"Contraction Property of thermal density matrix" in the Maldacena's paper of A Bound of Chaos

Question

In the paper, https://arxiv.org/abs/1503.01409 (Maldacena, et al. “A Bound on Chaos.”) in equation (24), the authors write an inequality, $$ Tr( y^{1+\eta} V y^{3-\eta} V ) \leq Tr(y V y^2 V) $$Where $y$ is one-fourth of thermal density matrix $\rho$, i.e $$ y = \frac{e^{(-\beta/4 \hat{H})}}{Z^{1/4}} = \hat{\rho}^{1/4}$$ and call it "the contracting property of $y$". Though I didn't find any appropriate literature for this but, I think the idea is to reduce the powers of $\rho$ in the expectation value and hence reduce the damping factor to get this inequality. Assuming the hamiltonian is lower bounded and discrete spectrum, and $V = v_{ij}$ some hermitian matrix, I can express $$ Tr (\rho^{a/\beta} V \rho^{b/\beta} V)= \sum_{i,j} e^{-a E_i} e^{-b E_j} |v_{ij}|^2/Z^{a+b} $$

When the spectrum of the hamiltonian is positive, the above inequality is straightforward, but my question is how to prove it for the general case, i.e there are finite number of negative energy states ? Do we assume some form of $v_{ij}$ or is there some mathematical identity to prove this ?

Answer 1

First write $\rho, y $ in the eigenbasis of the Hamiltonian,

$$ \rho = \sum_n p_n |n\rangle \langle n| \\ y = \sum_n q_n |n\rangle \langle n| $$

with $q_n^4 = p_n$, in particular $0\le q_n \le 1$. In this basis the inequality becomes

$$ \sum_{n,m} |V_{n,m}|^2 q_n^{1+\eta} q_m^{3-\eta} \le \sum_{n,m} |V_{n,m}|^2 q_n q_m^2 $$

At this point the inequalities $q_n^{1+\eta} \le q_n$ and $q_m^{3-\eta} \le q_m^2$ are easily verified for $0\le \eta\le 1$, implying the claim.