Vtyjkyuk

The von Neumann entropy is defined as $S(rho)=-Tr(rho ln rho)$, where $rho$ is density matrix.

http://en.wikipedia.org/wiki/Von_Neumann_entropy

In the above article it says:

S(ρ) is invariant under changes in the basis of ρ, that is, S(ρ) =
S(UρU†), with U a unitary transformation.

How can we prove this statement?

We have that the trace is independent of the choice of basis in which the matrix $rho$ is expressed:
$$Tr(rho)=Tr(U rho U^dagger)$$
But in the case of the von Neumann entropy we have the $ln rho$, so a change of basis for $rho$ gives:
$$Tr[U rho U^daggerln (U rho U^dagger)]$$
How is this equal to $Tr(rho lnrho)$?

I am not fully sure of the notations used here but the argument is standard for matrix computations.

Given that the von Neumann entropy can also be written as
$rho = -sum_j eta_j log eta_j $

where, $eta_j$ are the eigenvalues of $rho$, the only thing that remains to be proved is that eigenvalues are invariant under a change of basis. In other words,

$ Urho U^dagger = sum_j eta_j U|j><j|U^dagger$

is an eigendecomposition with the same eigenvalues $eta_j$. Therefore,

$ S(rho) = S(Urho U^dagger) = - sum_j eta_j log eta_j$

To prove that the von Neumann entropy (defined as $S(rho)=-mathrmTr(rho ln rho)$ with $rho$ being the density matrix) is invariant under unitary change of basis, one should first realize what $ln(rho)$ stands for.

The logarithm of a Hermitian matrix $rho$ is defined as

$$ ln(rho) = V ln( V^dagger rho V) V^dagger,$$

where $V$ is the unitary transformation that diagonalises $rho$, namely, $rho_D=V^dagger rho V$ is diagonal. See also the Wikipedia page on the logarithm of a diagonalizable matrix.

Therefore, the von Neumann entropy reads

$$S(rho)=-mathrmTr(rho V ln (rho_D ) V^dagger) = -mathrmTr(V^dagger rho V ln (rho_D ) ) = -mathrmTr(rho_D ln (rho_D ) ),$$

where the last step relies on the fact the trace is invariant under cyclic permutations, see the related Wikipedia page.

Note also that both $rho$ and $Urho U^dagger$ are associated with the same diagonal matrix $rho_D$ (unitary transformations do not change the eigenvalues).

Finally, because the expression of $S(rho)$ depends only on the diagonal expression of the density matrix $rho_D$, we conclude that $S(rho)$ is invariant under unitary changes of basis.

From a physical point of view, this means that the von Neumann entropy of an isolated system remains constant in time because of its invariance under unitary time evolution.

The matrix log is defined by

$$exp(A)=B implies A = log(B).$$

If $B=B^dagger$, then $B=UDU^dagger$ is diagonalizable by unitary matrices. We find that

$$exp(A) = UDU^dagger implies A = log(UDUdagger).$$

Using $exp(A)=sum_nA^n/n!$ in the first equality below, we can rearrange the above to give

$$U^daggerexp(A)U = exp(U^dagger AU) = D implies U^dagger AU = log(D).$$

Therefore, the unitary matrices "pop" out of the logarithm as desired:

$$A = Ulog(D)U^dagger.$$

Now the result for the entropy should be clear. The state matrix $rho=rho^dagger$ is hermitian, and so it is diagonalizable $rho = UDU^dagger$. Then the entropy is given by

$$S = -Trleft( rho logrhoright) = -Trleft(UDU^daggerlogleft(UDU^daggerright)right) = -Trleft(UDU^dagger Ulog(D)U^daggerright)$$

$$=-Tr(UDlog(D)U^dagger) = -Tr(Dlog(D)U^dagger U) = -Tr(Dlog(D)).$$

The cyclic property of the trace is used in the last line. Finally, one can say that

$$S = -sum_k rho_kklog(rho_kk)$$

where $rho_kk$ is the $k$-th diagonal matrix element of $rho$ in the diagonal basis.