# Does approximate equality of quantum states imply operator inequality in a large subspace?

Let the trace norm of $$X$$ be

$$\Vert X\Vert_1 := \operatorname{tr} \left(\,(X^\dagger X)^{1/2}\right)$$

and let the operator inequality $$A \leq B$$ denote that the operator $$B-A$$ is positive semidefinite.

If the quantum states (finite-dimensional Hermitian positive semidefinite matrices that have unit trace) $$\rho$$ and $$\sigma$$ are close to each other in the trace norm, $$\Vert \rho - \sigma \Vert_1 \leq \epsilon$$, for some small $$\epsilon > 0$$, does there exist a projector $$\Pi$$, such that

\begin{aligned} \Pi \rho \Pi &\leq (1+g_1(\epsilon)) \sigma \\ \operatorname{tr} (\Pi \sigma) &\geq 1- g_2(\epsilon) \end{aligned}

for some small functions $$g_1(\epsilon)$$ and $$g_2(\epsilon)$$, i.e., both $$g_1(\epsilon)$$ and $$g_2(\epsilon)$$ tend to $$0$$ as $$\epsilon \to 0$$? Note $$g_1$$ and $$g_2$$ should be independent of the dimensions of the matrices. Some observations:

1. This is true for classical states (matrices which commute) and also true when we also sandwich $$\sigma$$ in the operator inequality (proofs are here).

2. It seems that this is also true for randomly selected matrices in small dimensions (program and description are here; the Jupyter notebook also contains some more observations).

So far, I have not been able to come up with a way which avoids having a projector on $$\sigma$$ as well. It seems that using the assumption $$\Vert \sqrt{\rho}- \sqrt{\sigma} \Vert_2 \leq \epsilon$$, which is equivalent (upto the exponent of $$\epsilon$$) to the original assumption, is far more easier, since there does not seem to be a lot of ways one can manipulate the trace norm. I have been trying to use a strengthened version of Gresgorin Theorem (Corollary 6.1.6 of Horn & Johnson's 2nd edition of Matrix Analysis) but no luck so far.

Any help or ideas are appreciated.

• In convex algebraic geometry, the set of positive semidefinite matrices that have unit trace is called the spectraplex. Jun 29, 2023 at 6:23
• @Noel : "It seems that this is also true for randomly selected matrices in small dimensions": Can you describe mathematically how you construct the projector? I cannot read your code. Jun 29, 2023 at 15:20
• @IosifPinelis In the code, $\sigma$ is WLOG assumed to be diagonal, so I basically search over all possible diagonal projectors (projectors over all combinations of the eigenspace of $\sigma$) and see if they satisfy the conditions. This is done in a slightly more efficient manner
– Noel
Jun 29, 2023 at 15:50

Let $$\sigma$$ be represented by a PD matrix $$A$$ and $$\rho$$ by $$A+B$$. Note that $$|B|\ge B$$ (in the sense of PSD matrices) and has the same $$1$$-norm. Also $$\Pi |B|\Pi\ge \Pi B \Pi$$, so to dominate $$\Pi B\Pi$$ is always easier than to dominate $$\Pi|B|\Pi$$. Thus, we can replace $$B$$ by $$|B|$$ and assume that $$B$$ is PD as well, so its trace norm is just its trace.

Now let's take $$a<1$$ and consider the spectral decomposition of $$H$$ in which we unite all the eigenvalues of $$A$$ between $$a^{k+1}$$ and $$a^k$$, i.e., we write $$H=\oplus H_k$$ such that $$H_k$$ is an invariant subspace of $$A$$ and $$a^{k+1}I\le A\le a^k I$$ on $$H_k$$. Let $$d_k$$ be the dimension of $$H_k$$ and let $$P_k$$ be the projector to $$H_k$$.

Now we will construct our projector $$\Pi$$ as the sum of $$\Pi_k P_k$$ where $$\Pi_k$$ is a projection within $$H_k$$, i.e., if $$H\ni x=\sum_k x_k$$ is the orthogonal decomposition of $$x$$ with $$x_k\in H_k$$, we'll have $$\Pi x=\sum_k \Pi_k x_k$$. Notice that then $$\Pi A\Pi\le a^{-1}A$$ regardless of the choice of $$\Pi_k$$ and, if you think a bit, you will realize that this is essentially almost all you can do to ensure that inequality.

Now take a single $$H_k$$ and consider the operator $$P_kBP_k$$. Let its trace be $$\mu_k$$ so that $$\sum_k\mu_k=\varepsilon$$. Take $$\lambda>0$$ and remove from $$H_k$$ the eigenvectors of this operator with eigenvalues greater than $$\lambda a^{k+1}$$. The dimension of the removed space will be $$\le \frac{\mu_k}{\lambda a^{k+1}}$$.

On the remaining subspace $$V_k\subset H_k$$, we have $$\langle Bx_k,x_k\rangle\le\lambda \langle Ax_k,x_k\rangle$$. Let now $$x=\sum_k x_k$$, $$x_k\in V_k$$. Then, choosing some integer $$K$$, we can write $$\langle Bx,x\rangle\le \sum_{k,m} |\langle Bx_k,x_m\rangle|= \\ \sum_k \langle Bx_k,x_k\rangle +2\sum_{k,m:k-K\le m\le k-1} |\langle Bx_k,x_m\rangle| \\ +2\sum_{k,m:m< k-K} |\langle Bx_k,x_m\rangle| \\ =\Sigma_1+\Sigma_2+\Sigma_3\,.$$ We have $$\Sigma_1\le\lambda\langle Ax,x\rangle$$. Also by Cauchy-Schwarz and the positive definiteness of $$B$$, we have $$\Sigma_2\le 2K\Sigma_1$$. Thus, the sum of the first two terms is at most $$(2K+1)\lambda\langle Ax,x\rangle$$. The question is, of course, what to do with $$\Sigma_3$$. And the answer is that we will just kill it entirely by further reducing $$x_k$$ to the intersection of the kernels of the corresponding $$P_mB$$ on $$H_k$$. Since $$P_mB$$ is an operator of rank $$d_m$$ at most, that will reduce the dimension of $$V_k$$ by at most $$\sum_{m:m.

Then we shall have $$\Pi B\Pi\le (2K+1)\lambda \Pi A\Pi$$ and $$\Pi(A+B)\Pi\le [1+(2K+1)\lambda]\Pi A\Pi\le [1+(2K+1)\lambda]a^{-1}A\,.$$ On the other hand, the codimension of the final $$V_k$$ in $$H_k$$ is at most $$D_k=\frac{\mu_k}{\lambda a^{k+1}}+\sum_{m:m, so the trace lost is at most $$\sum_k a^k D_k=\sum_k \frac{\mu_k}{\lambda a}+\sum_{k,m:m because $$\sum_m d_m a^{m+1}\le \operatorname{tr} A=1$$.

Now one can just play with the parameters to balance. Suppose we want the domination with the constant $$1+C\delta$$ and the trace loss at most $$C\Delta$$ with $$C$$ being some absolute constant (say, $$5$$). Then we are forced to take $$a=1-\delta$$ and $$(2K+1)\lambda\le 3\delta$$. We also need $$\frac{a^K}{1-a}<\Delta$$, which calls for $$K=\delta^{-1}\log\frac{1}{\delta\Delta}$$ and $$\lambda=\delta^2 \left(\log\frac{1}{\delta\Delta}\right)^{-1}$$, so our $$\varepsilon$$ should be less that $$\Delta\delta^2\left(\log\frac{1}{\delta\Delta}\right)^{-1}$$ to make the conditions compatible. If $$\delta=\Delta$$, then we get them both around $$\sqrt[3]{\varepsilon\log\frac 1\varepsilon}$$. This is, probably, not optimal, but you requested just some speed of tending to $$0$$ with $$\varepsilon$$ and the power bound is not terribly bad, so I'll stop here.

• Thank you, this is great. This answer is still pretty non-trivial to me and those who work with me. I do remember looking at this vector space decomposition but obviously could not go too far. I have two soft questions: 1. if I use this result in my research, how would you like to be cited (you can dm me). 2. if you do indeed think this is a simple problem, has a similar technique been used before and if so where?
– Noel
Aug 1, 2023 at 15:00
• @Noel 1) Just cite MO; they need citations more than I do :lol: 2) Yes, this technique of "geometric stairway decomposition" (the name is mine, I'm not sure what it is officially called, if anything) is pretty standard in analysis though I haven't seen it applied in linear algebra much. The key is that the "short support" functions/sequences cannot interact much with the "long support" ones (what that means depends on the context), so you can localize to same size entries, which often makes life simpler. I picked the idea from Bourgain's paper on $\Lambda(p)$-sets, but it is quite ubiquitous. Aug 1, 2023 at 15:24
• Hi again @fedja, what exactly did you have in mind when you said that you can bound $\Sigma_2 \leq 2K \Sigma_1$ because I was writing this out for myself and I can only do something $O(K^2)$
– Noel
Sep 21, 2023 at 18:10
• @Noel Each diagonal sum ($m=k-\ell$ with fixed $\ell=1,\dots,K$) is bounded by $\Sigma_1$ by Cauchy-Schwarz: $|\langle Bx_{k-\ell},x_k\rangle|\le \frac 12[\langle Bx_{k-\ell}, x_{k-\ell}\rangle+\langle Bx_k, x_k\rangle]$ ($B$ is positive definite). Sep 21, 2023 at 18:31

Edit : the following argument does not answer the question and actually appears already in the OP's linked note.

Let $$H = 1_{[0,\infty)}$$. Let $$\Pi$$ be the projector defined by functional calculus as $$\Pi=H((1+\sqrt{\epsilon})\sigma - \rho)$$ and $$\Pi'=1-\Pi$$. Observe that $$\mathrm{tr}(\Pi'(\rho- (1+\sqrt{\epsilon})\sigma)) \geq 0$$

Now write $$\epsilon \geq \|\rho-\sigma\|_1 \geq \mathrm{tr} (\Pi'\rho) - \mathrm{tr}(\Pi'\sigma) \geq \sqrt{\epsilon} \ \mathrm{tr} (\Pi' \sigma)$$ so that $$\mathrm{tr}(\Pi \sigma) \geq 1- \sqrt{\epsilon}$$. By construction, we have $$\Pi ((1+ \sqrt{\epsilon})\sigma - \rho) \Pi \geq 0$$ and therefore $$\Pi \rho \Pi \leq (1+\sqrt{\epsilon})\Pi \sigma\Pi$$.

• I think your $\Pi$ is not in general a projector, even when $\rho$ and $\sigma$ are diagonal. Jun 29, 2023 at 15:25
• You get $\Pi (1+\sqrt{\epsilon})\sigma \Pi \geq \Pi \rho \Pi$, how do you transform this to $(1+\sqrt{\epsilon})\sigma \geq \Pi \rho \Pi$? It doesn't seem to me that $\Pi$ and $\sigma$ commute. This seems to me to be the same argument as Theorem 2.2 in my notes (github.com/goforashutosh/CloseStatesImplyNiceProjector/blob/…)
– Noel
Jun 29, 2023 at 15:56
• @Noel oops, you are right (and I should have read your notes better) Jun 29, 2023 at 16:14
• @IosifPinelis applying a $\{0,1\}$-valued function to a self-adjoint operator always produces a projector. Jun 29, 2023 at 16:18
• @GuillaumeAubrun : Right. Somehow I perceived $H$ as $\max(0,\cdot)$. Jun 29, 2023 at 16:25