Separable decomposition of states near the maximally mixed state

Question

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.

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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, $H = \sum_{i,j=0}^3 \omega_{ij} \sigma_i \otimes \sigma_j$.

Answer 1

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$^3$.

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$^1$ 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.

Answer 2

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 $\sum_i (1-\alpha_i)^2 \le 1$.