Vtgyjfy

$defT{operatorname{Tr}}$
$def1{mathbb{1}}$
Let $H=H_1otimes H_2otimes H_3$ be a finite dimensional Hilbert space, and let $rho_{123}$ be a self-adjoint matrix with $rho_{123}geq 0$ (positive eigenvalues) and $T{rho_{123}}=1$, so a density matrix.

Furthermore let $i,j,kin{1,2,3}$ different, and define $rho_i=T_{jk}(rho_{123})$ and $rho_{jk}=T_{i}(rho_{123})$, so for example if $rho=A_1otimes B_2otimes 1_3$ then $rho_{1}=A_1$ and $rho_{13}=A_1otimes1_3$.

I was reading this paper - https://aip.scitation.org/doi/pdf/10.1063/1.1497701?class=pdf (IV, equation 30) - and there is one property that it takes for granted, but I don't find it at all obvious. Even though it doesn't explicitly say it, I think it's implied there's an equality here:

$$
Tbig(e^{log{rho_{12}}otimes1_3+1_1otimeslog{rho_{23}}} (log{rho_{12}}otimes1_3+1_1otimeslog{rho_{23}})big) = T_{12}big({rho_{12} log{rho_{12}}}big) + T_{23}big({rho_{23} log{rho_{23}}}big)
$$

I am trying to prove the following generalization of the above property $(A_{12} = log(rho_{12}))$

$$
T_3( e^{A_{12}otimes 1_3 + 1_1 otimes B_{12}}) = e^{A_{12}}.
$$

If this is true then the above proposition follows since

$$
T(A_{123}cdot(B_{12}otimes 1_3))= T_{12}(T_3(A_{123})cdot B_{12}).
$$

I have a feeling like the generalization I am trying to prove is false.

Any different proofs of
$$
Tbig(e^{r_{12}+r_{23}}(r_{12}+r_{23})big) = T(e^{r_{12}}r_{12}) + T(e^{r_{23}}r_{23})
$$
are welcome, and any help is appreciated.
Thanks.

EDIT:

If the following is true, it solves all my problems
$$
S(e^{A_{12}otimes 1_3 + 1_1 otimes B_{23}}) = S(e^{A_{12}} otimes e^{B_{23}})
$$
where $S$ is the Von Neumann entropy, since
$$
S(e^{A}otimes e^{B}) = S(e^{A}) + S(e^B)
$$
and this leads to the desired result.

Note that LHS matrix operates on $H$, while the RHS one has a greater dimension. I don't know how to apply the known formula
$$
e^{Aotimes 1_2 + 1_1otimes B} = e^{A} otimes e^{B}
$$
in this case since this equation requires $1_1$ and $A$ to operate on the same space, and this is not true in my case.

Having looked at the original, here's what I think is going on:

The paper instructs:

A second, more powerful subadditivity inequality...
$$
S(rho_{123}) leq S(rho_{12}) + S(rho_{23})
$$
under the constraint $operatorname{tr}(rho_{123}) = 1$. To prove this, choose $A = rho_{123}$ and $B = e^{logrho_{12} + log rho_{23}}$ in Klein's inequality to obtain
$$
-S(rho_{123})+ S(rho_{12}) + S(rho_{23}) geq 1 - operatorname{tr} e^{logrho_{12} + log rho_{23}}
$$

For reference,

Klein's inequality:
$$
operatorname{tr}A(log A - log B) geq operatorname{tr}(A - B)
$$

With that, we apply Klein's inequality as follows:
$$
operatorname{tr}A(log A - log B) geq operatorname{tr}(A - B) implies\
operatorname{tr}rho_{123}(log rho_{123} - log [e^{logrho_{12} + log rho_{23}}]) geq operatorname{tr}(rho_{123} - e^{logrho_{12} + log rho_{23}}) implies\
operatorname{tr}rho_{123}(log rho_{123}) -
operatorname{tr}rho_{123}(logrho_{12} + log rho_{23})
geq operatorname{tr}(rho_{123})-
operatorname{tr}(e^{logrho_{12} + log rho_{23}}) implies\
-S(rho_{123}) -
operatorname{tr}rho_{123}(logrho_{12} + log rho_{23})
geq 1 -
operatorname{tr}(e^{logrho_{12} + log rho_{23}}) implies\
-S(rho_{123}) -
operatorname{tr}rho_{123}logrho_{12}
- operatorname{tr}rho_{123}log rho_{23}
geq 1 -
operatorname{tr}(e^{logrho_{12} + log rho_{23}})
$$

With that in mind, it seems to me that the equality being implied is
$$
operatorname{tr}rho_{123}logrho_{12} = operatorname{tr}rho_{12}logrho_{12}
$$
Or, more explicitly,
$$
operatorname{tr}[rho_{123} log[operatorname{tr}_3(rho_{123}) otimes 1_3]] = operatorname{tr}[operatorname{tr}_3(rho_{123}) log[operatorname{tr}_3(rho_{123})]]
$$
and similarly for $rho_{23}$. It seems to me that this equality indeed holds; let me know if you would like a proof of it.