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This question concerns the neighborhood of separable states around the maximally mixed state in a bipartite system; I will restate the theorem as it appears in Watrous' The Theory of Quantum Information (Theorem 6.13, page 321):

Let $H$ be a Hermitian operator on $\mathcal{X} \otimes \mathcal{Y}$ for complex Euclidean spaces $\mathcal{X}, \mathcal{Y}$ satisfying $\lVert H \rVert_2 \leq 1$. Then,

\begin{equation}\tag{1} \mathbb{I}_\mathcal{X} \otimes \mathbb{I}_\mathcal{Y} - H \in \text{Sep}(\mathcal{X}:\mathcal{Y}) \end{equation} where $\lVert A \rVert_2 = \sqrt{\text{Tr}(A^\dagger A)}$ is the Schatten 2-norm. An operator $S$ is here said to be in $\text{Sep}(\mathcal{X}:\mathcal{Y})$ if it admits a form \begin{equation}\tag{2} S = \sum_k p_k A_k \otimes B_k \end{equation} for a probability vector $p$ ($\lVert p \rVert_1 = 1$), and the sets $\{A_k\}$ and $\{B_k\}$ contain positive operators on $\mathcal{X}$ and $\mathcal{Y}$ respectively. In plain language, this says (up to normalization) that applying an arbitrary perturbation $H$ to the maximally mixed state results in another state that is not maximally mixed but is still separable.


My question is, can someone describe how to construct such an operator $S$ where both the perturbation $H$ and the decomposition of $\mathbb{I}_\mathcal{X} \otimes \mathbb{I}_\mathcal{Y} - H$ according to Equation (2) are known? I'm especially interested in the case of a pair of qubits, $\mathcal{X},\mathcal{Y}=\mathbb{C}^2$ where $H$ has a known decomposition in the Pauli basis, $H = \sum_{i,j=0}^3 \omega_{ij} \sigma_i \otimes \sigma_j$.

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    $\begingroup$ I took the liberty of clarifying the scope of the square root in the formula for the Schatten 2-norm (aka Frobenius norm). I also confirmed that the new formula agrees with the book (see eq $(1.179)$ on page $33$). Unfortunately, norm symbols like $\|.\|_2$ are ambiguous. $\endgroup$ Dec 26, 2021 at 1:59

2 Answers 2

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The following construction works for a pair of qubits and small perturbations with $\|H\|_2\le\frac15$. Remarkably, in this case we can choose operators $A_k$ and $B_k$ independently of $H$.

Basis

Begin by defining pure states

$$ \begin{align} |\psi_0\rangle &= |0\rangle& |\psi_2\rangle &= \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac23}e^{\frac{2\pi i}{3}}|1\rangle\\ |\psi_1\rangle &= \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac23}|1\rangle& |\psi_3\rangle &= \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac23}e^{\frac{4\pi i}{3}}|1\rangle. \end{align}\tag3 $$

The positive operators $P_i = |\psi_i\rangle\langle\psi_i|$ form a basis$^1$ of the real vector space $L_H(\mathbb{C}^2)$ of Hermitian operators on $\mathbb{C}^2$, so we can write $S=\mathbb{I}_\mathcal{X}\otimes\mathbb{I}_\mathcal{Y}-H$ as a linear combination

$$ S=\sum_{ij}\alpha_{ij}P_i\otimes P_j\tag4 $$

of separable states $P_i\otimes P_j$ with coefficients $\alpha_{ij}\in\mathbb{R}$. We will prove that if $\|H\|_2\le\frac15$ then $\alpha_{ij}\ge 0$.

Dual basis

Operators

$$ \begin{align} \tilde{P}_0 &= \frac12\begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}& \tilde{P}_2 &= \frac{1}{\sqrt8}\begin{pmatrix} 0 & -1-i\sqrt{3} \\ -1+i\sqrt{3} & \sqrt2 \end{pmatrix}\\ \tilde{P}_1 &= \frac12\begin{pmatrix} 0 & \sqrt{2} \\ \sqrt{2} & 1 \end{pmatrix}& \tilde{P}_3 &= \frac{1}{\sqrt8}\begin{pmatrix} 0 & -1+i\sqrt{3} \\ -1-i\sqrt{3} & \sqrt2 \end{pmatrix} \end{align}\tag5 $$

form a basis dual$^2$ to $P_i$, so we can compute the coefficients in the expansion of any operator $T\in L_H(\mathbb{C}^2)$ in the basis $P_i$ using

$$ T=\sum_i\langle T,\tilde{P}_i\rangle_{HS} P_i\tag6 $$

where $\langle A,B\rangle_{HS}=\mathrm{tr}(A^\dagger B)$ is the Hilbert-Schmidt inner product.

Coefficients

Finally, $\tilde{P}_i\otimes\tilde{P}_j$ is a basis dual to $P_i\otimes P_j$, so the coefficients in $(4)$ are

$$ \begin{align} \alpha_{ij}&=\mathrm{tr}[S\tilde{P}_i\otimes\tilde{P}_j]\\ &=\mathrm{tr}[(\mathbb{I}_\mathcal{X}\otimes\mathbb{I}_\mathcal{Y}-H)\tilde{P}_i\otimes\tilde{P}_j]\\ &=\mathrm{tr}(\tilde{P}_i\otimes\tilde{P}_j)-\mathrm{tr}(H\tilde{P}_i\otimes\tilde{P}_j)\\ &=\frac14-\mathrm{tr}(H\tilde{P}_i\otimes\tilde{P}_j)\\ &\ge\frac14-|\mathrm{tr}(H\tilde{P}_i\otimes\tilde{P}_j)|\\ &\ge\frac14-\|H\|_2\|\tilde{P}_i\|_2^2\\ &=\frac{1-5\|H\|_2}{4} \end{align}\tag7 $$

where we used the facts that $\mathrm{tr}(\tilde{P}_i)=\frac12$ and $\|\tilde{P}_i\|_2=\frac{\sqrt{5}}{2}$ for all $i=1,\dots,4$. Thus, if $\|H\|_2\le\frac15$ then $\alpha_{ij}\ge 0$. We conclude that, up to normalization, $(4)$ is the desired representation of $S$ as a mixture of separable states.

Generalization

The construction above can be generalized to stronger perturbations and higher dimensions. In order to accommodate stronger perturbation one needs to abandon the fixed basis $(3)$ and look for a spanning set of operators $P_i$ that yields small values of $|\mathrm{tr}(H\tilde{P}_i\otimes\tilde{P}_j)|$ in $(7)$. The choice of $P_i$ will generally depend on $H$.

Generalization to higher dimensions is conceptually straightforward, but potentially computationally difficult as the size of a spanning set grows exponentially with Hilbert space dimension. Also, the upper bound on $\|H\|_2$ that admits a fixed basis depends on dimension$^3$.


$^1$ The operators $P_i$ form a simple example of a SIC-POVM. Note that the above construction works for other sets of operators $P_i$ that are not SIC-POVMs. An alternative choice convenient when $H$ is expressed as a Pauli sum is the set of single-qubit stabilizer states

$$ P_0=|0\rangle\langle 0|\quad P_1=|1\rangle\langle 1|\\ P_2=|+\rangle\langle +|\quad P_3=|-\rangle\langle -|\\ P_4=|{+i}\rangle\langle {+i}|\quad P_5=|{-i}\rangle\langle {-i}|.\tag{3'} $$

$^2$ For a SIC-POVM in Hilbert space of dimension $d$, we have $\tilde{P}_i=((d+1)P_i-I)/d$ (thanks to @Danylo Y for pointing out this simple formula!). More generally, we can compute a dual basis using frame theory. For example, a frame dual to $(3')$ is

$$ \begin{align} \tilde{P}_0&=\frac13\begin{bmatrix}2&0\\0&-1\end{bmatrix}&\quad\tilde{P}_1=\frac13\begin{bmatrix}-1&0\\0&2\end{bmatrix}\\ \tilde{P}_2&=\frac16\begin{bmatrix}1&3\\3&1\end{bmatrix}&\quad\tilde{P}_3=\frac16\begin{bmatrix}1&-3\\-3&1\end{bmatrix}\\ \tilde{P}_4&=\frac16\begin{bmatrix}1&3i\\-3i&1\end{bmatrix}&\quad\tilde{P}_5=\frac16\begin{bmatrix}1&-3i\\3i&1\end{bmatrix} \end{align}\tag{5'} $$

leading to the same bound $\|H\|_2\le\frac15$.

$^3$ In particular, when using a SIC-POVM in Hilbert space of dimension $d$ the coefficients $\alpha_{ij}$ are non-negative if $\|H\|_2\le\frac{1}{d^2+d-1}$. Note that the existence of SIC-POVMs in arbitrary dimensions is currently an open problem.

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    $\begingroup$ It's probably worth to add that the set $\{P_i\}$ forms a SIC-POVM in dimension d=2. And, in general, $\tilde{P}_i = ((d+1)P_i - I)/d$. This gives a boundary $1/(d^2+d-1)$ for the norm of $H$ (if a SIC-POVM exists in dimension $d$). $\endgroup$
    – Danylo Y
    Dec 26, 2021 at 15:29
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    $\begingroup$ this is really nice! You are showing that all states close enough to the identity admit a separable decomposition using elements from a (most likely any choice of) SIC-POVM. Is this from the book, or somewhere else? Do you have a more general understanding of this result that explains why it holds? $\endgroup$
    – glS
    Dec 27, 2021 at 9:06
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    $\begingroup$ also more generally: if instead of a SIC-POVM you use another (linearly independent) set of positive semidefinite operators, you'll still get the result, only with a different bound depending on the what is the smallest L2 norm of the dual basis elements. Which of course makes sense: for $S=I$ any frame whose dual elements have positive trace will do the job, so it stands to reason that the closer to the identity, the more frames will work. That naturally brings to the question: can we say that the SIC-POVM provides a separable deomposition for the largest open ball around the identity? $\endgroup$
    – glS
    Dec 27, 2021 at 9:14
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    $\begingroup$ Could you please provide a bit more detail on how you computed the dual basis here? Did you explicitly construct the frame operator as a linear operator and then compute its inverse? $\endgroup$
    – forky40
    Dec 27, 2021 at 16:39
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    $\begingroup$ @gIS Regarding SIC-POVM providing the largest ball: I don't think this is the case, e.g. the six Pauli eigenstates work under the same bound $\|H\|\le\frac15$ on the perturbation. I updated the answer to include my initial approach using frames before I switched to SIC-POVM to have a basis. With hindsight, I suppose I should have stuck with the Pauli eigenstates or better yet describe constructions in both frames. Also, this can almost certainly be generalized to some nice condition on the frame (such as $\sum_{i=1}^m P_i=\frac{m}{d}I$), but I didn't have time to dig more into it. $\endgroup$ Dec 28, 2021 at 9:19
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In general, the problem of finding a separable decomposition (or even deciding if it exists) is NP-hard. I don't know if it's that hard for operators of the form $S = \mathbb{I}_\mathcal{X} \otimes \mathbb{I}_\mathcal{Y} - H$ where $\lVert H \rVert_2 \leq 1$, but if you want just examples it's easier to go backwards. You can start with a separable $S$, say, $$S = \alpha_{00}|00\rangle\langle00| + \alpha_{01}|01\rangle\langle01| + \alpha_{10}|10\rangle\langle10| + \alpha_{11}|11\rangle\langle11|,$$ and pick coefficients in such a way that $H = \mathbb{I}_\mathcal{X} \otimes \mathbb{I}_\mathcal{Y} - S \ge 0$ and $\lVert H \rVert_2 \leq 1$. In this example, it's equivalent to $0 \le \alpha_i \le 1$ and $\sum_i (1-\alpha_i)^2 \le 1$.

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