Nielsen / Chuang: Quantum Computation and Quantum Information
chapter 11.3.3 Measurements and entropy
Then it goes on to calculate the entropy of the state after measurement:
$$S(\rho^{'})=-tr(\rho^{'} \log\rho^{'})$$
I will now show that this is equivalent to the entropy that I have calculated in
post #7 (except Boltzmann's constant).
The state before measurement is ##\Psi=\alpha|0\rangle+\beta|1\rangle##, so: $$\rho = | \Psi \rangle \langle \Psi | =
\begin{bmatrix}
\alpha^2 & \alpha\beta \\
\alpha\beta & \beta^2 \\
\end{bmatrix}
$$
The projectors ##P_i## are: $$
P_0 =
\begin{bmatrix}
1 & 0 \\
0 & 0 \\
\end{bmatrix}
, P_1 =
\begin{bmatrix}
0 & 0 \\
0 & 1 \\
\end{bmatrix}
$$
Then:
$$\rho^{'}=\sum_{i} P_i \rho P_i =
\begin{bmatrix}
\alpha^2 & 0 \\
0 & \beta^2 \\
\end{bmatrix}$$
Since the matrix is diagonal, its logarithm is:
$$\log\rho^{'} =
\begin{bmatrix}
\log(\alpha^2) & 0 \\
0 & \log(\beta^2) \\
\end{bmatrix}$$
So, the entropy is:
$$S(\rho^{'})=-tr(\rho^{'} \log\rho^{'}) = -\left[ \alpha^2 \log(\alpha^2) + \beta^2 \log(\beta^2) \right]$$
which is equivalent to what I have calculated before.