# Pauli projector for multiple qubits

When operating on a single-qubit, any two-outcome measurement can be viewed as measurement of the Pauli observable

$$v \cdot \sigma \equiv v_1 \sigma_1 + v_2 \sigma_2 + v_3 \sigma_3$$

The projectors onto the eigenspaces of $$v \cdot \sigma$$ are

$$P_{\pm} = \frac{1}{2}(I \pm v \cdot \sigma)$$

and so knowledge of either of $$\langle \psi| P_+ |\psi\rangle$$ or $$\langle \psi| P_- |\psi\rangle$$ for any three orthogonal vectors $$v$$ is sufficient to reconstruct $$|\psi\rangle$$.

For systems of larger numbers of qubits, it is well known that reconstruction of $$|\psi\rangle$$ generally requires the measurement probabilities for all possible combinations of Pauli matrices, e.g. $$I \sigma_1$$, $$\sigma_1 \sigma_1$$, $$\sigma_2 \sigma_3$$, etc.

Presumably it is also possible in this case to parameterize all of these measurements by something analogous to $$v \cdot \sigma$$, which also has eigenvalues $$\pm 1$$. The most obvious choice, taking for example a 2-qubit system, would be $$(v_1 \cdot \sigma) \otimes (v_2 \cdot \sigma).$$

Expanding this out you find that all of the $$\sigma_i \sigma_j$$ combinations are indeed represented, except for those involving the identity.

What is the equivalent of $$v \cdot \sigma$$ for multiple qubits, and how can one write the projectors onto its $$+1$$ and $$-1$$ eigenspaces?

• I feel like you mostly answered the question yourself, or maybe I'm misunderstanding you. $\sigma_x,\sigma_y,\sigma_z$, or any other triple of operators obtained by rotating these, form a basis for the set of traceless Hermitian 2x2 ops. Adding the identity gives you a basis for the set of all Hermitian 2x2 ops. For $n$ qubits, you can simply take the tensor products of the Pauli operators on the different qubits. I.e., if $\{\sigma_i\}_i$ is a basis for one qubit, then $\{\bigotimes_j \sigma^{(j)}_{I_j}\}_I$ gives a similar basis for many qubits. You might notice how the eigs remain $\pm1$ – glS Dec 11 '18 at 17:56