I want to give you a general argument for the negative result: A general $n$-qubit density matrix cannot be contructed from this type of measurements.

The following observation will be needed:

Let $G$ be a compact Lie group acting on a vector space $V$ via a representation $\pi$. Let $\rho$ be a density matrix on $V$.

Consider the quantum characteristic function:

$\phi(g) = \mathrm{Tr}(\rho \pi(g)), \ \ g\in G$,

When $\pi$ is irreducible, then the density matrix can be reconstructed by virtue of the Peter-Weyl theorem:

$\rho_{ij} = \int \phi(g)\pi_{ij} (g) d\mu(g)\ \ i,j = 1,.,.,.,\mathrm{dim}(V)$

However, in the reducible case, there exists a basis in which all matrix representatives are block diagonal.Thus, in this basis, the matrix elements of the density matrix between basis vectors of different irreducible factors cannot be reconstructed, even given the quantum characteristic function at all points of the group manifold.

To apply that to our case consider the group $G = \bigotimes_{i=1}^{n} U(1) \bigotimes SU(2)$.

Where every Abelian factor acts on a different qubit by phase multiplication and the $SU(2)$ factor acts diagonally on the qubits.

Observing that an SU(2) group element acting on a qubit can be parameterized in terms of the Euler angles:

$g = e^{i\frac{\psi}{2}}P_1(\theta, \phi) +e^{-i\frac{\psi}{2}}P_0(\theta, \phi)$.

Then the quantum characteristic function can be obtained from the series of measurement expectations by a Fourier transform with respect to the Abelian coordinates and $\psi$ and vice versa.

But the group $G$ acts reducibly on the $n$-qubit, since every Abelian factor acts reducibly and $SU(2)$ acts irreducibly only on symmetric powers of qubits (thus reducibly on the full tensor power), which completes the argument.

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