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I think my question breaks down into two parts.

Let's say you have a two qubit system and you can perform projective measurements.

Each round of measurements will consist of results looking like below.

\begin{array}{|c|c|c|c|} \hline & \sigma_x \otimes \sigma_x& \sigma_x \otimes \sigma_y & \sigma_x \otimes \sigma_z& \sigma_y \otimes \sigma_x &\sigma_y \otimes \sigma_y&\sigma_y \otimes\sigma_z&\sigma_z \otimes \sigma_x&\sigma_z \otimes \sigma_y&\sigma_z \otimes \sigma_z \\ \hline round 1 &\uparrow \downarrow &\uparrow \uparrow &\downarrow \uparrow &\uparrow \uparrow&\downarrow \uparrow&\downarrow \downarrow&\uparrow \downarrow&\downarrow \uparrow& \uparrow\uparrow\\ \hline \end{array}

You repeat multiple rounds of this sort of measurements.

So you have experimentally acquired values of $\langle \sigma_i \otimes \sigma_j \rangle$ (expectation values of products of Pauli matrices) with i and j being x,y, and z, not including the identity.

Given what I found in this paper, I think you can construct the density matrix of the two-qubit system if you have expectation values of all possible products of Pauli matrices, including the identity matrix.

So you need to know $\langle \sigma_I \otimes \sigma_I \rangle$, $\langle \sigma_I \otimes \sigma_x \rangle$, $\langle \sigma_I \otimes \sigma_y \rangle$, $\langle \sigma_I \otimes \sigma_z \rangle$, $\langle \sigma_x \otimes \sigma_I \rangle$,$\langle \sigma_x \otimes \sigma_x \rangle$,$\langle \sigma_x \otimes \sigma_y \rangle$, $\langle \sigma_x \otimes \sigma_z \rangle$,$\langle \sigma_y \otimes \sigma_I \rangle$ ,$\langle \sigma_y \otimes \sigma_x \rangle$, $\langle \sigma_y \otimes \sigma_y \rangle$, $\langle \sigma_y \otimes \sigma_z \rangle$, $\langle \sigma_z \otimes \sigma_I \rangle$, $\langle \sigma_z \otimes \sigma_x \rangle$, $\langle \sigma_z \otimes \sigma_y \rangle$, $\langle \sigma_z \otimes \sigma_z \rangle$, for full tomograhy.

  1. Can I just get away with measuring expectation values of the products of Pauli operators without the identity matrix to figure out expectation values of all 16 products of two Pauli operators?

  2. Can I construct the full density matrix of a two-qubit system if I know the expectation values of all 16 products of two Pauli operators?

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  • $\begingroup$ Did you count parameters? $\endgroup$ – Norbert Schuch Jan 31 at 23:48
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You need all $16$ expectation values to totally reconstruct the density matrix. Knowing these values, the density matrix is simply written as $$\rho=\sum_{i,j=1}^{4} \langle \sigma_i\otimes\sigma_j\rangle \,\,(\sigma_i\otimes\sigma_j).$$ More generally, given any complete orthonormal set of ($16$) Hermitian operators $\{\sigma_i\}$, you can write the density matrix as $$\rho=\sum_i\operatorname{Tr}(\sigma_i \rho)\sigma_i.$$

Can I just get away with measuring expectation values of the products of Pauli operators without the identity matrix to figure out expectation values of all 16 products of two Pauli operators?

No. To see it consider for example the expectation value $\operatorname{Tr}\big((\sigma_x\otimes I)\rho\big)$. All products of two Pauli operators give zero contribution for this value, so you really need to know the value of $\langle \sigma_x\otimes I\rangle$ to have a complete description of $\rho$.

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