# Non-lockability of quantum max-entropy

Lockability and non-lockability are explained in this paper. A real valued function of a quantum state is called non-lockable if its value does not change by too much after discarding a subsystem. The max-entropy of a quantum state is defined as

$$H_{\max }(A)_{\rho}= \log \operatorname{tr}(\rho_{A}^{1 / 2})$$

For a bipartite quantum state $$\rho_{AB}$$, I would like to know if the max-entropy is non-lockable i.e. is there any relationship of the form

$$\text{tr}(\rho_{AB}^{1/2}) \leq \text{tr}(\rho_{A}^{1/2})\cdot|B|$$

which, after taking logs on both sides, would yield

$$H_{\max}(AB)\leq H_{\max}(A) + \log|B|$$

A couple of easy numerical examples suggests this might be true but I have not been able to prove it.

## 1 Answer

Let $$D_{\alpha}(\rho\|\sigma):= \frac{1}{\alpha - 1} \log \mathrm{Tr}[\rho^\alpha \sigma^{1-\alpha}]$$ be the Petz-Rényi divergence for $$\alpha \in (0,1)\cup(1,\infty)$$. Note that for $$\alpha \in (0,1)\cup(1,2]$$ this quantity satisfies the data processing inequality $$D_{\alpha}(\rho\|\sigma) \geq D_{\alpha}(\mathcal{E}(\rho) \| \mathcal{E}(\sigma)),$$ where $$\mathcal{E}$$ is any CPTP map.

Now let $$\rho_{AB}$$ be any state and take $$\mathcal{E}$$ to be the partial trace over the $$B$$ system. Then for $$\alpha = 1/2$$ we have \begin{aligned} -2 \log \mathrm{Tr}[\rho^{1/2}_{AB}] &= D_{1/2}(\rho_{AB}\|I_{AB}) \\ &\geq D_{1/2}(\mathrm{Tr}_{B}[\rho_{AB}]\|\mathrm{Tr}_{B}[I_{AB}]) \\ &= D_{1/2}(\rho_{A}\||B|I_{A}) \\ &= -2 \log \mathrm{Tr}[\rho^{1/2}_{A} |B|^{1/2}] \end{aligned}

Rearranging you get $$\log \mathrm{Tr}[\rho^{1/2}_{AB}] \leq \log \mathrm{Tr}[\rho^{1/2}_{A}] + \log|B|^{1/2}$$ which is actually a slightly stronger inequality with the exponent of $$|B|$$ being only $$1/2$$.

• It seems like you can also use data processing on fidelity under partial trace. $\text{Tr}(\rho_{AB}^{1/2}) = F(\rho_{AB}, I_{AB}) \leq F(\rho_A, |B|I_A) = |B|^{1/2}F(\rho_A, I_A) = |B|^{1/2}\text{Tr}(\rho_A^{1/2})$. – rnva Sep 24 '20 at 14:37