Let the [trace norm](https://en.wikipedia.org/wiki/Matrix_norm#Schatten_norms) 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](https://en.wikipedia.org/wiki/Matrix_norm#Schatten_norms), $\Vert \rho - \sigma \Vert\_1 \leq \epsilon$, 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](https://github.com/goforashutosh/CloseStatesImplyNiceProjector/blob/master/ClassicalProof.pdf)).
2. It seems that this is also true for randomly selected matrices in small dimensions (program and description are [here](https://github.com/goforashutosh/CloseStatesImplyNiceProjector/blob/master/CloseStatesImplyNiceProj.ipynb); 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](https://www.cambridge.org/us/universitypress/subjects/mathematics/algebra/matrix-analysis-2nd-edition)) but no luck so far.


Any help or ideas are appreciated.



Does approximate equality of quantum states imply operator inequality in a large subspace?
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 $\sum\_{m:m<k-K}d\_m$.


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 $D\_k=\frac{\mu\_k}{\lambda a^{k+1}}+\sum\_{m:m<k-K}d\_m$, so the trace lost is at most
$$
\sum\_k a^k D\_k=\sum\_k \frac{\mu\_k}{\lambda a}+\sum\_{k,m:m<k-K}a^k d\_m
\\
=\frac{\varepsilon}{\lambda a}+\sum\_m d\_m a^{m+1}\frac{a^K}{1-a}
\\
\le \frac \varepsilon{\lambda a}+\frac{a^K}{1-a} 
$$
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.


