1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $\pm1$ $\endgroup$ – glS Dec 11 '18 at 17:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.