Von Neumann entropy vs Shannon entropy for a quantum state vector Is there some kind of information about a quantum system, which we might derive from von Neumann entropy, which is impossible to deduce from Shannon entropy?
Let's say, that we have a bipartite system consisting of two qubits $A$ and $B$. These qubits might be in an entangled state $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) = \frac{1}{2} (1\ 0\ 0\ 1)^T$. We can obtain a density matrix representing this system as
$$ \rho^+ = |\Phi^+\rangle \langle \Phi^+| = \frac{1}{2}\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \\ \end{pmatrix}.$$
Then, we can get the reduced density matrix of the subsystem $A$:
$$ \rho_A^+ = Tr_B(\rho^+) = \frac{1}{2}\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}.$$
Now, we might want to calculate the von Neumann entropy:
$$ S_{vN} = -Tr(\rho_A^+ \log_2 \rho_A^+) = 1.$$
But, if we just treat the squares of coefficients $c_i$ of a state $|\Phi^+\rangle = \sum_i c_i |i\rangle$ in a $\{|00\rangle, |01\rangle, |10\rangle, |11\rangle\}$ basis as probabilities (which is the case for a quantum state), we get exactly the same value of Shannon entropy as von Neumann entropy:
$$ S_S = -\sum_i |c_i|^2 \log_2 |c_i|^2 = 1.$$
 A: Thanks to the comment of @Norbert Schuch I finally understand, that Shanon entropy calculated from a state vector represented in the Schmidt basis, is the same as von Neumann entropy.
Let's consider a state $|\psi\rangle = \frac{1}{4}|00\rangle + \frac{1}{\sqrt{2}}|01\rangle + \frac{\sqrt{3}}{4}|10\rangle + \frac{1}{2}|11\rangle$. If we treat the squares of the coefficients as probabilities, we get Shanon entropy:
$$ S_S = -\frac{1}{16} \log_2\frac{1}{16}   -\frac{1}{2} \log_2\frac{1}{2}   -\frac{3}{16} \log_2\frac{3}{16}  -\frac{1}{4} \log_2\frac{1}{4} \approx 1.18. $$
By creating density matrix from the state vector as $\rho = |\psi\rangle \langle \psi|$, and computing entanglement entropy as described in the initial question, we get von Neumann entropy:
$$ S_{vN} \approx 0.148. $$
Now, to represent initial state $|\psi\rangle$ in the Schmidt basis, we have to do an SVD of this state. We begin by representing the state vector as a matrix:
$$ |\psi\rangle = \begin{pmatrix} \frac{1}{4} \\ \frac{1}{\sqrt{2}} \\ \frac{\sqrt{3}}{4} \\ \frac{1}{2} \end{pmatrix} \rightarrow M_\psi = \begin{pmatrix} \frac{1}{4} & \frac{1}{\sqrt{2}} \\ \frac{\sqrt{3}}{4} & \frac{1}{2} \end{pmatrix}. $$
$SVD$ gives singular values equal to $0.98286148$ and $0.18434563$. These are the only non-zero entries in the state vector written down in the Schmidt basis. Now, if we compute the Shanon entropy of squares of these singular values we get
$$ S_S' = -|0.98286148|^2 \log_2 |0.98286148|^2 - |0.18434563|^2 \log_2 |0.18434563|^2 \approx 0.148, $$
which is exactly the same as calculated previously von Neumann entropy.
