I'm looking for some guidance calculating the following trace. The trace I want to calculate is: $$ S = \text{Tr}(\rho \ln \rho) $$ Where $\rho$ is the density function in statistical mechanics. The homework problem I'm working on is:
Let's look at a density matrix for a quantum system which is well described by a basis with three states. An example is the electron of a hydrogen atom that is prepared with spin up and in a linear combination of the three $2p$ states. Two standard bases for the hydrogen atom are the Cartesian states given by $\big | 2p_x \big >$, $\big | 2p_y \big >$, $\big | 2p_z \big >$, and the good $L_z$ angular momentum states, $\big | 2p_{+1} \big >$, $\big | 2p_0 \big >$, $\big | 2p_{-1} \big >$. They are related by a unitary transformation: \begin{align} \big | 2p_{\pm 1} \big > &= \mp \frac{1}{\sqrt 2} [\big | 2p_x \big > \pm \big | 2p_y \big > ] \\ \big | 2p_0 \big > &= \big | 2p_z \big > \end{align} You can prepare a beam of hydrogen atoms so that with probability $1/4$ an atom is prepared in the $\big | 2p_x \big >$ state, with probability $1/4$ it is prepared in the $\big | 2p_y \big >$ state, with probability $1/4$ it is prepared in the $\big | 2p_z \big >$ state, and with probability $1/4$ it is prepared in the $\big | 2p_{+1} \big >$ state.
Where my $\rho$ equation is: $$ \rho = \frac{1}{4} \big | 2p_x \big > \big < 2p_x \big | + \frac{1}{4} \big | 2p_y \big > \big < 2p_y \big | + \frac{1}{4} \big | 2p_z \big > \big < 2p_z \big | + \frac{1}{4} \big | 2p_{+1} \big > \big < 2p_{+1} \big | $$
I'm simply not familiar enough with evaluating traces to know what to do with the logarithm. My initial attempt was to plug in my equation for $\rho$ and to see what happens when I try and simplify, but I kept getting stuck with logarithm terms in my trace function and got stuck.
I just need some help to get this calculation started. Thank you!