Von Neumann Entropy of a Qubit

$H(\frac{1+\overrightarrow{r}}{2})$, where:

• $\overrightarrow{r}$ is the Qubit $\rho =\frac{1}{2}(I+\overrightarrow{r}\cdot \overrightarrow{\sigma })$ represented as in the Bloch Sphere
• $H$ is the Binary entropy function

Proof

The Von Neumann Entropy for a state $\rho$ is defined as $-\mathrm{Tr}[\rho \log \rho ]$. This can be calculated by diagonalizing $\rho$ and then summing $\lambda \log \lambda$ over the 2 eigenvalues $\lambda$. For a 2x2 Matrix $A$, we calculate the eigenvalues through

$$\begin{array}{rlc}0& =\det (A-\lambda I)& \\ & =\left|\begin{array}{cc}a-\lambda & b\\ c& d-\lambda \end{array}\right|& \\ & =(a-\lambda )(d-\lambda )-bc& \text{Plugging in }A=\rho =\frac{1}{2}\left(\begin{array}{cc}1-r_3& r_1+i{r}_2\\ r_1-i{r}_2& 1+r_3\end{array}\right)\\ & =(\frac{1-r_3}{2}-\lambda )(\frac{1+r_3}{2}-\lambda )-\frac{1}{4}(r_1+i{r}_2)(r_1-i{r}_2)& \\ & =(\frac{1}{2}-\lambda {)}^2-\frac{1}{4}(r_3^2-r_1^2-r_2^2)& \\ & =(\frac{1}{2}-\lambda {)}^2-(\frac{1}{2}|\overrightarrow{r}|{)}^2& \\ & \Downarrow & \\ \lambda & =\frac{1\pm |\overrightarrow{r}|}{2}& \end{array}$$

Hence we can now calculate

$$\begin{array}{rl}-\mathrm{Tr}[\rho \log \rho ]& =-\left[(1+|\overrightarrow{r}|)\log (1+|\overrightarrow{r}|)+(1-|\overrightarrow{r}|)\log (1-|\overrightarrow{r}|)\right]\\ & =H(\frac{1+|\overrightarrow{r}|}{2})\end{array}$$

Where $H(q)=-q\log q-(1-q)\log (1-q)$ is the Binary entropy function, following the distribution below