consider range $\theta \in [0, \pi / 2]$ \[fig:qerr\_entropy\]](./eps/ent.pdf "fig:")

Then the entropy of binary code is calculated. Here, eigenvalues and angle representation are used instead of a variance-covariance matrix, and only the two-dimensional case is treated. In this representation, elements of the variance-covariance matrix are described as follows; $$\begin{aligned}
 \label{eq:r-theta_representation_of_2d_gauss}
\sigma_{11} = \bar{\lambda} + \lambda \cos &2\theta, \quad \sigma_{22} = \bar{\lambda} - \lambda \cos 2\theta, \quad 
\sigma_{12} = \lambda \sin 2\theta, \\
&\bar{\lambda} = \frac{\lambda_1 + \lambda_2}{2}, \quad \lambda = \frac{\lambda_1 - \lambda_2}{2}, \nonumber\end{aligned}$$ where $\lambda_1$ and $\lambda_2$ are eigenvalues of variance-covariance matrix ($\lambda_1 \ge \lambda_2$). $\theta$ means the angle between $x_1$-axis and the longer axis direction of the gaussian ellipse (Fig. \[fig:ellipse\]).

From the symmetry of gaussian distribution, it is enough to get probabilities of binary code $(1,1)$ and $(-1,1)$. It is possible to analytically calculate these probabilities as follows; $$\begin{aligned}
&p_{(1,1)} = \int_0^{\infty} \!\!\! dx_1 \int_0^{\infty} \!\!\! dx_2 \ \frac{1}{2\pi} \frac{1}{|\Sigma|^{-1/2}} e^{-\frac{1}{2} \bm{x}^T \Sigma^{-1} \bm{x}}= \frac{1}{2} - \frac{1}{2\pi} \tan^{-1} \left( \frac{2}{\gamma \sin 2\theta}\right) \nonumber \\
&p_{(-1,1)} = \frac{1}{2} - p_{(1,1)} = \frac{1}{2\pi} \tan^{-1} \left( \frac{2}{\gamma \sin 2\theta}\right) \qquad \qquad \theta \in [0, \frac{\pi}{2}],\end{aligned}$$ where $\gamma$ is defined as $\sqrt{\lambda_1 / \lambda_2} -\sqrt{ \lambda_2 / \lambda_1}$, which is 