Considering pure QR algorithm (without shifts and preliminary tridiagonal reduction) are there sufficient conditions for algorithm to converge to quasi-diagonal form?

For the the following matrix $$ A = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right). $$ with eigenvalues $\lambda_1 = 1$ and $\lambda_2 = -1$ it apparently does not. QR decomposition produces the following $$ A = QR = \left(\begin{array}{cc} 0 & 1\\ 1 & 0\end{array}\right) \left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right), $$ which generates stationary sequence $\{A^{(k)}\}$.