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To repeat the titular question: "What is the largest absolute value attainable by an off-diagonal real or complex component of a $4 \times 4$ density matrix?"

Paul B. Slater
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Consider a $2\times2$ Hermitian matrix $ A \equiv \begin{pmatrix}a & b \\ \bar b & c\end{pmatrix}.$ For $A$ to be positive semidefinite (PSD), we clearly need $a,c\in\mathbb R$ and $a,c\ge0$. The eigenvalues are $$2\lambda_\pm = (a+c)\pm\sqrt{(a-c)^2+4|b|^2}.$$ For $A\ge0$ we thus need $(a+c)^2 \ge (a-c)^2 + 4|b|^2$ i.e. $ac\ge |b|^2$ and $|b|\le \sqrt{ac}$.

Suppose now $\rho$ is some $n\times n$ density matrix. A matrix is PSD iff all its principal minors are. For any $i<j$, we must thus have $|\rho_{ij}|\le \sqrt{\rho_{ii}\rho_{jj}}$.

This reduces the problem to figuring out the maximum value of $pq$ over all $p,q\ge0$ such that $p+q\le1$. This is achieved by the choice $p=q=1/2$, and thus $$\max_{\rho\in\mathrm D(\mathcal H)}|\rho_{ij}| = \frac12,$$ for $i\neq j$, where $\mathrm D(\mathcal H)$ denotes the set of density matrices over a space of dimensions $\ge2$.

In short, the maximum value of any coherence is $1/2$, achieved with a state of the form $|+\rangle\!\langle +|$ (with $|+\rangle$ the Bell state defined over the appropriate subspace).

glS
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  • OK, very good, thanks! Notationally speaking, the last equation might indicate $i \neq j$. – Paul B. Slater Dec 28 '20 at 15:52
  • In my preprint https://arxiv.org/abs/2012.09152, I sloppily claimed the the off-diagonal real and complex components have to lie in [-1,1], rather than as glS shows in $[-\frac{1}{2},\frac{1}{2}]$, but the underlying argument that random sampling from the interval is computationally infeasible--if one seeks to generate numerous density matrices--still apparently holds. When the preprint is replaced, I will make the indicated change. – Paul B. Slater Dec 28 '20 at 16:45
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    @PaulB.Slater well in fairness, if you consider the expectation value of a coherence term (e.g. $\langle \sigma_x\rangle$ or $\langle\sigma_y\rangle$), which might often be a more physically relevant quantity, you get the $[-1,1]$ bound – glS Dec 28 '20 at 16:56