Remember that angles on the Bloch sphere are doubled so orthogonal states are an angle $\pi$ different on the Bloch sphere, i.e. opposite directions along the same axis.)

The eigenvalue (or, more precisely, the relative angle between the two eigenvalues) is the angle of rotation around that axis.

$\endgroup$
0
$\begingroup$

A state $\rho$ with Bloch sphere coordinates $\newcommand{\bs}[1]{\boldsymbol{#1}}\bs r\equiv (x,y,z)$ has the form $$\rho = \frac{I + \bs r\cdot\bs \sigma}{2}\equiv \frac{I+x\sigma_x + y \sigma_y + z\sigma_z}{2}, $$ with $\sigma_x,\sigma_y,\sigma_z$ the Pauli matrices.

Computing the eigenvalues (eigenvectors) of $\rho$ thus amounts to computing those of $\bs r\cdot\bs\sigma$. Observe that $$\bs r\cdot\bs \sigma=\begin{pmatrix}z & x-iy \\ x+iy & -z,\end{pmatrix}$$ and thus the eigenvalues are $\lambda_\pm = \pm\sqrt{-\det(\bs r\cdot\bs \sigma)}=\pm\|\bs r\|$. The corresponding eigenvectors are then seen to be $$\lvert\lambda_\pm\rangle = \frac{1}{\sqrt{2\|\bs r\|(\|\bs r\|\mp z)}}\begin{pmatrix}x-iy \\ \pm \|\bs r\| - z\end{pmatrix}.$$ The vectors in the Bloch sphere corresponding to $\lvert\lambda_\pm\rangle$ have coordinates $$\begin{cases} x_\pm &=& \pm x/ \|\bs r\|, \\ y_\pm &=& \pm y/ \|\bs r\|, \\ z_\pm &=& \pm z/ \|\bs r\|. \end{cases}$$ In other words, the eigenvectors of $\bs r\cdot\bs\sigma$ correspond to the two unit vectors in the Bloch sphere along the same direction as $\rho$.

The eigenvectors of $\rho$ are then clearly the same as those of $\bs r\cdot\bs \sigma$, while its eigenvalues are $(1\pm\lambda_\pm)/2$.

$\endgroup$
2
  • $\begingroup$ In glS's response they say: "Computing the eigenvalues (eigenvectors) of ρ thus amounts to computing those of r⋅σ." how come you can just ignore the identity matrix part of the bloch vector? $\endgroup$
    – user13760
    Nov 9 '20 at 17:49
  • $\begingroup$ because the identity is diagonal in every basis. I'm not ignoring it, I'm just observing that the only nontrivial part of solving the problem is diagonalising the non-identity components of the expression $\endgroup$
    – glS
    May 2 at 10:55

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.