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I understand that the Pauli operators and the identity matrix span the space of complex 2x2 matrices.

Let's say you have two qubits and you can perform projective meausurements.

So you can measure spin states of the two qubits along $\sigma_i^1$ and $\sigma_j^2$, with the superscripts 1 and 2 indicating which of the two qubits and the subscripts being any of x,y, and z.

I am under the impression that from such measurements, you can construct expectation values of all two qubit correlators, consisting of the Pauli operators and the identity matrix.

So I have two questions.

  1. Is it true that I can construct expectation values of all two qubit correlators, consisting of the Pauli operators and the identity matrix? (All of $\langle \sigma_I \otimes \sigma_I \rangle$, $\langle \sigma_I \otimes \sigma_x \rangle$, $\langle \sigma_I \otimes \sigma_y \rangle$, $\langle \sigma_I \otimes \sigma_z \rangle$, $\langle \sigma_x \otimes \sigma_I \rangle$,$\langle \sigma_x \otimes \sigma_x \rangle$,$\langle \sigma_x \otimes \sigma_y \rangle$, $\langle \sigma_x \otimes \sigma_z \rangle$,$\langle \sigma_y \otimes \sigma_I \rangle$ ,$\langle \sigma_y \otimes \sigma_x \rangle$, $\langle \sigma_y \otimes \sigma_y \rangle$, $\langle \sigma_y \otimes \sigma_z \rangle$, $\langle \sigma_z \otimes \sigma_I \rangle$, $\langle \sigma_z \otimes \sigma_x \rangle$, $\langle \sigma_z \otimes \sigma_y \rangle$, $\langle \sigma_z \otimes \sigma_z \rangle$ although I can only perform $\sigma_x, \sigma_y, \sigma_z$ projections.)

  2. In general, for a single qubit, how would you extract the expectation value of the identity matrix from projective measurements $\langle \sigma_I \rangle$?

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  1. Yes. The space of two-qubit observables ($4\times 4$ Hermitian matrices) has dimension $4^2$ (more generally, the vector space of $n\times n$ Hermitian matrices has dimension $n^2$). Any orthonormal set of $n^2$ Hermitian matrices can therefore be used as a basis. You can easily verify that this is the case for the set of $16$ operators you mention. Regarding their being all local observables, this is not surprising because you can always find a basis for a tensor product space $\mathcal H\otimes \mathcal H$ that is made up of product vectors of the form $v\otimes w$.

  2. The expectation value of an observable $A$ over a pure state $\lvert\psi\rangle$ is defined as $\langle A\rangle=\langle \psi\rvert A\lvert \psi\rangle$. It follows that the expectation value of the identity is $\langle I\rangle=\langle\psi\rvert\psi\rangle=1$.

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