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In the book of From Classical to Quantum Shannon Theory, in exercise 11.8.1, there is a property of logarithm of a tensor products of two matrices, defined as follows: $$\log ( A \otimes B) = \log(A) \otimes I + I \otimes \log(B),$$ where $A,B$ are positive semi-definite operators. So I tried to use it on the following examples. Suppose $A$ and $B$ are density matrix defined as: $$A = |0\rangle \langle 0 |,\\ B = \omega_0 |0\rangle \langle 0 | + \omega_1 |1\rangle \langle 1|,$$ where both are constrained to density matrix requirement such as positive semi-definitess, same as the requirement in the exercise. When I apply the property for $A \otimes B$, I got different results as follows: $$\log(A\otimes B) = \log(\omega_0) ( |0\rangle \langle 0 |_A \otimes |0\rangle \langle 0 |_B) + \log(\omega_1) ( |0\rangle \langle 0 |_A \otimes |1\rangle \langle 1 |_B).$$ Meanwhile, for the RHS, I obtained: $$\log(A) \otimes I + I \otimes \log(B) = I \otimes \log(B) = \log(\omega_0) ( I \otimes |0\rangle \langle 0 |_B) + \log(\omega_1) ( I \otimes |1\rangle \langle 1 |_B),$$ since $\log(A)$ should be zero since its eigenvalue is 1. By inspecting their matrix alone, these two are not equal. I am wondering if I made any mistakes or the property has additional requirements?

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    $\begingroup$ You do you define $\log A$ if (at least) one of its eigenvalues is $0$? $\endgroup$ Commented May 17 at 9:08
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    $\begingroup$ ...so the text is imprecise: It should read: "positive-definite operators" or so... $\endgroup$ Commented May 17 at 9:36
  • $\begingroup$ @TobiasFünke I didn't think about it carefully, but from a quick look the formula might very well still be correct if one assigns $-\infty$ to the zero eigenspaces of the operators. And it might well be that this is how the book defines the logarithm. $\endgroup$ Commented May 17 at 9:55
  • $\begingroup$ @NorbertSchuch Well, but then $\log A$ is not well-defined operator; $\infty$ is not a number (in standard analysis; i.e. to make sense of this you have to consider the extended real number line or so), and if $A|\varphi\rangle=0$, then $\log A|\varphi\rangle$ is un-defined, e.g. it has no finite norm etc. $\endgroup$ Commented May 17 at 11:54
  • $\begingroup$ @TobiasFünke As long as you treat this from a theoretical physicist's point of view, rather than mathematical physics or math, I'd say all is fine. $\endgroup$ Commented May 17 at 17:32

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The problem is that you are using the logarithm incorrectly. If you are willing to allow for $\log(0)$, i.e., zero eigenvalues of your operators, then the correct definition must be $\log(0)=-\infty$. (While you use $\log(0)=0$, which is wrong regardless.)

These $-\infty$ terms will then show up in both of your equations. E.g., the first equation will read \begin{align} \log(A\otimes B) & = \log(\omega_0) ( |0\rangle \langle 0 |_A \otimes |0\rangle \langle 0 |_B) + \log(\omega_1) ( |0\rangle \langle 0 |_A \otimes |1\rangle \langle 1 |_B) \\[0.5ex] & \qquad\qquad\qquad+ \log(0) ( |0\rangle \langle 0 |_A \otimes |0\rangle \langle 0 |_B) + \log(0) ( |0\rangle \langle 0 |_A \otimes |1\rangle \langle 1 |_B) \\[1.5ex] & = \log(\omega_0) ( |0\rangle \langle 0 |_A \otimes |0\rangle \langle 0 |_B) + \log(\omega_1) ( |0\rangle \langle 0 |_A \otimes |1\rangle \langle 1 |_B) \\[0.5ex] & \qquad\qquad\qquad+ \log(0) ( |0\rangle \langle 0 |_A \otimes I_B)\ , \end{align} and correspondingly for the right hand side. If you set $\log(0)=-\infty$ and use that $-\infty + x =-\infty$ (for finite $x$),you should find that both sides of the equation are the same.

Note that in the end, this is just a "classical" problem about numbers, since you can always work in the eigenbases of $A$ and $B$: Then, it just reads $\log(xy)=\log(x)+\log(y)$, which indeed also holds if $x$ or $y$ are zero if you adopt the convention $\log(0)=-\infty$.

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One problem is that the logarithm $\log(A)$ of an operator $A$ is singular if $A$ has zero eigenvalues. E.g. $$\log(|0\rangle\langle 0|)~=~\log(1)|0\rangle\langle 0|+\underbrace{\log(0)}_{=-\infty}(I-|0\rangle\langle 0|). $$ OP should as a minimum include these singular contributions in their calculation.

(More generally there might also potentially be issues with the branch cut of a complex logarithm: However in OP's case $A$ is assumed to be a positive semi-definite operator.)

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