How do I calculate the von Neumann entropy of a pure one-qubit density matrix?

Question

Let's say I have a pure state of the form: $$\psi = \sqrt{\frac{3}{9}} \lvert 0 \rangle + \sqrt{\frac{6}{9}} \lvert 1 \rangle$$

Then the density matrix representation would be: $$\rho = \psi \otimes \psi' = \begin{bmatrix}.3333&.4714\\4714&.6667\end{bmatrix}$$

Now, what would be the Von Neumann entropy of this matrix? I saw that the equation is:

$$S(\rho) = -\text{trace}(\rho \log \rho) = 1.3455.$$ But it's supposed to be 0, isn't it?

Answer 1

It turns out to be a novice mistake. I was using matlab and this log is elementwise, as @Ahusain pointed out. We must take the matrix logarithm in Matlab which is denoted by $logm$. Then the calculation becomes: $$-\text{trace}(\rho \log m (\rho)) = \text{NaN}.$$ The reason is, we have to define $0 \times \log (0)$ as $0$ instead of $\text{NaN}$ which is the default behavior of Matlab. Another way to calculate it is the following:

As this density matrix is representing a pure state, it would have a diagonalization. I.e. it can be written as:

$$\rho = \sum_{j} \lambda_j \lvert \phi_j \rangle \langle \phi_j \lvert$$

Where, $\lambda_j$ are it's eigenvalues and $\lvert \phi_j \rangle$ are it's distinct eigenvectors. In which case, the von Neumann entropy is simply the shannon entropy of it's eigenvalues:

$$S(\rho) = H(\lambda_1, \lambda_2) = -0 \times \log(0) - 1 \times \log(1) = 0$$Using the above definition of $\log(0) = 0$.