I use this decomposition all the time, but I have never read a paper solely devoted to the topic. From my experience a complete characterization of the constraints on $t_{i_{1}, t_{2},..t_{n}}$ is tricky, and so if you want to be sure $\rho$ is physical you should calculate the density matrix and its eigenvalues.

However, there are a lot of necessary conditions that have a useful form in this decomposition. For example, for a positive unit-trace Hermitian operator $\rho$ is follows that

$|t_{i_{1}, i_{2},.. i_{n}}| \leq 1$

$tr ( \rho^{2} ) =\frac{1}{2^{n}} \sum_{i_{1}, i_{2},.. i_{n}} t_{i_{1}, i_{2},.. i_{n}}^{2} \leq 1 $

The above condition tells us that if we think of $t$ as a vector in a real vector space, then the physical states live within the unit sphere. This is a bit like the Bloch sphere for 1 qubit but for many qubits we have some other constraints that take the form of hyperplanes. For every $\vert \psi \rangle$ expressed in the same form
$\vert \psi \rangle \langle \psi \vert = \frac{1}{2^{n}} \sum_{i_{1},i_{2},... i_{n}} Q_{i_{1},i_{2},... i_{n}}\sigma_{i1} \otimes \sigma_{i2}... \sigma_{in}$ we require that

$\langle \psi \vert \rho \vert \psi \rangle \geq 0 $ and so
$\sum Q_{i_{1},i_{2},... i_{n}}t_{i_{1},i_{2},... i_{n}}\geq 0$
which defines a hyperplane.

The problem is you have a hyperplane for every $\psi$ so that requiring $t$ to satisfy every inequality one of the infinite hyperplanes is impossible to check by brute force. If you want sufficient conditions for positivity of $\rho$ I suspect you have to calculate eigenvalues.

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