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Assume $A,B$ are noncommutative positive operators on $\mathbb{C}^n$, it is easy to see the following trace equality $$\text{Tr}ABBA = \text{Tr}BAAB.$$ Do we have $$\text{Tr}(ABBA)^\alpha = \text{Tr}(BAAB)^\alpha$$ for any positive real number $\alpha$?

The background of this question may be irrelavant. It comes from the definition of sandwiched version of quantum Renyi divergence, which takes the following different forms: $$ D_{\alpha}(\rho||\sigma) = \frac{1}{\alpha-1}\text{Tr} (\sigma^{\frac{1-\alpha}{2\alpha}}\rho\sigma^{\frac{1-\alpha}{2\alpha}})^\alpha \\ D_{\alpha}(\rho||\sigma) = \frac{1}{\alpha-1}\text{Tr} (\rho^\frac{1}{2}\sigma^{\frac{1-\alpha}{\alpha}}\rho^\frac{1}{2})^\alpha $$ where $\rho$ and $\sigma$ are density operators and $\alpha\in(0,1)\cup(1,+\infty)$.

Here I want to confirm the two expressions are equivalent. Take $A=\sigma^{\frac{1-\alpha}{2\alpha}},B=\rho^{\frac{1}{2}}$ to recover the above original question.

Jiawei Wu
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  • It's straightforward to convince yourself this is true when $\alpha \in \mathbb{N}$. To extend to the general case I would look at the power series expansion of $X^\alpha$ and use the linearity of the trace together with the observation for $\alpha \in \mathbb{N}$. – Rammus Aug 13 '21 at 13:53
  • another approach is $\big(ABBA\big)$ has the same eigenvalues as $\big(BAAB\big)$ so assuming your mapping with $\alpha$ is well defined (it doesn't make sense in many cases) then you have $\sum_{k=1}^n \lambda_k^\alpha = \sum_{k=1}^n \lambda_k^\alpha $ – user8675309 Aug 13 '21 at 17:07
  • @Rammus This method seems to work only for limited range of the eigenvalues of $A,B$ due to the convergence radius of power series, though it should be enough for me. – Jiawei Wu Aug 16 '21 at 09:25
  • @user8675309 This may be stupid, but how can I see that $ABBA$ and $BAAB$ have the same eigenvalues? – Jiawei Wu Aug 16 '21 at 09:28
  • Indeed, @user8675309 method is much cleaner and more useful. For cyclic invariance of eigenvalues see this question. – Rammus Aug 16 '21 at 09:42
  • @Rammus The answer to this question only shows that the spectra coincide as sets. For the current question one has to take multiplicities into account (the statement is still true with multiplicities). – MaoWao Aug 16 '21 at 09:45
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    @JiaweiWu ref https://math.stackexchange.com/questions/3623345/products-of-matrices-in-either-order-have-the-same-characteristic-polynomial/ . The key insight is $\text{trace}\Big(\big(AB\big)^k\Big)=\text{trace}\Big(\big(BA\big)^k\Big)$ for all $k$ working over $\mathbb C$. From here you can use many tools-- Newton's Identities is the standard choice though that requires a bit of machinery. I give a more direct proof in that link. – user8675309 Aug 16 '21 at 17:42

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