# Prove that the trace distance is upper-bounded by the Hilbert-Schmidt distance

In (Haah et al. 2015), in the third page, second column, the authors use the following result: given a pair of states $$\rho,\sigma$$, we have $$\|\rho-\sigma\|_1 \le 2\sqrt{\min(\operatorname{rank}(\rho),\operatorname{rank}(\sigma))} \|\rho-\sigma\|_2,$$ where $$\|A\|_1\equiv \operatorname{Tr}|A|$$ is the trace norm, and $$\|A\|_2\equiv \sqrt{\operatorname{Tr}(A^\dagger A)}$$ is the Hilbert-Schmidt norm.

I haven't encountered this fact before, and I can't find a reference pointing to its source in the paper. What's a good way to prove this result (or equivalently, what's a reference discussing such result)?

• Lemma 11 of this paper might be helpful. Oct 6, 2022 at 6:01
• This also seems to be a specific case of eq. $1.170$ from Watrous' TQI, although it is not proven there.
– JSdJ
Oct 6, 2022 at 9:08
• @narip relevance with this question aside, that's a really nice paper I somehow missed, thanks for the reference!
– glS
Oct 6, 2022 at 14:28

Let $$P - Q = \rho - \sigma$$ be a Jordan-Hahn decomposition, meaning that $$P$$ and $$Q$$ are the unique positive semidefinite operators that satisfy that equation and have orthogonal images. We'll start by collecting some basic facts about this decomposition:

• It is the case that $$\|\rho - \sigma\|_1 = \operatorname{Tr}(P) + \operatorname{Tr}(Q)$$. Moreover, because $$\rho$$ and $$\sigma$$ are density operators, we have that $$\rho - \sigma$$ is traceless, and therefore $$\operatorname{Tr}(P) = \operatorname{Tr}(Q)$$. This implies $$\|\rho - \sigma\|_1 = 2\operatorname{Tr}(P) = 2 \operatorname{Tr}(Q).$$

• By the Pythagorean theorem, we have $$\|\rho - \sigma\|_2^2 = \|P\|_2^2 + \|Q\|_2^2$$, and so $$\|P\|_2\leq\|\rho - \sigma\|_2 \quad \text{and}\quad \|Q\|_2\leq\|\rho - \sigma\|_2.$$

• Because $$\rho$$ and $$\sigma$$ are positive semidefinite, it is the case that $$\operatorname{rank}(P) \leq \operatorname{rank}(\rho) \quad\text{and}\quad \operatorname{rank}(Q) \leq \operatorname{rank}(\sigma)$$ (in essence, by counting positive and negative eigenvalues).

Now we have what we need to get the inequality from the one that narip and JSdJ referred to in their comments (which follows from Cauchy-Schwarz for operators): $$\begin{gathered} \|\rho - \sigma\|_1 = 2\operatorname{Tr}(P) \leq 2 \sqrt{\operatorname{rank}(P)} \,\|P\|_2 \leq 2 \sqrt{\operatorname{rank}(\rho)}\, \|\rho - \sigma\|_2\\ \|\rho - \sigma\|_1 = 2\operatorname{Tr}(Q) \leq 2 \sqrt{\operatorname{rank}(Q)} \,\|Q\|_2 \leq 2 \sqrt{\operatorname{rank}(\sigma)}\, \|\rho - \sigma\|_2 \end{gathered}$$ (and then get the minimum by chosing whichever is better).

• great, thanks! When you say that $\operatorname{Tr}(P)\le 2\sqrt{\operatorname{rank}(P)} \|P\|_2$ "follows from CS", are you thinking of the inequality $\|ST\|_r\le \|S\|_p \|T\|_q$ for $1/r=1/p+1/q$, taking $T=I$ (or better said, projection onto the suppor of $S$), $r=1$, $p=q=2$? Or is there a more direct way?
– glS
Oct 6, 2022 at 15:15
• @gIS Cauchy-Schwarz inequality is indeed a special case of Hölder's inequality with $p=q=2$, but here we only need Cauchy-Schwarz: $\mathrm{tr}(P)=\langle T,P\rangle_{HS}\le\|T\|_2\|P\|_2=\sqrt{\mathrm{rank}(P)}\|P\|_2$ where $T:=I_{\mathrm{supp}(P)}$ as you described. Oct 6, 2022 at 17:52