Density operator and natural log/trace

[t0pquark]

Homework Statement

The average value of the Von Neumann entropy is associated with $tr( \rho *log(\rho))$, where $log$ is the natural log and $\rho$ is the density operator. Simplify this expression.

Relevant Equations

$\rho = \vert \psi \rangle \langle \psi \vert$
$tr(\rho) = 1$

I feel like I'm going around in circles trying to do something with the expression $tr( \rho *log(\rho))$. I thought about a Taylor expansion, but I don't think there's a useful one here because of the logarithm. We learned the Jacobi's formula in class, but I don't think I want a derivative here either. I saw on Wikipedia a corollary that $tr(log(A)) = log(det(A))$, but the equation is $tr( \rho *log(\rho))$, not $tr(log(\rho))$, and ordinarily $tr(x*y) \neq tr(x)*tr(y)$.

 

[Gaussian97]

Have you tried to apply the spectral theorem to the density matrix?

 

[t0pquark]

Have you tried to apply the spectral theorem to the density matrix?

I just looked it up, and I guess the spectral theorem says that any Hermitian matrix is diagonalizable (so it would be true for the density operator). The trace of the density operator is 1, so the sum of these entries on the diagonal would be 1. Also, if the density operator was a diagonal matrix, then its natural log would simply be the natural log of the entries on the diagonal. I can see this would be helpful, but the end isn't clear for me.
Also, I'm not sure how you would notate this so it could be later multiplied by $\rho$?

 

[Gaussian97]

I just looked it up, and I guess the spectral theorem says that any Hermitian matrix is diagonalizable (so it would be true for the density operator). The trace of the density operator is 1, so the sum of these entries on the diagonal would be 1. Also, if the density operator was a diagonal matrix, then its natural log would simply be the natural log of the entries on the diagonal. I can see this would be helpful, but the end isn't clear for me.
Also, I'm not sure how you would notate this so it could be later multiplied by $\rho$?

Yes, you forget an important property, that is that the diagonal base is orthonormal, with all that you can simplify this expression to a very famous one.

 

[t0pquark]

Yes, you forget an important property, that is that the diagonal base is orthonormal, with all that you can simplify this expression to a very famous one.

Not sure I see what you are talking about?
I'm going to try writing out what I know here.
Applying the spectral theorem to $\rho$, we can get something i.e. $\rho = U A U^\dagger$, where $A$ is a diagonal matrix.
If this goes into the original equation:
$tr(\rho*log( \rho )) = tr(U A U^\dagger * log(U A U^\dagger))$
With log rules, $log(U A U^\dagger)) = log(U) + log(A) + log(U^\dagger) = log(A) + log(UU^\dagger) = log(A)$?
Original equation is now
$tr(\rho*log( \rho )) = tr(U A U^\dagger * log(A))$
I'm not sure how to use the fact that the diagonal base is orthonormal.

 

[Gaussian97]

You should be careful, let $A$ and $B$ be hermitian matrices, then $\log{A}$ and $\log{B}$ are well defined matrices. Also $\log{AB}$ is well defined and thus
$$\log{AB} = \log{A}+\log{B} = \log{B}+\log{A} = \log{BA}.$$
Now, taking the exponential of both sides
$$AB = BA, \qquad \forall A, B \text{ hermitian}$$

Clearly there's something wrong

 

[t0pquark]

Clearly there's something wrong

Yikes, you're right there
So I'm back to $tr( \rho*log(\rho)) = tr( UAU^\dagger*log(UAU^\dagger))$?

 

[Gaussian97]

Ok, how would you compute the logarithm of a density matrix in general?

 

[t0pquark]

Ok, how would you compute the logarithm of a density matrix in general?

Well, it would be easiest if it was already diagonalized; just take the logarithm of the entries on the diagonal.
If not, I guess I could either diagonalize it like above or write it as a sum of projections and then try to do something with that?

 

[Gaussian97]

Well, it would be easiest if it was already diagonalized; just take the logarithm of the entries on the diagonal.
If not, I guess I could either diagonalize it like above or write it as a sum of projections and then try to do something with that?

Yes, fair enough, in other words, to compute the logarithm of a non-diagonal matrix you must first diagonalize the matrix and then if $\rho = UAU^\dagger$ you can compute $$\log \rho = U \left(\log A\right) U^\dagger$$.

 

[t0pquark]

Yes, fair enough, in other words, to compute the logarithm of a non-diagonal matrix you must first diagonalize the matrix and then if ρ=UAU†ρ=UAU†\rho = UAU^\dagger you can compute

logρ=U(logA)U†​

Ahh, so then I would have $tr( \rho * log(\rho)) = tr(UAU^\dagger*log(UAU^\dagger)) = tr(UAU^\dagger*U (log(A))U^\dagger) = tr(UA (log(A))U^\dagger) = tr(A *log(A))$ ?

 

[Gaussian97]

Yes, in general, you can prove that the trace of a matrix is base independent. But you can still simplify more writing this with the eigenvalues.

 

[t0pquark]

Yes, in general, you can prove that the trace of a matrix is base independent. But you can still simplify more writing this with the eigenvalues.

Starting with $tr(A * log(A)) = tr(diag(a_1, ..., a_n)*diag(log(a_1), ..., log(a_n))) = tr(diag(a_1*log(a_1), ..., a_n*log(a_n))) = a_1*log(a_1) + ... + a_n*log(a_n)$??

 

[Gaussian97]

Yes, maybe you can use the $\Sigma$ notation, but I think this is the result that you should get.

 

[t0pquark]

Yes, maybe you can use the $\Sigma$ notation, but I think this is the result that you should get.

Thank you so much for your help!

 

[Gaussian97]

Anytime