Proof: Time independence of the entropy under unitary time evolution

[barcodeIIIII]

Homework Statement

The unitary time evolution of the density operator is given by
$$\rho(t)=\textrm{exp}(-\frac{i}{\hbar}Ht)\,\rho_0 \,\textrm{exp}(\frac{i}{\hbar}Ht)$$
General definition of entropy is
$$S=-k_B\,Tr\,\{\rho(t) ln \rho(t)\}$$
Proof: $$\frac{dS}{dt}=0$$

Homework Equations

I am not very clear how does the chain rule works in matrix derivative, also how do I differentiate 'ln' of a matrix?
$$\frac{d}{dt} ln\,\rho(t)=?$$
Any insights would be great, Thank you

The Attempt at a Solution

$$ \frac{dS}{dt}=-k_B \,Tr \, \{\frac{d}{dt}[\rho(t)\,ln\,\rho(t)]\} $$
$$\frac{dS}{dt}=-k_B\,Tr\, \{ \frac{ d}{dt} \rho(t) ln\,\rho(t)+ \frac{d}{dt} [ln\, \rho(t) ] \rho(t) \}$$
where $$\frac{d}{dt} \rho (t)=-\frac{i}{\hbar} [H, \rho(t)]$$

 

[mfb]

How is the logarithm of a matrix defined, or how can you express it in other ways? You should be able to find the time-derivative of this.

The trace is linear in the matrix components, exchanging it with derivatives is not an issue.

 

[Birrabenzina]

I got this proof in my last quantum statistical mechanics exam, I remember it pretty well.
So there are two ways, one using $$\frac{\mathrm{d}\rho}{\mathrm{d}t}=\frac{1}{i\hbar}\left[\mathcal{H},\rho\right]$$
Where $\mathcal{H}$ is your Hamiltonian. (My professor suggests this as it should be easier but imho it's really not).
The other one is an elegant proof using diagonalizations and diagonal operators' invariance under unitary transformations.
So, since
$$\rho=\rho^{\dagger}\longrightarrow\exists!U\in U(n)\text{ s.t. }\tilde{\rho}=U^{\dagger}\rho U$$
And your new ρ is diagonalized. Now we can define Entropy in terms of this diagonalized density operator
\begin{align*}
&S=-Tr\left(k_B\rho\log(\rho)\right)\\
&S=-Tr\left(k_B\rho U\log(\tilde{\rho})U^{\dagger}\right)
\end{align*}
Now we can use the main property of unitary transformations, that is
$$U^{\dagger}U=UU^{\dagger}=Id$$
And we can then stick it inside our previous equation
$$S=-Tr\left(k_BUU^{\dagger}\rho UU^{\dagger}U\log(\tilde{\rho})UU^{\dagger}U^{\dagger}\right)$$
Simplifying everything and substituting the definition of our diagonalized density matrix we get
\begin{align*}
&S=Tr\left(k_BU\tilde{\rho}\log(\tilde{\rho})U^{\dagger}\right)\\
&S(\rho)=-UTr\left(-k_B\tilde{\rho}\log(\tilde{\rho})\right)U^{\dagger}=US(\tilde{\rho})U^{\dagger}
\end{align*}
Since by definition time evolutions are unitary operators and diagonalized operators are invariant under unitary transformations we get
$$U(t)^{\dagger}S(\rho)U(t)=S(U(t)^{\dagger}\rho U(t))=US(U(t)^{\dagger}\tilde{\rho}U(t))U^{\dagger}=US(\tilde{\rho})U^{\dagger}=S(\rho)_{\square}$$

 

[blue_leaf77]

I am having trouble trying to understand the quantity $S(\rho)$, is it a scalar or matrix and how does it differ from $S$? Moreover since the thing in the middle of

\begin{align*}
-UTr\left(-k_B\tilde{\rho}\log(\tilde{\rho})\right)U^{\dagger}
\end{align*}

is just a number, shouldn't it be possible to write it as
\begin{align*}
-UTr\left(-k_B\tilde{\rho}\log(\tilde{\rho})\right)U^{\dagger} = -UU^\dagger Tr\left(-k_B\tilde{\rho}\log(\tilde{\rho})\right) = Tr\left(k_B\tilde{\rho}\log(\tilde{\rho})\right)
\end{align*}

 

[Birrabenzina]

S isn't just a scalar, that's a function of the density operator, hence you want to diagonalize that operator through a transformation, in order to say that S is invariant through time transformations